ML19211D360

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Engineering Evaluation of Structural Ability of Kenasco Corp Radwaste Container,Drawing 6901-002B USA DOT 7A,Type B, to Meet Special Tests of DOT Rules & Regulations,Paragraph 173.398
ML19211D360
Person / Time
Site: 07106104, 07106144
Issue date: 09/23/1969
From: Haelsig R, Susdin S
MECHANICS RESEARCH, INC.
To:
Shared Package
ML19211D359 List:
References
15038, NUDOCS 8001180099
Download: ML19211D360 (40)
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AN ENGINEERING EVALUATION OF THE STRUCTURAL ABILITY OF KENASCO CORPORATION'S RADIOACTIVE

. MATERIAL WASTE CONTAINER, DRAWING NO. 6901-002B

. USA DOT 7A, TYPE B, TO MEET THE SPECIAL TESTS OF THE DEPARTMENT OF TRANSPORTATION'S a.ULES AND REGULATIONS PARAGRAPH 173.398 C-2295,. Task 4 January 21, 1969

  • Indicates Revised Paragraph Revised September 23, 1969 Prepared for:

Kenasco Corporation Prosser Washington Prepared by:

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Sherman A. Suridin hq$ESSl

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Approved by:

No.16538

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Y R. T. Haelsig, P. E. /

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Clytt Revision

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Approved by:

R. T. Haelsig, P. E./

l Mechanics Research, Inc.

15 North Broadway Tacoma, Washington 98403 15023 MECHANICS RESEARCH INC.

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TABLE OF CONTENTS Page

1. 0 INTRODUCTION I
2. O CONTAINER CONFIGURATION 2
3. 0 CONTALNER DESIGN FEATURES 4

3.1 Shock Tube Design 4

3. 2 Inner Container Seals 4

4.0 EVALUATION OF SPECIAL TEST REQUIREMENTS 7

4.1 Thirty-foot Drop Test 7

4. 2 Puncture and Penetration 14 4.3 Reduced Pressure 14 4.4 Vibration 14
4. 5 Water Spray 14 4.6 Fire Test

........16

5. 0 WEIGH T C ALC ULA TIONS 36 5.1 Inne r Box 36
5. 2 Outer Box 36
o. 0 REFERENCES 37 4

I765;214 MECHANICS RESEARCH INC.

1. O INTRODUCTION The Department of Transportation requires that hazardous materials be transported in containers capable of surviving normal shipping and handling environments and that will remain intact after severe accident conditions. Reference 1 describes the applicable rules and regulations.

This report is an engineering evaluation of the ability of the Kenasco Corporation's Radioactive Material Waste Container, Drawing Number 6901-002B, to structurally survive the various test conditions of Reference 1.

This is a Type B container.

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2. 0 CONTAINER CONFIGURATION Drawings describing the construction and configuration of this nuclear waste shipping container are enclosed as Kenasco Corporation shop drawing No. 6901-002B, and the Bill of Material No. 6901-002C. The container consists of a one-inch thick steel outer box and a two-inch thick steel inner box. The inner and outer boxes are separated by a set of shock tube s.

The shock tubes are a special design feature of this waste c ontaine r.

These devices are intended to absorb the energy of impact and to support the inner box so that it will remain intact throughout severe accident conditions. The design characteristics of the shock tubes are described in Section 3. 0.

The cover of the outer box is secured with 461-1/4 inch diameter high strength bolts, type ASTM A490. The cover of the inner box overlaps the box sides by six inches. This overlap will ensure that the' container contents will always remain totally enclosed by the inner box.

The container is completely constructed of mild steel. This will enable the container to survive severe impact conditions by absorbing impact energy through plastic deformation. The edges are welded with full penetration of the material thickness or with an equivalent fillet area.

The empty container weighs approximately 37,000 pounds and is intended to carry a maximum payload of 5,000 lb. The payload contents will consist of fifteen Type A, 55-gallon drums.

The container is sealed by two lengths of Garlock single lip closure seals, Type ZI. The seal pressure is maintained by finger type stainless steelsprings which force the seal lip against the container wall.

The seal is rated for intermittent use up to 250 F.

