ML19337A260

Forwards Licensed Operator Training Lessons for Fluid Flow Heat Transfer & Thermodynamics,In Response to NRC .Training Re Mitigation of Severe Core Damage Per Inst for Nuclear Power Operations Guidelines Will Be Developed

ML19337A260
Person / Time
Site:	Comanche Peak
Issue date:	08/11/1980
From:	Gary R TEXAS UTILITIES ELECTRIC CO. (TU ELECTRIC)
To:	Collins P Office of Nuclear Reactor Regulation
References
NUDOCS 8009090387
Download: ML19337A260 (184)

Text

{{#Wiki_filter:' .c-TEXAS UTILITIES GENERATING COMIMNY 200i earm inwsa. oat.us. rEXAS 75201 (( - h August 11, 1980 50 N R. J. GARY 8 ABCts?tvt v.C8.etS@t=T ^ ~ ~ " ~ ~ ' " ~ ' ' ' THIS DOCUMENT CONTAINS P00R QUAUTY PAGES Mr. Paul F. Collins, Chief Operator Licensing Bureau Office of Nuclear Reactor Regulation U.S. Nuclear Regulatory Commission Washington, D.C. 20555

Dear Mr. Collins:

This is in response to Harold R. Denton's letter of March 28, 1980, " Qualifications for Reactor Operators," Enclosure 1, " Criteria for Reactor Operator Training and Licensing," Sections A.2.c, A.2.d, and A.2.e, respectively. " Training programs shall be modified, as necessary, to provide

1) Training in heat transfer, fluid flow and thermodynamics.

2) Training in the use of installed plant systems to control or mitigate an accident in which j

the core is severely damaged.

3) Increased emphasis on reactor and plant transients.

Effective date: Present programs have been modified in response to Bulletins and Orders. Revised programs should be submitted for OLB review by August 1,1980." CPSES is using the enclosed training lessons for heat transfer, fluid flow and thermodynamics for licensed operator training in addition to training provided by contract for cold license candidates. CPSES will develop training for recognizing and mitigating the consequences of severe core damage using Institute of Nuclear Power Operations (INPO) guidelines of June 30, 1980. 8 8009090 8/ /

e' ?) ' CPSES. training is providing increased emphasis on reactor and plant transients. Cold license certified candidates have completed additional simulator training conducted by General Physics Corporation which emphasized transient and accident response to the degree of experiencing multiple equipment failures, solid plant-and natural circulation operation. As part of the program, the operators were subjected.to operating circumstances which duplicated the TMI-2 accident. " Training Center and facility instructors who teach systems, integrated responses, transient and simulator courses shall demonstrate their competence to NRC by successful completion of a senior operator examination. Effective date: Applications should be submitted no later than August 1, 1980, for individuals who do not already hold a senior operator license." CPSES has a licensed operator training program, the content and conduct of which is supervised by an individual who is senior' licensed on a similar facility. We propose to hot ' license all instructors who teach systems, integrated responses, transient and simulator courses. " Instructors shall be enrolled in appropriate requalification programs to assure they are cognizant of current operating history, problems, and changes to procedures and administrative limitations. -i Effective date: Programs should be initiated May 1, 1980. Programs should be submitted to OLB for review by August 1, 1980." i CPSES instructors will be enrolled in appropriate requalification programs after initial qualification. Very truly yours, R. . Gary i RJG:mkg Enclosures x..

.i CPSES TRAINING LESSON NOTES PAGE 1 0F 36 J. EICHELBERGER LESSON: HEAT TRANSFER AUTH: C. MEYER REV: No. 1 PROGRAM: CPSES OPERATOR TRAINING

REFERENCES:

REVISCD BY: R. WIRKKAL. NUCLEAR POWER SYSTEMS, C. D. GREGG KING, CHAPTER 14 AND 15. NOTES: TE:G:

1.0 INTRODUCTION

• One of the unusual aspects of a nuclear reactor is that there is no theoretical upper limit to the rate of energy release from fission. However, the maximum operable power U

1,evel is heavily dependent on the rate at which this energy can safely be removed. Therefore, the design of the reactor core is dependent not only on the nuclear considerations, but the heat-removal aspects as well. Heat is the movement or transfer of energy by means of a tem-perature difference. When a substance rece>.ves heat, its molecules begin to vibrate more rapidly. This increases their kinetic energy and this increased kinetic energy is passed along to other molecules. Hence it can be seen that the flow of heat is dependent on the temperature difference. 1.1 TYPE 3 0F HEAT TRANSFER There are 3_ general means by which heat is transferred from one point to another conduction, convection, and radiation. q._

• \\..

) e ) 1 PAGE 2 0F 36 REM /RKS LESSCN NOTE (CCNTINUATICN SHEET - EVEN) .l.1.1 Conduction Conduction is the transfer _ of energy by forces between adjacent molecules,of a body. The proceis may be pictured as a. series of impacts between vibrating molecules in wnich the net energy flow is frcm the mc.e rapidly vibrating molesc[les to the less rapid. Conduction may occur between parts of a single body or between bodies in centact. This is the type of heat transfer that we will be most concerned with. An example of conduction is two metal bars, one hot 'and one cold, placed in contact side by side. The heat travels frem the hot-bar to the cold bar by the hot bar's molecules impacting and losing energy to the cold bar's molecules. 1.1.2 Ccnvecticn Convection is the transfer of energy in a fluid body by a lbr5 ./ movement or ficw of fluid mass between regions of high and !~ Icw temperature. The ficw of the fluid may be entirely the result of differences in density due to differences in the tercerature of the fluid, as in natural convection; er ficw may be produced by mechanical means (i.e. a pump), as in forced convection. During ccnvection, heat transfer energy is also being transferred simultaneously by molecular l condaction. So in essence, convection is merely conduction ( with fluid (mass) ficw. An example of convection can be found ( in an autcmobile's engine. A water pump circulates a forced j - fluid ficw;- the fluid ficws through the hot engine bicek and removes heat from the cylinder walls. This is an example of forced convection. 1 L y ,,p.. w 4 g

n e t PAGE o CFic LESSCri fl0TE (CCriTItiUATICl SHEET - CDD) REf4 ARKS 1.1.3 Radiation Radiation is the transfer of energy between a hot and cold body. by the emission and absorption of electromagnetic It has been showr5 that heat transfer by conduction waves. and convection requires a medium for transmission, but heat transfer by radiation does not. A relevant example of heat transfer by radiation is the sun, which transmits all of its ei..rgy to earth by radiation. This can be easily shown since the energy travels through the vacuum of space, where there are no molecules to serve as a medium of transfer as needed'by conduction or convection. 2.0 CCriCUCTICri THROUGH A PURE SUBSTAtiCE There will be a transfer of energy (called heat) between two L points in a substance whenever there is a temcerature s-difference between those points. How much heat is transferred depends on the size of the temperature difference (the driving force) and a property of the substance called thermal con-ductivity (k), which describes how much heat (Stu/hr) can be transferred per unit thickness (1/ft) of substance and per degree of temperature difference (1/*F) across that thickness. The general expression for finding how much heat is transferred across a substance fecm point x) to point x2 due to a temperature difference from T) to T2 is: (T)-T ) 2 0"M g -X j 2 a 3 g23[r:fF1 g tu), e Stu. 'F nr nr r: t j 4

b i PAGE 4 0F 36 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 2.0 (cc,,) / 7 T 3 g / X g X2 HEAT CONDUCTION GEOMETRY Without including the cross sectional area for heat flew, 2 the expression yields heat flux (Btu /hr ft ), Consider heat conduction across a steel pfate; a graph of tem-perature versus distance can be drawn. l l 7 l l rew.cnatune acmess a nurc l The resulting graph can have different slopes; it could be a straight er curved line, but the general trend of the curve is to decrease to the right. The sicpe of this line at any point represents the temperature gradient (h). It is assumed that the material is hemogeneous, therefore for each substance 'k

=a s PAGE 5 0F 36 {" ,. LESSC!l NOTE (CCNTIllUATIC!l SHEET - CDD) REMARKS t 2.0 (Cent.) the thermal cenductivity constant (k) can be found. The larger.the value of k, the better of a heat conductor it is. The "olicwing is a list of scme thermal conductivity constants: TABLE OF AVG. CCNDUCTIVITIES Matarial - 5tu He ft *F CORK 0.025 FIBER INSULATI!!G BOARD 0.023 MAPLE OR OAK WOOD 0.096 BUILCING BRICX 0.4 WI!!DCW GLASS 0.45 3 CCNCRETE 0.19 sy 1% CARSON STEEL 25. z 1% CHROME STEEL 35. ALUMI!!UM 113. CCPPER 223. SILVER 235. e 4' Y,I2 I, I 2 4 9

a -( PAGE 6 0F 36 REMARKS LESSON NOTE (CCNTINUATICN SHEET - EVEN) 2.0 (Ccnt.) Examole: A glass window pane,1/8" thick, as shewn in the above figure is installed in a heme. On a cold day in winter the cutside surface temperature is O'F and the inside w face temperature is 68'F. Find the rate of heat ficw out of the pane and the temperature in the glass 3/32" frem che inside surface. Solution: A) Using: Q=XAh=(.45)(3x4) g 0') = 35,307 Stu/hr.

3) The rate of heat ficw is censtant throughout the pane,

[T-T therefore: Q = XA T)-T2 3 X -A A-A 2 l l 35,307 = (.45)(12) jh-0) Sc1ving for T: T= 68 - (35,307) (.0078) s ( 45) (12) T = 17'F. 2.1 CCNCUCTICN THROUGH CCMFOSITE BODIES If more than one material is present, as in a multi-layered wall, each substance has a different themal conductivity constant. The resulting temperature versus distance graph would be as follows: d l N. I i r. To A e c o 2 1 3 2 3 8 4 X S ~ I Xw FIGURE 2.1-1 Trwca ATURC VS CISTANCE M A COWCSCC NTH

e s PAGE 7 0F u LESSON NOTE (CONTINUATIO*: ""EET - 000) REMARKS 2.1 (Cont.) Notice that the rate of heat flow is steady, therefore the same amount of heat is conducted through each individual layer per unit of time. So: T -I AT., AT AT 2 l C O 3 a[3 O"M A ax C 4A D an A C D Solving there equations simultaneously yields: Q= A 17 tI 3x g, for "n" layers where ATteal g { ""{ is the total temcerature drop IK across the "n" layers. %m Examcle: A side panel of a refrigerator (above figure) is made of an inside sheet of aluminum 1/16" thick, a layer of cork insulation, and a 1/8" sheet of 1% carbon steel. The interior surface temp of the refrigerater is kept at 40*F and the outside at 68'F. Hcw thick must the cork FIGURE 2.1-2 layer be to limit the ficw of heat through the wall to [ j[ 50 Stu/hr? Solution: Using: Q= AAT N /[ ~ a5 ad ad 9 +4+1 / s / S t ! !'. ] 3 / ,ss - s f ,cej6l3 28 - 50 stu/hr 'S;;u" ,cie. lla G [/ / 1 ~ 4y L/ 4* t/8* [ ~

a i PAGE 8 0F ?6 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 2.1 (Cont.) Solvtng for X2 = [O 8 M28)_,,_0052,.0104] (.025) (50) 118 25 X2 =.252' X2 = 3.02" 2.2 CCNDUCTICN IN CYLINORICAL GECMETRY Although conduction thrcugh flat platas and walls is the most direct application of the conductivity equation, in the PWR a great deal of heat transfer takes place across cylindrical walls. In this type of geometry, where heat is moving from a smaller diameter to a larger diameter, the area increases as temperature decreases, thus complicating the problem. For this a:: plication the conductivity expression takes en a different fom. g, 2:L X (T;-T ) h 2 in tr,lr 1 7 The furm 2:L represents area in cylindrical geccetry and the (3.T/in r /f ) represents the temperature gradient across the 2 i wall. Just as with the flat plato, a heat flux per unit length can ~ be obtained by not including the length factor, L. Examole: A 10 ft length of steam pipe made of 1% carbon steel with rj = 1", r = 1.25", carries steam at 100 psig. o The heat ficw rate is 30,000 Stu/hr. The surface temp of the pipe.is 250*F. Find the interior surface temperature. Solution: UsingQ=2.iKL(T-7) 3 2 In(r I"ll 2 l Therefore, Tj = Q in (rg r )3* l 4:XL

• I2 f

T) = (30,000) in u.2}") + 250 2a25)(10') n u g,,, l-T) = 254.25*F. ~(

a = ( PAGE o 0F M LESSON NOTE (CCNTINUATION SHEET - 000) REMARKS 2.3 CCNDUCTION WITH' INTERNAL HEAT GENERATION Ccnsider a red with internal heat generation (such as a fuel red in a nuclear reactor and the fission heat generated, or that of an alectrical wire ar.d the heat generated due to resistance when current is passed through it). Once again it is not a simple prcblem. However, with the basic assumptions that heat is only conducted in a radial direction and that the rate of heat generated per unit volume is constan one can derive a set of. equations. But first examine a graph of 1 T vs. r FIGURE 2.'; TEMPERATURE P9eFIL2 1M C A cveunomicAL war sounct This shcws that maximum temp rature occurs at the gecmetric center of the red. To find the temperature at any radius in the rod the derived equation is: 2 T -Tg= g where 3 r = radius from center to measured point (ft) T) = max tamp (at center) (*F) Tg = temp at radius r (*F) 3 G = rate of heat generated per unit volume (Stu/hr-ft ) K = thermal conductivity constant (nr F ) 4 .g. .. g .m a w

a b PAGE 10 0F 36 ^ REMARXS LESSON NOTE (CCNTINUATION SHEET - EVEN) 2.3 .(Cent.) and to find Q at the surface of the rod: Q = 4HKL (T) - T ) 2 L = length of rod (ft) T2=tempatsurfaceofred('F) These two equations will enable a basic understanding and problem solving capability. Examole: A 15" heating coil is imersed in a centainer of water. The resistance of the coil is.24 c/ft and it draws 100 amps. The radius of the coil is 1/4", and the surface temp is 200*F. If the thermal ccnductivity constant ,is 10 Stu/ft hr *F find the max temp of the coil, the rate of heat generation, and the ficw rate of heat to the water. h Solution: a) Q= "{$g, = pcwer = 1 R = (100)2(.2k) (1.25 ft) 2 = 3000 watts converting: Q = (2000 watts)(3.413 Bj e ) = 10,239 Sti2/hr b) Assuming that heat is generated at a uniform rate over the coil's cross section, then Q = GHLr2 = 10,239 Btu /hr 0 10.239 6 Stu therefore G = 2 a(1.25)(.0208')2 = 6.0.x 10 3 aLr-f y c) T,,x would be fcund at the gecmetric center of the coil using X(T; - T )

• 4 2

2 7I = Gr +T 4K 2 T) = 265'F l l 'D t I

PAGE 11 0F 36 LESSCN NOTE (CCNTINUATION SHEET - CCD) REMARXS 3.0 CCNVECTION Ccnvection is generally listed as a mode of heat transfer along with conduction and, radiation. Strictly speaking, however, convection is really conduction combined with mass transfer. This transfer of c olant mass is the characteristic feature of convection, whether the coolant be a gas, liquid, or liquid metal, A' an example, cor,si/~,' a heating element submerged in a centainer of water. Heat is generated within the element by electrical resistance and the temoerature of the element increases. When the temperature of the surface bec:=es greater than that of the water contacting it, the potential exists for heat transfer. However, as soon as seme thermal energy is transferred to the water, that water undergoes a decrease in density and is prceptly displaced by e r-f water of greater density. The warmer water rises because of natural buoyant forces acting on it. The rising water cen-tacts cooler water and its slightly higher temcerature causes the extra heat energy to be dissipated in the bulk ccolant. The heater element, meanwhile, is still surrounded by c:aler water, and will be until the whole bulk temperature gradually increases. This increase in bulk temperature is caused mostly by the heat energy carried by movino fluid, as opoosed to pure conduction A clore icok at the water immediately adjacent to the heater surface would shcw a temperature gradient from the surface into the' water, just as exists in conducticn. A thin layer of water remains in contact with the surface because of ad-hesive forces, and heat is transferred thrcugh this relatively stable "bcundary layer". Because of the variables affecting this layer (surface roughness, shape, size, fluid viscosity, 6 f

e PAGE 12 0F 36 REMARKS LESSC1 fl0TE (CC;1TIllUATIC!i SHEET - EVE?l) 3.0 (CCliT. ) ~ temperatures and temperature differences) it is extremely difficult to quantify. Rather, the rate of heat transfer through the layer is detemined for a range of conditions, and this heat transfer coefficient is used for design purposet. The coefficient, designated h, expresses the amount of heat (Stu) which can be transferred acrcss a solid / liquid interface 2 per unit area (ft ) per unit time (hr), per degree difference between the surface and the bulk ccolant (*F). The equation for the rate of convective heat transfer is similar to that for conducticn: q = hAa7 ( ") = ( 8tj ) x (ft ) x (*F) hr it- 0F h The equation works for heat transfer in either direction, frca surface-to-liquid, er frem liquid-to-surface. 3.1 FCRCED CClVECTIC{ The initial discussion on convection emphasi:ed the fact that heat. energy was transported and dispersed because of fluid movement. In that case the movement was caused by buoyant forces, and the fluid motion was inherently slow and diffuse. The amcunt of heat which can be removed frem (or added to) a surface can be significantly increased by forcing the water to =cve past the surface with some centrolled velocity. This is analagous to the difference between a hot radiator depending on natural circulatien of air to heat a reem or having a fan move air across t!ie radiater to speed and improve the peccess. Just as in natural ccnvection, many facters affect heat removal capability, including the ficw characteristics, i.e., -L

(, PAGE n CF y LESSON NOTE (CCNTINUATION SHEET - 003)' REMARKS 3.1 (Cent.) Is the ficw pattern laminar or turbulent? Is the flow being forced alcag the length of a heated surface, or perpendicular to the surface? Is the ficw contained inside a channel (pipe, duct) or flewing outside heated surfaces? For nany of these different conditions, special correlations have been developed which yield the appropriate heat transfer coefficient. These '~ cmcirical chrrelations generally require information abcut fluid properties and ficw conditions, as well as tne solid surface preperties. Temperature differences between bulk i coolant and wall will vary as a function of distanca along a closed ficw path, so heat transfer rate will vary accordingly. ' For this particular case of forced ficw along a c1csed heated flow path, an " effective" temperature difference can be used _(__. to describe the tctal effect of the entire channel. This leb mean temoerature difference (ai ) is given by the expres-m sien T I"),g eutlet AT = a aT in(.Iinlet) cutlet where AT = temperature difference at the inlet inlet AT = temperature difference at the outlet oug),g w ] ~ 'En( l ( ) Iil ' L. FIGURE 3.1 1 iLLusinAnoH or log WCAN TIVPERATURt CtFFERENCE i

PAGE lACF 36 REMARKS LESSC!i NOTE (CCNTI?tUATIC!i SHEET - EVE?i) 3.1. (Cent.) - Use of this AT, ailens evaluation of the total heat transfer _ capability of the heac exchanger, which will be discussed next: q = UAAT m 3.2 THE OVERALL COEFFICIE?iT The discussion of heat transfer by cenducticn concerned itself with the treatment of ecm::osite bodies, where the cenduction path censisted of different media, each with its own themal conductivity. A very cc::ran application of convective heat transfer is the heating of the fluid by another, where one of the fluids is moving inside a flew channel. Typical exam-ples in a F',iR are the steam generator, condenser, MSR, and ..,.h feedwate: heaters as well as all other heat exchangers. For these components, heat is transferred thrcugh one boundary layer to the channel (tube) wall, conducted through the wall, and then transferred through another bcundary layer to the seccnd fluid. For this gecmetry an overall heat transfer coefficient is defined which accounts for all the heat transfer characteristics. r. , n. a n,; \\ ,/ ,..-, n j

i.. i {. '9*7'\\

1 i i =.. ,tuio r, ,,/ 3 l r, l s 's, -.. h' / r, l \\~ i u ~.. - r 4... i a 8008:044f LAYERS l I I GECWETRY-AND TIuPCRATURE ORCP THMCuGH pygggg 3,2 A TU8E WAL! wlTH TWO 80VNCARY t.AYERS l'

O e b PAGE se OF sc LESSON NOTE (CCitTI!!UATION SHEET - 000) REMARXS 3.2 (Cont.) Since the most ce=en application is the tube shown in the diagram, that is the for n of the equation which will be deve-Assume a given am' unt of heat, q, is being transferred loped. o frca FLUID 1 to FLUID 2. For the boundary layer of FLUID 1: q = h aT)A) j For the tube wall: (some length L-) , 2:L K aTwall r in(f) 2 For FLUID 2 boundary layer q = h A 0T 22 2 (;.; Since the same amount of heat is being transferred in all three cases, all the above equatiens are equal. To condense the problem, each equation will be rewritten as a function of its AT, and the aT's added 'l " O d. n)A) r wall = 9 I" ( ~3) AT 2 lnLA AT =" A22 in( ) 3 .aij + AIwall + AI2* I + 2 + ) nA n, A 11 2:LA c2 AI = Tfluid 1 - T,3)) (r ) l 2 AIwall *'Iwall (#2) ~ Iwall ("3)

1. I2 = T,))(r ) - Tfluid 2 y

3

PAGE 16 0F 36 i REMARKS LESSCri tJOTE (C0tiTItiUATI0tt SiiEET - EVEil) 3.2 (Cont.) r Tfluid 1 - Tfluid 2

• I

} ~ n ^2 n; j 2 2:LA Ti-T2 1 "l l

• In( )*hA A

22 2 2:.A This equation is of the same form as that presented for heat conduction through composite macerials. The dencminator re-presents the total resistance to heat movement frcm FLUID-1 to FLUID-2. It is simpler to treat the 2 boundary layers and tube wall as a single entity and thus define an overall heat transfer coefficient, U, such that k*UA(T)-T) 2 'Jhere A is the total tube surface area involved, and (T;-T ) 2 is the total temperature drop between bulk fluids. Examole: In a steam generator, the total tu'be surface area 2 Stu is 48,000 ft. Given that U is 1402.9 and CPSES hr ftd 'F full power conditions, what is the total heat transferred in the S.G.7 s.G. primary inlet = 618.2 F, outle- = 557.8 F Solution: At full pcwer, primary Tyyg = 588 F, steam pressure =1000 psia ~ T 4000 psia) = 544.6 F SAT So AT = 588.0 544.6 F = 43.4 F 2 h = UAAT = 1402.9( 8tj ) x />8,000 (ft ) x (43.4)*F hr ft" *F 5 q = 2922.5x 10 Stu/hr -x -

+ ( PAGE 170F u LESSON NOTE (C0tlTINUATION SHEET - 000) REMARKS 4.0 HEAT EXCHANGERS A heat exchanger is an application of forced convection heat transfer. Since fluids inside and outside tubes are involved, an overall heat transfer coefficient can be used. And since temperatures between the fluids vary with position, a loo.mean temeerature difference is also required. There are two general categories of heat exchangers. The first type is called parallel-row because the two fluids travel in the same direction through the exchanger boundary. The diagrams indicate the Fluid 2 temperature increases as Fluid 1 temperature decreases. This leads to a very large AT at the heat exchanger inlet, and a relatively small AT at the cutlet. Since heat transfer always depends on AT, the rate of heat trar.sfer decreases significantly frem inlet to cutlet. Another obvious constraint in this type of heat exchanger is tnat the Fluid 2 temperature can never exceed Fluid 1 temperature anywhere. PLUID Ilu 9 7.. ~ l l r- ~ 'T

• f**"".

l h '%$ 2 f ~ I I r--- F FLU 10sOut t FLUID I TEWP FLU 10 2 14 CUT p{g*ggg f,,Q.] PAR ALLck FL.0W MEAT EXOHANCr.R

( PAGE 18 0F 16, ) REMARKS LE550:1 NOTE (CCitTINUATION SHEET - EVEN) 4.0 (Cont.) A second type of heat exchanger is called cbunter. flow because the fluids move in opposite directions tnrough the exchanger boundary. In the counter-flew design, the maximum temperatures of both fluids occur at the same end of the exchanger and the ai remains relatively unifom thrcughout the heated length at a high level. It is possible to have a FLUID 2 temperature higher than a FLUID 1 temperaturt in a counter. flow design e S = j -4 I i c-{ 7302 'Nr l [: I i i 6._.......-............ t stuio e our PLL101 TtWW FLU 10 2 in our rIGURE 4.0-2 ccmtamcw wcAr EXCHANGER i l i. l

PAGE 19 0F 36 , ' E.SSC:4 tiOTE (CO?tTI!lllATICil SHEET - C00) REMART,5 4.0 (Cont.) In general, a counter-flow heat exchanger is more efficient than a parallel flow design under the same conditions. The total rate of heat transfer in a heat exchanger (where no change of phase is involved) can be determined from fluid stream properties for either fluid. q=mCp(T -Tin) EITHER FWID cut 8 q = total heat transfer rate (h ) m=massficwrate(fr) ( 'out ' 'in) = temperature change for FWID (*F) C - specific heat for fluid, (Stu/lb 'F) p Exaeole: The shaft of the reactor coolant pump is (._ sealed to prevent leakage of main ecolant frem the primary system. The seal is normally provided by high pressura makeup water with a controlled leakoff rate. If this source of fluid is lost, primary fluid will leak up thrcugh the seals. Since the high tamperature of this water (cold leg temperature s 558'F) might seriously damage the seals and bearings, a thermal barrier heat exchanger is an integral part of the leakage path. This HX is effectively a single pass counter ficw device as shown in the diagram. If primary leakage througn the HX is SGFM, and the maximum allewable temperature of the bearing and seals is 235'F, what is the ai of the c =ponent cooling water inside the tubes if flow is 40 gpm and inlet temperature is.105 F? e l

b CF 7g PAGE 79 RE.MARXS LESScri fl0TE (CCtlTI?tUATICil SHEET - EVEll) PRlWARY W,4TER CUT 4.0 (Cont.) 2350+r Max a T p -ceuneucur p ceCuno m 1050F V --m CO W pC N cNT COCLaG Out FIGURE 4.0-3 SHAFT THERWAL BARRtgn HEAT EXCNANGER PRIMAbY WATER IN g/-- 558 F Scpu Solutien: For this case, an overall heat transfer coeffi- ~ cient is not available, but the total heat removed frem the I primary fluid can be calculated (using steam table pro-perties). HEATRE"0VED=$1C AT 3 P ft

• 55 lbm
• 1.0 Stu x (558-235)*F
• 9E"
• 7.48i gal 3

ica "F fg = 11,873 Stu/ min g l This same quantity must be added to the ccmponent cooling l water. 3 Stu = 0 gpm x

