ML16256A276

Revision 309 to Final Safety Analysis Report, Chapter 6, Engineered Safety Features, Appendix 6.2C

ML16256A276
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Site:	Waterford
Issue date:	08/25/2016
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WSES-FSAR-UNIT-36.2C-1Revision 10 (10/99)APPENDIX 6.2CDECLASSIFICATION OF FAN COOLERDUCTWORK:HEAT AND STEAM REMOVALAND CONTAINMENT MIXING WSES-FSAR-UNIT-36.2C-2Revision 10 (10/99)ObjectiveAfter a Design Base Accident (DBA), the fan cooler ductwork may not survive if it does not meet therequirement of Safety Class II and Seismic Class I. This investigation is to illustrate the capability of the fan cooler system to carry out the function of containment heat and steam removal without the consideration of ductwork.AssumptionsThe air steam mixture discharged from the fan cooler will be considered as submerged turbulent jet; theanalytical treatment of the jet with regard to mixing and momentum transfer is based on the jet theories [1, 2, 3,]*. Furthermore, the containment spray enhances the mixing process by the wake action which will also be included in this analysis. Also assumed is a complete severance of the duct at the junction of the classified to nonclassifed ductwork following the fan exhaust.Inputs [4]

Q=Fan cooler capacity after accident = 44000 CFMA=Discharge opening area = 8 x 8 = 64 ft 2 R f=Radius of containment space for unobstructed jet flow = 40 ft R c=Primary containment radius = 70 ft R d=Direct distance from the fan discharge to the containment wall = 98 ft Q w=Containment spray flow rate = 1750 gpm d w=Spray drop size = 700

µV w=Droplet terminal velocity = 8.36 fps C d=Drag coefficient = 0.5=Angle between jet flow and spray flow = 45

°=Spray coverage factor = 0.95AnalysisWe can assume a round jet at the discharge even if the initial cross-section is square [3]. So we have:

r o=equivalent round jet radius, ftSince A r o=2 thus r o=A ft==64451..*Reference numbers WSES-FSAR-UNIT-36.2C-3Revision 10 (10/99)We define:

x H=length of the initial region, distance along the jet axis from the discharge plane to the pointbeyond which the jet velocity on the axis begins to change.

x Hcan be calculated from [1]:

x r H o=8or x H==8451361xft... which is within the unobstructed zone.Also we define:

veolictity jet initial o U=axis jet the on velocity m U=section cross jet entire the over averaged mass the on based velocity c U=with:°°=A d U A d U Uc A A2where:=densityU=velocityAt the end of the initial region where the main region is assumed to begin, the universal velocity profile isvalid and that leads to: [1, 2,]:

U c=.52 U mIn the initial region U m=U owith fps;11.46 60 x 64 44,000 A Q o U===accordingly, at the end of the initial region WSES-FSAR-UNIT-36.2C-4Revision 10 (10/99) fps.5.96 11.46 x 0.52 c U==Due to the fact that the momentum and heat content of the jet remain unchanged, the following relation isobtained for turbulent jet [1]:T T U U c o c o=also from the similarity of heat and mass transfer [1]:X X T T c o c o=whereTTTooa=TTTcca=XXXooa=XXXcca=with T o= initial jet temperature T a= ambient gas temperature T c= mass averaged temperature over the jet cross-section X o= initial concentration of certain species of the jet X a= ambient concentration of the same species X c= mass averaged concentration over the jet cross-section T c and X c are defined in a manner similar to U c:°°=A A c a d U a d T U T°°=A A c A d U A d X U XwithT = temperature X = concentration We then conclude:X X U U c o c o=Since the containment atmosphere is made up of air and steam, the initial density of the jet differs fromthat of the surrounding medium by the effects of cooling and condensation through the cooler. The percentage decrease in density is found less than 15 percent which is based on the output from the containment analysis. This value represents approximately the initial steam concentration difference:

WSES-FSAR-UNIT-36.2C-5Revision 10 (10/99)X0.15=the minus sign indicates defect. At the end of the initial region, one can find:XX U U0.52 c o c o==Within the free space zone, the concentration deficiency is down to half of what it was initially. This is truealso for the temperature defect. The effect of containment spray is investigated next.B.Containment Spray The interaction between the jet and the spray is studied here. Defining N = drop number density N can be found by the continuity relation between spray flow rate and pump rate:

w Q 2 c R w V w d 6 N=or2 3 6 c w w w R V d Q N=0.95 x 70 x 8.36 x 3.28 x 10 x 700 6 60/0.13368 x 1750 2 3 6== 5032 drops/ft 3The interaction between the jet and the droplet motion can be found through the momentum exchange;the momentum flux gain by the jet in the transverse direction must be equal to the total drag force to thedroplets in the same direction. The result is [2]:

()=4 d N Cos V 2 C V 2 w 2 w d2 s fps 0.427 45 cos x 8.36 x 1/2 4 2 3.28 x 6 10 x 700 x 5032 x 2 0.5 s V=°=where WSES-FSAR-UNIT-36.2C-6Revision 10 (10/99)

V s= transverse velocity gain by the jetHowever, the momentum transfer is dissipative, namely, it contributes to the mixing of the jet with itssurrounding through the wake of the drop. V s represents approximately the turbulence fluctuationsupplied to the jet by the spray.VVfps ss 20427.where=V s 2 turbulence fluctuation in the jet contributed by the spray; we also define:=2 transverse turbulence fluctuation of the jet; it is observed that [1]:V U m 2 01.In the initial region, U m = U otherefore fps 1.15 11.46 x 0.1 V 2=The ratio of the turbulence fluctuation due to spray to that due to the jet alone isV s 20427037..2Thus, the spray increases the mixing rate about 37 percent . At this point, it is felt that the mixing in themain region of the jet should also be considered since it can extend beyond the initial region. If the jet can reach to containment wall directly, the jet center velocity is found to be [1]:

d ro o m R U U 4.12=or Um = 12.4 x 4.51 x 11.46/98 = 6.54 fpsthe turbulence fluctuation isVxfps 2016540654...and the ratio of fluctuations 0.652 0.6545 0.427'V's V 2 2=The spray can increase the mixing rate by 65.2 percent at this distance. Without the spray, theconcentration ratio is WSES-FSAR-UNIT-36.2C-7Revision 10 (10/99) 0.285 U m 0.5U U Uo o c o c===with the spray, the concentration ratio can be reduced to:

0.653 1 0.285 Vs V 1 spray no ococ2+=+2= 0.172or ()0.03 0.026 0.15 x 0.172 c=we can conclude that the concentration difference is reduced below 3 percent. This situation should beconsidered sufficiently mixed.ConclusionThe fan cooler without ductwork will assure adequate mixing in association with the spray inside thecontainment. Interference between the fan cooler discharge and solid object in its path will result in further enhancement of mixing through wake action. Any form of natural circulation within the containment will also increase the mixing process. The arrangement of the fan cooler discharge without ductwork ring headers and the strength of the jet will assure the ejected cool air to be mixed well with the ambient air before it is drawn in the cooler inlet. We can conclude that the fan cooler will perform the heat and steam removal functions as designed.References1.Abramovich, G.N., "The Theory of Turbulent Jets", MIT Press, 1963 2.Schlichting, H., "Boundary Layer Theory," Mc Graw Hill, 1968.

3.duPlessis, M.P., Wang, R.L. and Kahawita, R.,"Investigation of the Near-Regionof a Square Jet," Journal of Fluid Engineering, Trans. ASME, Vol. 96, 1974.4.Drawings [LOU-1564] G-146, G-854, G-855, G-856 (6/15/76).