After shock tube deformation, there can be a maximum of 3. 5 inches of clearance between the inner container and outer wall. The 6 inch deep inner container cover skirt will insure that the inner cover always 1765 216 MECHANICS RESEARCH INC. r

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4 remains in place. The outer seal on the cover skirt is about 4 inches down from the inner container edge so that it will always remain effective.

The outer surfaces of the inner container and inner cover are coated with aluminum foil to reduce the heat radiated into the inner

' container and payload.

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3. 0 CONTAINER DESIGN FEATURES 3.1

_ Shock Tube Design A shock tube is a thin wall tube with a diameter generally about the same as its height. It is designed to collapse at a stress somewhat greater than the material yield stress, rather than buckle. The shock tubes used with this container are shown in Figure 1.

Typical force-defleccion curves for shock tube collapse are shown in Figure 2.

The characteristic shock tube energy absorption is shown in Figure 3 in nondimensional form. For given shock tube dimensions and material properties, the amount of energy which a particular shock tube can absorb by collapsing up to 80% of its original height can be predicted by this curve.

The maximum acceleration of the arrested body may also be calculated by noting that initial shock tube collapse occurs at approximately the calculated ultimate strength of the shock tube.

The shock tube is welded to the inner box. A small gap is left between the tube and outer box texcept for a few of the shock tubes which are tack welded to the upper and lower sides of the outside box to provide stability during normal container shipping and handling). This single attachment is intended to prevent buckling of the shock tube by relative inner and outer box sliding motion.

  • 3.2 inner Container Seals The seal on the inner container, as shown on ~Kenasco Corporation Drawing 6901-002B, is a single lip split closure seal, Type 23, made by Garlock, Inc., Palmyra, New York. The seal section dimensions are shown below. Nominally, each seal is compressed.1/8. " to 3/16'.'. Under extreme conditions when the inner container sidewall is compressed against the inner cover lip, the outer seal will still be compressed from 1/16" to 5/16?.

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In general, impact in any drop orientation will tend to compress the cover against the inner container. However, in an extreme case, the f

cover shock tubes could be crushed up to a maximum of 3-1/2" and subsequent bouncing could tend to keep the inner cover away from the inner container body. The wiping depth of the outside sealis therefore designed to be a minimum of 3-1/2".

The seal is capable of withstanding intermittent temperatures up to 250 F.

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4. 0 EVALUATION OF SPECIAL TEST REQUIREMENTS The test requirements for which this container is being evaluated are given in paragraph 173.398 of the Department of Transportation Hazardous Materials Regulations (Reference 1). Since this is a Type B container, the standards of both subparagraphs b and c apply.

4.1 Thirtv-foot Droo Test The principle of the conservation of energy requires that when a falling body impacts against an unyielding surface, the impact force multiplied by the distance over which it acts must equal the kinetic energy of the falling body. The practical significance of this principle to the evaluation of an impacting container is that some deformation must occur to absorb the energy of motion.

The principle of the conservation of energy is expressed a.

2 Fd 5 = 1/2 mv For the impact of a freely falling body 2

2 gh v

=

so that Fd 6 = Wh where W is the container weight and h is the drop height.

The inner container and payload weigh approximately 26,000 lb.

The kinetic energy of the inner box and payload for a 30 foot drop condition is Wh = 26,000 x 30 x 12 = 9.4 x 106 in -lb.

The ultimate strength of the 6-inch diameter shock tubes are Fu6 * %A = 75,000 x 5. 58 = 4. 2 x 105 lb.

where e is the material ultimate strength and A is the shock tube area.

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Similarly, the ultimate strength of the 8-inch diameter shock tube is 5

Fu8 = 75,000 x 7. 27 = 5. 5 x 10 lb.

For a container impact against the top, bottom.~or sides (see Dwg. 6901-002B the total shock tube deformation required to absorb the kinetic energy of the inner box can be calculated from Figure 3 to be 6

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where N is the number of shock tubes and h is the original shock tube height.

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and 5 m = 3.4 inches The maximum inner box deceleration is 5

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= 167 g's

2. 6 x 104 1765 224 MECHANICS RESEARCH INC..