• 1.0 Stu 55 iha 1

ft 11,873 min 7.481 gai lem "F f;3 40 F = ai = T - 105 F 145 F=T cut l

~ 21 7 26 oAGE 0 LESSCN NOTE (CCNTINUATION SHEET - CCD) REMARKS 4.0 (Cent.) Most heat exchangers combine the characteristics of parallel-ficw and counter-ficw designs and also include flow perpen-dicular to heater tubes by means of baffles and divider plates. In the CPSES plant, most of the heat exchangers are of the U-tube design; this type of construction is favored for high pressure applicaticns because of its simplicity of construc-tien and ability to accept large ther=al transients since each tube is free to expand as needed. Generally, the higher temperature / higher pressure fluid is inside the tubes, and tne icwer temperature /lcwer pressure fluid cccupies the space between the tubes and shell. Flow within the shell is directed and reversed by baffle plates to assure maximum exposure to

• ube surfaces. For this type of heat exchanger where heat transfer coefficients and bounefarv laver AT's vary widely, the use of an overall heat transfer coefficient, U, and the log-mean temperatures for AT is advisable.

l COCL FLutD IM w I H1 l ' J,, L COOL FLulo CUT ,,~, MOT FLUa0 14 MCT FLUID CUT fig'd,qE 1,0..t TYPICAL (VERTICAL) SHELL AND U-TU5E HE AT EXCH ANGER SHOWING FLUID FLCw PATHS y

'C-PAGE 22 0F 36 REMARKS LESSC:1 tiOTE (CONTINUATION SHEET - EVEN) 4.0 (Cent.) Examole: The RHR HX's are of the "shell" and vertical U-tube type. Primary coolant on the tube side has a design 6 AT from 140*F to 119.40F at400 psig and 1.9x10 1b/hr. Component cooling water en the shell side has a design aT frcm 105'F' 6 to 115.2 F at150psig and 3.8x101b/hr. How much heat is this HX designed to remove from the primary ccolant; to what fraction of full pcwer is this equivalent? Solution: Heat removed = rs C aT-p 6 =, ' l 9 x 10 lbm/hrx.9975hp x (140 119.4) F 3.9 x 10 (Btu) = ,g, nr With both RHR trains in operation, total heat removal is 7 u 7.8x10 , which corresponds to 0.67% of full rated pcwer 6 (11690 x 10 Stu/hr). This is the level of decay heat presen scme 20 hours after shutdcwn frem extended full-pcwer cperation. 5.0 HEAT TRANSFER WITH CHAliGE OF PHASE Many of the heat exchangers in a PWR are involved directly in thermodynamic processes and axperience a change in phase of

• ane or both of the fluids involved. The steam generatcr and ccndenser are prime examples of this mcde of heat exchange.

l For these cases, the expression g=bC AT p i dces not work since the temperature of a boiling or condensing liquid will not change and also C changes drastically with the phase change. To evaluate these HX's, a simple energy balance using enthalpies I is required en the boiling fluid. The following diagrams show C

{~ PAGE 23 0F 36 LESSON NOTE (CONTINUATION SHEET - OCD) REMARKS ~ 5.0 (Cent.) the temperature /enthalpy relationships for heat exchangers with change in phase. The t.H for both fluids is the same in mag-nitude (no energy is lost) and again, the counterflow design canyieldahighertemperature(highersaturationpressure)in the boiling fluid than the parallel flow design. Obviously, for these T-h diagrams to be valid, the hot fluid must be subccoled (under a greater pressure), and the amount of heat ~to be transferred must not exceed the capacities of the design. This concern will beceme more relevant after a discussion of general boiling heat transfer. YT PARALLEL FLOW 'gcyNTER FLcw MCT PLyl0 HOT FLulO \\ l e so1 LING FLut3 soiums FLuio l= y l: j An an HE AT EXCH ANGER PERFCRidANCE WITH CMah3E IN DMASE FIGURE 5.0 i

O 4 r PAGE 24 0F 36 .) RE.'iARKS LESSCN NOTE (CONTINUATION SHEET - EVEN)' 5.1 THE BOILING CURVE Consider the typical example of a heated tube submerged in - a saturated water coolant. As the tube surface temperature rises above that of the saturated water, small bubbles form at nucleation sites on the tube wall. These bubbles increase in si:e and break loose from the surface because of' buoyant forc'es, thus disturbing the boundary layer and resulting in a signi-ficant heat removal capability for a small AT. As the aT increases due to a higher wall temperature, the bubbles formed are larger and appear at a greater rate, so that heat removal is furcher enhanced. But beyond this point, a higher ai will cause the bubbles to coalesce, and form patches of vapor on the tube surface, before the bubble mass is replaced by the liquid film re-establishing itself. 'ihile the vapor patch is covering the surface, heat removal is significantly reduced in that area because conduction through steam and radiation at icw temperatures are not very efficient heat transfer mechanisms. If the wall temperature is increased still more, the vapor blanket on the tube surface stabilizes and heat transfer is severely restricted. This condition is referred to as "dryout" and may be regarded as a potential failure condition for the heat exchanger. The tube _ wall wi11 increase in temperature to that of the hot fluid, and will be extremely susceptible to thermal shock when the liquid film again wets the outside sur-face. The diagram below shows the heat transfer capab'ility frca a heated surface to water with boiling effects included. l

{- PAGE 25 0F 36 ~ LE550!!tl0TE (CCitTItiUATICtl SHEET - 000) REMARKS 5.1 (Cont.) e l FILM 10000 - SOILING i l .t n A i CONVECTION l l l00 ~ l 2 8 STU/HR FT 'F 8 S. 1 NUC L EAT E l-SOILING l e I to ico 1000 AT [* F] (p FIGURE 5.1 HEAT TR ANSFER COEFFICIENT TO WATER AT S ATURATION TEMFER ATURE The curve is generally referred to as the "boiline" curve because the folicwing regimes of boiling can be identified: A. In this range, heat transfer is by pure convecticn. Superheated liquid will Mse to the surface of the water pool and evaporation will take place. 3. In this range, bubblas fom at the surface of the heated element (nucleation sites). At the icwer end of this range, the bubbles collapse in the superheated bulk liquid after breaking loose. At the top end of the range, bubbles rise to-the surface. C. Because of the high surface temperature, vapor patches form on the surface and heat transfer becomes less efficient. As surface tamperature increases, a stable film is formed and heat transfer capability decreases further. At scms:

[ pAGE 25 0F 26 RE.". ARKS ' .LESSCN NOTE (CONTINUATION SHEET.- EVEN) 5.1 . Cl -(Cont.) i point, however, surface temperature beccmes so high that~ radiation heat transfer through the vapor film beccmes a significant mechanism, and heat transfer capability begins l to increase. Eventually, a high rate of heat removal can be established, but now radiation is the mechanism, and surface temperature is. extremely high. Generally, if a high rate of heat transfer is required, such as when the heated surface contains a heat source, the temperature at the surface will exceed the melting point of the material. i 5.2 DNB '4hile the boiling curve as presented above is applicable to the icw. pressure side of all heat excnangers, it is also applicable i to heat transfer frcm the fuel reds'in 'the core. ' For this application, the "h" axis will be relabeled the "q", or heat flux axis, since new a certai t rate of heat removal is recuired l for both power generation ar i fuel ecoling. Again the same regions of the curve cL-N identified, only new the transition to film boilidg takes on new significance. 6 DN8 10g j ______________,8 5 l l 10 l l l l l ^ 4 HEAT FLUX-l STU/HR FT2 l { l l i l l l l l l l l l' t 10 10 0 10C0 AT FIGURE 5.2 FUEL ROO HEAT FLUX VS, CL AD/ COOL ANT CELTA-T

PAGE 27 0F 36 LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 5.2 (Cont.) This point is-referred to as the Departure from Nucleate Boiling point, or DNB_ point, and the corresponding heat' flux is referred to as the DNB heat flux. Since the curve represents the behavior of a physical system, the operating point must lie on, i the curve.. If some heat flux higher than the CNB heat flux is produced in a fuel rod, then the operating point shifts from DNB over to point ~B on the curve, and the corresponding surface temperature rises to several thousand degrees F. This f high temperature will probably melt the fuel rod cladding and create a potential for release of radioactive fission products into the coolant. Obviously the CNB heat flux is to be avoided, and accordingly most cores operate at some point "A" on the curve. Just how far below the DNB point a plant operates, and W how close it approaches during plant transients, is a major portion of core protection analysis. This will be discussed in detail later. 5.3 THE CCNDENSER Thus far, the discussion has involved a change of phase frcm liquid to vapor, but the reverse change is equally important. In the main condenser, a low temperature and high quality steam is converted back to a liquid with a resultant decrease in specific volume. One pound mass of saturated stoam at ICO*F occupies 350 cubic feet, while the same Ibn of saturated liquid at 100*F occupies only.016 cubic feet. This drastic decrease in specific volume is the mechanism used to maintain an extremely . low pressure in tae condenser. In the CPSESplant at full power, the condenser pre sure is significantly below atmospheric. t y . ~,,.,, ,.v

E PAGE 23 0F 36 REMRKS. LESSON NOTE (CONTINUATION SHEET - EVEN) 5.3 (Cent.) ~ Typically, the " pressure" (or vacuum) can be expressed either as 3.5 inches Hg, 29.9 - 3.5 = 26.4 inches of vacuum, or 29j = 1.7 psia. This vacuum is extremely important 14.7 x since it represents the pressure condition at the exit of the. Icw pressure turbine. This exit pressure, cccmenly called back pressure, helps determine the driving ap across the turbine, and thus directly, the amount of work which can be done. The heat t; nsfer process frem the steam to the c'irculating water inside the condenser tubes is described by a heat transfer coefficient just as when heat is ficwing cutward. The -initial steam condensing en the tube surface forms a liquid film which acts as a beundary layer. This layer builds up to some thickness which depends on the rate of condensation, ad-hesive forces, and gravity. The layer acts as a heat transfer buffer because it produces a temperature drop and thus raises the temcerature which the vapor sees. The' smoother the tube finish, the thinner the liquid layer will be, and the better heat transfer and condenser capability will be. Another factor which can seriously degrade condenser perfomance is the presence of ncncendensible gasses mixed. in with the vapor. As the vapor-gas mixture comes in contact with the colder tube wall, the vapor condenses to a liquid and forms part of the liquid film. The gas, hcwever, does not condense and since it physically blankets the cold tube, it acts as a barrier for other vapor contacting the tube. Since the thermal conductivity of gasses is very Icw, the presence of the tube is actually masked frem the vapor. This obviously represents a i~ sericus degradation in condenser performance, and justifies e efforcs made to remove non-cendensible gases (air) frca 'C

b pAGE 29 0F 36 LESSON NOTE'(CONTINUATICN SHEET - CDD) REMARKS 5.3 (Cont.) the cendenser. The principle of operation of the vaceum pumps used for this purpose will be covered in a later lesson. 5.4 FOULING A major concern in the operation of heat exchangers where a change in phase is involved, is ten::ed fouline. This refers to the deposition and buildup of mineral deposits on tube walls as the water in which they were dissolved boiled off. Obviously, the majcr location in a PWR where fculing is important is the steam generator. Not only does a layer of deposits greatly reduce heat transfer (remember conductivity through composite walls), but the concentration cf some of these substances also accelerates corrosion of j =C th'e surface on which they deposit. This concern for fculing is responsible for the stringent chemistry requirements for secondary water, and also explains the attentica paid to : ndenser tube integrity since this is where most impurities can enter the secondary water. 6.0 NATURAL CIRCULATION the phenomenon of natural c:nvecticn was described in an early section on convective heat transfer. The motive force rescen-sible for the movement of fluid was the buoyancy of the lower density, heated fluid. A lcwer density fluid naturally tends to rise, while a higher density, colder fluid tends to =cve ~ dcunward to replace it. In an open can'tainer with an electric heater element, convective currents are established with water rising as it is heated, and moving dcwnward as heat is lost to the air and container walls.

C pAGE 30 0F 36 REMARKS LESS03 NOTE (CCitTIllUATIC:1 SHEET - EVEll) 6.0 (Cont.) In the primary icop of a pWR, a phenc=enon called natural ~ circulation is observed which functions in the same manner. The driving force for Tiis circulaticn is the buoyancy of heated water moving through the core. Physcially, the core is located in a low part of the coolant loop, so the water does have a higher elevation to move to. Just as with the naturai ccnvection example, heat is removed in the steam generators at the high elevatica in the Icop. It then returns thrcugh the c'oid leg to the bottom of the core.where the loop is repeated. This mechanism can provide several percent of nomal forced circulation flow in the absence of pump power (an abnormal condition).. The heat source is nomally the decay heat present in a shut-down core, which can be as much as several percent 'g of normal full pov:er. This natural circulation is extremely desirable since it transports decay heat to an available sink rather than allcwing it to accumulate in the core vicinity, where temperatures would increase to unacceptable levels. The process is enhanced if pressure in the primary system is maintained well above saturation to prevent void formation, and if the steam generator secondary water level is maintained above the tube bundle (auxiliary feed is recuired). Up to several percent power, the ficw/ power ratio is greater with natural circulaticn than at full pcwer operation, so the indicated core AT will be less than the full-pcwer value. Also, since heat is being reaoved by steam formation in the steam generators, the indicated S.G. water level will remain relatively constant even with aux. feed ficw present. The absence of natural circulation will be indicated by rising core outlet temperatures and decreasing steam pressure as S.G. water level increases due to continuing auxiliary feed with no heat removal. ,e

6 PAGE'31 0F 36 ( L ~ L a / 0_0_0_9 j NATURAL CONVECTICN CURRENTS CVER A HEATER CCIL FIGUTtE 6.0-1 J '/ -3 + STEAW sen.

e rr_p

+ h im

/y tn, i lll = $v E. .,.7bN p a ad j!:rT 7, 2 i 4 .![ l !""c!! $ 3 1 i e ! >T t, f. y + o o 1A .~%fl.4.5.% 1 l PL ANT ELEVATICHS FOR NATURAL C:RCULATICH FIGURE 5.0-E

PAGE 32 0F 25 i PREHEAT STEAM GENERATOR FLOW SCHEMATIC I 6 Dry and saturated steam A leaves vessel F.,7 St,eam water , ; y, q - mixture enters jj/ sg h secondary separator fg Jl] j if m v

\\ A 'ih-Water level 3'g g '.

rculated - Steam water )s e mixture leaves / tube bundle KL {!l l h l aIhs f .n a ut li,l 'li ii l iti I~l l 'I Il I ij pg Preheater Feedwater inlet Recirculated %7' ~ water enters e 0""20"' boiling section Primary fluid outlet Primary fluid inlet FLCW PATHS FOR FEED S STEAM FIGURE 7.1 -{

J' . w 1 ( pAGE 33 0F 26 LESSC t NOTE '(CONTINUATION SHEET - 000) REMARKS 7.0' PLANT TRANSIENT RESPONSE -Because all: the systems responsibl'e for' producing shaft power in the turbine are. intimately connected together, it is nearly impossible to make a change anywhere in 4 the plant without scme effect being felt everywhere. A major aspect of plant operations is the ability to recognize (i.e. understand) the normal response of the many systems to a single change. The automatic control systems in the PWR are designed to respond to changes and to produce the necessary actions to accomodate that change. A few examples'will clarify the point } 7.1 STEAM GENERATOR SHRINK AND SWELL In discussing heat transfer in the steam generators, it was D menticned that the system was saturated, so that bubbles were rising through the liquid and breaking through the surface into the steam space. An automatic level controller adjusts feed flow (and pressure) to assure a constant level as progr,amed for the operating power level. Realize that steam pressure is designed to decrease as power level and steam ficw rate increase, i In the event of a rapid load increase, steam ficw is suddenly increased and as a result, pressure decreases. Althougn steam flow exceeds feed flow, the steam generator water level does not fall as might be expected, but rather rises. This unex-pected result is.due to a phenomenon called swell and is caused _by the pressure reduction effect on the number and size of the bubbles ri ing through the (now) super-saturated liquid. With 1 pressure.recuced. the num er and size of the bucbles increases, and a greater volume of water is displaced by the butbles. This causes the level of the two-phase mixture (water plus bubbles) I -,e a

'(? PAGE 34 0F 26 - REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 7.1 (Cent.) to rise, although the total mass of water does not change. - Also, since the pressure has just decreased, and correspond-ingly saturation temperature has decreased, bubbles are being formed further down in the riser (heated) section at slightly. . lower temperatures than before the load increase. The total effect of all the additional bubble formation is a-significant volu=e increase in the riser section which exerts a higher backpressure en the feedwater entering that section at the bottcm of the wrapper..Since feedwater flow has not j ' changed (feedwater enters thrcugh the feed inlet nozzle into the pre-heat section)the water level in the downcccer section--between the wrapper and shell, will increase. Steam generator level is measured in this downccmer region, so the indicated level -{ _. will increase. The higher level in the downcemer produces a greater driving head to move more water into the riser section. Feed ficw autccatically increases to match, steam ficw, thus maintaining the higher level. Conditions will stabilize with a higher indicated steam generatcr water level and a higher steam ficw rate. i A e

4 PAGE 35 0F 35 LESSON.N0TE (CONTINUATICN SHEET - 000). REMARKS . 7.1 (Cont.) -Just the opposite sequence takes place for a lead decrease. ~ Steam ficw decreases, steam pressure increases, the bubble-si:e and production decreases', the backpressure on the downcomer decreases and'dcwncemer level decreases. A lower water level is required to produce the head necessary for ficw to the riser matching the reduced steam ficw. i Thus the following transients can be expected to naturally occur due to lead changes: For a power decrease, steam generator water level will decrease (SHRINK). For a pcwer [ increase, water level will increase (SWELL). 7.2 PRIMARY SYSTEM TEMPERATURE CHANGES Wher. a sudden pcwer level rise produces increased steam ficw ki7 and reduced steam pressure, the effect is felt through the steam generator tubes in the primary ccolant also. The reduced saturation temperature produced by the lead increase causes a larger temperature gradient between primary and secondary bulk coolants. A larger AT results in greater heat transfer, so more energy is removed frca the primary water as it passes through the steam generator tubes. The subccoled primary water exhibits this greater energy less as-a temperature reduction. When the water passes through the core at this lower temperature and the same accunt of heat is added tzt the core (ignoring for new the reactivity effects of ccolar water), the primcry coolant ex;eriences a net energy loss. More energy is being removed from the coolant than is-being added to it, and it: temperature will continue to drop. In a very shcr: time, if. nething in crvenes, this ccnditions will be sensed by the reactor protection system as. a potential accident, and protective +-ww- ~,m g e., en<~ g m 3 g

n PAGE 36 0F 36 m ..) REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 7.2 (Cont.) measures initiated. In fact, the core is designed te operate with a negative temperature coefficient, so the cooler primary water represents -a reacitivty addition, which will serve to raise core power. ' This higher pcuer will increase the core AT, but because of the reduced cold leg temperature, T or the primary ecolant AVE will remain at a reduced level. (The reduced T is necessary AVE to provide positive react 0 try to offset the negative feedback frem the core pcwer increase.). So without automatic action, core pcwer level will increase to the new (power / steam) demand with an accompanying drcp in primary coolant temperature; the magnitude of the drop is determined by the magnitude of the coderator temperature coefficient. If this condition is allowed to stabilize, steam pressure and temperature would be well below their design values, and mois-ture carrycver in the steam might well be excessive. For this reason, the automatic red control system will add more excess reactivity to the core by pulling control rods, and thus force primary coolant temperature to increase. Total core power output will remain at whatever the current power (steam) demand requires, and the reactivity insercion will be offset (via the negative mcderator temperature coefficient) by an increase in primary coolant (moderator) TAVE

• The opposite process will occur in respenn to a load decrease.

Control rods will autcmatically be insated into the core to reduce the excessive Tgyg. \\

r / CPSES TRAINING LESSON NOTES 37 1 0F PAGE LESSON: PWR THERMODYNAMICS - FUNDAMENTALS AUTH: C. MEYER-CPSES Operator Training ggy: 1 PROGRAM:

REFERENCES:

Westinghouse Lesson Notes Nuclear Power Systems General Physics Text, " Heat Transfer" NOTES: TEXT *

1.0 INTRODUCTION

TO POWER PLANT FUNCTION Significant attention has already been given to the nuclear fission process and to nuclear core performance. The unique __z (.: feature of nuclear fission which has prompted its development is the large (200 MEV) amount of energy produced by each fission process. Through proper design, nuclear cores are presently capable of producing trillions of watts of thermal power. In this respect, they are not different from oil or coal-fired boilers, or even advanced solar collectors. The processes used to exploit these concentrated energy sources and ultimately to convert that energy to the more useful form of electricity are common to all central power stations. Al-though the following=sectiins will concentrate on the comoonents and processes present in a PWR powar plant, similar equipment i and processes can be identified in any power plant. 1.1 FLOW PATHS A single-line flow diagram for a PWR is shown in Figure 1.1; it consists of two separate, closed loops and one open loop. In the primary loop, water'under high pressure is circulated through the core where heat is added. It then moves through ( 4 --~n-

,.s- .. m. b PAGE 2 0F 37 c - NOTES LESSON NOTE (CONTINUATION SHEET - EVEN) " hot leg" piping to the ' steam generator where heat is removed. A short run of piping from the steam generator to-the pump-Lis called the " crossover". leg. And the return piping from the fpump back to the core is ' called-the " cold leg." The total ~ - mass of waterscontained in'the primary loop is. effectively constant. Transients:or perturbations to the system are accom-modated by surge space in the pressurizer. The secondary loop represents a more complicated flow path. In the steam generator, heat from the primary system water converts secondary water, at.a lower pressure, to steam. This steam flows to and through a turbine where the heat energy is removed and converted to shaft energy.which turns the. generator. The: steam'is cooled back to a liquid (water) in the condenser, and this water ir pumped back to the steam ' generator to repeat the cycle. There are several.more components in1the secondary flow path whose importance will become apparent 1 in future-lessons. It can be assumed that during steady-state operation, the total mass of water and steam in the secondary -loop is also.1early. constant. (h_? The third loop moves. water through the condenser-to remove heat from the spent steam. -The heated water is rejected to some large,1 renewable reservoir. A major objective of-the following sections is to develop an ' understanding and appreciation of the flow processes taking place in both the primary and secondary flow loops. l + l-t (1 b. I T v w ~-

fki, q ',

J v ,m- - - - --. -. mm, r ) STEAM ^'- j-PRESS. p Ft.0W Ji t GEllERATOR d ,', TURalllE H_iF5RQ - STE A!.i (Y., NEbhIbEkhiVN 3 GEllERATOR b r., ,ms, y H ' ! p?~T: R7r i-~~ p .3 Fl.OW N COilDEllSER Y{A}ER / ( CORE -%g h MiP5755ETR g. D I 1 Jh,';,~g PUMP WATER FID W i. p -14123.7 dy. t .4 i.., 1 't PUMP y ...... ~. _... ~...... - nw'i risure i.i SINGLE LlHE PWR FLOW DIAGRAM t S 4

. - ~ i l STEAtt FUEL ROD G EllER ATOR CLADDiljG TUDES fl0ZZLES SifAFT l 1 l l l 1 l l 1 1 l l 1 I i l g --- > S E C 0 tjD A R YWATER/--> T URDINEDL PRit.1ARY FISSoll l 'M W ATER - r l l STE Al.t l C0llDEllSER TUDES V LAKE WATER riuore 1.2 __ SINGLE LINE Els!ERGY FLOW DI AGRAM _ fi? u r-

) PAGE - 0F r LESSCl fl0TE (C0tlTI?!UATICll SHEET - CCD)' REMARKS 1.2 HEAT TRAriSFER PATHS Unlike the ficw paths presented in the previous section, the ' movement of Mat energy is ver/ simply diagrammed. Figure 1.2 illustrates hcw energy frem fissien is transferred from fuel reds into the primarf water and frca primary water across steam generator tubes into secondary water converting that water into steam. The energy present in the steam is c:nverted to the mechanical energy of a rotating shaft in the turbine and scme residual steam energy is dissipated across condenser tubes into a large icw-energy sink. The energy pre:ent in the rotating shaft is used to turn the generator rotor, producing finally electrical energy for further distributien and censum: tion. Tne mechanisms by which the nuclear fissien energy is transferred across intarfaces' and chanced in fem will be discussed in ._._. f W e> cuch greater detail. 1.3 ELECTRICAL CUTPUT The electrical energy produced by the generator represents a much more useful fem of energy than the fission heat produced in the fuel rods. Unfortunately, real-world pewer plants experience complications which do not show up on simple flew . diagrams. The productica of electricity is not a very efficient process, even with the best equipment available today. A-final objective of this section will be an understanding of overall pcwer plant efficiency. 2.0 CEFIllITIO!!S To pr0cced with the discussion of how a pcwer plant functions, it -is first necessarf to establish a vocabularf of basic concepts. 1 \\ w ---

4 pAGE 60F 37 ' REMARKS' LESSON NOTE (CCNTINUATICN SHEET - EVEN) 2.1 THERM 00YNAMICS Therecdynamics is the study of energy and its transfor=ations and of the relationships between the precerties of substances. S;:ecifically not included in themedynamics is the relativistic behavior of matter and the interchangeability of mass and energy. A basic fundamental of this science states that all mattar. possesses scme internal energy. That energy may exist. in different foms, and may be increased frem, or lost to other "cutside" lumps of matter. The cbject of thermodynamics is to identify and acccunt for the various accunts of energy and mass involved in a particular situatien. In order to do ( this, the boundaries of a prcblem must be specified l 2.2 SYSTEM The regien in which mass and energy are to be studied is called a system. It is defined by physical or imaginary bcundaries, and everything cutside the system is referred to as its surrcundings. All exchanges of mass and energy take place at the boundaries of a system. The boundaries are not necessarily rigid, depending en whether the system is allcwed to expand er centract. For example, the system defined as the centents of' an air-filled balleen might change si::e and shape depending en j pressure and temperature variaticns ~in the surrcundings. Two distinct categories of systems can be identifed. A closed system centains a constant mass, and only different fems of energy may cross ics boundaries. A sealed jug of-het er cold water is an example cf a closed system. l lo l s F

• e PAGE 7 0F 37 (2 :)