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The cover is secured in place by 461-1/4" diameter ASTM A490 high strength bolts. Each bolt is capable of withstanding a tension load of 150,000 lb. The bolt pattern is located such that at least four bolts are adjacent to each shock tube.

The peak force exerted by each cover shock tube is 4. 2 x 105 lb. and since the four bolts together can withstand at least 6. 0 x 105 lb., the cover will remain in place during impact conditions where the shock tubes react against th: outer cover surface.

The inner container surface, opposite the impact surface, is loaded like a plate during flat impact against a side. See Figure 4.

Assuming a uniform deceleration across the surface of 163 G's, the inner container cover stress and deflection, for each of the 1" thick plates, may be calculated from the table given in Roark, page 246. The factor 4

4 wb 163 x 0. 283 x 75 4

6 Et 30 x 10 xl<

= 50 For this factor, the maximum deflection is about, 6

1. 2 5 = 1. 2 x 1

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2 c d = 3. 8 Et

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=

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75 c d = 20,000 psi The maximum stress, c, due to tension and bending combined is:

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=

2 2

b 75 c = 53,000 psi MECHANICS RESEARCH INC. 1765 225

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r The ultimate strength of the steel from which this container was fabricated ranges from 60,000 to 72,000 psi. Therefore, the container panels will not rupture.

The 1. 2 ' deflection is well within the 3" clearance between t

the cover and the top of the 17H drums.

Most of the shock tubes are arranged on the inner container to react directly against the container edge and do not appreciably load the container surface. On the top, bottom and long inner container sides, two shock tubes have been placed toward the center of the panel to provide sup-port for the deceleration of the container panel plus payload. (See Figure 4)

The total downward load P, for a top or bottom panels g

plus a 5,000 lb payload is:

Pd = 163 (2 x 0. 283 x 126 x 75 + 5,000) 5 Pd = 16. 8 x 10 lb.

The upward reaction of the two shock tubes, P, is:

u P = 2 x 4. 2 x 10 = 8. 4 x 105 lb u

Since the stress and deflection of a load concentrated near the center of a plate is approximately equal to that caused by 1. 5 times the same total load if it were uniformly distributed, the action of the center shock tubes will iargely cancel the stress and deflection of that caused by deceleration of the panel plus payload.

The inner lip of the inner container cover must prevent the inner container sidewalls from excessively bending inward. There are two shock tubes which directly load the inner lip and two more which contribute to the load. Assuming that the force against the inner lip is 1765 227 MECHANICS RESEARCH INC.. __..._.

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equivalent to the peak load of three shock tubes, the peak force is, F = 3 x 4. 2 x 105 = 1. 26 x 106 lb p

If this force is uniformly distributed along the inner lip, the shear stress e, is:

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1. 26 x 10

10,000 psi g

=

A 124 x 1 The shear strength of the weld is about 20,000 psi so that the inner lip will not tear away from the cover.

4. 2 Puncture and Penetration A series of experiments were performed, as described in Reference 5, to determine the puncture resistance of flat panels. The re-sulting plot of puncture resistance for various skin thicknesses and weights is reproduced in Figure 5.

Plotting the container total weight and outer skin thickness on these curves indicates that a 42,000 lb. container with a one-inch thick skin panel will pass this test without skin penetration.

The penetration test consists of dropping a steel rod weighing 13 lbs. through a distance of 40 inches onto the package surface. The energy of this 6

condition, 520 in-lb, as compared to puncture test energy,1. 7 x 10 in-lb, in-dicates that if the skin panel will survive the puncture test it will easily survive the penetration test.

4.2 Reduced Pres sure The inner container seal is rated as being capable of withstanding a 7 psi pressure. The inner container walh can, of course, easily withstand this pressure differential.

4.4 Vibration Vibration normally incident to transportation will not affect this c ontaine r.

4. 5 Water Sorav Since its construction is entirely of metal, the container is exempt from this requirement.

MECHANICS RESEARCH INC. i765 228

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4.6 Fire Test The container has been analytically subjected to the required 1475 F fire test for a 30 minute period, using a transient electrical network analogy program (reference 8). The results of this analysis show

. for the modified design that the temperature of the payload at no time exceeds 160 F.

Figure 6 illustrates the thermal and time history response of the container. The - key feature of the modified design is the applicatien of the thermal control surface to the outer surface of the inner box. This thermal control surface consists of bonded aluminum foil. Figure 7 illustrates the response of the original design without this thermal control c oating. The temperatures of the payload in the unmodified original design approach 240 F.

The aluminum foil thermal control coating is bonded to the steel inner box using a contact cement, 5cotchgrip Rubber Adhesive - 1300, which maintains adhesive integrity up to 300 F.

Figure 8 illustrates both the analytic model used in this analysis anc the equivalent electrical analog. The outer box wall has been subdivided into three nodes representing the thermal capacitance and conductance through the thickne.as of the outer box wall. The inner box has been idealized as four. nodes, again representing both conductance and capacitance through the thickness of the inner box wall. The assumed emissivities for this problem are as follows:

Heat Source:

c=.9 Outer Container Walls:

c=.8 Inner box Walls Outer wall:

c=.2 1765 230 Inner wall: c =. 8 Payload:

c=.8 The model includes the following special features:

Coupling between the outer box wall and the inner box wall e

consists of a radiant term, a temperature dependent air conduction term and a shock tube conduction term.

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Coupling between the inner box wall and the payload consists e

of a radiant term, and a temperature dependent air conduction term.

Heat input consists of a programmed source temperature e

equal to 1475 F acting through a radiant coupling term to the outer box wall. This heat source at 30 minutes drops 0

to a temperature of 70 F.

At thirty minutes, a free convection and air radiation term is e

introdaced coupling the outer box wall to an air temperature of 70 F.

At 3-1/2 hours into the test sequence artifical cooling on the outer box wall is introduced as noted in Figure 6 This reduces the outer wall temperature to approximately 70 in about 15 minutes.

The details of this analysis are summarized in the following calculations:

Nocal Caoacitances A.

Outer Wall Nodes - 2,3,4 Total Weight = 13,860 lbs., c=.11 for steel C, 3,4 =.11 x 13,860

= 508. 2 Btu / R 2

3 B.

Inner Wall Nodes - 5,6,7,8 Total Weight = 23,120 lbs; c =.11 for steel C, 6, 7, 8 =.11 x 23,120 = 63 5. 8 Btu / R 5

4 C.

Payload Node - 9 Total Weight = 5,000 lbs A s sumed Specific Heat =. 33 Btu /lb-R C9 = 5,000 x. 33 = 1650 Btu / R 1765 234 MECHANICS RESEARCH INC.. _ _ _

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Evaluation of Conductive Resistors ij Ra..j = Aij kij A.

Outer Wall Resistors - 2,3 2

Area = wt

= 13,860 48,975 in

=

pt

(. 283)(1) 2 340.11 ft

=

= 1/3 inch /12

=. 02777 ft.

k = 25 Btu /hr-ft OF (1% carbon)

R,3

.j2777 ft (3600 sec /hr) 2 2

( 340.11 ft )(25 Btu /hr-ft-F)

=. 011757 R /B tu/ see B.

Shock Tubes - 4 2

A rea = 54 3 in,

5 in 4

(5 /12)

R 15.911 R/B tu/s ec

=

g 543j(25) 144 C.

Inner Wall Resistors, 7,8,9 2

Area = wt

=

23,120 lb

= 40,848 in pt

(.283)(2)

= 283. 667

=

2

=

. 041666 ft 4.12 R, g, 9 =

.041666 (3600) =. 021148 R/ Btu /se c 7

(283. 7)(25)

MECHANICS RESEARCH INC.. - -.

o Radiation Resistors Radiation Resistors assume conservative values for emissivities consisting of c =. 8 for bare metal surfaces and c =. 2 for aluminum foil surfaces.

The conservatism of these assumptions is verified by Table I presented on page 4-110 of reference 6.

1 Ri '

  • 3 Kij [(Ti)2 + (Tj)d] [Ti + Tj]

CAi Kij =

(c1 1

Ai I

-1) + (F j ) + Aj

( c; - 1) i i

c =.1714 x 10-8 B tu /hr-ft2o 4 g

=. 4761 x 10-12 Btu /sec - ft 2 4

R A.