LESSCN NOTE (CCNTINUATION SHEET - 0D0) REMARKS The second type of system is logically called an ocen system. This.is one in which mass can ficw either into, or cut of, across the boundaries, and in which mass is not necessarily_ ccnstant in time. A kettle of water boiling on a stove is an example of an cpen system; hedt energy is being added to _the water, and steam is leaving the system. A special type of open system which is of particular interest is the steadv flew svstam. In this system, all points remain in a particular conditien, or periedically return to seme par-ticular condition. An cperating automcbile engine radiator is a system which experiences steady ficw; all points remain at constant conditiens. An autemebile engine (internal combustien) represents a steady ficw systam in which conditions return periodically to. their initial values, and mass ficws in both directions across system boundaries. (Gasoline and air flow hg in, exhaust gases ficw cut.) 2.3 PROPERTIES OF A SYSTEM Once a system is defined by specifying its beundaries, the characteristics of that system can be determined. These charac-teristics, cbservable or otherwise measurable, such as mass, temperature, volume, color, electrical charge, etc., are called crecerties. Each preperty has specific units, such as pcunds-mass, degrees, cubic inches, volts, etc. Those preperties which deel with characteristics within the system are desig-nated as internal crecerties, to distinguish them frcm external crecerties of the ent. ire system such as its position or velocity. Jhe stata of a system is its condition at one point in time as described by its properties. ~

pAGE 8 0F 37-N i RE'4ARXS LESSC:1fl0TE (CCITI?tUATIC?( SHEET - EVEll) A crecess occurs whenever a system changes frem one state ~ to another. The gradual ecoling of cof#ee in a thermos jug 't is a prccess because it,results in a changing property of the system, namely, temp vature. In addition to designating a process by the prcperty which changes, prccesses can be characteri:ed.by the fact the certain prcperties de not-change. An isother al crecess is one in which tamperature remains constant. A process is designated isebaric if pressure remains constant, isemetric if volume remains constant, and adiabatic if no heat is transferred. 2.4 CYCLE A cvela is a series of crecesses whica results in a final state of a system which is identical to the initial state of the system before the series of processes was begun. h 2.5 WORKI?tG FLUID Ther cdynamics is concerned with the transformaticn of heat energy into more useful forms or to perfom work. Often it is

necessary to store energy, or to perform several transfers to different systams to achieve these objectivas. A substance l

called a workinc fluid is used to perfom the desired functicns. l In many ccoling systems, water or freen is the working fluid used to transport heat. The substance used to transport, trans-( fem or store energy in a themedynamic system is called a l workinc fluid. 2.6 EilERGY A great deal has already been said about energy and everybcdy probably has a general feeling for wnat it is, but in order to -have a ce=en basis for discussion, energy will be defined here as "the ability to do work where work is the applicatica l l-i i

- 4, pAGE 9 0F 37 s ~~ LESSCN NOTE (CONTINUATICN SHEET - OCD) REMARKS of a force over a distance. In other words, ener.gy can be expended to raise a weight to scme height, or, if the energy is not expended,it still exists as the latent capability to do that work. It is extremely important in the evaluation of thermodynamic systems to be able to identify and quantify the different.for=s of energy present, both within the systems or moving across system boundaries. 2.5.1 Stored Energy latent or stored energy is categorized as. either internal or external' depending on whether it affects internal or external properties. Internal enercy exists as chemical (in the strength of chemical bends forming polyatemic molecules), molecular (in the intens'ity of moiecular vibration), or nuclear (binding energy between nuclecns) energy forms. According -igfly to the basic definitien, these forms of energy represent the capability of exerting a force over a distance. The units of work would logically be expressed as force x distance, with units of lb x ft, ccanonly called "fcot-peunds." By selec-f ting scma arbitrary :ero point which would correspond to zero energy, the tctal stored energy of any system could be expres-sed as some number of "fcot-pounds." A larger unit called the British Thermal Unit (BTU) has proved much more convenient to use, so that the'following relationship is cc=menly applied: 1 BTU = 773.16 foot-pcunds The conversien factor is designated "J"~where j g, 773.16 ft-lbg oiU Mr e "a e

r 1 PAGE 0 op 37 REMARKS LESSON NOTE (CONTINUATICN SHEET - EVEN) External stored enercy exists by virtue of a system's position - ~ in a gravitaticnal field or its motion. These forms are designated, respectively, potential and kinetic energy. Potential enercy exists 'because of sc=e vertical distance (Z) over which gravitaticnal force (g) can be exerted on a systam'smassQe). The prcduct (g x fc) has units of pcunds-force and (Z) is in feet, so potential energy also has units of fcot-pounds (or, using the conversien facter "J", units of BTU. The conversion facter, gc, is used to relate force, mass and acceleratien units, gc = 32.2hwp #seC ). 2 Kinetic enercy exists because a system is mcving relative to a fixed reference point outside the system. A force had to be exertad over scme distance to get the systam moving at its present velocity. The integral expressien for the total energy supplied is: 'S Kinetic energy = f ye where V is system velocity. The units for kinetic energy are also fcc -pounds (or BTU). 2.5.2 FLOW ENERGY Up to this point, the forms of energy have been asscciated with " closed" systems. For the case of "cpen" systems, still another form of energy can be identified which exists because of the fluid mass entering and leaving the systam. The fluid moves because of a certain applied pressure (P) per given volume (V) and so centains an amount ci ficw enercy, Flcw energy = PY (ft-lb ) = h (STU) f Even in ~a ncn-ficw system, the product PxV will not be :ero, and this form of internal energy may be regarded as the energy applied to the system to get it to its present stata. PxV will always be a part of the total stored energy. j E. l l 4

f. PAGE 110F 17 LESS0:1 NOTE (CCriTI?!UATION SHEET - 000) REMARKS 2.7 HEAT The general understanding of what is meant by " heat" has already been discussed under the name of stcrstd molecular energy. In thermodynamics, heat, (designated as Q) is energy moving across system boundaries. Temperature in this centext, is not a measure of heat, but of stored molecular energy, and heat exists because' of differences in stored energy - or differences in temperature. Heat is measured in Stu just as are all other forms of energy. A given quantity of heat w.ll have different effects en different substances. The amcunt of temperature increase any given sub-stance will exhibit for a given heat addition depends..cn a material prcperty called " specific heat" designated "c", and the total mass of substance involved (m). TE Q = cc (T -T ) l 2 ~ where Q = amount of heat (energy) transferred (Stu) m = mass of substance (lbm) c = specific heat (Stu/lba *F) Tj = initial temperature of substance (*F) T2 = final temperature of substance ('F) To further ccmplicate matters, the value of specific heat for a substance is not constant, but can vary depending on conditions. For examsle, it takes more heat to raise the'temcerature of a given mass of aluminc= frcm 600*F to 700*F than to raise it C frcm 100*F to 200 F, and it takes more heat to raise the tem-m. w -.,..<. .S perature of a pcundof water by 1*F than to raise a pound of s of steam by-l'F. 4

PAGE 12 0F 37 REMARKS LESSC;l fl0TE (CC:ITI?iUATIC:1 SiiEET - E'/E?l) Heat is transferred, in respense to a temperature difference, by two basic mechanisms: cenduction and radiation. With conducticn, energy of molecular vibraticn is " passed alcng" tnrcugh seca substance 'frem a high temperature regien to a 1cwer temperature region. Within the conducting substance (or substances in centact) a cradient in temperature exists. With radiation, a bcdy emits heat to its surroundings withcut being in centact with them. In fact, all bedies are radiating' energy and absorbing energy all the time. It is only when a bcdy is at a higher temperature than its surroundings that a net transfer of energy takes place. The driving force for radiative heat is a temperature difference. It might seem strange that convecticn is not included as a mechanism of heat transfer. In fact, convecticn is an extremely -( im;crtant ac!anism for transporting energy, but it explicitaly ..(:; involves a transfer of mass also. The real mechanism in cen-vecticn is conduction into and cut of the moving fluid. Mere will be said abcut the details of heat transfer in a later lessen. 2.8 WORK Thermodynamics was defined partially as the study of energy and its transformations, and energy is the ability to do werk. So the connecticn between energy and work has already been established. Work (W) is energy transferred across the boundary cf a system because of a difference in a prc;erty other than tem:erature between the system and its surroundings. This definitien it very similar to that for heat in that it describes energy in transit. After the work is dene, no more work exists, only energy in a different place or fom. Our L

a-( PAGE 13 CF 37 LESSCN NOTE (CCNTINUATICN' SHEET - CCD) REMARKS ccmmen perceptien of this phencmenen is mechanical work. where energy is transferred by scme applicaticn of force through a distance, frem one systen'sboundary to its sur-rcundings.or to another system. Typical examples are rotating shafts, reciprecating linkages, gear trains, belts and pulleys. A simple water wheel (a steady ficw system) converts the poten-tial-energy of falling water to rotating shaft energy. A rocket engine converts the chemical energy of a prcpellant to heat energy, the heat energy to a pressure (PxV) energy of the expanding ccccustien gases, and finally to kinetic energy by ficw through a no::le. The reacticn to this kinetic energy increases the velocity of the rocket. Work has energy units cf ft-lb or Stu. f Power is cc==cnly interpreted as the a cunt of work which can r ;; be done. The correct definiticn of pcwer is the rate of doing v-work; so it has units of werk per unit time such as Stu/sec or ft-lb /sec. Larger units of pcwer are the horsepewer (HP) and f the kilcwatt (E1) 1 HP = 2544 Stu/hr = 550 ft-lb /sec f 1 KW = 3413 Stu/hr = 738 ft-lb,/sec When a V-8 aute=cbile engine is rated at 3C0 HP, it means that the engine (a system) is capable of converting chemical energy (gasoline) to reciprecating and then rotationa'l =echanical energy and then transmitting it as work via a shaft across ft-lb. the system boundary at a maximum rate of 300 x 550 r n. sGC 200 x 2544 3 nr. V )

• O O

s PAGEi4 0F 37 REuARKS LESSCN NOTE (CCNTINUATICN SHEET - EVEN) 2.9 ENERGY

SUMMARY

The discussicn of energy has covered several different forms of energy. Stored enerb,y can be either internal (U) usually referring to molecular vibration or - *ef d, manifesting itself as potential energy (pE), kinetic e.cergy (XE) or ficw energy (W ). f pE = 9 gZ (ft-lb ) = "gw "'Z (3tu) sc f I XE = V2 (ft-lb ) = }2 (Stu) c f f=pV(ft-lb)=h(3tu) W f Each of these energies represent the total inventary of an entire systam. Generally, ccnditiens are uniform thrcughout a system, so that evaluations of energy can be dcne for a unit mass, such as one ocund mass and when the evaluatiens are ccmplete, the result can simply be multiplied by the tctal number of pounds-cass in the system. One applicatien of this precedure has already been mentioned--remem::er specific heat was(Stu/iba'F). Any other property dealt with en a unit-mass basis is also called " specific," such as specific volume 3 (ft /lbm), specific internal energy (3:u/lba) or specific heat .(Stu/lbm). Specific quantities are generally represented as the Icser case letters of the prcperty, so speciffe internal energy is (u), specific heat is (q), specific volume is y, and so on. Also menticned in the discussicn of energy (E) were the concepts of heat (Q) and work (W). Heat and work are not prcperties of a system and only exist when some precess is taking place. They describe a transfer cf energy into, or out of, a system acrcss its bouncary. The cemen designaticn and signs are as folicws:

(r pAGE 15 0F 37 LESSCN NOTE (CCNTINUATICN SHEET - CDD) REMARX3 -aQ is heat frcm (cut of) system +aQ is heat into system -aW is work into system +aW is work out of system 3.0 THE FIRST LAW OF THERMODYNAMICS 3.1 INTRCCUCTION At this point, it prcbably seems as if little progress has been made in meeting the original cbjectives of the ther=edynamics section - understanding hcw a power plant functicns. Certain aspects of the connecticn between raw fissicn energy and final electrical energy might be gaining subs-tance, but new a major step forward can be taken. The First Law of Thermcdynamics, otherwise kn~cwn as the Law of Conservation of Energy very simply statas that: .5... _, Energy can neither be created nor destroyed, but caly converted frcm one form to another. Now the labericus precess of defining a system in order to do an energy balance and defining the various forms of energy and precasses which can take place starts to make sense. Once all the energy in a system is acccunted for, we don't have to worry about any new energy shcwing up anywhere - or disappearing any-where. The First Law, which, so far, has never been violated, states that this is so. 3.2 THE GENERAL ENERGY EQUATION _The study of any thermodynamic system is new reduced to a simple acccunting prcblem. With all the forms of energy in a systam inventoried ' then what the system can put out (heat or work) or er what can be done to the system (heat or work) by any precess will simply shcw up on cur invencery sheet. The final balance e

PAGg 16 op 37 RE:GRXS LE55C1fl0TE (CCitTIlUATIC?i SHEET - EVE?i) can be expressed very simply: Any change in total energy within a systems boundaries must be equal and opposite to a change in the systems surroundings (outside the boundaries). The change will be due to some " process" taking place, and the exchange of energy across system beundaries is either in the fem of heat or work. This balanca can be expressed mathe-matically for scme process whfen changes a system fr m state 1 to state 2 as: FEj + XEj + U) + W ) + Q -2 = PE2 + KE2+U2 + "f2 + W -2 7 l i This expressi,cn for the First Law is called the ceneral enercy ecuatien, and can be solved for any desired c:=penent if the others are kncun. Reali:e that scme of the c:=penents of this equation are themselves fun:tiens of other variables, so that the expression can be mcdified as follcws: ' ( _. 2 PV h2h + hh I 0 -2 U; 1 + + + = h9h + 2 d "I-2 + + 2 2 (If chemical energy is involved in the system, then those tems c:uld be added to each side of the equation also.] Figure 3.2 THERMCDYN AMIC SYSTEM SHCWING APPLICAT10N CF THE GENERAL ENERGY ECUATICN PACCIS: PATH =- r --- - - 7 P--- 7 l peg ] l PE2 l [ xci I i xcz l 1 vi I

• v:

I A l u: I -(C l ui L._ _ _ _ _ _ _.a L. _ __ ._ J 1 Q w i

PAGE 17 0F 37 1.ESScil NOTE (COlTIllUATION SHEET - CDD) REMARKS For many practical problems in thermodynamics, several of the terms are unchanged by a particular process so can simply be removed frem the expressicn. For a stationary, clcsed system the equation beceres: U) + Q -2 = U2*N1-2 l If the system censists of a sealed container of water which, is being " heated," the expression becc=es: U) + Q -2 " U which l 2 excresses what shculd be obvious. The internal energy of the system after the heating process is equal to the original ~ energy plus the heat added. If the system were a particular volume of air, such as in a diesel engine cylinder and the volute of air were reduced by a pisten moving upward, then S + @J g,,PY22 U

0 c.

4 For an open system cc posed of a pump casing where the pressure of a fluid is increased, but not its density or temperature (internal energy) P)V) P Y 2 l-2=(P)-P)h d d + "l-2 OI W 2 (Since P is greater than P), Wl-2 2 is negative and frem the given sign ccnvention, this represents work dene to a system.) Obviously.then, this general energy equatica is a = cst ver-satile tcol. It can be used to evaluate hcw much work will be done by a certain peccess, how much heat needs to be added to effect a de'stred change, or hcw much ccnditiens inside a system change as a result of sc e process. 'l

PAGE 18 0F 37 REMARKS LESSC1 tiOTE (CC iTINUATIO?I SHEET - EVE?i) 3.3 ENTHALPY Secause the properties U, P, V, of a system often shcw up 'tcgether in the energy equaticn, a new parameter has been defined which ccebines them. Enthalov, designated as H (or h for specific enthalpy) is defined as: H=U+PV(Stu) T Since H is made up of prcperties, it also is a preperty, and will be used in all future expressiens for total stored energy. 3.4 EXAMPLES OF STEADY STATE, STEADY FL 1 SYSTEMS At this point, several of the major ccm;:enents in the FAR will be regarded as systems, and the varicus energy parameters studied so far can be asscciated with them. (Fig. 3.4.) Each of these individual cc=penents can be regarded as a system, cr the entire secondary system in a FWR, of which they all are a par: can be censidered as the system. For the total system, the boundary censists of all surfaces centaining the working fluid, and the energy balance would relate heat additions and icsses to work additiens and losses across system boundaries. To evaluate changes within the system, at least the water part of the system, it is generally assumed that water is effectively ince:pressible. That is, the density dces not change signi-ficantly over a wide range of a:: plied pressures. For this type of fluid, the Law of Cencinuity applies, which states that the product of cross-sectional ficw area and fluid veiccity is ccnstant. Or, frem simply associating units, that volumetric ficw (i) is ccnstant for steady ficw conditions. f*3 2 A(ft)V(j;)=Ccnstant'3;c) i l L1.

PAGE 19 CF 27 e 9 5!*'.! ..e l o c e G = l[ a c. = E % = 1 W J 4' ~ r a. .= a of ~ m.. d* B g W i.; a 5 .3 emf +- -P e 5 a b:, O IJJ g N. s H = M m ? 1 E m 2 5.$. J m w = 6 2 a _.j w m t8* J =- wl -l% - h, wJ ~ s 2* = W C _.U f, El e ca "O m y a t m *+ e c. j .J 3 U _%3 %v-5 S M 3 s U Q u o e e e a w y f E = C* J c -4 f 2 n. 2c' r 4 g 2 s w o e. J. m \\ o 2-at > g k 4 = w. 3 4 l Q f 8 y ): = =w 3 = 1~ ~ ? 5 E U-. ~ o q-f M c W Z g g a 4 "q w 3' .tu H C, =r 6 y-p 9" C at 4 bl 3 m % _a E o O w 7_

SEAL.ED AtJD lj0RI.t AL lilGil EVACUATED V A C U U f.t AT f.10SPilER E PRESSUR E Yr, .Y r Yg mi 1. '( j 1 p j

)

b I b.') }

)

)

r C J L i' 9 \\. 4 i b i (; 50" f 29.92" j I, t LL 4 I 1 (f:.$. 7 y } l i L y f, jd j 3 }' f 4., ..,,I [3 E PRESSURE EFFECTS ON A SIMPLE MERCllRY BAROMETER EI 8 o Figure 3.6 ] ~

([ PAGE 21 0F 37 LESSON NOTE (CClTI?!UATION SHEET - 000) RE!4 ARKS This simple expression allcws evaluation of the change in ficw velocity when pipe diameter changes throughout the system. For example, if water is ficwing through a 8-in diameter pipe at 10 ft/sec, what will the velocity be where the pipe necks dcwn to 6-in diameter? What is the vclumetric ficw rate (i)? Using the law of continuity, A)V) = A Y22 {$8)n 2 f' (6)fn2xV x = 2, e 2 7x10N*Y2 = 17.8 ft 8 s 6 sec ,.3 - Volumetric ficw yjc)Y '" k5H f t)'

• 3 g,.sy=3.49ft go 3

sec ( The same relationships can be used to solve ice required pipe .-f diameter changes, and using the fluid density, for mass ficw rate, m (lbm/sec). 3 A(,1b) = N ft ) x o(1,bym) sec sec 3. ,g Still another expression is available for evaluating steady-ficw systems. 3.5 .BERNCULLI'S EQUATICN For the special case where a ficw system experiences no tempera-ture change (change in stored internal energy), nor any heat or work effects, then the General Energy Equation reduces to the folicwing form: ~ P.E. + X.E. + Vf = Constant h9 A ~+ +f=Ccnstant(Stu) g es 4

~ PAGE 22 07 37 REMARKS t.ESSON tiOTE (CCNTINUATICli SHEET - EVEN) .This expressicn is kncwn as Bernoulli's equation for incem-pressible fluids, and states that in the absence of energy ~ p crossing-the flow systam boundaries, the total energy in the fluid d es not change; only the fons of the energy may shift. An example using this equation is as folicws. Water is ficwing in a 12" diameter horizcntal pipe at 15 ft/sec and 25 psia. The pipa makes a dcwnward turn for 16 feet, then levels cut again and increases in dicceter to 15". If the final mass flow rate is 40015m, what is the final pressure? 3 see (use o = 62.4 lbm/ft ), The problem requires seme relaticnship between flew rate, elevatten (PE) and p'ressure (ficw work, PV) so is a natural application of Bernoulli's equatien. Scme conversien of para-meters is necessary, and a " specific" calculatien will be dene. 3 , 400 t Final velocity, V2, x 2 2 3 2=5.22he V ~ '.016 ft3 Specific volume, y = 1 = 1 ft 3 = 0 62.4,1ba. Iba 2 P;V)hq1;# 1 2,py + _g-Z Ib I _y1_ 22 gc 2 + 7 GC sc gc Since pr ure is given in psia (lb.) it will be multiplied sn by 144 g: to get units censistent with the other energy tems. in <+3 25(Theg:) x 144 (pi)2 x.016(g)+g(lb)x16(ft)+ 32 g_ 2I lbf lx15 (sec-) sec = 2 32 ;cmit jf,+0+7(5.22)2f*2 2 x.016'f*3 1 sec lbf P x 144 2 sec 32 lemft 57.5 #tIh# + 16 #tIh# + 3.5 ft1bf, ~ Ica Icm lbm [ -h 2.2P 425 76.7=2.3P(.yb:') 33.5 = P (P 3f;)

m_ f PAGE 9, OF 17 LESSON NOTE (CONTINUATICN SHEET - CDD) REMARKS 3.5 ICEAL. GAS SEHAVICR With the law cf centinuity and Bernculli's equation, most basic problems concerning liquid behavior can be solved. This leaves a definite void in the treatment of steam and air which are gases. The deficiency will be eliminated by the use of ideal cas relationships for these very non-ideal gases. Sefere describing ideal gas behavior, it is necessary to say a few words about absolute temperature and pressure. Our c = men temperature scales are really relative measurements, based en the freezing and boiling temceratures of water under nor=al at=cspheric pressure. The use of a temperature as a property of a particular working fluid is a relative measure of the internal energy of molecular motion. In exploring the behavior of gases under ' ideal condi'4cns, early experimenters found that _[, _ the. temperature dependence fit an absolute scale. That means, MG# at a tem:erature of absoluta zero, all molecular motion ceases, and most of the world as we knew it operates well up en the absolute temperature ~ scale. The folicwing table shews tne . absolute temperature scales correscending to cur cc:=cn scales. Freezing Boiling Scale /conditicn Absolute Zero Point Point Farenheit ('F) -460 32 212 Rankine. ( 'R) 0 492

672, Centigrade ( 'C)

-273 0 ~ 100 . Kelvin ( 'K) 0 273 373 Note that the absolute scales use the same si:e increments as the corresponding relative scales: one Farenheit degree =

ne Rankine degree; one Kelvin degree = cne Centigrade degree.

' As mentioned before, ideal gas behavior is always expressed as a function of absolute temperature, f a

pact 24op37 REMARKS LE550:1 fl0TE (CC!iTIliUATI0il SHEET - EVE!!) Similarly, when cbserving ideal gas behavice, the early experi-menters found the pressure dependence also required an casolute _ scale. Most pressures in mechanical applications are measur'ed relative to cur nemal' atmospheric pressure. Hcwever, at=cs-pheric pressure (due to scme 50 miles of at=osphere held dcwn by gravitatienal forces) varies with elevation, and with weather c:nditions,so an absolute scale would see'n very con-venient. The bottem end of this absolute pressure scale wculd correspond to a perfect vacuum, which can be approached in l outer space or under centrolled laboratory conditions. In a perfect vacuum, a unit volu=e would c:ntain no gas molecules. As the density of gas molecules increases, the effect of mole-cular vibratien is sensed as a pressure force on the walls of j any container, until at nomal at=cspheric conditions, a 2 l pressure of 14.7 lbf/in is exerted. Just as with the temcera- '+ l ture scales, this "ncmal" pressure is used as the zero point for a pressure scale called'"cace." Pressures relative to at : spheric are =easured in units of csic. Pressures belcw at= spheric are referred to as vacuums, and scmetimes are exp-ressed as psiv. Cc==enly, pressures are expressed as inches of mercury (in. Hg), since the mercury baremeter was one of the original pressure measuring devices. It censists of a icng U-tube c:ntaining mercury; cne end of the tube is evacuated l and sealed, the other is left cpen to the at csphere. Since atmospheric pressure acts en ene side of the mercury, and no pressure at all acts en the other, the mercury column is forced j upward en the evacuated side. In fact, at=cspheric pressure will su: port a column of mercury 29.92 inches high! Realize that if the open side of the tube were exposed to a perfect i vacuum, both legs of mercury would be the same height. (Fig. 3.5) t i

s E" pAGE ?s0F 27 LESSCN NOTE.(CCNTINUATICN SHEET - CCD) REMARKS .Thus any pressure expressed in inc..es of mercury is an ~ absolute measurement, since it corresponds to a measurement relative to a perfect vacuum. Scmetimes a vacuum condition (pressure less than atmospheric) is expressed as " inches of vacuum" which simply means a certain number of inch'es of mercury belcw 29.92 in. INCHES OF GAGE ABSCLUTE MERCU:.Y INCREASING. PRESSURE. + + + t NORMAL ATMOSPHERE O PSIG 14.7 PSIA 29.92 in Hg + VACUUM 4 PERFECT VACUUM -14.7 PSIG 0 PSIA 0 in Hg Examola: Shcw that 29.92 inches of mercury correspends to 14.7 psia. 'sj; ~ Solu *ien: Use a tube containing mercury with an inside area of one square inch, and filled.to a height of 29.92 inches. 3 Thus the total volume of mercury is 29.c2 in, so the Octal 3 12xik"x12in3*E8)* 1 ** weignt of mercury is 29.92 in x ft " = 14.7 lbm. The force exerted by this " weight" is just the mass x gravi-tational acceleration. F=mfxg 14.7 lbm x 32 ft/sec2 = 14.7 ibf Itm gc oz<;,z737 Since this force is acting on the ene square inch of surface 7 area, it can be expressed as 14.7 lbf/in or 14.7 psia. Examole: Express the pressure 51.8 in Hg in other units. Solution: Since 29.92 in Hg corresponds to 14.7 psia, a simple ratio will give the answer. P(psia) = 14.7 osia x 51.3 in Hg = 25.45 psigo' 29.92 in ng Or, since 14.7 psia = 0 psig, this new pressure correspencs to 26.45 _14.7 = 10.75 psig.

PAGE 25 0F 37 ) RE. MARKS LESSCN tt0TE (CCitTIllUATI0tt SHEET - EVEM) Exaanle: Express the pressure 5 psia in other units. Soluticn: Since this is less than 14.7 psia, it ccrrespends to a vacuum. Again,using a ratio: P(in Hg) = 29.92 in Ha x 5 psia = 10.18 in Hg 14.7 psia This cculd also be expressed as 14.7 psia -5 psia = 9.7 psiv, or 29.92 in Hg -10.18 in Hg = 19.74 inches of mercury vacuum (or inches of vacuum). ficw that absolute temperature and pressure have been discussed, the general behavior of ideal gases can be expressed very simply as: 11=PY PV 22 .'l .'2 'Where: schscrt;t I refers to same initial state and subscript 2 Y s.m refers to a final state after scme precess has occurred, and ~~ pressure and temperature are in absolute units. This expressicn makes sense when analy:ed qualitatively. If a fixed vclume of air is heated, pressure will increase:T2*I,P2

• f1' Y2*Y*

I l If ~a certain quantity of air is heated and allcwed to expand so 2

• Y, and T2*I.