Inner Wall to Outer Wall Coupling Resistor - 6 2

Ai = 283. 67 ft Aj = 340.11 ft~

Fij = 1. O ci =. 2 (1), =. 8 (2) cj=.8

(. 4761 x 10'IZ) (283. 6 7)

(II K6*

(1 -1) +1 +

(1 -1) 283.67

.2 340.11

.8

= 25. 932 x 10'I Btu /sec - R4

(. 4761 x 10-12) (283.67)

(2)

K

=

(I- -1) +1 +

(L,l) 283.M

.8 340.11

.8

= 92. 63 x 10' Btu /sec - oR4 1765 236 MECHANICS RESEARCH INC..

-e-wa==

j

~

~

B.

Payload to Inner Wall Coupling - 11 A s sume K33=

92631 x 10-10 Btu /sec - R4 C.

Radiant Source to Outer Wall - 1 Assume A

= 1. O 3

A 2

(.

x

(

(1 600)

K =

(

- 1 ) + 1 + (1. )

(

-1)

.8 9

=.118976 x 10-9 Btu /sec - R4 D.

Space Radiation - 12 12 = (.1714 x 10-8) (340.11) (. 8) (1/3600. )

K

~9 4

=.129543 x 10 Btu /sec - R Temnerature Denendent Air Conduction Resistance The conductivity assumed for air uses the imperical relation given in section 16 of reference 7. Table 20.

i Ri_

i 1

Akii Ai ki where k = k 492 + C

[T T 1. 5

.2

, Btu /hr/ F/ft T+C (492)

For Air:

K.

0.155 3

C

= 225 A.

Wall Airgap - 5

((A/

-3 5/12

= 1. 468 x 10

fft,

=

\\

5 283.667 MECHANICS RESEARCH INC.

1765 237 B.

Payload Airgap - 10

= 2 /12

= 5875 x 10-3/ft A o 283.667 R

ki [B ru /hr-ft-F]

1/ki

[ft-sec-R/ Btu]4 6

530.

.013715

. 262486 x 10 6

810.

.018902

.190456 x 10 1050.

.022646]

.158968 x 106 1310.

.026213

.137336 x 106 6

1550.

.029176

.123389 x 10 1810.

.032113

.112104 x 106 6

1950.

.033593

.107149 x 10 Note: See following page Combined External Air Convection and Radiation Following application of the simulated 1475 degree fire the outer box wall is assumed to be cooled by combined air convection and radiation. The coefficient utilized for this combined equivalent convection loss has been taken from the data presented in Table 11, pages 4-106 and 4-107 of refe rence 6.

q = (he + hr)A aT h = (he + hr) @ 150 AT = 2. 40 (Avg. ) B

/ft -hr / F R=

1

=

1 (3600 sec /hr) hA (2. 4 btu /ftz-hr / F) (340.11 ftd)

= 4. 41 F/ Btu /sec This resistor switches in @. 5 br.

Sample output from the computer program, reference S, is included in the following pages and to substantiate the plotted results, Figures 6 & 7.

1765 238 n!ECHANICS RESEARCH INC.

~-

~

1..E % [*,.
  • i.

1.0ADER 6M CORE

^----- T [

~

-Z ggd" T(R) =53P.0 T(F) = 70 0 M(IN) = 0 86456E+00 M(FT) = 0 137.15E-01*

i 1... *

  • TTXI"755070 T(F) = 90 0 M(IN) = 0 16949E+00 M(ff) = 0 34824E-Ul E*

T(R) =570 0 T(F) = 110.0 MCIN) = 0 17432E*00 M(FT) = 0 14527E-01 T(R) =590 0 T(F) = 130 0 MCIN) = 0 17907E+00 M(FT) = 0 14923E-03 T(R) =610 0 T(F) = 150 0 MCIN) = 0 38374E+00 M C F T) = 0 15312E-01 1

T(R) =630 0 T(F) = 170.0 MCIN) = 0 18834E+00 MCFT) = 0 15695E-01 T(R) =650 0 T(F) = 190.0 M(IN) = 0 19287E+00 M(FT) = 0 16073E-01 T(R) =670 0 T(F) = 210 0 MCIN) = 0 19733E+00 M(FT) = 0 16444E-01 T(R) =690 0 T(F) = 230 0 M(IN) = 0.20172E+00 M(FT) = 0 86810E-03 T(R) =710 0 T(F) = 250 0 MCIN) = 0 20605E+00 MCFT) = 0 17371E-01 T(R),=730 0 T(F) = 270 0 M(IN) = 0 21032E+00 M(FT) = 0 17527E*01 T(R) =750 0 T(F) = 290 0 M(IN) = 0 23453E+00 M(FT)

=.0 37876E-01 T(R) =770 0 T(F) = 310 0 MCIN) = 0 21866E+00 M(FT) = 0 38224E-01 T(R) =790 0 T(F) = 330 0 M(IN) = 0 22278E+00 M(FT) = 0 16565E-01 T(R) =610 0 T(F ) = 350.0 KLIN) = 0 22682E+00 MCFT7 = 0 18902E-01 TCR)7 63070 T(F) = 370.0 M(IN) = 0 23082E+00 "R(FT D 'O.19235E-Ol*

TCR) =850 0 T(F) = 390 0 MCIN) = 0 23476E+00 M(F T) = 0 19563E-01 T(R) =870 0 T(F) = 410.0 M(IN) = 0 23665E+00 M (F T) =

'90 i

0.19666E-Ol'*

T(R) =890 0 T(F) = 430 0 MCIN) = 0 24250E+00 MCFT) [a 0 20206E-01 T(R) =980 0 T(F) = 450 0 MCIN) = 0 24630E+00 McFT) p 0 20525E-01 TCR) =930 0 T(F) = 470 0 M(IN) = 0 25006E+00 M(FT) = 0 20636E-01 T(R) =950 0 T(F) = 490 0 MCIN) = 0 25378E+00 M(FT) = 0 21148E-01 T(R) =970 0 T(F) = 510 0 M(IN)

  • 8 25745E*00 M C F T) = 0 21454E-01 CD T(R) =990 0 T(F) = 530 0 MCIN) = 0 26108E+00 M(FT) = 0 21757E-01 T(R) 4010 0 T(F) = 550 0 M(IN) = 0 26468E+00 MCFT) = 0 22057E-01 T(R) 9030 0 T(F) = 570 0 M(IN) = 0 26623E+00 M(F T) = 0.22353E-01 T(R) 1050.0 T(F) = 590 0 M(IN) = 0 27175E+00' M(FT) = 0 22646E-01 "ICAT"4070.h T(F) = 610 0 M(IN) = 0 27524E+00 MCFT) = 0 22937E-0!*

T(R) M090 0 T(F) = 630 0 M(IN) = 0 27869E+00 M(FT) = 0 23224E-01 TCR) =110.0 T(F) = 650 0 KCIN) = 0 28210E+00 M(FT) = 0 23509E-01 T(R) =130 0 T(F) = 670 0 M(IN) = 0 28548E+00 M(FT) = 0 23790E-01 T(R) =150 0 T(F) = 690 0 M(IN) = 0 26663E+00 M(FT) = 0.24069E-01 CD T(R) =170 0 T(F) = 710 0 MCIN) = 0 29215E+00 MCFT) = 0 24346E-01 T(R) =190 0 T(F) = 730 0 MCIN) = 0 29544E+00 M(FT) = 0.24620E-01 T(R) =210 0 T(F) = 750 0 M(IN) = 0 29670E+00 M(FT) = 0 24691E-01 T(R) =230 0 T(F) = 170 0 M(IN) = 0 30192E+00 M(FT) = 0 25160E-O!