The equation reduces 2 " ?, then V that P I I l to simoler forms for certain types of precesses. Isothermal 71=T2 PV1i=PY22 P P 1 2 Isemetric V3=V2 V1 V., Iscbaric P3=P2 p 1 2 Although real gases behave scmewnat differently than this " ideal" mcdel, these ecuations will be sufficient fer the accuracy required by this 'essen's cbjectives. ~

r~

Pt.GE 27 07 37 LISSCN NOTE (CCNTI!!UATICN SHEET - 000) REMARKS 3 Examle: The steam hubble in a pressuri:er occupies 780 ft 0 under saturation conditions (2235 psig, 652 F). A slew 3 insurgo reduces the volume to 500 ft withcut'signi ficantly changing the temperature.' Ignoring condensation or sub-ccoling effects, what is the new pressure? Solution: Since T2=T,PV1y=PY' y 22 Pressures must be absolute, so 2235 psig g 2250 psia. P2*PY11 V 2 3 P, = 2250 psia x 780 ft E0 rt-P2 = 3510 sta = 3495 psig 4.0 HEAT EtiGItiES 4.1 INTRCCUCTION With the previcus infer =a:icn abcut energy and its different forms, and the equivalence of energy, heat, and werk, it seems like a logical next step to figure cut scme way of converting the internal energy of a working fluid into useful work. A heat engine is a device which does precisely that; it is a closed or steady ficw system which converts heat to work. It always cperates in a' cycle, so that it can produce werk continuously. This means (remember the definition of cycle) that conditiens in _ the system', i.e. enthalpy, return periodically to an initial value. So for the entire cycle, within the system, AU = 0 Energy in = Energy cut G

PAGE 28 CF 37 RE.'tARXS LESSCN :10TE (CC:lTI!WATIC t SHEET - EVEN) Obvicusly, heat can be added to cur heat engine frca different sources and in different ways, and work can be removed in different foms, but all heat engines use a cycle with three fundamental / stages. 1. Heat is added at a high temperature, (a temperature gradient is required to transfer energy as heat). 2. Scme of the heat added is cenverted to work at an inter-mediate tamperature (depending on engine type). 3. The remaining heat is rejected frca the engine to a lcwer tamperature sink. (This final step is necessary to return the ccnditions within the system to their initial values.) Cbvicusly, the more heat which is ccnverted to work, the less that needs to be rejectad to return the systam to its initial conditiens. This aspect of heat engine perfemance is ccvered in the next section. 4.2 THER:GL EFFICIE ICY By definition, the thermal efficiency (n) of any ther cdynamic cycle is the ratio of werk cut;:ut to heat input. 4

• g " Oin - Ceut W

,Cr the Cycle i gg Several factors centribute to making this ratio less than 1.0. Probable the most obvious is friction which converts some of the werk output back into heat which is generally lost frca the system. Another cc::=on-inefficiency is simply heat loss by conduction or radiation. The same temperature / energy gradient which drives the heat engine, also causes heat to be lost to the surrcundings. The exotic insulatici, and attention to lubricants used in large pcwe plants represents a direct effort L e

,{ [ PAGE .m CF,, LESSC1 NOTE (CClTIt:UATIO:t SHEET - 000) REMARY.S to reduce losses and increase thermal efficiency. Perhaps a greater appreciation for thermal efficiency can be prcmoted by expressing it as a profit ratio. If a person cwned-a heat engine and was able.to sell al1 the work it could put cut, and had te buy.all the heat to drive the engine, then the return en his investment'in heat would bec- $ profit = 5 received frca work output - 5 spent for heat input. .If both sides are divided by 5 spent for heat input, S crofit, S received -1=nx -1

1. iC8 Cf * "k 5 spent a spent ces: cr neat For the hypothetical case of fixed ccsts and prices, the way to maximize the profit ratio is to increase thermal efficiency.

Obvicusly then, it is desirable to increase efficiency. There is nothing in the First Law of Ther:cdynamics which restricts full conversion of heat to work. So what is the difficulty? 4.3 Reversibility and the Second Law of Themedynamics All around us we cbserve precesses which only go in one direction: water ficws dcwnhill, heat ficws frca a high temperature body to a icwer temperature body, mountains erede, time passes. The r9varsa of these precasses is never observed - they are irreversible. ,There is an analascus concept in thermodynamics which restricts hcw efficiently a heat engine can perforn. l Think for a minute cbout the different foms of energy which can exist. The First Law treats all forms as equal and interchangeable -as icng as the total a: cunt is censerved. Yet our experience cen- _sistently demonstrates that scme forms of energy transform to others care easily then the reverse process. A speeding bullet strikes a wail and all of its kinetic energy is transfomed into heat. But the. heat cannot be col'ected to send the bullet speeding

PAGE 30 0F37 REMARKS-LESSC1 10TE (CC;1TI!iUATICl SHEET - EVE?l) off again. A mountain climber accidentally drops his sunglasses and they fall to the bott:m of the mcuntain. Their potential energy was quickly conve,rted to kinetic energy. When they hit the grcund, the kinetic energy was converted to heat, but the same heat cannot be collected to move the sunglasses back up the mountain. A machipist grinds a new edge en his tcol bit, the rotaticnal energy in the grinding wheel is converted to frictional heat in the tool bit and then transferred to a quenching fluid. The heat cannot be collectsd to rotate the grinding wheel again. It bec:mes cbvicus' that heat is the ultimate ehergy. All other forms end up as heat and it is very difficult to c:nvert heat back to other fems. Usually the final fem of heat is very diffuse "lew grade" energy. The Second Law of Thermodynamics - {7-n. rec:gni:es this directienality of energy transfomations. The fact that heat ficws frem high temperature to icwer temperature bcdies, and that heat engines can never be perfectly efficient are applications of the Sec nd Law. To better explain the Secend - Law, scme new definitions are helpful. 4.4 AVAILABLE E?tERGY Several examples have been given to illustrate the general trend of energy to bec:me heat. Consideratien of similar examples t should make us reali:e that scme foms of energy are better able to be converted to work than other forms. Also, once a system is in equilibrium with its surroordings, even though it still possesses definite foms 'of snergy, it is incapable of doing work. This is expressed by stating that scme forms of energy are unavailable for doing work. And also that available enercy is that pcrtion of energy which could be c nverted to work by ideal precesses which reduce the system to a state in equilibrit.m

31 F 37 PAGE 0 LESSON NOTE (CONTINUATION SHEET - CDD) REMARKS with its environment. Obviously, available energy is the only form that shows up as work in the equation for efficiency. The unavailable energy is a necessary loss to the pcwer cycle to return conditions to their initial values. 4.5 ENTROPY New that the reality of unavailable energy has been presented, along with the c ncepts of " directional" energy transfer, we need se e "=easure" of energy availability which will tell us hcw well a given process will transfer energy. We knew that heat ficws frem a.high tem:erature scurce to a lcw temperature sink, but temperature alene dces not give any indication of hcw much energy will be unavailable for heat transfer. From the Second Law, we knew that any process which results in sc=e conversion of energy to heat will not be reversible because energy -In the form {33E' of heat is not c:=pletely recoverable to other fer=s. It seems that seme c:=oination of heat transfer and temperature will pro-vide the desired parameter for measuring energy availability (and-unavailability). The new quantity is called entreov(s) and is definel in tares of a change in entropy which takes place during a process, aa a Btu a

• jr-332 g )

U is = change in specific entropy (Stu/lbm R) ao u heat transferred (Stu/lbm) T = absolute temperature ( R = F + 460) 'According to the definition. if heat is added to a substance 1at s0:e temperature, then the entropy change is positive and the molecular activity of the substance will increase. If heat is removed daring a process, the entrepy change will be negative and molecular activity will decrease. By extrapolating the 4 4

PAGE 32 Of 37-REMARKS. LESSC:1 fl0TE (CC;ITINUATIC?f SHEET - EVE?l) relationship between entropy and molecular activity back to absolute zero temperature, whert there is no mole'cular activity, there is no entropy. So entropy is a measure of change in molecular activity, which ccrrespends to the amount of heat added or removed from a substance during a process; and since heat is the energy for s respcnsible for irreversibility, entropy 4 is also a measure of unavailable energy. This property called entrcpy turns out to be a very useful waj of evaluating precesses when other methods fail. 'If heat is being conducted to a pot of water, the amcunt of heat transferred is given by: q= meat Su: when the water turns to steam, the value of "c" changes drastically, plus there is no way of acccunting for the latent energy asscciated with the phase change (liquid to vaper).

i Also, if the system is doing scme work at the sue time heat is being added, then the aT is not a measure of the heat transfer.

In both cases, the entrecy chance of the syste'n is a true measure l of the amount of heat transferred. In the heat engine discussed previously, heat was added to the working fluid at a high temperature and rejected at a 1cw tempet sture. The working fluid will probably experience scme i variatien in entrepy while work is being produced by the engine. This change in entropy is a measure of hcw much heat energy is not converted into work by the v.gtne. The greater the change in' entropy over a cycle, the less work is produced frem heat - remember entropy measures unavailable energy. -Now we have ! delightfully cenfusing situation where a prcperty of the sys' tem measures what it can't do. Perhaps a few applications will clarify "entrcpy."

e PAGE 33 0F 37 LE550:1 t40TE (C0!iTItNATI0tl SHEIT - CCD) REMARKS Imagine two bodies A and B at temceratures T and T, respec-A B tively. If T is greatar than T, heat will be transferred g g frcm A to B because of the temperature gradient between them. Sedy A will lose energy in the form of heat and so will experi-Sody B will receive ence a decrease in entrcpy ASS = heat energy and thus experience an increase in entropy AS, = n 4a0 ]. ficw if both bcdies are lumped tcgether in a system, the total change in er.trcpy will be: AST0TAL " ASA * #33* A w Since T is greater than i, g g $S- > dE. 'S 'A and the AS g7.g for the heat transfer prccess is cesitive. } Reali:e new that no heat citered cr left the. system, but a property of the system changed. Another interesting thing has hacpened: the heat energy which initially existed at temperature T. new exist: at T, which is a 1cwer temperature. Heat will B n net flew back up a temperature gradient, so the heat transfer precess dcwn a temperature gradient is definitely irreversible and is characteri:ed by an increase in system entrcpy. It app' ears obvicus in this case that entrepy measures irreversibility. If a heat engine were added to the system, which cculd ccnver heat energy te work, then the amount of heat reaching bcdy 3 would be significantly reduced. Scme of the heat energy frcm body A would leave'the system as work. But a heat engine must -reject scme heat, se aS will increase by scme amcunt. The 3 increase in entr'py of the system is a measure of the heat e transferred,'and thus a =easure of irreversibility. It is ccnvenient te regard the icwer temperature, T, as scme atmes-- g pheric, or recm, temperature which is not suitable for doing ff'

PAGE 3d CF 37 REMARKS LE55C l 10TE (CC ITI!iUATIC( SHEET - EVE?l) work by driving a heat engine. The processes described above can be plotted en temperature and entrepy axes. Frem the definition of entropy, ac a5 = p ag = TAS The accunt of heat transferred, aq is the area underneath the precass path T x aS. The heat engine precesses are also defined by lines, and the work done is the area between the lines. Figure 4.5 ENTROPY EFFECT3 ouRING A HEAT TRANSFER Proc 33 SYSTEM Al.205 I ~ ~ Esr7, ~ ~ 9 I 1 %4 -.,---, r. I I r.>r, I l t____ _ _ __ _a la8 l. lase l [ _ / r:n fut cast matne I (/4=...l.. I.i.,l 1 ,y = , = =. t =~j r, - 777 7 i.7,. ...f, / /H A 434 433 l';; l < V!.=l

...I<I,..l

. $ I SYSTtW D C

pAcg 35 o-37 LESSCN' NOTE (CCNTINUATION SHEET - 000). REMARKS 4.6 THE CARNOT CYCLE Back in 1824 when thermcdynamics was a new science, a Frenchman named Carnot was develeping these same concepts of irrever-sibility and cycles and heat. engines, and he invented a cycle which censisted of the folicwing steps: 1. Heat is added at a constant temperature. 2. Work is done at censtant entrepy. 3. Residual heat is removed at a censtant (1cwer) temperature. 4. Work is dene en'the systam (ccmpression er pumping) at constant,entrepy (since no heat was added) to return cendi-tiens to their initial values. This cycle can be plotted on temperature-entrepy exes to get a ce=menly used T-s diagram. Y. a TEMPERATURE-ENTRCPY DIAGRAM FCR A ' CtRNOT CYCLE t 'k T Ti o b W.--=.o " w~Jr d C 72 gan Y ~ s, s: s-- Figure 4.5

EC PAGE 16 0F 17 r REM. ARKS LESSCM NOTE (CCNTINUATICN SHEET - EVEN) On this diagram, the heat addition process is frem a + b, the work process is frcm b - c, the heat rejection precass frem c - d, and the work addition process frem d - a. No work is dene while heat is being transferred, and no heat is being transferred while work is being dene. So the cycle is not very realistic, but is cartainly represents 'a desirable situation frem a thermal efficiency standpoint. The entire area within, the curve represents available energy. U

• heac accea " Oin - Cout work cut Qin I

7 08 ~ I #8 I2~I l l*I~F y, ~2 l T as T 2 2 2 For this unique Carnot cycle, thermal efficiency depends only on the source and sink temperatures. In fact, this cycle has ' YJ the maximum thermal efficiency of any cycle cperating between T and T. Observe the processes during which entrepy does g 2 not change (as = 0). Energy is being added and removed frem the system in scoe form other than heat, and as a result tie tam;erature changes, but entropy does not. Remember thct entrcpy requires a change in heat, aq-, in crder to be ncn-zero. Although this Carnet cycle represents a maximum in terms of - thermal efficiency between two temperatures, it is not alway s achievable cr even desirable in real-world applicaticns. L2t's examine what a real-world situation icoks like. O 9 4 -V

Figure 4.6a TEMPERATURE-ENTROPY DIAGRAMS ( FOR THE CARNOT CYCLE o = b a =- g d c 7 a e 7 y LN <c> S S G =D C D T d c T c c e M <c, S S e %3

( CPSES TRAINING LESSON NOTES PAGE 10F 42 LEss0N: PWR THERMODYNAMICS - APPLIED 3973. C. MEYER/D.HUBBAF PROGRA>1: CPSES OPERATOR TRAINING REV: 1

REFERENCES:

Westinghouse Heat Transfer Notes General Physics Heat Transfer Course Nuclear Power Systems NOTES: TEXT: 1.0 STEAM POWER ENGINES

1.1 INTRODUCTION

+

"75 Now that the basic laws of thermodynamics have been presented, ^~' along with the definitions of the system properties enthalpy ? and entropy, the workings of a basic steam power cycle can be analyzed. Such things as changes in system properties, work output, and thermal efficiency will be the most important applications of past discussions. Since we will be using water as the working fluid, and its vapor phase, steam, it is convenient at this time to study the properties of water before putting it to work. 1.2 PROPERTIES OF WATER' ~ ,Like many other fluids, H 0 can exist as a solid, liquid or vapordependingonthepr$ssureandtemperatureofitsenvironment. Different phases can co-exist at several different conditions (ice water, wet steam) and all three phases can co-exist at a single set of conditions called the triole point. The different regimes for the phases car be represented grapnically on a pressure-temperature diagram. s

• a W-s

PAGE ? 0F 37 REMARKS LESSCM NOTE (CCNTINUATION SHEET - E'/EN) -- - - --- - - - - - - - - - -- -- - ------- 7 o'[E' E sace.2 e.i. l UCul0 i ~ n.T,,,. - = - l j sous PRES $UNC (P) e VAPCM / i I 31*F 212*F 7 0 3.34*F TEnsPERATUnt (71 1.2 1 PRES 3uRE-TEMPERATURE OfAGRAu FoR WATER Notice that at a given temperature, there is only one pres,sure' at which.a phase change will take place. Above the critical C point (32C5.2 psia, 705.3 F), the characteristics of ifquid and vapcr beccme identical so that separate phases cannot be dis-tinguished. At the triple point, all three phases can exist. together in equilibrium. Sublimation, the direct transitten frcm solid to vapor, can onTy eccur belcw the triple point. 5b Since we are not concerned with ice, further discussion will concentrate on the liquid and vapor regions. The transition of phase is called boiling, and can occur at a wide range of temperature-pressure conditions. For a given pressure, the l temperature at which boiling cccurs is called saturation l temoerature for that pressure. At a given temperature, the pressure lat which boiling occure is called the saturation pressure for that temperature. A liquid which is saturated is at the boiling temperature and the next addition of heat j will cause vapori:stion. The vapor prcduced will be saturated t- - vacer, that is, at the boiling temperature. A saturated system is one in which liquid and vapor both are present at I saturation conditions. l l e l ~) l t

PAGE 3 0F 42 LESSCN NOTE (CCNTINUATION SHEET --0DD) EUdARKS Liquid which' is belcw the-boiling temperature is called subecoled licuid, and vapor which is above'the boiling temperature is called superheated vacer, at a given pressure. These three states lof cur working fluid, below, at, or above, the boiling temperature are cc= enly presented graphically as a function of pressure and corresponding specific volume as shewn belcw. Y ~ f.-e:m. j e u-..-. I u m.. / The different characteristic: of constant temperature lines (isotherms) are shcwn on the diagram. Nctice that in a satu-rated system, pressure and temperature are constant. The change in specific volume at the constant pressure and tem- . perature is due to the constant addition of heat energy. The heat required to change liquid into vapor at a ccnstant tam-perature is called the latent heat of, vacerization. In a saturated system, the tamcerature cannot increase above the boiling point until all liquid is convertad to vapor because -of this latent heat requirement. 4 /

a i PAGE 4 0F 42 l ,) REMARKS LESSCN NOTE (CCNTINUATION SHEET - EVEN) The relative amount of liquid and vapor in a saturated steam mixture is designated by a factor called quality (x) where x = mass of vaoor expressed in %. i mass or mixture If the quality of a " wet" vapor is 80%, then for one ibm of mixture,.8 lba is saturated vapor and.2 lba (1-x) is satu-( rated liquid, prcbably in the fora of tiny dreplets. 1.2.1 Steam Tables Because the steam-pcwer engine which is the object of this section uses a working fluid which could be in any part of this p-v diagram, it is necessary to have tables which give the l themedynamic properties of tne fluid over the entire possible range. Because all of the calculatiens used to evaluate ther-medynamic systems are based on changes in properties rather than absolute values, it is not necessary that the tables be absc-lutely correct. It is sufficient that the properties are given relative to scme censistent reference point. i One characteristic of these property tables is the use of the ' subscripts "f" to designate ifq fd or fluid, and "g" to desig-l nate vapcr or gas. Therefore v is the specific volume of a f fluid, s is the specific entrepy of a gas, and h is the g gg .enthalpy associat:H with the phase change frca liquid to vapor. For the two-phase mixture, the specific properties can be deter- -mined using the definition of cuality, i-e e p i 4

PAGE.5 0F 42 LESSC1 tiOTE (CC?iTItiUATI0:1 SHEET - 000) REMARKS h, (for seme quality x) = Xh + (1-X)h g f x (for scme quaH ty d = Xv + (1-X)v v g f x (for seme quality x) =,Xs + (1-X)s s g f If the parameters with subscript "fg" are available hg = hf + Xhfg f

• IVig v
• V x

' f + Xsfg s =s x For subcooled liquids and superheated vapors, both tamperature and pressure are necessary to define a statapoint. For satu-ration conditiens, pressure and temperature are not independent (cne can change only if the other changes) so only one is neces-sary to define a statepoint. Generally tables of saturated ( values are given for each parameter separately. The follcwing are typical examples of information contained in steam tables: TEMPEPATURE TABLE (CRY SATURATED STEAM) ABS TEiP PRESS SP. VOLUME EITHALPY E!!TROPY O F-osia LIO. EVAP VAPOR LIO. EVAP VAPOR L10. EVAP VAPOR T P V "V V h h h s s s f fg g f fg g f fg g 32 .088 .016 3306 3306 0.0 1076 1076 0 2.19 2.19 100 .949 .016 350. 350. 68.0 1037 1105 .13 1.86 1.98 200 11.53 .016 33.6 33.6 168 978 1146 .29 1.48 1.77 _212 14.7 .017 26.8 26.8 180 970 1150 .31 1.44 1.75 300 67.0- .017 6.4 6.5 270 910 11S0 44 1.20 1.54 i

i PAGE 6 0F 42 REMARXS LESSC 1 NOTE (C0tiTINUATION SHEET -- EVEN) PRESSURE TAELE ~ (DRY SATURATED STEAM) ABS INTERNAL PRESS TEMP SP. VOLUME . ENTriALPY ENTROPY ENERGY O psia F LIO. VAP LIO. EVAP VAP LIO. EVAP VAP LIO VAP P-T V V h h h sf s s u u f g f fg g fg g f g 1.0 101.7 .016 344 69.7 1035 1106.13 1.85 1.98 69.7 1044 14.7 212.0 .017 25.8 180 970 1150.31 1.44 1.75 180 1077 100. 327.8 .017-4.4 298 899 1187.47 1.13 1.60 298 1105 200 382 .018 2.3 355 843 1198.54 1.00 1.54 355 1114 1CCO 545 .022 .45 542 659 1191.74 .65 1.39 538 1109 SUPERHEATED STENi ABS TEMP. O 5 PRESS F s psia-U U 0 0 (sattemp) 200 F 500 F 800 F 1100 F 1 v 452 572 750 929 (101.7)h 1196 1288 1433 1585 s 2.11 2.22 2.35 2.46 s 14.7 v 30 39 51 63 (212) h 1193 1237 1432 1585 s 1.82 1.93 2.06' ~2.17 100 y 5.6 7.4 9.3 (327.8) h 1279 1429 1583

1. 7.1 1.84 1.95 2U0 v

2.7 3.7. 4.6 (381.8)h 1259 1425 1581 s 1.62 1.77 1.88 300 y 1.8 2.4 3.1 (417) h 1258 1420 1573 s 1.57 1.72 1.83 y -r

l.

PAGE 7 0F o LESSCN NOTE (CCNTINUATION SHEET - ODD) REMRKS E:tamole: Hcw much heat must be added to 1 lbm of water at saturation temperature at 200 psia to reach 80% quality? Solution: From the pressure table at 200 psia. = 1198 (Stu/lba). hf = 355, hfg = 843,-hg For 80% quality ( 8 lbm steam,.2 lbm water) Btu h (80%) = 355 +.8 (843) = 1029.4 lem or h (80%) =.8 (1198) +.2 (355) = 1029.4 {t[. Since the liquid is saturated, h = 355 B/lbm so 1029 - 355 = 674 Stu of ener9y must be added. Examole: One iba of water at 200 psia is boiled completely to steam, and the steam is superheated 120 degrees F. Hcw s much heat energy was required for the superheat? What would the density of the superheated steam be? Mcw much 0 water at 212 F, 14.7 psia could the superheated steam c,.}.. s.J convert to steam? Solution: Frca the superheat table, saturation temperature U U for 200 psia is 382 F; 120 F superheat raises the temperature U to 502 F '(assume 500 F) where h = 1258 Stu/lbm. Frcm the pressure table, h for saturated steam is 1198 Stu/lbm, thus heat added is ah = 1258 - 1198 = 60 Stu/lbm. 3 Also frem the steam tables, V = 2.7 ft /lba, which is just the inverse of the desired density. Density = 1 = 1 3 .37 lbm/ft, = V-4./ The maximum amount of new steam whicti can be raised by the seperheated steam would be realized if this stealn were bubbled -through the saturated liquid. In this way, the superheated steam would be condensed to liquid at 14.7 psia, 212 F. .*W

PAGE 8 0F 42 REMARXS LESSON NOTE (CONTINUATION SHEET - EVEN) U h (sat liquid at 212 F,14.7 psia) = 180 btu /)bm U fg (212 F, 14.7 psia) = 970 Btu /lbm h Total energy available in superheated steam = 1233 Stu/lbm - 180 B/lbm = 1078 Btu /lbm Amount of steam raised = 1078 = 1.11 lbm new steam 910 lcm of superneat steam. Obviously, for extensive use of the tables, much more detail is available, and interpolation between data colu=s is scmetimes Because of the manipulations involved in using the necessary. tables, a graphical form was devised by Mollier which creatly simplifies the analysis of many ther cdynamic pr: cesses. Its ~ c: ordinates are enthalpy and entropy, and lines of constant temperature, pressure, and moisture (superheat) are pictted. It c: vers a wide range of superheat and moistures dcwn to saturated liquid. ,..] / d SUPERHEAT 9 REGION / / / /= v CONSTANT e gS4TO; PAY -h TEMPERATtJRE v CO E ? ky,O A Y f Apog Q h Olsy q. y [+N a WET .y 6 REGION k4 l' $ o*a h' 1.2-3 MC L!ER DIAGRAM

.- 3 PAGE 9 0F 42 LESSCN NOTE (CCNTINUATION 'Sh'EET - 000) REMARKS Obvicusly, on.the chart, any prccess in which a property remains constant can be folicwed very easily. Constant pressure processes move along the isebars. Isothermal processes folicw the constant temperature lines. Isentrepid processes move vertically on the chart. The Mollier diagram allcws direct evaluation of those processes in which many properties change at one time. 1.2.3 The T-s Diagram Still one mere diagram is in ec=cn usage for presenting ther-medynamic processes. The ulti= ate application of these processes is in a pcwer cycle, whcse thermal-efficiency depends on entropy (available and unavailable energy) and temperature. A basic T-s diagram was shewn in the presentation on entropy, and a similar Q property _ diagram.fer cur working fluid is shewn belcw. CRITICAL POINT \\ j / SUPERHEAT D COMPRESSED REGICN w LICUID \\ C LINES y [ WET REGION \\ 4 c:: d SATURATED 2 I W VAPCR -/ Y / LINE SATURATED CONS AN._ i i L!CUlO PRESSURE L I N ~=. LINES ') ENTRCPY (s) 1.2.a TEMPERATURE-ENTROPY CIAGR AM FCR STEAM er e.. g ,r-