T(R) =253 0 T(F) = 790 0 MCIN) = 0 30512E+00 MCFT) = 0 25427E-01 T(R) =270 0 T(F) = 810 0 M(IN) = 0 30830E+00 M(FT) = 0 2569tE-01 T(R) =290 0 TCF) = b33 0 k(IN) = 0 31144E+00 M ( F T) = 0 25953E-01 T(R)

T(F) = 650 0 M(IN) = 0 31456E+00 "f(RT"=1310 0330 0*

T(F) = 870 0 M(IN) = 0 31765E+00 M(FT)

'k(FT,m,0 26213E-p; 0 26471E-01,

) =

T(R) =350 0 T(F) k 890 0 MCIN) = 0 32072E+00 M(FT) = 0.26726E-01 T(R) =370 0 T(F) = 910 0 MCIN) = 0 32376E+C0 MCFT) = 0 26960E-01 T(R) =390 0 T(F) = 933 0 M(IN) = 0 32678E+00 k(FT) = 0 27232E-Ol T(R) =410 0 T(F) = 950 0 MCIN) = 0 32977E+C0 M(FT) = 0.27461E-Cl T(R) =430 0 T(F) = 970 0 MCIN) = 0 33274E+00 M(FT) = 0.27729E-01 T(R) =450 0 T(F) = 990 0 M(IN) = 0 33569E+00 M C F T) = 0 27974E-01 T(R) =470 0 TCF) =1010 0 M(IN) = 0 33862E+00 M(FT) = 0 28218E-01 TCR) =490 0 T(F) =1C33 0 M(IN) = 0 34152E+00 M(FT) = 0 26460E-04 T(R) =510 0 T(F) =1050 0 MCIN) = 0 34440E+00 MCFT) = 0 287COE-01 T(R) =530 0 T(F) =1070 0 MCIN) = 0 34727E+00 M(FT) = 0 28939E-01

~T(R) 4550 0 T(F) =1090 0 MCIN) = 0 350llE+00

,MCFT) 0 29176E-Ol

=

TCR) =570.5 T(F) all10 0 MCIN) = 0 35293E+00 M(FT) = 0 29411E-01 TCR) =590 0 T(F) =1130 0 MCIN) = 0 35573E+00 MCFT) = 0 29644E-01 T(R) =610 0 T(F) =1150 0 M(IN) 0 35853E+00 MCFT) = 0 29876E-01

=

TCR) =630 0 TCF) =1170 0 M(IN) = 0 36128E+00 M C F T) = 0 30106E-01 T(R) =650 0 T(F) =1190 0 M(IN) = 0 36402E+00 MCFT) = 0 30335E-01 T(R) =670 0 T(F) =1210 0 MCIN) = 0 36675E+00 M(FT) = 0 30562E-01 TER) =690 0 T(F) =1230 0 M(IN) = 0 36946E+00 M(FT) = 0 30768E-DI T(R) =710 0 T(F) =1250 0 M(IN) = 0 37215E+00 MCFT) = 0 31012E-01 T(R) =730 0 T(F) =1270 0 MCIN) = 0 37482E+00 M(FT) = 0 31235E-01 T(R) =750 0 T(F) =1290 0 M(IN) = 0 37748E+00 M(FT) = 0 33457E-01 TCR) =770.0

'T(F) =1310 0 M(IN) = 0 38012E+00 M(FT) = 0 31677E-01

~

TCR) s790 0 T(F) =1330 0 M(IN) = 0 38274E+00 MCFT) s.0 31695E-Cl T(R) 1E!0 0 T(F) =1350.'O MCIN)'s 0 38535E+00 M(FT) = 0 32113E-01*

""TCR)"=830'.0' T(F) =1370 0 M(IN) = 0 36794E+00 KLFT) = 0 32329E.Ol T(R) 4E50 0 T(F) =1390 0 M(IN) = 0 39052E+00 M(FT) = 0 32543E-01 T(R) 0870.0 T(F) =1410 0 MCIN) = 0 39308E+00 MCFT) = 0 32757E-01 TCR) ze90 0 T(F) =1430 0 M(IN) = 0 39563E+00 M(F T) = 0 32969E-01 T(R) s1910.0 T(F) =1450.0-M(IN) = 0 39816E+00 M(FT) = 0 33180E-01 TCR) 4930 0 T(F) =1470.0 MCIN) = 0 40067E+00 MCFT) = 0 33390E-01 T(R) 0 T(F) =l490.0 M(IN) = 0.40318E+00 M(F{L=,0,.,3159%~0k TYPE G TO CONTINUE.P"T? "EXI T.

X C_1 EXIT

'C s

e

. COST CONNECT S2 26

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s, "

COMPUTE $1 78

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