PAGE 10 CF 42 REMARKS _LESSCil NOTE (CCNTI!iUATIO!! SHEET - EVEN) Building upon previous discussions of entropy and the pcwer cycle,- this diagram will be used to evaluate the steam pcwer cycle. - 1.3-THE RANKINE CYCLE The Rankine Cycle is actually another name for the steam-pcuer cycle and is the themedynamic cycle used in a steam-pcwer plant. Co not lose sight of the fact that the ultimate cbjec-tive of' the pcwer plant is to prcduca electrical energy frcm themal energy. The intarmdeiate step is conversien of themal energy to mechanical energy, or werk, which turns the generator shaft. _The more efficienciv'we can conver: heat energy to work, the less the final product will cost. The more heat energy which recains available to do work, the less heat will have to be rejecte'd. So it beccmes cbvious that a study of the entreov behavice of the cycle will give a very goed indication of its R. t peric mance. The basic ficwpath of cur working fluid along with foms of energy which cross the system boundary is shewn belcw. Fcr this study, the system will be defined by all surfaces centaining the working fluid. STEAM - 2 T UR 8 th E w 0UT 'g STC1W CENEA ATC A 3 CONOCNSCM w ' CUT FEE watC8t t-

• !M

!.3 1 S A SI FLCw7ATM FCR THC 4 ANUNC CT LC wm

PAGE 11 0F o LESS0tt it0TE (C0ttTI!!UATI0tt SHEET - 000) REMAP,xS The precesses which take place are: 1. Heat is added to-the working fluid at a high temperature in the steam generator. 8,ecause the system is satnrated, the temperature and pressure of the water / steam mixture are not independent. The steam temperature will be saturation temperature at system pressure. Since a significant amount of heat is added to the working fluid, in the form of latent heat of vapori::stion, its entrcpy is increased. Steam leaving the steam generator has 99.75% quality. 2. Work is removed frca the steam in the turbine. The exact mechanism of ccnversion of internal energy to kinetic energy, and then to mechanical energy will be discussed later. The staam increases in moisture as pressure and temperature decrease and specific volume increases. Ideally, no heat is transferred in this process, so entropy does not change. 3. Heat is rejected frcm the cycle in the condenser. This process-of removing the latent heat from the steam cenverts it back to liquid, with a resulting large decrease in specific volume. This is responsible for the vacuum back-pressure (pressure at the exit side), on the turbine. It also implies i large decrease in specific entrepy. 4. The feedwater/ condensate at a very Icw pressure must be returned to the steam generator which is at a much higher pressure. The required energy to do this is added in the form of mechanical work in a pump. Ideally no heat is added by the pump, so no entrcpy change takes place, but the internal ~(PV) energy of the fluid increases, so the resulting tem-perature also increases slightly. .w

m PAGE 12 CF 42 REMARKS LESSC?l ?l0TE (CC?tTI!(UATICtl SHEET - EVE?i) At this point, the cycle is ccmplete. The working fluid has been returned to its initial conditions and is ready for heat to be ~ added again. - A cer=cn questien at this' point is '"Why not take the exhaust steam frem the. turbine and return it directly to the heat scurce for reheating?" The difficulty with this preposal is that the ' fluid has to be returned to th,e steam generator at a high pressure. This means that a great deal of pump work is required, both because of the large volume of steam to be moved, plus the ccm-pression which will subecci the liquid phase. In fact., much less work is required to pressurize a liquid so that the candenser/ pump ccmbination represents a significant savings in work which must be done to, ccmplete the cycle and move the working fluid. This simplified and ideali:ed Rankine cycle can be plotted on the T-s diagram for water / steam as shcwn belcw. 1. Heat is added at ccnstant pressure to a subccoled liquid. Once the liquid ten:perature increases to saturation, the quality increases acrcss the " mixture" region to the saturated vapor line (effectively) where it leaves the steam generatcr at point 2. Entropy has increased significantly. JL T -2 d l 4 3 = C I. *2 RasesiNC gyg*,g CM A TCidPCRATUAC *CMTRCPY CIAGRAnd y-3

PAGE p 0F o LESSC:1 NOTE (CCiTIllUATIC:1 SHEET - CDD) REMARKS 2. Steam enters the turbine and internal energy is converted to work. The expansion process is rapid and no heat is transferred, so temperature decreases at constant entropy to point 3. 3. Heat is reccved in the condenser, converting the icw pressure / temperature steam back to a liquid. Because the system is saturated, pressure and temperature remain c:nstant as heat is removed and entropy decreases. The final state is saturated liquid at point 4. 4. The pressure of the liquid is increased back to steam gene- . rat:r presture without heat being added. Entropy remains c:nstant but fluid temperature increases because of the increased internal energy. The feedwater ccmpletes the cycle back to point 1. Obvicusly, the heat energy rejected frem the system is not available w to do work. Just as in previous discussions, this 1ount of heat is numerically equal to T as. The 1:emainder of heat and work c energy added to the system is then totally available to do work. This is represented by the total area inside the corve describing the cycle. T

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//~///

) s ---== I 3*I ' Av&4LASLC CNCRGY IN THC ICCAL mehKINC CTCLC

( PAGE 14 0F 42 REMARKS LESSCit NOTE (CCtiTItlUATIClI SHEET - EVEM) Just looking it the diagram, the thermal efficiency corresponds to the relative si:e of the two areas: AVAILAELE ENERGY thermal efficiency = AVAILAELE + U W/AILABLE ENERGY Anything whi'ch can' be done to increase the amcunt of available energy and/or decrease the amount of unavailable energy will increase the thermal efficiency of the cycle. So increasing steam tamperature will help to increase efficiency; hcwever, the correspending increase in pressure in the steam piping represents an offsetting cost facter. The maximum steam temperature is reached by heat transfer in the steam generater frcm primary fluid at a slightly higher temperature. Because the primary fluid must remain subccoled, this maximum tamperature determines the pressure which must be maintained, and thus the strength of the primary system bcundaries. Again a cost facter .is involved. Decreasing the condensate temperature also will improve efficiency by increasing available energy and_ decreasing unavailable energy. This temperature is slightly above the temperature of the final heat sink, which is usually fixed, or in scme cases seasonal. For a given engineering ccmprcmise on values of T373;;4 and TCOMOENSATE, seme additional improvements can be made in effi-ciency by using the areas to the right a'nd lef t sides of the l [ present ideal cycle, that is, the superheat and subccoled regicas of the diagram, h i l I NCw TsTCiu< TsTCiu......... _..... l N t' NN h s\\ 3 TC. D ,,e,r.,//1ZZZZd.LZZZ/i c i' 3 ' 1.b - %CT !%CPC ASC ;M AvatL ABLC CSCMGT ST W A11:, G STCJW TEWPCR AILAC, . CACPPima CCmCCh1 ATE TChPC A ATORC

/I PAGE 150F 42 LESSCN NOTE (CCNTINUATION SHEET - CDD) REMARKS J Notice that the quality of the steam leaving the turbine is well within the 2-phase region. This means that it contains a significant amcunt of moisture, scme of which exists in the feria of dreplets. ' The small,'but dense dreplets of water can and do causa, considerable damage (pitting, erosion) to the turbine blading in the last stages near the exhaust. So there is another gced reason for mcVing the right hand side of the cycle diagram cver closer to the saturated vapor line. This is done using ccmponents called moisture secarater reheaters (MSR's) which take steam after it has passed through several stages of expansion (high pressure. turbine) and add more heat to actually superheat it befcre it re-enters the turbine (1cw pressure stages). The effect of the MSR's en the T-s diagram is shewn belcw. d / / / / / / / T s' o' / p' / / / / Ts. s' e' /, r ACCITIONAL AValLASLE ENERGY / FROM REHEAT [ -'C -- t t 3 I 3 8 I l 1 t S ) 1.3-5 THE EF ECT CF REHEAT CN AN IDEAL RANKINE CYCLE

a \\ PAGE 16 0F 42 REMARK 5 LESSCN NOTE (CCNTINUATICN SHEET EVEN) ~ Not only is a significant amount of available energy supplied by the MSR, the quality of the steam in the turbine remains higher threughout the expansion process. Frcm the diagram, and also in reality, the improvement in steam quality alone justifies the cost of the MSR's. The heating source for the MSR's is live steam which bypassed the high-pressure turbine. No new heat is added to the cycle, hcwever the steam ficw rate through the HP turbine is scmewhat reduced. Another big difference between the Rar. kine and Carnot cycles shcws up on the subccoled liquid side of the cycle diagram. A I'.s........_.............../ r.. 3 M AMKIN C .g_.gggggy s --puw, wenz r.,. !.U."L.".2*.T I c Y 1 l lt s-. C!FFERENCES,8,ETWEEN CARNOT ANO R ANKtNC 1.3-4 The Rankine cycle applies pump work to a saturated liquid and subccols it while the Carnot cycle pumps,(cr ccmpresses) its fluid directly back to a saturated liquid at steam temperature. Thus the Rankine cycle must add a significant amount of heat at icwer tro;:erature than T37g;g; a lock at the relative available/ unavailable energies of the two cycles shcws that this will decrease relative ther.al efficiency. Obviously, the higher the tom;:erature at which heat is added to the cycle, the greater s b

,] PAGE 17 0F 42 LESSON tt0TE (C0!iTIrlUATIO;l SHEET - 000) REMARKS will be the amount of available energy, and thus thermal efficiency. Using the definition of thermal efficiency, work out ut heat accec total wort , total worx v neat reJectec I = 1,(heat re7ectec) total werx as the ratio of total work to heat rejected increases, so does themal efficiency. For a fixed amount of heat rejected, (ie fixed T as) the higher the temperature at which heat is added, c the more work can be done. Ts._______________________ TC----- ---s------- Sg S2 31 S2 1.3 7 INCREASED AVAILABLE E?iERGY FROM AD0 LNG HEAT AT A HIGHER TEMPER ATURE

PAGE 18 0F 42 REMARKS LESSCN NOTE (CC:ITI!iUATICM SHEET - EVEN) ~ Generally, this process is called regenerative heating or feed-water heating and makes use of steam extracted f' rem the turbine to heat the feedwater. Since the heat ccmes frem within the cycle, it does not. affect the formula for efficiency directly. Mcwever, since the feedwater new enters the steam generators at a higher temperature, heat will be added at a higher tempera-ture, and thermal efficiency will im;, rove. On a T-s diagram, this is shewn as a small frac *fon, y, of the steam ficw leaving the turbine and giving up its heat to the feedwater. The fraction, y, is selected so that the temperature rise to satu-ration is split equally between the feedwater heaters and the main heat scurce (steam generators). Ccmparing this cycle with the unmedified Rankine cycle. (1-y) lba of steam rejects heat frca the' cycle, rather than 1 lbm, so thermal efficiency is improved. l C x-- =- ; / su .. -..,, m.... -. The different heat additicns can be quantified by ccmparing the specific enthalpies of the fluid through the cycle, assuming 1 ihm of fluid enters the S.G. at point 1 and' leaves at point 2. C N, d [ I / im

m...,..t3m..or n.

L

'} PAGE 19 CF 42 LESSC:1 NOTE (CC:iTI;UATIC:( SHEET - CCD) REMARKS aq; = heat added in S.G. = h2-h1 5 - h )(1-y) ac = heat rejected to condenser = (h 4 r 2 - h ) + (1-y)(h3-h) AW, g = total-work done = (h 3 4 g - h )(1-Y) y - h ) + (h AWin = w rk into system = (b 5 7 The diagram shows the y lbm of regenerative steam mixing back with the (1-y) lbm of feedwater at point 7. In actual practice the path is mere ccmplicated, and the mass is usually reccmbined at some ficw point further back, closer to the condenser. Mcwever, the heat-ficw paths are similar to what has been shown. Thus an ideal Rankine cycle with one stage of reheat and cne stage of regenerative heating (feedwater heating) lcoks as fc11cws en a T-s diagram. A T v 939 - S.G. WOUT -HIGH PRESSURE TURSINE \\> _ _ _ _ _ _ _ - - _ N ST E A M EXTRACTIO IDEbNE - FEEDWAT ER HEATING ?/ 4 9 0UT - CONCENSER S IDEAL RANKINI CYCLE WITH REHEAT AND FEEDWATER HEATING ) .. -10 1 i i

~ PAGE 20 0F 42 REMARKS LE550;I NOTE'(CCNTI?iUATIC:( SHEET - EVE?i) 1.4 INDIVICUAL COMPONENTS ~ New that the overall, ideal, Rankine cycle has. been presented, scme 'of the physical components of the real FWR will be examined individually. 1.4.1 Steam Generator The steam generator is extremely efficient at transferring heat frcm primary water to secondary water. The vertical u-tubes which centain the higher pressure primary water are contained in a cylinder called a tube wrapper; this wrapcer g6fdes the secondary water along and across the heated tubes. The water level inside the wrapper is well 'above the tube bundle, so that boiling takes place at the steam-water interface, and also en tube surfaces within the bulk coolant. So ner= ally, bubbles are rising through the liquid and breaking thrcush the surface. This is the mechanism which lifts a great deal of moisture into the steam, which in turn necessitates the mechanical moisture separaters _lucated in the tcp portion of the steam generator shell. Because no ficw work is-done in the S.G., all the energy added to the working fluid shcws up as internal energy (or 6 enthalpy since =o). The general energy equati' n for the S.G. a as a constant ficw systen reduces to the simple ferm: l q =hout - hin ~ i q = heat added to the system per lbm of steam l-l hin = specific enthalpy of feedwater l .hcut = specific enthalpy of steam leaving S.G. (requires kncwledge of quality) l' l L; l L ~

^ r PAGE 210F 42 TEXT:(Continuation sheet-odd) NOTES: (* Examole: At full power, feedwater flow into each S.G. is 0 3.785 x 10 lbm/hr at 440 and 975 psia. Steam leaving has 99.75". quality. How much heat is added per lbm? How much total heat is transferred per S.G.? Solution: Using the steam tables h(sat _ liq at 975 psia) = 538.7 BTV/lbm _h(sat. vapor at 975 psia) - 1193.8 BTU /lbm h = (.0025 x 538.7) + (.9975 x 1193.8) = 1192.2 BTU /lbm out hin (subcooled water 440, 975 psia) = 419 BTU /lbm q=h -h = 1192.2 - 419 = 773.2 BTU /lbm out in Q = qin = 773.2 BTU 3.785 x 106 lbm = 2927 x 106 BTU lbm hr hr Note that 2927 x 106 BTU - 1 hrKw =.858 x 106 Kw - 858 MW i hr 3412 BTU and the plant has 4 S.G. 858 MW x 4 = 3432 MW which is the approximate thermal output for one unit of CPSES. 1.4.2 Turbine

.g The Turbine is the machine which actually converts the heat energy which was added to the working fluid into mechanical work.

Ideally, no neat is transferred in this conversion process, so aq is zero and subsecuently as is zero. The expansion process in an ideal turbine is an isentropic (constant entropy) process. The internal energy of the steam (obvious from its high tempera-ture and pressure)- is first converted to kinetic energy by nozzles inside the turbine. The steam, now moving at high speed, is directed against blades attached firmly to a moveable shaft. By design, the energy of the mcVing steam exerts a force on the bladiag, causing the shaft to rotate. To minimize the forces on any single set of blades, this steam energy is shared between many sets of blades. The driving force, which continues to move steam through the consecutive sets of nozzles and blading, is the pressure difference between steam inside the turbine and the large vacuum space inside the condenser at the turbine exhaust.

9 s PAGE 22 0F 42 iJ REMARKS LESSON NOTE (CCNTI?iUATION SHEET - EVEM) The general energy equation reduces to N =hin - hout cut W = specific work,done by steam out hin = enthalpy of entering steam hout " enthalpy of exhaust steam Because the energy loss through the high pressure turbine is substantial, the exhaust steam at that stage is very wet, abcut 14 percent moisture, or equivalently, 85 percent quality. In the moisture separater reheaters (MSR's) seme 10 percent of the total steam ficw is removed as liquid; the remainder is super-C heated e,cre than 100 F. This "new" steam is expanded thrcugh two. parallel lcw pressure turbines and then exhausted to the cen-denser. Obviously, evaluation of turbine work is corplicated by this staging and reheat, not to mention the several differen: points where extraction steam is bled off. And in reaI turbines, there will be scme heat losses just because of the physical size of the machine and scme fricticn losses in the moving parts. This - implies that not all of the energy loss as calculated by enthalpy decrease will be converted to work. There will be some mechanical efficiency for the turbine, designated n, which su=ari:es g is always }ess than 1. effectiveness in energy conversion. n t j (hin - hout) real real work ~ "t " (n n me reucal won in out4 ene retical 4 The difference between the ideal and real turbine processes are easily shewn on a Mollier diagram: t =

==. / A I ... ?.?= -{ I 1.4 1 tCC AL VS. AC AL ttJRSINC C17tN3tCh1 CM & WC(t:ER c:Assau

C PAGE 23 0F d2 LESSCN NOTE (CCNTINUATION SHEET - 000)' RE. MARKS Both cases exhaust at the same pressure. The ccmplete turbine process can also be presented on a Mollier diagram as follows. b 2/5 h,. - m-I- I,' e S r,t' n / c n / ], / /- ./ tape %

• • ~~~.

/ . -q - y el 5,- Y EYamp ,~ 's, Igy. -ee r S 2.4.g C0uPLETE TURSINC PRCCES3 Of4 A WOL'lER 01AGRAW A Ccnditions at turbine inlet A-B Slightly non-isentropic expansion B Exhaust HP turbine B-C MSR adds superheat at ccnst. pressure C Enter LP turbine C-D-E Non-isentropic expansions in stages E Lcw pressure, high moisture exhaust to ccndenser. G 5 l 1

PACE 24 0F 42 NOTES: TEXT:(Continuation sheet even) 'Examole: A main turbine uses saturated steam at 975 psia h and exhausts-to the condenser at 1.7 psia. Turbine efficiency is 85".. Calculate the amount of work done per lbm of working fluid entering the condenser. Solution: The ideal process involves an-isentropic expansion between the given pressures. Using the constant entropy value at 1 psia allows evaluation of-quality, and from this, enthalpy. ~ Actual exhaust _enthalpy is obtained using the definition of efficiency. From the steam tables Sf(975' psia) g 1.3940 Following the constant entropy line down to 1 psia reveals a constant moisture percent of g30% or a steam quality of 70%. h (1.7 psia) =.3 x hf +.7 x h = (.3 x 80) + (.7 x 1108.5) g = 800 ideal work = hin - hout g (715 psia) - 800 =h = 1194 - 800 = 394 Btu /lbm Actual. work =.85 x 394 = 334.9 Btu /lbm Actual h = 1194 - 334.9 = 859 Btu /lbm out (Compare the work done by one ibm of steam (334.9 Btu) with the heat added in the S.G. of the previous example-(773 Btu). Obviously, a significant amount of energy remains in the exhaust steam. f S D . r

.e. PACE 250F 42 ' TEXT:(Continuation sheet-odd) NOTES: 1.4.3. Condenser Like the steam generator, the condenser is an extremely efficient heat transfer. device. By removing energy from the steam and condensing it back 'to a liquid, it significantly reduces the specific volume of the working fluid and thus produces an extremely low pressure (1-2 psi-a is typical). Its behavior can be expressed by an equation timilar to that for a S.G. -q = hin - hout q = heat removed per lbm of fluid h = specific enthalpy of exhaust steam in h = specific enthalpy of condensate out Examole: From the exhaust steam of the previous example, condensate is produced at a temperature of 120 F and 1.7 psia. How much heat is removed by the condenser per lbm of fluid? Solution: h = actual turbine h = 859 Btu /lba in out h = 80 Stu/lbm out q'= 859 - 80 = 779 Btu /lbm 5: (since only 52 percent of the S.G. flow actually passes through ID* the condenser, total heat loss is 3.785 x 106 x 4 S.G. x Btu .52 x 779 x = 1.797 x 106 kw = 17 f$, or 34 2 u about half of the plants rated thermal output.) 1.4.4 Pumps In the early discussion on cycles, it was pointed out that in-order to complecea cycle, it was necessary to return the working. fluid to its initial conditions. Since this discussion started out with feedwater entering the steam generators, that is the condition which must be achieved. It was also pointed out earlier that.it is much easier to raise the pressure of a liquid than that of a gas. So the final component of the cycle, the pump, has the function of adding sufficient work to the fluid

PACE 260F 42 NOTES: TEXT:(Continuation sheet even) ~ to raise 'its pressure to S.G. pressure, or 975 psia. -Again in ( the ideal case no heat is transferred in the process, so it is an isentropic compression. But again friction and turbulence losses result in some loss in efficiency, so more work must be done by the pump motor than is actually added to the fluid. w rk actually done on fluid Pump efficiency n work supplied to pump = p and pump work, or flow work is given by: pump work, Wp = for incompressible flow. Examole: Water from the condenser hotwell at 1 psia and 90 F must be raised to S.G. pressure of 975' psia. How much work is added to the fluid? If pump efficiency is 52",, what horsepower motor must be provided? Solution: From the steam tables, specific volume of the condensate, y %.0161 ft /lbm and will be assumed constant. 3 Pressures are already in absolute units. ft3 I" Wp " -VAP .0161 lD* (975-1) x 144 J in2 ft2 ft Ibf 788 Btu Wp = -2.87 To calculate total work, it will be assumed that effectively all the feedwater requires the same work. Total feedwater flow = 15.1 x 106 1bm/hr. Wp = -2.87 -x 15.1 x 106 -43.4 x 10s B u r HP B x 103 =- hr 2543 Btu /hr s

17,028 HP total work added to fluid If6 Required pump'HP

= 32746 HP. = (This _ total required pumping power is divided among several stages of pumping and parallel pumps at each stage.) L e s

m 9 PACE 27 0r.42 TEXT:(Con'tinuacion sheet-odd) NOTES: T/. + 1.4.5 Feedwater Heaters The initial discussion of the Rankine cycle pointed out the desirability of using energy from within the cycle to raise the feedwater temperature. This allows external heat to be added to the cycle at a consistently higher ~ temperature. From the examples, it is obvious that a significant amount of heating is required. The heating is done in six stages with the lowest energy feedwater being in the sixth stage, near the condenser. Steam is extracted from'various stages of the LP turbine to provide heat for the 6th, 5th and 4th stage feed heaters. As the feed temperature gets higher, higher energy steam is required so extraction is takcn from the HP turbine exhaust and various extraction points in the HP ~ turbine to supply the 3rd, 2nd and 1st stage feedheaters. Drains from the moisture separator reheater as well as the 'd5 condensed extraction steam from the feed heaters are also collected and used to preheat lower energy feed flow (remember, heat always has to flow downhill, hotter to colder). The net result of this seemingly complicated _ process is to reduce the amount of heat rejected in the condenser (since this water and steam never gets there) and produce feed water with U temperatures within 100 of the S.G. saturation tempera-ture. While it is true that this steam is lost ~ to do work in the turbine, the net effect is to raise the curve closer to the ideal and increase overall efficiency of the cycle. To examine the very large amount of energy represented by -this process, we'll examine the flows through one of the i feed heaters. e 9 A

PACE 2 0F R NOTES: , TEXT:(Ccntinuation' shoot even) (' Examole: Feed heater 5 receives condensate at a temperature of 150 F and a flow rate of 9.8 x 106 lbm/hr. 6.21 x 10s lbm/hr extraction steam is supplied from the LP turbine at 20 psia and 4% moisture. Condensed extraction' steam from heater 4 is also drained into-heater 5 to add its heat to the feedwater. The drains consist of 5.76 x 10s 1b'm/hr water at 226 F. What will be the final temperature and entaalpy of the feedwater when it leaves the No. 5 feed heater? Solution: Using Steam Tables Initial feed temp = 150 F Initial h = 118 Btu /lbm .04 (h ) +.96 (h ) =.04(196) +.96(1156) Extraction steam h = f g y = 1117.6 BTU /lbm Drains h2 (by interpolation) = 194 BTU /lbm gs added by steam = h - h = 1117.6 - 118 = 999.6 BTU /lbm g y gd added by drains = h - h = 194 - 118 = 76 BTU /lbm 2 Q total added =,qsms + qdhd = 999.6 6:21x105fD + r 76 5.76 x 105 = 6.65 x 108 r r To determine the final feedwater enthalpy, find the specific s gtotal 6j'i x lo BTU /hr = 67.8 BTV/lbm heat added - m reed 9.8 x,10 lbm/hr =h add this.to the initial enthalpy - hinitial + hadded final 118 + 67.8 = 185.8 BTU /lbm 0 Assume 1 BTU will raise 1 lbm water 1 F in temperature. 0 67.8 BTU /lbm = 67.8 F increase 150 + 67.8 = 217.8 F So, the final temperature of the feed is 217.8 F and the final enthalpy is 185.8 BTU /lbm. In actual plant operation approximately 2.46 x 106 lbm/hr of steam is extracted from the HP turbine and 1.63 x 106 lbm/hr is extracted from the LP turbine for the purpose of feed heating. (

p..

= ge

e -e PAGF. 290F 42 TEXT:(Continuation sheet-odd) NOTES: ( 1.4.6 -Moisture Separator Reheaters -The purpose of the MSRs is to add energy to low pressure steam and decrease its moisture conent so that more work may be extracted frem 'it. The device physically performs two functions: the moisture separation is done mechanically, similar to the process used in the steam generators. The heatjng functicn uses steam directly from the steam generators which is condensed to a saturated liquid in the process of adding its heat to the low pressure steam. The MSR is located in the flowpath between the HP turbine and the LP turbine. The HP turbine exhaust is treated as just described and then enters the LP turbine as dry, superheated steam. Examole: The exhaust steam enters the MSR at a mass flow rate of 11.3 x 106 lbm/hr, at a pressure of 160 psia and a moisture content o f 14*.'. The steam leaves the moisture separator portion with a moisture content of IL What mass -of moisture is removed from the steam ficw? 11.3 x 106 lbm/hr x w-~ (.14 .01) = 1.469 x 106 lbm/hr Steam remaining = 11.3 x 106 - 1.469 x 106 = 9.83 x 106 lbm/hr What is the specific enthalpy cf the steam leaving the moisture separater? h at 160 psia = 336 BTU /lbm h = 1195 BTU /lbm f g h = (.01 x 336) + (.99 x 1195) = 1186 BTU /lbm out The steam leaving the MSR will be superheated nearly 150 and have an enthalpy of 1279 BTU /lbm. How much reheat steam (975 psia) is supplied tothereheaterioaccomplishthis? Amount of enery needed is = h -h out in = 1279 BTU /lbm - 1186 BTU / ibm - 93 BTU /lbm Q total = inah = 9.83 x 106 lba/hr. (93 BTU /lbm) = 9.14 x 108 BTU m Enthalpy of steam used to reheat = h at 975 psia = 1193 BTU /lbm Assume reheat steam gives up heat and is condensed to a saturated liquid at 975 psia, then energy transferred would be: I-i&

n.- PACE 300F 42' NOTES: TEXT:(Continuation sheet even) The steam leaving the MSR will be superheated nearly 150 and have an enthalpy of 1279 BTU /lbm. How much reheat steam (975 psia) is supplied to the reheater to accomplish this? Amount of energy needed is = h -h out in = 1279 BTU /lbm - 1186 BTU /lbm = 93 BTU /lbm Q ' total = hah = 9.83 x 106 lbm/hr. (93 BTU /lbm) = 9.14 x 108 BTU Enthalpy of steam used to reheat = h at 975 psia = 1193 BTU /lbm Assume reheat steam gives up heat and is condensed to a saturated liquid at 975 psia, then energy transferred would be: h -h = energy transferred 1193 - 538 = 655 BTU /lbm g f hfg = 655 BTU /lbm Since +Q (heat gained) must equal -Q (heat given up) then Id(steambeingheated)ah=m(reheatsteam)hfg 9 there3 ' ' heat steam) = 5B /1 mass. aw of reheat steam = 1.395 x 1.06 lbm/hr h 1.4.7 Summary The components presented in this section are major pieces of equipment in the PWR steam plant. The presentati'on has been scmewhat simplified to emphasize the most thermodynamic processes involve basic concepts. This simplification results in some 1.nconsistencies in the ' total heat balance but the numbers are fairly close to the actual situation which will be found at CPSES. The basic concepts of conservation of mass and energy frequently appear in these problems and are basic to our u,nderstanding of closed system thermodynamics. 2.0 THE PWR POWER PLANT At this point, the basic thermodynamic function of the power cycle has been presented. The primary loop of the PWR represents the main heat source for the modified Rankine cycle. The power output of the core is a function of the fission rate and is governed by reactor physics effects ' studied previously. The ( fission heat is removed from the core by flowing water and serves to raise the internal energy (enthalpy) of the coolant. This increased enthalpy is measured as an increase in coolant temperature

PAGE 310F 42 TE.TI:(Continuation sheet-odd)

0TES

across the core. Core power = Flow rate x Enthalpy rise '(BTU) - ( 1 bn.) * (BTU) (sec) (sec) (lbm) Effectively, all this heat is transferred across 'the steam generator tubes.to the working fluid in the Rankine power cycle. to the working fluid = from fission - directly and in-directly. The inclusion of a time factor in evaluating energy production now makes the equation a power balance. 2.1 THE REAL RANKINE CYCLE The thermodynamic balances done on individual components in section 1.4 are reasonable examples of the evaluations done for real PWR's. The appropriate equations are summarized below: Steam Generator k=b) (h -hfeedwater) 3 steam t " 't s2 (hsteam in - hsteam out) Turbine W Condenser q ' Jns3(hsteam in - hcondensate) Pumps w =5 p s4 In these expressions ' 6 - rate of heat addition to cycle + rate o.f heat rejection from cycle +w - rate of work output from cycle -w - rate of work input to cycle hs1 = mass flow rate of steam out of S.G. hs2 = effective steam flow rate through turbine Es3 = mass flow rate of steam entering condenser s4 = mass flow rate of feedwater through pump (s) m As mentioned earlier, the turbine work and pump work are divide'd into several stages, each with its own mass flow rates, so the analysis only gets more_ complicated. ~ .c .o. 1

PACE 320F 42 NOTES: . TEXT:(Continustion sheet even) The evaluation of overall thermal efficiency is done'using the original expression: net work cerformed real turbine work - real pump work Uth heat added heat added in S.G. Note that the effects of steam reheating and regenerative feedwater . heating do not enter into.the expression. The following specific values are taken from section 1.4. real turbine work = 334.9 BTU /lbm r'ns2 g.78 x 15.1 x 106 lbm/hr = 11.8 x 106 lbm/hr ' Total turbine work = 3.95 x 109 BTU /hr Real pump work = 2.87 xh=4.4 I ins 4=15.1x106[(assumed) total pump work = 6.64 x 107 BTU /hr B heat added = 2927 x 10s B x 4 = 11.7 x 109 h 3.95 x 109 -.066 x 109 Uth

• 11.7 x 109 nth =.332 g

The thermal efficiency of the real Rankine cycle presented earlier is about 33 percent. Figure 2.1-1 shows a more detailed mass distribution sequence than the one analyzed above, including several other applicati6ns of steam flow. Notice that some of the steam is used to drive boiler feed (feedwater) pumps. This usage reduces turbine output, but also reduces external pump work inputs!, Figure 2.1-2 shows the actual CPSES heat balance which exactly details the various flow paths throug' the cycle. It shows the feed going through the six stages of feedheating as well as the points where drains and extraction steam are added back into the cycle. Careful study of a heat balance such as this reveals the many interrelations which exist in a complicated real cycle.

il Y y {0 V C0KDEit3 ATE a 35.5% onAlH5 s O A 16.5% "ExusCII0: 10% stoist. n $1[ Alt ((ED li1R. [I]3 ACilCM 64.5% l O n 31[ AH CEsti'.A10R$ l j 11015f.164.51 63% l L.r.wn.l52;5% Coxornu a 1 it.r. ,g 1001 I luitD. I 1 5tr.lggigai p.l l l l i i 1 3 9% R[tt[AI Slit. k M [11RAtliON Ak 1.5% soitts.. siran v-rito ititrS Tito :lla 10. 5% 4 L~ -U i m a om ~ DIAGRAM OF MASS FLOW TilROUGli STEAM PJ 2'1'1 AND FEEDWATER SYSTEMS ZPS 7 'il 2/27/75 g a 4

PAGE 34 0F 42 1 ,C. -. .....'\\ ~' ,= = "; r......_.....- T......... ( ".'.: *.';'.."J1 r.... [ ','.T.* '.'.' .~ /h%.....'...~ f._. g =,.,7 ...,.n. / m......,....... / O.,... .. 2 4 N";,~;,',,* **.- ~ e,.r.c., ietw.n Rt.NKINE CYCLE 'b 2-3 ,e.,,e,st:w a e.ew e = c.:ar c a' sic,".. a wt sys;tw := a nat:suai::s c~. i ww

.} pAGE 35 0F 42 LE5501 fl0TE (CC:ITIllVATICit SHEET - 000) RE?GRKS Figure 2.1.3 shews the actual Rankine cycle as it would shcw up en a T-s diagram for water. 2.2 THE PRESSURIZER Although the pressuri:er is not associated with the secondary side of the plant at all, it is similar in that it represents a saturated systam; as such it eerits a short discussion of its equilibrium state. The pressuri:er is a fairly large vessel (1800 ft") which acts as a surge tank for volumetric changes in the primary coolant. The total available volume in the primary flew system is fixed, that is, the beundaries are censidered as rigid. Because water is effectively ince pressible, sc e surge volume is necessary to accc=cdate changes in the volume of primary water caused by [. temocrature changes. The steam space in the pressurizer provides this surge volume via a " surge line" connected to cne of the hot' leg pipes. The pressure existing in the steam space of the pressurizer exists throughout the entire ccolant system, although the liquid tecperature in the pressuri:er is considerably higher than T in the ccolant. The change in pressure caused by an AVG insurge or outsurge of water ebvicusly depends on the si:e of the surge space available. Conditions in the pressurizer can be regarded as a Ecn-steady ficw system and can be expressed using the general energy equatien. For the insurge/cutsurge conditions, changes in PE and KE can be disregarded, and no mechanical work is done so W is zero. Likewise, no heat is added so AQ is :ero. (The effect of heaters and spray is con-idered.later.), s I .I

PAGE 3s: OF 49 f REMARKS LESSCtl NOTE (CCitTIllUATICil SHEET - EVE?l)

• A"o o
• E"f f + m u ]after amtj+[muf f + m u ]before h

u h g gg am = mass of watcr entering (i) or leaving (o) pressurizer m,m = masses of liquid and vapor in pressurizer f g = internal energy of liquid and vapor >f, ug h = enthalpy of water entering or leaving Considering an insurge first AU ht 9 + [m uf f + m u ] initial

• E"f f + m u ] final u

gg gg ^"i + (*f + "g initial

• Im * "g) final l

f l l The equations merely state that no mass or energy disappears l in the insurge.. The final state of the system will probably have a higner pressure because of the insurge, so that scme of the initial steam will have cendensed back to liquid. Also, since the expanding coolant which added to the liquid invent:ry 'h was cooler than the saturation temperature existing in the pressuri:er, the total liquid temperature decreases slightly. I=ediately after the insurge, the volume available to the j steam is decreased, therefore its pressure is increased, much l as the c mpression stroke in a diesel engine. The increased l pressure in fact.does PV work on the steam and as a result its enthalpy and temperature are increased. Thus we have two systems really, with'a superheated high pressur,e steam in contact with a subcooled liquid. Obvicusly, the pressurizer is not in equi-librium, and scme of the steam energy will be. transferred as heat to the liquid because of the temperature difference. Evan-l l tually, the two phases will,ccme into scme equilibrium saturation condittens. The follcwing T-s diagrama shew the thermodynamic precasses taking place: I L ' b, e 1

PAGE 37 0F 42 LESSON NOTE (CCNTI!WATIC1 SHEET - CCD) 'EMARKS

1. 2 s P l

/ / I' /, ll/l/ / T --- h s.~~ I/ .- T S - ---.. s --..-- b' 's ~ SATURATED UCUlO UNC SATURATE 3 VAPOR UNE 3 3 2.2 PROC:*:. E3 CCURRtNG QURING A PRESEURIZER INSURCC The pressurizer liquid experiences a heat transfer to the eccler insurging liquid and at the same time experiences the pressure ] increase caused by the steam ecm::ressien 1-2, P 'P ' I *I

• 2 1 2 1 Saturated steam is isentropically compressed, frem point 1 to point 2 and its temperature and pressure increase.

It is tem-perarily.superheated. As both phaces seek equilibrium, heat will be transferred frem the steam to the liquid; hcwever this is an extremely slew prccess. Both systems will experience changes in temperature, entrcpy, and i pressure, until eventually conditions reach equilibrium at sene- .new saturaticn pre'ssure P. The final accuats of liquid and 3 vapor at this new pressure are given by the initial mass / energy censervation equations. e 'Av 1 ,/ w

~ ~PAGE 28 0F 42 REMARXS LESSC;1 NOTE (CONTI?iUATICN SHEET - EVEN) Examole: Zicn Unit 1 experiences a load transient which U 0 ~ to increase frcm 550 F to 565 F at causes the primary Tgy --2235 psig. The 1800,ft pressurizer was half ful'1 of water ini tially. What will the final pressure and pressuri:er conditiens be? Solution: For this problem, the intermediate statas are not required, so the final cenditions can be calculated frem the mass / energy balance. Change in primary volume = 556300 lbm , 5563C0 lbm = 89.3 ft 3 IU 3D 3(560%) 45.86 3 (565 F) 46.20 ft ft This is the volume entsring the pressuri:er and corresponds to 3893 lbm of fluid at 595 F since it surges frem the het leg. The liquid and steam in the pressurizer are at saturatien tem-U perature at 2235 psig, 652.7 F so liquid mass

• 900 ft x 37.07 ID = 33353 lbm and steam mass =

3 3 ft SCO ft x 6.37 N = 5733 lba, 3 ft" The total energy of fluid in the pressurizer after the insurge is t.m h9g+ muff +mu. This energy will not change in mag-gg nitude, but will be redistributed as cenditions stabilize. The final conditiens must be censistent with the initial totsi mass 4 and must represent a saturated system. However a significant time may be required to reach the final equilibrium. The short l tem conditions are of interest because of limits en safety valve settings.. Assuming the ccepressien of the steam bubble U is isentrepic, the entropy remains constant at 1.2523 Stu/lba R. L l l- ~k ~T

pAGE 39 07 42 { LESSC1 NOTE (CCNTI!1UATIOl SHEET - 000) RE." ARKS Also, before the steam condenses back to liquid, its density is 3 increased because the available volume is reduced frem 900 ft to 3 3 3 810.7 ft. The new density is 7.07 lbm/ft up frem S.37 lbm/ft, This ccmbination of entropy and specific volume (density) are c fcund in the steam tables at about 2550 psia and about 682 y, seme 11 F superheat. Notice the pressure transient is quite .large and definitely-an undesirable shock to the system. Auto-matic pressure controls will be discussed later. After exter.sive interpolations in the steam tables, the final equilibrium state is fcund to re saturatien at a pressure slightly belcw the original 2235 psig. This is to be expected since the insurging liquid was much ccider than the original centants in the pressuri:er. A significant amcunt of heat was transferred frem the steam (and liquid) to raise the insurging mass up to the new satur$1 tion cenditiens. Starting frem the same initial conditions, a decrease in primary system average temperature will result in an cutsurge and an isentrepic expansion of the steam bubble. Using the steam tables again, the short term pressure transient reaches nearly 2100 psia, which results in a highly superheated liquid and violent boiling and flashing. The equilibrium state, again frem extensive inter-pelatien in the steam tables, is just under 2200 psia. Again, the large pressure transient represents a significant shock to the system, and thus is undesirable. During large pressure transients, autc=atic centrols initiate eithe'* pressuri:cr spray or heater action. The spray acts like -an artificial outsurge, only the change in mass is negligible. 'The colder spray water (frem scoops in the cold leg piping) premotly removes heat frem the pressu'i:er by condensing some r steam back to liquid. Obviously.after an insurge, the pressure will increase in proportien to the decrease in available steam volume. Cendensing seme of the steam back to liquid is an W 4

J PAGE 400F 42 - NOTES: TEXTi(Continuation' sheet even) extremely efficient way to reduce the mass of steam and thus reduce the pressure. (Compare this to the hours required to reach equilibrium after.an insurge with no automatic actions.) Heaters act to increase the pressure in the pressurizer after an outsurge has caused pressure to decrease. By adding heat to the water (already above saturation temperature) and converting it to steam, more steam mass is forced into the available steam volume, resulting in a pressure /. increase. Of course, once the pressure has increased, the liquid is subcooled and heat must be added to bring it back to saturation tempera-ture at the new higher pressure. The entire inventory in the pressurizer thus reaches new saturation conditions uniformly. 3 Example: The 1800 ft pressurizer is half full of water at 2235 psia. With all heaters on, how long would it take to raise pressure to 2285 psig (ie + 50 psi)? Solution: The total mass in the pressurizer will not change, however some water will be convarted to steam to increase the pressure. The total initial mass in the system can be obtained from an earlier sample problem as 39096 lbm. Determine the specific volume from the steam tables of the steam at the new conditions. V =.02718 V =.15349 f g ~ Knowing that the total mass remains the same (39096 1bm) and 3 that the total volume is 1800 ft, we can set up a simple algebra formula and determine the relative masses of steam and water. Let X = lbm of water 3 .02718 ft /lbm (X) +.15349 ft /lbm (39096 - X) = 1800 ft .02718 (X) .15349 (X) + 6001 = 1800 .12631 X = -4201 X = 33259 lbm of water 39096 - 33259 = 5837 lbm of steam Recall the original masses of steam and water from the previous problems, and subtract. Stater = 33363 lbm ' Msteam = 5733 lbm We find that we now have 104 lbm less water and 104 lbm more steam. I

e PACE 410F 42 TEXT:(Continuation sheet-odd) NOTES: (E Now the problem reduces to finding how much energy was required to 1) Raise the temperature of the water and steam to the new ' saturation conditions and 2) how much energy was required to turn 104 lbs of water to steam at this pressure. The total will be the amount of heat which the pressurizer heaters must add. Since we are interested in total heat added, we can use total system enthalpy (rather than just internal energy U) to find the answer. So, Q added = H -H and H = dbtm hsts + Mwater hwater 2 y The whole formula then is this: 9 " ( stm stm +" ( stm sta + "C# "E# wtr wtr 2 2 y = {33253 (705.4) + 5837 (1114.5)} - {33363 (695.5) + 5733 (1118.1)} 6 0 = 29.965 x 10 BTU - 29.61 x 10 BTU 5 Q = 3.57 x 10 BTU If total heat capacity is 1800 kw (which it is) then the heat ,[l-addition rate would be: 6 1800 kw

• 3413 BTU = 6.143 x 10 BTU /hr kv.hr Now, since we know the BTU's needed, and the BTU addition rate, we can simply divide to find the time.

6 3.57 x 10' BTU 6.143 x 10 BTU /hr =.058 hr = 3.48 minutes In summary, the automatic pressurizer pressure control system utilizes heaters and spray to provide fast response to primary system pressure (volume). transients. The pressurizer is a saturated system in equilibrium, but responds as a closed surge tank in the short term. Return to equilibrium in the absence of automatic controls can be very slow. '2.3 CALORIMETRIC The total quantity of heat added to the Rankine cycle is determined by a calorimetric analysis. All of this heat, in turn, is provided 'by both the nucicar fission process in the core, and a small amount from the main RCS pumps. e

PAGE 420F 42 NOTES: TEXT:(Contint.ation sheet even) The required inputs for a calorimetric analysis are: total steam flow, steam pressure, and feedwater temperature. Knowing the steam pressure. allows a steam table lookup of saturation enthalpy. Similarly, feed-water temperature allows lookup of feedwater enthalpy. The result of the calculation is total NSSS output and has units of (energy / time). NSSS output = 6, (h,g,,,- hfeedwater} { hr} Conditions should be measured at steady state, and losses are some-times factored in, if knoun. Figute 2.3 is a computer-generated calorimetric evaluation for all four steam generators in Zion Unit 2. (The Barton DP is a measurement proportional to feedwater flow.) Core power may be measured directly if the total coolant flow is exactly known. ""#(B ) = (h (Tg ) - h (Tcold The enthalpies of the coolant are evaluated at the measured tempera-tures at system pressure.

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e CPSES TRAINING LESSON NOTES PAGE 1 0F en AUTHOR: B. TUMBLIN LESSON: FLUID FLOW REVISION NO: 3 PROGRAM: CPSES OPERATOR, TRAINING WesEinghouse Heat Transfer Notes General Physics Heat Transfer Course Nuclear Power Systems REMARKS LESSON NOTES 1.0 FLUID FLOW

1.1 INTRODUCTION

Frictional forces exist between all real substances. The flow of any real fluid in a system produces tangential friction forces called viscosity forces. These forces give rise to a' coefficient of viscosity which is dependent en temperature, pressure, and the type of substance. ~ The forces acting in a moving fluid and the velocity of a fluid determines the type of flow that exists in a system -- laminar or turbulent. The t3pe of flew can be determined by calculating or knowing the fluid's Reynolds Number (Re). We can produce perturbations in a flowing fluid by placing obstacles in'the fluid's flow path. These perturbations can be useful in a steam plant system for energy conversion ce I measurements such as turbine nczzles, venturi, orifices, etc. f

r e o PAGE 2 0F 59 ~ REMARKS LESSON NOTE (CONTitiUAT10t1 SHEET - EVEN) 1.0 .(Cont.) .We need a device called a turbine to obtain useful work from the energy stored in steam. The turbine converts steam energy to work output. We will look at the various types and classifications of turbines. Anoth.er device used to convert steam energy to useful work is an air ejector. We will examine its design and principle of operation. A driving force is needed to force a fluid through a system. This driving force is supplied by a pump, either centrifugal or positive displacement. As the fluid is forced through'a system by a pump, friction forces restrict the fluid's flow. Losses in pressure and flow rate are encountered called head losses. We will examine some of the effects of pump laws and head loss characteristics in a fluid system. 1.2 VISCOSITY When two substances are in contact, one or both moving, frictional forces will cause a relative velocity difference between them. These tangential frictional forces are called viscous forces. The motion of a fluid past a stationary wall can be thought of as an infinite number of very thin layers. Each layer is in contact with another layer with some frictional force between them. They slide over one another with some velocity as shown in Fig. 1.2-1. Vy g'X _7 _7 H 7 _/ V:o-f //// / / / / / / / / / / / / / / // DIPE WALL FIGUE. l.1-1 FLUID VELOCITY INSIDE A PIPE

e, 'PAGE 3 0F.59. LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 1.2

(Cont. ) '

The fluid layer next to the pipe wall will stick.to the wall and have exactly the same velocity as' the wall. This is known as the'non-slip condition. Each of the other layers will have some velocity relative to one another up to some velocity V max

• An example of a fluid velocity profile _is a deck of. cards on

- top of a table with a hori:ontal force applied to the top card. Frictional forces between the cards cause the cards below the top card to move with the top card. Due to slippage, the top card does not fully displace the bottom card. As each pair of cards is analyzed, we can see that the upper card displaces the lower card some distance due to frictional - forces, but not completely due to' slippage. When we reach the' bottom pair of cards, we find that the last card has stuck to the table top. The last card sticks due to insufficient force to overcome the adhesive frictional force between the card and table top..(Fig. 1.2-2). I F0mOC g/ g 1 t/ l 4/ t l I 6/ I t/ f v .//s///// / ///////////// FIGURE 1.2 'accx oF CARoS* FRICTtoN ANALOGY This property lof internal friction in fluids is called viscosity. From Newton's theories of fluid flow and subsequent experinents, .the actual velocity distribution of the fluid layers relative to one another is not: linear, but parabolic in nature. :The viscous s ._m.t__-----_m_.

PAGE 4 0F 59 _ REMARKS-LESSON NOTE (CONTINUATION SHEET - EVEN) 1.2 ,(Cont.) forces in the fluid flow were found to be proportional' to the distance from the boundary wall. (See Fig. 1.2-1)' / Frictional Force = F = h The magnitude of the frictional force is also proportional to the area being considered (A). (Fluid area,.;ot pipe area). In order to calculate this frictional force, we need a constant of proportionality which we will call u. This constant is called the c_oefficient of dynamic viscosity. Now we have for _ viscous frictional force (F) F = uA h . The coefficient of viscosity is dependent on the type of fluid, temperature, and pressure of the fluid. It has units Ibf-sec of ft or in CGS unit system dyne-se_c-and is called. a 2 Cm l-poise or a centipoise. (1 poise = 100 centipoise.) lbf - sec dyne - sec I = 08.8 ft cm j The table below shows the dependence of the viscosity coefficient on temperature, pressure and fluid type. p l

• -e

W 59 PAGE 5 0F. LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 1.2 -(Cont.) TABLE I COEFFICIENT OF DYNAMIC-VISCOSITY-FLUID TEMP (*F)/ PRESS (CENTIPOISE) AIR 32*F 0.0171 -150*F 0.0201 MERCURY 32 F 1.70 150*F 1.35 STEAM 600'F/0 PSIA 0.0208 (SUPERHEATED) 1000*F/0 PSIA 0.0286 600*F/500 PSIA O.0310 1000 F/500 PSIA 0.0388 WATER 32*F 1.792 100*F 0.679 212*F 0.284 The more viscous a fluid becomes, the gr. eater the frictional forces in the fluid and the more difficult it will be to make the fluid flow. From Table 1, it is apparent that for a liquid, as temperature-increases, viscosity (u) decreases; but, for a gas, as temperature increases, viscosity increases. Many-times it is convenient to calculate the friction force per unit area, F/A. This is called the shearing stress, given ths. symbol T. So from the friction force equation: F = uAh .F/A = t = u Y c--- .;m

PAGE 6 0F 59 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 1.2 (Cont.) .Since h is not linear, it is extremely difficult to evaluate mathematically. Therefore, to simplify we will assume that h(velocityprofile)isastraightlineorlinear,andfrom Fig. 1.2-3 we can see that F/A = r = p( ma )"" ,0 wax H [ FIGURE 1.2 3 veo ////////////////// wtLL LINEAR FLUID VELOOITY ASSUMPTION The shearing stress is in units of pressure (psi), and acts parallel to the wall. The shearing stress is the force necessary to overcome the fluid's internal friction and start moving. 1.3 TYPES OF FLOW There are two types of flow that occur in a bound system. One is laminar flow and the other is turbulent flow. Laminar Flow is characterized by the example of the deck of 1.3.1 cards on the table top; the fluid levels are sliding over one another. In,this type of flow, the viscous forces pre-dominate. At the boundary wall, the rluid velocity is zero while at midstream the velocity increases to V as shown max in Fig. 1.3-1.

59 PAGE 7 0F LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 1.3.1 (Cont.) ///////V*O ///// N ~1]v.x v =l 2 / / /// / V=0// // FIGURE 1.3-1 L AMIN AR FLOW VELOCITY PROFILE Laminar flow is uniform and steady. It can be easily analyzed to find frictional forces, temperature gradients, etc. Laminar flow is typical where low fluid velocity and/or a high viscosity fluid is present. 1.3.2 Turbulent Flow is characterized by irregular, fluctuating velocities in various directions. In this type of flow, the inertia forces predominate. Although the analysis of this type of flow is difficult and incomplete, the velocity profile can be characterized by three general conditions. (a) The velocity at the boundary wall is zero. (b) 'There is a region around the boundary wall where the flow is laminar. (c) The turbulent flow occurs in the central region of the bound system and produces a more uniform velocity profile. See Fig.1.3-2. +

PAGE 8 0F so REMARKS 1.ESSON NOTE (C0!iTINUAT10!i SHEET - EVEN) v'O 1.'3.2 (Cont.) ,,,,j yyygm .h*.".'.".'."'.'.*.".... -'\\ vunn FIGURE 1.3-2 "'~ ~ """] taww = rue. ///////////////// v=0 TURBULENT Flow vtLoctTY PROFILE f Turbulent flow is typical where high fluid velocity and/or low viscosity fluid is present. The effects of velocity on a fluid were made by Osborne Reynolds using the equipment shown in Fig. 1.3-3 FIGURE 1.3-3 i k=fra j)Q f l Water was stored in a tank and the velocity of the. water was controlled in a glass tube (B) by adjusting valve (A). Dye ~ stored in a small tank was introduced into the flow of water in the glass tube. At low velocity, the dye moved in a -straight line parallel to the glass tube with no mixing . occuring, just a thin filament-along the length of the tube. This he called laminar flow. i i r. w .n. ~

m. PAGE 9 0F 59 . LESSON NOTE (CONTINUATION SHEET - 000) 9EMARKS 1.3.2 (Cont.). At high velocity, beyond a certain range, a violent mixing and swirlir.g of the dye occurred. This,he called turbulent flow. No gradual change occurred between the two types of flow; instead, a sudden change was observed. Reynolds derived an expression between the inertia forces and the viscous forces in the fluid to determine which force was dominant. He called this number the Reynold's Number (Re). pV 0 ave Re = u 9e 3 where: o = fluid density lbm/ft . sec u = fluid viscosity ft V = average fluid velocity ft/sec ave D = characteristic length which depends on the geometry associated with the flowing fluid in feet (i.e. pipe inside diametar). lbm. ft gc = 32.2 ibf. sec Reynolds found for any flow with Re less than 2000, laminar conditions exist. For flow with Re greater than 3000, turbulent conditions exist. For Re between 2000 and 3000, no exact condition can~be predicted. Thus, observation of the flow must be utade since the change from laminar to turbulent. is sudden with Re in the 2000 to 3000 range. 2.0-FLUID FLOW THROUGH-N0ZZLES AND CRIFICES In or' er for a turbine to use the-energy stored in the d steam, it must'be, converted'to a more usable form-- kinetic energy. ' This is.the function performed by a nozzle. ( There are two types o'f nozzles used depending on their application. One type is the. convergent no::le, characterized by k

r. -

PAGE to 0F 59 - REMARKS LES50ti fl0TE (C0t4TitiUATI0tl SHEET - EVEft) 2.0 (Cent.) a constantly decreasing cross-sectional. area along its axis (Fig.2.1)..The other is the convergent-divergent nozzle, or' simply divergent nozzle, characterized by a cross-sectional area that first decreases then increases along its axis (Fig. 2.2). / ws ms V / / / CONVERGENT Nozzle CONVERGENT-OlVERGENT Nozzle FIGURE 2.1 FIGURE 2.2 flozzles convert the_ energy of the steam in the form of enthalpy to kinetic energy in the form of steam with a relatively high velocity. From the general energy equation I + Q = PE2 + XE2+U2,PY PE) + KE) + Uj + P;V 22 +W J J The potential energy terms are negligible. The steam passes .through.the nozzle so fast that Q is zero and there is no way 'for work to be removed. i e

t..

.i .s' p [' PAGE ll0F 59 L LESS0tl NOTE (C0tiTIt!UATI0ttSHEET - ODD) REMART,5 $2.0~ (Cont.) Therefore the general energy equation becomes ~ KE; + U) + P)V 2, + U2+EY 1 = KE 22 or J J 2 2 V V 1 2 2g d 29d 2 e c Where g

• I ft - Ibf c

J = 778 2, BTU '1bf - sec If the steam is considered to be zero velocity as it would t-l .in a steam chest, then h, = V2 J + h), where h is called the " stagnation" g enthalpy. 2 h =V2 2g d + h2 g c y 2, 2g 50 h) c 0 2 L 'V2 " Y2 9c (h -h)= 223.7Y(h - h ) '- J g 2 g 2 If the steam is considered to have some initial velocity as it would moving down the steam pipe, then 2 V V 2 + hg j 2g g.+ hj =' 29c e-V "29 d (h);- h ) + Y1 2 c 2 i If V) is~ unknown, then use the continuity equation as follows. AV I P j j = gA Y2 2- .A V) AY-j 22 l.- L- ,i >2 p

i.

y e

o o s PAGE 12 0F 59 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) i 2.0 (Cont.) V A "l 2' 2 V) = A;v2 2, (A "l)2 2 2 y y 1 A v2 2 j Then: V

• 29 J (h) - h ) + (

) Y 2 c 2 2 V ~{ ) Y = 2g J (h) - h ) 2 2 c 2 y 2, 2g d Ih - h ) c l 2 1 - (A v2A"1)2 2 1 29 J (h) -h) y2,, c 2 A "1} 2 )I 2 A)v2 3.0 N0ZZLE DESIGN Nozzles are designed for specific applications where the pressure at the inlet and discharge, desired steam velocity. -and specific volume of the' steam are known. As the steam moves axially through the nozzle, there will be a definite pressure, velocity, and specific volume at each point. The . length'of the' nozzle has.no effect on the convergent section. If the' nozzle is short, then the transition'from high to low- -pressure is accomplished quickly. If the nozzle is longer. then the transition from high to low pressure is_ longer; 'but, the same pressure, velocity, and specific volume are produced-at.the same crus-sectional area along the-travel

~ PAGE 13 0F 59 LESS0:1 NOTE (CONTINUATI0tt SHEET - 000) REMARKS 3.0 (Cont.) path. The nozzle-shape should be similar to Fig. 3.1. l THROAT s 8, APFRGulMATc l cur auet. inau::co i crit Anac l Y'EcY FIGURE 3.] DEStGN FEATURE oF A Nozzle i The shape of the nozzle is not critical at the entrance since the velocity is low. The only requirement is that the cross-sectional area decreases gradually. To reduce friction, this section can be made short with no significant effects. .The divergent part must be treated more carefully since the velocit'ies here are very high. The throat must merge into the divergent section with no discontinuity a *%t no disturbance will be created in the high velocity fluid stream. To_ prevent disturbances in the divergent section, the walls must curve gradually an'd become parallel at the exit. The maximum included angle must be small (<20*) or turbulence will be generated as the steam is expanding. ~*~

PAGE 14 0F 59 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 3.0 (Cont.' ) For applications where a small pressure drop is expected, a convergent nozzle is most appropriate as lower velocities are expected at the throat exit. For large pressure drops where extremely high velocities of steam' are expected, a convergentdivergent nozzle is most appropriate to minimize turbulence at the nozzle exit-Lower turbulence minimizes energy losses going to the turbine blading. 3.1 N0ZZLE FLOW RATE As steam moves through the nozzle, pressure decreases, velocity and specific volume increase, (Fig. 3.1-1) \\ ~ l ,n.o* h# I ~~'[0 ;' .....-7,..., encssunc "t,U,unc,oo 7, 33 i i = ' ri.ow l-i // \\ FIGURE 3.1 1 Flow CHARACTERISTICS oF A Nozzle ~

PAGE 15 0F 59 LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 3.1 -(Cont.) As the steam begins to-flow, the velocity increases rapidly

as the pressure drcps. As the-downstream pressure decreases, the velocity at the throa,t continues to increase until a maximum velocity is achieved; this velocity is sonic velocity.

In the converging section of the nozzle, no further increase in velocity can be ac',ieved. The mass flow rate through the nozzle reaches its maximum value at the throat when it achieves sonic velocity. The pressure at the throat, at this point, is called the critical pressure (P*). This gives rise to 'the critical pressure ratio r *. p P r = p-p o Where P is the pressure at the nozzle entrance. For saturated g steam,-r = 0.577 or s 0.58. Therefore, if pressure at the p

• tnroat is above the critical pressure, the velocity will be less than sonic velocity. But, if the pressure at the throat is below the critical pressure, the velocity will reach sonic velocity and no higher.

(see Fig. 3.1-2) For divergent nozzles, where the steam is allowed to expand at a controlled rate in the divergent section, the velocity can achieve supersonic velocity. These supersonic velocities are downstream of the throat. MAI I i de 5 j.;- o 10 PPES$unt R ATIO ' 'p VARI ATION oF pass-FLCW RATE < FIGURE ' 3.1-2.- wiTH THacAT-rRessuat ratio m-m._

15 PAGE 0F 59 REMARKS LESSON NOTE (C0tlTINUATION SHEET - EVEN) 4.0 VEitTURI Th'e converging-diverging passageway is characteristic of both a nozzle and a venturi. A pressure at the exit slightly /' lower than the pressure at the entrance causes a greatly decreased pressure at the throat. p t.0 ' 'l A e e- .e~ Same venturi 4 different flow I.'".----------i-rate > A flow g FIGURE 4.1 rates. B flow - rate. l .2" e PRESSURE CONDITIONS IN A CONVERGENT-OtVERGENT Nozzle i LENGTH ! ll i ENTRANCE E XIT P 'P _ p g l THRO AT /. / In the convergent section, expansion proceeds with the achievement of high velocity and low pressure at the throat. (Fig. 4.1) In the divergent section, compression occurs because the velocity is progressively reduced while the pressure is increased. This action continues as long as the throat pressure is greater than or equal to the critical pressure. P f:

PAGE 17 0F 59 LESSON NOTE (CONTIt!UATION SHEET - ODD) REMARKS 4.0 .(Cont.)- .The venturi is constructed such that pressure in the annular space at the inlet is measured -(in the form of a piezometer.) / The inlet pressure is compared to the throat pressure where there is another series of radially drilled holes. This series of holes ensures an average pressure is being measured and guards against loss of the instrument in the event one hole gets plugged C ne. oeb .1...... '2 .g... [ FIGURE 4.2 l The differential pressure between the inlet and the throat ~ is proportional to the velocity squared of the fluid as it passes through the throat (see Fig. 4.2). We can see this from the general energy equation. Here, the potential energy terms are negligible. The temperature of the fluid is relatively constant and there is no heat added or work done 'by the' fluid as it passes through the throat. Thus we are left with: P PV KE + g)v)-,E2+7-22 K j Then. 2 V) pV 2 1I

2.y+

=Y 9 2 PV g 22 + 29J .J-c

o. o. PAGE-18 0F* -59 L , REMARKSf . LESSON NOTE (CONTINUATION. SHEET-- EVEN) '4. 0 '.

(Cont.)'

'or Y P v) .P V2 2, (V2 1 j 2g c Looking at Fig. 3.11, we can see that the specific volume fromthe'inlettothe.throathasanegligiblechange,sovj =v. Alos, the. velocity of the fluid at the inlet is very 2 small compared to the velocity.at the throat, thus we can neglect the.V) term 2 l- .P) -'P2= y 2 2g"c Since 29 v is constnat,'then L V = kap 2 .or V. = / kap 2 From.the continuity' equation for steady flow + m.= pAV A is constant since the throat diameter does not change; ~ ' p 'is the density of the fluid; V is the velocity of the -fluid which is proportional to the square root of the pressure differen'ce between the inlet and throat. i In = c AV = Kp / AP - [; Therefore, by measuring the pressure drop frcm the nozzle Linlet to the throat, we can measure the flow rate through a fluid: system by-knowing the density of the fluid. t M, ~ =. r l ~

'~~ + L'.- x-y y PAGE < l9 0F 59 LESSONLNOTE (CONTINUATION' SHEET - 000) REMARKS -4.0 . ( Cont. ) ~ 'The venturi is used in plant systems such as main steam ~ system where it serves two purposes. .(a) To measure the mass flow rate of steam;for the -steam generator' level control system. The density of the steam (c) can be determined from steam tables by -knowing the steam pressure. The venturi' nozzle area (A)-is fixed. By measuring the pressure drop, which is proportional to the velocity. squared of the steam as it passes through the venturi, the mass flow rate (m) of .the steam can be calculated. m = pAV (b) -Since the venturi is a nozzle, we know that the pressure at the throat has a limit; this is the critical pressure. In the. event of a main steam line rupture, the lowest pressure "seen" by the steam generator would be the .o critical pressure at the throat of the venturi

Thus, the venturi would act as a flow restrictor.

i-5.0 ORIFICES 'An orifice is installed in a system where a reduction or a I controlled flow rate is desired. The orifice most be carefully constructed if the desired flow rate is to be. attained. 'The orifice ~is installed as shown in Fig. 5.1 with the sharp i edge orJsquare edge.on the upstream sida and beveled edge on .i

the.down stream side.

l L '? b k i

? PAGE 20 0F 59 REMARKS. LESS0ft NOTE (C0t1 Tit 10ATION SHEET - EVEll) i 5.0 (Cont.) f FIGU N 5.1 FLOW ANo_ PRESSURE THRoUGH A SHARP-EoGEo ORIFICE l ncssunt 0 Any~ rounding of the sharp edge corners will decrease the jet contraction at the vena contracta thus increasing the flow area and the mass flow rate. The pressure in front of the orifice may be slightly higher than true pressure because of the impact of the fluid on the. orifice plate. As the fluid passes through the orifice, the e pressure drops abruptly. It reaches a minimum value at the ~ vena contracta or location of smallest jet diameter. At this point, all filaments of the fluid flow become parallel. As the fluid continues to flow beyond the vena contracta, the pressure increases as the fluid is slowed' down. Since no diffuser section ~is present, considerable turbulence is created causing a decrease in the downstream pressure. l By controlling the diameter of the orifice the maximum flow rate through the orifice can be determined as in section 2.0 where ~V2 = 2g J(h)-h ) c 2 AV 21 1-(A;v )2 2 l. i s -

( Q ;, I eV ~ PAGE 210F 59 LESSON NOTE (CONTINUATION SHEET - ODD) REMARKS 5.0' .. ( Co nt. ) - and the mass flow rate can be calculated by the equation ?- .m = pAV An example of an orifice used in the plant wo'uld be in the letdown system. The maximum flow rate through the system is J determined by three orifices (75 gpm, 75-gpm, or 45 gpm). l The actual flow rate can be control. led by regulating the pressure.down:tream of the orifices. 6.0 PUMP AND FLOW CHARACTERISTICS The effect of a pump oi a fluid was discussed previously in the. context of a thermodynamic cycle. Before proceeding on j to'a discussion of specific pump characteristics, it will be

beneficial to identify. the processes which occur in a closed fluid-system.

Just as with other. systems, the first law of thermodynamics can be used to suciarize those processes -- The energy within a system can be transformed from one type to another. However, energy can move into and out of the system as heat or work. This " conservation of energy" can be summarized by the general energy equation for any process which changes the system from state I to state 2. ~PE)+.KE)+(h))+.'U)+Q=PE2 + KE2+( ) +U2+W 2 This can be expanded'as follows for a unit mass. 2 2 Z V Pv Z V py +U 29J+ l

• 9 ".

+ 2g 2 c The work term in the-context of this discussion refers to l-pump work done on.the' fluid; it may also be expressed as a positive energy. addition, E (just like heat) on the other. p p side of the equiaticn. Since the forms of energy are commonly y .S. l 7

-t PAGE22 0F 59 REMARKS . LESSON NOTE (CONTINUATION SliEET - EVEN) 6.0 (Cont.) expressed-in mechanical terms the conversion from Btu to ft-lbs will be removed'and the equation can be rearranged as follows: 2 2 y y Z) + 2 + P)v)-+ E Z + PV2 2 + [J(U2 - U)) - Jq) p 2 + 2g = c The terms in brackets represent energy which raises the fluid temperature. This results from flow turbulence and friction as well as heat transfer. Once energy has degraded to " heat" it is generally lost to the system, so the term in brackets can be designated the loss term, and is commonly. referred.to as " head loss," h. t The other terms in the expression can be~ assigned specific names also, because of their. significance in the flow of incompressible fluids. (v) = v, specific volume is constant) 2 (P)-P )v = pressure head (ft-lbf) 2 lbm V 2_y 3 2 f t-lb 2g = vel city head ( lbm f) ~ c I2-Z)h=potentialhead--( l 1 2 ) c The potential head is the historical basis for the " head" designation.~ It referred to the height (in feet) of a water level in'a reservoir which was responsible for a particular flow rate at the t : 3in. The effect of gravity on a particular

height ~of water produced the driving energy to move water out the drain pipe.

Notice that all of. the " head" terms ~ i t w

x ~ PAGE '23 0F.59 LESSON NOTE (CONTINUATION SHEET.- ODD) REMARKS-6.0 L (Cont.) 'have units of ft-lb ,_and that a unit.of " feet" can be f Ibm obtained if the ib and ibm are cancelled. Although the f cancellation is invalid, it is still common to see flow 4 energies' referred to as " feet of head." In other.words, the equation now simply states that the sum of pressure head, velocity' head, potential head, and pump L head, must equal system losses in the steady flow system. The following sections will provide some f.nsight into the different mechanisms for adding pump work to a system, and the factors affecting head loss. 6.1 CENTRIFUGAL PUMP Acentrifugalpumpisitdeviceforincreasingtheenergyof ~ 4 , a fluid so that the fluid can overcome the energy losses in a fluid system. One of the main parts of a centrifugal pump is the rotating part called the impeller..The fluid enters the eye of_the impeller. The vanes of the impeller impart a velocity or kinetic energy to. tne fluid (see Fig. 6.1-1). The second main part-of the centrifugal pump is the volute, which is a~ passageway of.~ constantly increasing area. As the.hig'h velocity fluid from the periphery of the impeller enters the volute, the velocity decreases.as the area increases. lSince there is no-work added to the. fluid'in the volute, the energy of.a certain mass of fluid remains' constant. As the .veloc'+y decreases, the kinetic energy.of the fluid also Edecre'ase's. This implies that the-energy is converted to . another' form called flow energy' in the form of an increased - -pressure. I i

1 ~ t, s l-t- PAGE 24 0F 59 -REMARKS. ~LESSONNOTE(CONTINUATIONSilEET-EVEN) L '6.1 (cont.) b V ( V ( v3tV24VI $ 4 CCNSTANTLY CECREAS:NG I (,T - vEtociTv (xE t uv 2 v y' ' FIGURE 6.1-1 ( f-" 8 / N ' CROSS SECTioN OF A CENTRIFUGAL.' PUMP / ',v,^3 SHOWING DECREASE IN VELOCITY WITH (s

• ~

INCRE ASE IN FLOW ARE A. tuPELLER VOLUTE ,Vt,,4 A A, c A2 # A 4. A4(#$ 3 f Al CONSTANTLY INCRE A$1NG A%E A t Due to tolerences. required for moving parts of machinery, 'some of-the fluid pressure is lost. Thus, the pressure can j reach only a certain upper value. To increase this value of total head, multi-staging of the pump is necessary. By . arranging a group of impellers in series such that the fluid I leaving the discharge or volute of one impeller it fed to the suction or eye of the next impeller or stage, the pressure can be increased. There are a number of variables involved which &cem.ine the characteristics of a centrifugal' pump. They are: .Q = volumetric flow rate N = pump speed - E= total head a = fluid density d = diameter of impeller p= fluid viscosity By using dimensional. analysis, we can derive some interesting facts about centrifugal pumps. Letting M = mass, L = length, _T = time, then d = L' or ' L = d 3 l p = h o r M = '. 'p d > j._ .t L 1 1 l N = 7 or T = g L p l Using these: facts and continuing the dimensional analysis I '_for Q:and E, we find 'that l (. am

PAGE 25 OF 59 . LESSON NOTE (CONTINUATION SHEET 000) REMARKS'- 5.1 .(dont.) 3 3 Q*h"5 =" N 5 3 = pH d e, E=MgH=pd.fL.L=pd(1/Nh2' For calculating -hydraulic horsepower (this is the rate of energy addition to-the fluid while passing _ through the pump ~ which-is proportional-to the power from the prime mover ),we find that ft-lbf hydraulic horsepower = j. (550 "I" sec thus prime mover power = where n = centrifugal pump efficiency factor. Continuing the dimensional analysis, then 3 2 2 3 2 (Nd )(3 d')c gdc 1 power (prime mover) = ii 550 650n What we see is that the flow rate through a centrifugal pump is' proportional to the pump speed (N), the total head is 2 proportional to the. pump speed squared (N ), and that the power to drive the prime mover of the centrifugal pump is proportional to the pump speed cubed (N ) or 2 3 -Q = N, E = N, PWR = N. Using this data we.can arrive at pump curves for a centrifugal pump at a constant speed or for various speeds. Looking at Fig. 6.1-2 we-can see the effect of doubling the speed of a _ centrifugal pump. i s i ,tn _o. Y V' f

.O s PAGE 26 0F 59 l REMARKS LESSON NOTE (CONTINUATION SHEET.- EVEN) L 5.1 (Cont.) E, FIGURE 6.1-2 Na a2Ni 2 Q*N [*N Flow RATE LVs PUMP HEAD ' h N 2 FoR A CENTRIFUGAL PUMP k E C, NI i FLCw RbE a0 From the PWR equation we can also see that for doubling the 3 pump speed since PWR = N implies that the power to drive the pump would increase by a factor of eight. PWR2 = 8PWR) 6.2 POSITIVE DISPLACEMENT PUMPS Positive displacement pumps are unique in that they have a definite volume (determined by the pump size) that can be moved by the pump. Remember-3 3 L L 3 Q = 7- = 773 =LN Therefore the capacity or flow rate of the pump is' directly proportional.to the pump speed. Since the tolerances l designed in a positive displacement pump are small, there is little or~ no ' leakage past the pump internals. This means

that-the positive displacement pump will produce whatever pump-head is necessary to deliver a specific flow rate.

determined by tne pump speed. A typical head vs flow curve- ~ l for_a positive displacement pump would look like Fig. 6.2-1. i 1

__e. s-n PAGE 27 0F1 LESSON HOTE-(CONTINUATION SHEET - 000)_ REMARKS 6.2 (Cont.) FIGURE 6.2 ! e. es, s I Flow RATE VS PUMP H'E AD FOR A PostTIVE DISPLACEMENT PUMP The little fluctuation at the top of the curve occurs when the pressure (pump head) is so_large that the fluid will . leak past the pump internals resulting in a reduced flow rate. - 6. 3 ' PUMP. COMBINATIONS sometimes it. is more economical or more beneficial for system reliability to use a number of smaller pumps rather than one large pump. Centrifugal Pumps in Parallel By placing.two identical'. centrifugal pumps in parallel, we . find that the pump head'does not change. But, the flow rate will double. A resultant set of pump curves would look like Fig. 6.3-1 (i.e., two. feed pumps running in parallel). e <e L, ? x

PAGE 28 0F 59 - REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 6.3 (Cont.) A. [^. w n O $= N PUMP PUMPS A OR 8 ASB l-FLOW RATE : Q FIGURE 6.3-1 TWO CENTRIFUGAL PUMPS IN PARALLEL Centrifucal Pumos in Series By placing two identical centrifugal pumps in series, we find'that the flow rate remains the same but the total pump head :douvies. A resultant set of pump curves would louk like Fig. 6.3-2 (i.e., condensate, condensate booster pumps). _ %[ t,

-e- .o i PAGE 29 0F 59 - LESSON NOTE-(CONTIflUATIO!1 SHEET _- 000) REMARKS 6.3 (Cont.). 8 A w PUMPS k ABB E s .E PUM9 A oR 8 Flow R AT E s Q FIGURE 6.3 2 Two CENTRIFUG AL PUMPS IN SERIES 2 Centrifugal and Positive Displacement Pumos in Parallel If a minimum guaranteed flow' rate _is necessary but a larger -flow may be desirable, then placing a positive displacement and centrifugal pump.in parallel is advantageous (i.e., CVCS t system). By placing a positive displacement and centrifugal ' pump'in parallel, we find ~ a minimum flow rate is always achieved no matter what the pump head may be. A larger flow rate is attainable at lower pump heads as in Fig. 6.3-3. - n. .-A ..---.-.a

PAGE 30 0F 59 REMARKS LESSON NOTE _(CONTINUATION SHEET - EVEN) 6.3 (Cont.) 4 e 7-w = o 0 o z 4 A 2 4 o e, + FIGURE G.3-3 Flow R ATE : o POSITIVE DISPL ACEMENT AND CENTRIFUG AL PUMP IN PARALLEL l 6.4 HEAD LOSSES IN A SYSTEM Losses occurring in a piping system, (due to frictional processes resulting in the loss of useful energy by trans- ' formation into increased internal energy, and heat transfer-through.the system,) can be broken down into two catagories (a) ' losses due to friction in a pipe (b). losses due to friction in fittings, valves, and elbows. 6.4.l LOSSES DUE TO FRICTION IN A PIPE The head loss due to friction in a straight pipe may be expressed by the Weisbach equation:

lin ,s; s s pAGE 31' 0F 59 LESS0ii NOTE (C0hTINUATION SHEET - ODD): REMARKS 6.4.1 l(Cont.) L7= f b ' h D-where f = friction factor L'= length of pipe D = diameter of pipe V = velocity of fluid ~ Th'e term f (friction factor) is a function of the Reynolds number and the relative roughness of the pipe walls. 6.4.2 LOSSES DOE TO FITTINGS AND VALVES The losses due to ' pipe fittings (such as elbows, reducers, valves,z ect.) are often minor ~as compared to pipe' losses and may be neglected through long pipelines. However, for short piping systems, large ~ errors may result if they are not accounted for in the' flow computations. The losses resulting -from the fittings can be expressed by the equation 2 V .hl"KLk C-where K " loss: coefficient L V =~ fluid 1 velocity iThe' loss coefficient is ' dependent on the type of fitting and -i

must be ~ calculated for each individual' type of fitting.

6'.4.3 : SYSTEM FLO'.4 CHARACTERISTIC ' CURVES

The curve f for a particular system's flow -can be calculated

'using~ the;comoination of the headloss. equations.. In,. gene ra l, d ~' it williicok --like Fig. 6.4-h x -.4 L-e

s t, o-t PAGE 32 0F 59 REMNRKS LESSON NOTE-(CONTINUATION SHEET - EVEN) .t.4.3 (Cont. ) = i A L w _./ Y a 2 O FIrVRE 6.4-1 l FLOW RATC o TYPIC AL SYSTEM HEAD LOSS CURVE .The curve will be a parabola since the'headioss equations 2 contain a V term. By changing the flow characteristics, (i.e., open or cloSe a valve), the system flow curve will l change as in Fig. 6.4-2. l A Lg hq SHUT CPcN oi N agco.----... o l l l e. ca. o, FIGURE 6.4-2 i CH ANGES IN SYSTEM HEAD LOSS CURVES f n h.

W PAGE 33 0F 59 LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 6.4.3 ( Co nt. _) Assuming h is the original position of a valve producing a gg -particular. curve, then at a total head of E, a flow rate of g Q will be maintained.- By opening the valve,.less system g head loss is. experienced; thus, for the same head E, a g higher. flow rate will be produced, Q. If the valvc is 2 closed by a few turns,.then more restriction to flow would be experienced by the fluid. For the same head a lower flow rate, Q), would be produced. By combining the system flow curves with the pump curves, we can determine the amount of ficw in a system with a specific pump as.in Fig. 6.4-3. n, u FIGURE 6.4-3 _c____________ SYSTEM Flow CURVE WITH N - - - ~ ~ ~ ~ ~ ~ ~ i { N, A SPECtFIC CENTRIFUGAL PUyp 8 I l Go Gs FLOW RATE a Q From Fig. 6.4-3, we can see a centrifugal pump operating at some speed fl) with the initial valve setting h A flow Lo. rate of O with a head of E -is established. If the valve is g g opened some' amount, a new system curve, hg, is produced. A-new head, E, is produced resulting in a new flow rate, Q). j The same can be done for a positive displace: rent pump as in ~ Fig. 6.4-4

'PAGE 34 0F 59 REMARKS .LESS0ti tt0TE (C0tiTI?1UATI0t1 SHEET - EVEft) 6.4.3 (Cont.) \\ N, ke w n E FIGURE G.4-4 g c, g e SYSTEM FLOW CURVE WITH SPECIFIC POSITIVE DISPLACEMENT PUMo E Q FLOW RATc :Q i From Fig. 6.4-4, we can see a positive displacement pump operating at some speed ti). Regardless of the valve position, it produces the same flow rate, but changes its discharge head;to meet the head losses produced by changing valve positions. Changing pump speeds can be_ illustrated as in Fig. 6.4-5 and Fig. 6.4-6. J- "L L U "g N, S

• j j 5.'..............

0 ,s o ci -o u ~ l Os J8 0 Cy FLon RaTc s G . Flow RATES Q FIGURE G.4 5 FICURE 6.4 6 _ CENTRIFUG AL PUMP SPEED COUBLED PostTivE DISPL ACEMENT PUMP SPEED DOUBLE 0 J

e .) i PAGE 350F 59 LESS0tl NOTE'(C0tiTINUATION SHEET - ODD) -REMARKS 6.4.3-(Cont.')- Fig. 6.4-5 and Fig. 6.4-6 show that by changing the pump speed a new head and new flow rate will be generated. They can be~found by knowing the pump characteristics. Note that as flow increases the system flow curve does 'not change. l

Although. fluid velocity has increased, this -is accounted for by'the flow curve. Only a change in system characteristics (i.e., value position, pipe size, etc.) will change the system flow curves.

6.5 -CAVITATION Periodically one walks past a pump and hears a sound similar to gravel going through a chute. This phenomena is known as cavitation. When a fluid flows into a region where its pressure.is reduced to its vapor pressure, it boils and vapor: pockets develop. The vapor bubbles are carried along with the' fluid until they reach a region of higher pressure and suddenly collapse. As the vapor bubbles collapse and the fluid rushes into the vapor cavity, a high pres:ure pulse or wave on the: order of 200,000 psi results. For bubbles collapsing naar or in contact with the pump walls or impeller, the resultant pressure can cause pitting of the metal. This results in a weaker component which may later fail. 'In' order to minimize cavitation, a parameter known as net positive suction head -(NPSH) is calculated from the following. equation: NPSH = Pstatic # dynamic - saturation static is a combination of pressure head and potential head P (Refer to' Secticn 6.0). P is always a positive term. static -P is the velocity' head term. P is a negative dynamic dynamic term and becomes more negative-as the fluid velocity increases. m_

1 o-PAGE 36 0F 59 R5 MARKS LESSONTNOTE (CONTINUATION SHEET.- EVEN) From this ecuation, we can see that as long as Pdynamic + P is _ greater than the saturation pressure of the static fluid, then NPSH will be positive and no cavitation will This can be, accomplished by adding a b' coster pump to occur. increase the suction pressure for a specific pump (Psta tic)

• Another method would be to lower the saturation pressure.

.This requires lowering the fluid temperature. In a condenser, the lower the pressure (more vacuum) the better the efficiency. A pump drawing a suction from this condenser is more likely to cavi, tate since P is high compared to P and P sat dyn static

• To prevent lowering condenser vacuum (raising pressure) the only other alternative is to lower the fluid temperature.

Lowering the fluid temperature is called condensate depression. Condensate depression is the amount of subcooling_given.a fluid before it leaves the condenser. It is usually on the order of 1 - 2*F. Condensate depression lowers the overall system efficiency since more heat is being rejected from the system. However, the small loss in efficiency is more than compersated by the reduced amount of plant downtime for maintenance or replacement of pumps damaged by cavitation. 7.0 TURBINES We have seen the effect of a nozzle which converts the flow energy stored in the steam to kinetic energy of the 'steem. In order to utilize this kinetic energy, we need a device which will convert the kinetic energy of the steam to a useful work output. This device is called a Turbine. A simple turbine consists of a shaf t, a wheel, and some blades (sometimes referred to as buckets) arranged similar ~ ~to Fig. 7-1. Y f

s PAGE 37 0F 59 LESSON NOTE (CONTIt:UATI0!l SHEET - 000) REMARKS 7.0 (Cont.)- BLAoE WHEEL %/ \\ r '( ) SHAFT s, Figure 7 1 The principle of operation for a steam turbine originates f rom itewton's Laws of Motion. 1. Every particle remains in a state of rest or moves with a constant velocity in a straight line unless an unbalanced force acts on it. 2. The acceleration of a particle is directly proportional to the resultant force acting on it and inversely proportional to the mass, and the sense of acceleration is the same as that of the resultant force. 3. To every action there is an equal and opposite reaction. There are two types of turbines. One is the impulse turbine, the other is the reaction turbine. We will look at .each type. -7.1 IMPULSE TURBIriE An impulse force results whenever a jet stream is bent from ) its original course or dimini.shed in. velocity. The high velocity steam from the no :le-ic directed by the no::le 'onto'the turbine blading. flewton's'second law gives the force on the blade as .F =d t.t

o PAGE 38 0F gg -REMARKS ~ . LESSON NOTE'(CONTINUATION SHEET.- EVEN) 7.1 (Cont.) So the component of the force in the x-direction can be expressed'as Fx' 'Y ( I' x . Looking at Fig. 7.2, we can see that the force on the blade is- =b(V),-V2w) " (Y cos A - V cos B) F 1 2 x Viw Vi

1. A.

c.i. l WORK = Fs V2 Fig. 7-2 Velocity diagram 8/ For an impulse turbine blade H V 2w f G k +

i o. p I ~ ~ PAGE 39 0F 59 1.E:50N NOTE (CONTIfiUATION SHEET - 00D) REMARKS 1 7.1 -(Cont;) This force or impulse is the result of the high velocity steam from the nozzle striking the turbine blade, changing direction, and leaving the turM ne blade at some lower velocity. The force acting on'the turbine blade causes the blade to move some distance producing useful work as the turbine wheel and shaft rotate. Example: Steam flowing through a nozzle at 100 lbm/sec leaves the nozzle at a velocity of 500 ft/sec striking the turbine blade at a 30 angle moving the blade a distance of one foot. The steam leaves the turbine blade at an angle of 120' with a velocity of 250 ft/sec. How much work was done by the steam? Solution: Using the equation for force on the turbine blade F = In :(V) cos A - V cos B) 2 F=(100lbm/seck x ((500 ft/sec x cos 30 ) - (250 ft/sec x cos 120*)) r 32.2 lem. f 2 -lbf. sec F = (3.1 D ) x ((.866 x 500 ft/sec) - (.500 x 250 ft/sec)) ft lb F = (3.1 f. sec) x (433 ft/sec + 125 ft/sec) ft F = 1730 lb f Since work is equal ~ to a force acting through a distance W = Fs-W 1730 lbf x 1 ft W ='1730 ft - Ibf. ~

O o PAGE 40 0F 59 REMARKS LE550:1 NOTE (CONTINUATION SHEET - EVEN) 7.2 REACTION TURBINE Newtons third law states that for every action there is an equal and opposite reaction. A reaction force results wherever a velocity is generated. Unlike the impulse blade which is in the shape of a bucket, the reaction turbine blades form a shape which is similar to a nozzle. See Fig. 7-3. FIG. 7-3 3 TURBINE BL A0E SHAPES Y IMPULSE REACTION BL Ac!NG BL AclN G As before, from Newtons second law, the force on the blade is given as Y F = o*t l and the component of the force in the X-direction is expressed as F, = [n aVx Now, looking at Fig. 7-4, we can see the force on the blade is (Ylw ~ Y2w) * " (Y1 2 cos A - V cos B) F x

,F 4+% \\4A +/ %'+4 _ eEE A< EAT,em TEST TARGET (MT-3) 1.0 5 Ea EM y y El Il [m HE I.8 J.25 1.4 1.6 4 6" MICROCOPY RESOLUTION TEST CHART +//p 5*r%

1. 5+Q 4y>,,if k

hcy O n=

e y / 4k / . < >+ yy N'\\ IMAGE EVALUATION NNNN TEST TARGET (MT-3) i 1.0 !!EaDaa !ll NE I.I y,'" lE .2 i'.25 y.g g 4 6" 5 4 MICROCOPY RESOLUTION TEST CHART 4}5l%

• +%p

// p>?,)fjjp $)<3 f f r e~4-L. .._. 1.> ~l _j 0 e g

N s> - PAGE 410F 59 LESSON fl0TE (C0ftTIf10ATI0fi SHEET - ODD) REMARKS 7.2 (Cont.) 9 VIN k A l v, l @ fh = W = Fs /jy FIG. 7-4 l VELOCITY DIAGR AM V2 FOR A REACTION B l TURBINE BLADE. l*-- Vz w Since the velccity V) is usually very small, the force . acting on the turbine blade is due to the velocity of the steam leaving the blade. The velocity leaving the turbine blade is much higher than the velocity of the steam at the blade entrance. The velocity increase is due to the nozzle effect as the steam passes through the blading. The reaction force produced will cause the turbine blade to move some distance resulting in useful work as the turbine wheel and shaft rotate.

g 420F 59 PAGE REMARKS. ' LESSON NOTE (CONTINUATION SHEET - EVEN) 7.3 . TURBINE CLASSIFICATION. In-a pure impulse turbine, the steem pressure decreases and the velocity of the steam increases as the steam passes through the nozzle.. The steam coming from the nozzle enters the moving ~ turbine blading. Some of the steam's kinetic energy.is imparted to.the bicding, and the steam emerges from the moving blade at a lower velocity. No pressure drop occurs in the pure impulse turbine stage, enly a decrease in steam velocity. In a_ pure reaction turbine, the steam enters the moving turbine bitding at zero velocity. The steam decreases in pressure while passing through the turbine blading and generates a velocity. It is impossible to build a pure impulse or pure reaction turbine. All turbines have some impulse effect and some reaction effect occurring within the blading.s?.~urbines are classified depending on which of these effects predominate. There are four main ways to classify turbine stages by blading (a) Simple impulse staging .(b) Velocity staging (c) Pressure staging (d) Reaction staging 7.3.1 SIMPLE IMPllLSE STAGING Tht-simplest type of turbine. is one containing a single impulse stage called,a Delaval turbine. A stage consists of one set of nozzles and one set of' moving turbine blades. A pressure drop occurs'across the nozzles with a resultant velocity' increase in the steam. Since all of the pressure drop is across one' set of nozzles, the velocity of the steam is very high. The steam enters the turbine blading at high L

P; ,~ Ep. - - - ou" g. i h p. L p PAGE 43 OF 59 - LESSON NOTE' (CONTIfiUATION SHEET - ODL} REMARKS:. 7.3.1 (Cont.): - velocity,' gives.up some of its kinetic energy ~to. the turbine blade, and: leaves at a' lower velocity. The steam retains a considerable unused portion of its-kinetic energy. 'It can be shown-; theoretically that'the optimum blade speed for-an impulse turbine is approximately. one-half-the velocity of the' steam entering the turbine blading. Since the velocity of the' steam in the Delavel turbine is' very high, the l turbine 'must. operate at high velocity to achieve any reason-- able efficiency. The high velocity of the steam and turbine. creates-a large amount of friction. Thus, the~ efficiency of the Delaval turbine is very ~ poor and is.normally used to . drive small plant auxiliary equipment. l .It'is.often helpful.to look' at the pressure-velocity diagram for a. turbine. stage or set of turbine stages. These diagrams help to visualize what is happening in the turbine as steam flows through it. Fig 7-5 is a'P-V diagram for a simple -e irpulse stage. Steam pressure drops through the nozzle with 3 resultant increase.in velocity and specific volume. The pressure and specific volume of the steam remains constant-across-the turbine.blading. The steam velocity decreases due to'the turbine blading absorbing _some of the kinetic l energy of-the s,aam. .d l h i L [.- i' uc - :. =

4 L PAGE 44 0F 59 REMARKS LESS0ti t10TE (C0:lTIriUAT10ft SHEET - EVEft) 7.3.1 (Cont.) N ,a V MB-N = N'OZ Z LE. (j ! MB: MOVING BLADE O l

---

_q j,,,,__

PRESS. i l /l l l l ll l l / l ! VELOCITY i i/ FIG. 7-5 l/ l l PR ESSURE -VELOCITY

---

y l

l DIAGRAM FOR A 8 8 i SIMPLE IMPULSE STAGE TURBINE CASING 7 /MA% N / 8 m TURBINE SHAFT

q _ R. co: w PAGE 45 0F 59 l LESSONNOTEI(CONTINUATIONSHEET:-000) REMARKS ~ 7.3.2 VELOCITY STAGING-Velocit'y staging gets its name from the pressure-velocity relationship across the turbine blading. The pressure ~ across the turbine blades is constant while the velocity decreases across each set of moving blades. The velocity staged turbine, called a curtis wheel, consists of a nozzle, a set of. moving blades, a set of_ fixed or stationary blades, and another set of moving blades. (See Fig. 76.). All of -the steam pressure drop occurs through the nozzles resulting in a very -high steam velocity just as we observed in a-simple impulse turbine. However, in the velocity staged turbine, the steam enters one set of moving blades and has energy extracted rasulting in a lower steam velocity. The steam now enters a set of stationary blades. Here, the magnitude of the velocity remains constant, but the direction of the steam is changed and directed into a second set of ~ moving blades. Again, more energy is extracted from the steam in the second set of moving blades resulting in a lower velocity than before. The turbine speed for this type of staging does not'have to be as high as that for a simple impulse stage. There are more sets of moving blades to extract,the steam's energy making it more efficient. Fig._7-6 is a pressure-velocity diagram for'a velocity staged impulse turbine. m.

-g e 9 PAGE 46 0F 59 REMARKS -LESSON NOTE (CONTINUATION SHEET - EVEN) 7.3.2 (Cont.) TURBINE CASING-TURBINE SHAFT N = Nozzle FB MB MS = Moving Blade N MB V O FB = Fixed Blade I ~ j l l Fig. 7-6 _,0,LUME,,],_,,,,,_,4____ V p___________'! 8 l l Pressure - velocity diagram PRESS.i l, l l for a velocity staged impulse turbine. _j ,i i i i l l ll i I ll .l l I I 'l i e i ,t l VELOCITY i I e i I / I l l f,#,/- l l 1 i i i 3 I l l gf. _ __ _ _ s l l l l l

I+' s-p a pAGE 47 0F.59 LLESSON NOTE---(CONTINUATION. SHEET - ODD) REMARKS

7. 3. 3 -~- PRESSURE. STAGING,

Pressure' staged' tur' ines are known as rateau turbines. b The turbine consists of many-simple impulse turbines in. series. -Again, just-as we observed in-the simple. impulse turbine, the pressure. drop of the steam occurs through the nozzles. By;having many sets of simple impulse turbines in ~ series, the pressure drop.across-any one set is quite-small. The'small pressure drop will produce a lower steam , velocity from the nozzlas..This means that the turbine can ~ operate efficiently'at slower speeds. This increased ' efficiency is due'to loser internal and frictional energy losses -in the steam and turbine assembly, respectively. The advantage of increased efficiency is unfortunately off-set by.the increased construction cost. Fig. 7-7 is a pressure-velocity diagram for a pressure staged turbine. 9

g-PAGE 48 0F 59 -REMARKS LESSON' NOTE (CONTINUATION SHEET - EVEN) 7.3.3 (Cont.) TURBINE CASING \\ Fig. 7-7 / Pressure - Velocity N / / diagram for a /[/ / h \\_ Pressure - Staged Turbine. NNNNNNNNNNN NNNN'N NNN' TURBINE SHAFT N N N MB MB MB 'A U A V A V N = Nozzle i e i PRESS. ! l l l l l MB = Moving Blade I i g g n, l l l l l t /y_________+l l l '/ l: l l i l l i i, p__

______7, l

l i i i 3 f i l 'l l l i l l l l i jf----------f'l l l -j vstocir e e /, g i i f I I l l l-l l l l 4 I I l 1 I e (

a p: PAGE 49 0F 59 LESSON NOTE (CONTINUATION SHEET - 000) REMARKS 7.3.4 REACTION STAGING Reaction staged ~ turbines derive their. classification from the. type'of turbine blading used - reaction blades. (Refer to Fig. 7-3). In reality, this is actually an impulse-reaction staging becat ' the steam enterinq the blading has some intitial velocity producing an impulse effect. How- ~ ever, the reaction effect is predominant. This type of r turbine is called a Parsons Turbine. In'this type of staging, the stationary and moving blades are designed so that the passageway through the blades converge then diverge. Each set-of blades,' both stationary and moving, is actually a group of convergent nozzles. There is a pressure drop across both stationary.and moving blades as the steam passes through the blades. In the stationary blades, the p'ressure urops with a resulting increase in steam velocity. (See Fig. 7-8). The steam is then directed by the stationary blades into the moving blades. The moving blades absorb some of the steam's kinetic energy-(impulse effect) but also generate some steam velocity due to the nozzle shape of the turbine blades (reaction effect). The overall effect is a decrease in steam velocity across the moving blades. The steam leaves the moving blades and enters another set of stationary blades;. and the process continues. The parsons turbine is'most efficient when used with low pressure steem to utilize the low velocities produced. Fig. 7-8 is a pressure-velocity diagram for a reaction staged turbine. 1

9 PAGE 50 0F 59 - REMARKS LESS01 tiOTE-(C0tiTit1UATI0t1 SHEET - EVEN) TURBINE CASING '// /////////// \\\\ \\ FIG. 7-8 \\ ). ,/ N // \\ // N PRESSURE - VELOCITY DIAGR AM ' 1 1 1 FOR A RE ACTION - STAGED xNNNNNNNNNNNNN' TURBINE SH AFT FB MB FB MB FB MB FB = FIXED BLADE MB = MOVING BL ADE V V & SW' & thv' l l l f- - - - - - PRESSURE ! l i i i i 'r -l l l l l -l l l l l,' l l l l' l l l s ,'f l l l l l e n s a l l l s i l l l l . f, i l i i i, l l

,~

i l i .e s, s 'l l l l 4 i VELOCITY l i k i l i i i i i i l', -l l l VOLUME _________f j l l l l l l l 8 8 I i l i I i

t 0; ~ PAGE 51 0F 59 LESS0tl NOTE (CONTINUATION SHEET'- 000) REMARKS-7.3.5 -STEAM FLOW PATH IN A TURBIflE - Besides' classifying a turbine by blading, a turbine is also - classified by the flow path of steam in the turbine. Steam flow can-be in the radial direction or axial direction. Steam flowing in the axial direction can be in'one direction or two directions. Fig. 7-9 shows steam flow in the. radial direction. Fig. 7-10 shows steam flow in the axial direction. - STEAM Flow FIG. 7-9 R ADIAL Flow TURBINE } O v STEAM Flow STEAM Flow W I h -3 h 3 slNGLE ' AXIAL' Flow couBLE AXIAL Flow FIG 7-lo AX1 AL Flow TURBINE

p o F, 7: PAGE -b20F 59 REMARKS-LESSON NOTE-(CONTINUATION SHEET.- EVEN)- . 7.3.6 - STAGING COMBINATIONS We have looked at the variousitypes of turbine staging. depending on the designed use of the turbine, turbine j staging is often combined to achieve maximum' efficiency. The curtis wheel is more efficient-at high steam pressures. .The'rateau and parsons turbines are more efficient at lower steam pressures. There are numerous combinations of turbine taging that can be utilized. For. example, a curtis wheel has four times the energy absorbing capacity of a Delaval, turbine (Velocity staged vs. simple ~ impulse-stage). Fori - this reason, curtis wheels are often used as the first stage in combination curtis-rateau or curtis-parsons staged turbines. The curtis wheel absorbs a large amount of energy in one stage and limits the amount of the turbine subjected to high i initial steam pressure and temperature. This allows the .following stages to be built with less weight and at lower cost. The combinations depend upon steam pressures to be used, efficiency. desired, size of turbine required, and cost of turbine units. Turbines with various types of turbine staging are classified as pressure-velocity compounded, velocity-reaction compounded, etc.- Example: A turbine is classified as.a velocity-pressure-reaction compounded, single-axial-flow turbine. Make a drawing of this turbine, and show the pressure-velocity diagram for this turbine. Solution:. Look at Fig. 7-11 -1 i

e e PAGE 53 0F 59 LESS0!l NOTE (C0iTIt:UATIO!! SHEET - CDD) REMARKS 7.3.6 (Cont.) N MB FB MB N MB N MB FB MB FB MB V O V V bbb Av vnvnvOUeU l 1 l l l l l 1 i i I (---~ l j j l l l l l ,ji,, ' l PRESS. t i i i e i i l l l e i i i i g l-e i l l l l l l i i 'l i i i i l i i i i l I i,- l l l l l l l l l l .-l' l l i i i i i l i i i . -l-l l i i i i i i i i i i e i j i i i j i I I t I i, 3 f 9 i a g i-/ i I I g e i i I i / i i g I l l l l l <l,-----t, ' l l l l l i i i i i i i g i 1 e g i I e t i i l i 4 I l 3 e g l i i i i i i i i i i i i. i i i i i, i l l' l l l l i l ,p-- - ----4----<, l l l l l l l l i i i e i ,l i 1 1 I i ,l i t l i i i i i i i l i i i i ,i ,i i i i i i i l l l l l VELOC TY ' l l l i ,f i i i i I I I i I l i I i I g i i VOLUME 'e I i i i i i i e i i i i i i i i i i i i FIG. 7-ll i

4 PAGE 54 0F 59 ' REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 7.4 TURBINE DESIGN CONSIDERATIONS 'A typical turbine assembly consists of a number of parts arranged similar to Fig. 7-12. .m E ~ 2' STEN 4 r-r ~~' - BLADE CHEST N { '3 J L hb ~" N0ZZLE - CASING - DI APHRAf4 ROTOR u / l n ROTCR NSHAFT SHAFT ? SEAL SEAL. e, [] N - \\ / CASlfG \\ TURBINE WHEEL FIG. 7-12 TYPICAL TURBifE ASSET'BL'I i l I I

y ~ x y ~ PAGE 55 0F 59 .LESS0:1fl0TE (CONTIttVATI0tt SHEET - ODD) REMARKS -7.4 I(Cont.) The turbine wheel-and rotor may be forged as one unit such as small turbine units. For. larger units the turbine wheels are shrunk and pinned to.the turbine rotor. The edges of.the turbine wheel are grooved (Dovetail) to position the turbine blades and weld them in place. The nozzles are drilled in the turbine casing (sometimes cast). The diaphrages are positioned between stages to minimize stear. leakage and provide pressure boundaries.for each-stage. Each diaphragm forms a seal at the shaft (similar to the shaft seals in the casing) to minimize steam leakage. For velocity staged turbines, the.diaphragns have holes drillec' to equalize pressure between the two moving sets of blades since no pressure change is to occur. The shaft seals are parallel groeves between the turbine casing and the rotor. This type of seal is called a LABYRITH SEAL. (See inset of Fig. 7-12). Looking at Fig. 7-12, notice the size of the turbine blades is increasing in length. This also results in the turbine diameter increasing. The reason for the increasing size of clades and turbine diameter is -two-fold.

1. 'To equalize the torque along the shaft length.

If the torque.is not equalized, then the shaft or rotor would twist and, break. We can visualize the changing torque -as the. steam moves through each stage of the turbine. The steam is giving up more and more-energy. 'As the steam's energy decreases,'the amount of force the steam 1 can exert on the turbine blades becomes less and less. The energy in the steam is in the form of pressure, f4a thematically. Pressure (P) = f or F = PxA

j

u s

a PAGE 56 0F co-REMARKS LESSON NOTE-(CONTINUATIONLSHEET'- EVEN')- If we are to maintain a constant force, then the area of the turbine blades must increase. We are limited to the 'I' width of the bipdes, so the-length must increase. .2. To allow for the increasing volume occuppied by the steam. Fr'om the pressure-velocity diagrams, we can see-that for each pressure drop an increase in steam volume occurs dueLto expansion. In order ~ to minimize fluid = friction, we must increase the volume of each turbine stage to accommodate the expansion of the steam. In Fig. 7-2, we looked at the force component in the X-direction only. This force component gives ut useful work. But, what about the: force component in the Y-direction? The Y-component of force would try to push the turbine along its axis. If we design a turbine with single axial flow, then we must prevent the turbine from moving along its axis. We can prevent turbine axial movement with a thrust bearing which you will study in Phase 11. If we design the turbine for double axial flow, then the Y-components of force will cancel eac', cther. Another consideratien for' double axial flow is that twice as much steam can be-handled.by the turbine for a.particular blade size and turbine diameter. Double axial flow turbines are most useful as low pressure turbines using the steam leaving a high pressure turbine. The increased volume of the steam due to expansion leaving the high pressure turbine will not be wasted. 9

=,, 4 ~ PAGE57 ~0F 59 -LESSON NOTE (CONTINUATION-SHEE1 - 000) REMARKS 8.0' AIR EJECTORS Air ejectors are'often used.as vacuum pumps because of their -low cost, simplicity, and ' reliability. -The principle of

operation for an air. ejector depends upon the entrapment of -

a gas by a'high velocity fluid--steam. The res'ulting mixture of gas and-fluid is then compressed'in a diffuser and discharged at a pressure greater than the pressure in the gas chamber. Fig. 8-1. is a diagram of a single-stage air ejector and pressure graph. 1 l \\ I- '3 ~

n. ~ PAGE'58 0F 59 REMARKS LESSON NOTE (CONTINUATION SHEET - EVEN) 8.0 (Cont.) I I i STEN 1 l l i I l PRESSURE i l i i l ATt10SFHERE ]gg~~~~~ l tilXTURE l 8 l i i i l l l l l I l l l-c l }l l l n l STEN 4 i DISCHARGE I" # b,' i j - STEN 1 Yi f N + STEAt1 /, /pgggggga 5 IJ0Z2LE m A AIR O W4EER SUCTIOi1 l. l l FIG. 8-1 AIR EJECTOR NfD PRESSURE GRAPH 1

PAGE 59 0F 59 'LESSONNOTEf(CONTINUATIONSHEET-ODD) REMARKS 8.0 (Cont.)- Steam enters-the ' air ejector at high pressure and expands through-the convergent-divergent nozzle dropping in pressure and increasing in velocity. _The high velocity steam leaving the-nozzle passes through the air chamber and entraps-part of.the air to be removed. The mixture of steam and air-then enters a convergent-divergent diffuser that will recompress the mixture to the pressure at the discharge. It is necessary to have a convergent-divergent diffuser because the pressure in the air chamber is low and the velocity of the steam from the steam nozzle is supersonic. To recompress such a stream, the diffuser must first converge and then diverge as the velocity of the mixture drops below sonic speed. To create a low vacuum or to discharge to a high pressure system, several air ejectrrs can be connected in series. These are called first, second, third,'etc. stages. The last stage discharges directly into some system or to the a tmosphere. By using a small condensor between stages, called an intercondenser, the steam from each stage will be condensea. This will reduce the amount of gases to be compressed by the following stages. If we use feedwater or condensate as the cooling medium in the intercondenser, then the beat removed from the steam in the intercondenser can be used to preheat the feedwater, thus improving thermal ' efficiency. l J k l0. a}}