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prob.vii-1 - II lie‘ ii is"'J its-n iIiretil-li‘.1...

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Unformatted text preview: II - .- lie‘. ii ' is ."..'J' its-n. iIiretil-li‘. .1 at]. I_ :i'lillill .1]: r'- 'i|"- :..' “ii It in .F-éill'ailin-n . -;- i .3 t:- l l‘ i in: I _I Tau :l I'- - ”fies" t .: 1H 1-. 2- file“ . . F " - L‘I'1 seen? *I- ' in a that“: " '- .11, l .. 'H 'lis - lWehee program PROBLEM 7- 1' SOLUTION Containment Pressure Analysis The objectise of this problem is to find the peak containment pressure of the reaeter described' to Example ?- l for a large break LUCA with a St] lube rupture accident. We will determine the contairunent hamper-anus: and pressure at State 3 (me final equilibrium state after the SG tube rupture} using the results from State '2'. of Example ?-1.The following conditions Erotn State 2 of Example T-l are given below: 1-34TE-33 where the mass of water in are secondary can he found by: a? V1.5 to We as earns 1.34? E— llflm 1' kg— Fer the transition from State 2 to 3. we realte the following assumptions: I Neglect heat transfer from me containment to the structures (QM, = I Neglect decay heat [OHM = [1} Ir Neglect effects of water displacement of air an air partial pressure - Assume 56‘ s are solid Starting with Eq. [ii-13}: Vr = rum-[tr 3{T:.m} + .rat-r;1{Tr.:n}] ll [ -—-]-VTW:I{T1:¢'J MW] 'W'x ll: Thrill!) I}: From this expression. we are left with two unknowns, x3 and T3. Therefore. we need another equation relating these two variables. This equation exists as Eq. ['l-lS} which states: Rev member ! t. 1'“? Page I of2 panes h mw?*mWr‘i mans +m'3JU-a “‘ mep‘i'mwa ”t m‘w'f- WIT'afjlull . Problem ":51 Solution ilnwflE-r W‘IW UL}; "t mael5\-\(tilwttl1 ‘ iflw [mm + "MPH“: ' “2} + mull-la ' “113' + "lacy—{T3 ' T2} = ‘3 thrfl. I]: = ufE{T3JBt) + Kfiulgsfram) m +tu =2.1lEfl5kg u: = 1556 Lang m“, = :5. ease: kg us. = 1255 tang to. = SEEM kg C” = 719 Hike “K T1 = 415.5 “a: Reananging for as: x, = manamflmzm + rnee lemme m finial mwellngIfTflefli) + (mw:+mfivp}uffl {Tint} l3:- Mflnmmhnemmgn mi- 1255 area + {5. gen: 1531(le on -°on3__l.aflst {EIfiDElM kg}llfg3{T3l$al} + [2.1 lEDfi kglufgyflTlmL) From this expression. we are also left with the same two unknowns, x3 and T3. as in Eq. [7-13] above. Now that we have two equations and two unknowns, we may proceed. to solve for :3 and T3. Since the two above equations are complex. we must do this iteratively. Left out of the solution because of their [aherienr nanne. the iterations yield the following results: 5; = 15.45: T3 = 421.35 “a (148.? T) Since we know that the water vapor in the containment is samrated, we can find the pressure in the containment clue to the water vapor to the a.1.r from the denved temperature. This partial preasme due to water vapor ts; p“; = 0.45% MPa The' increase in temperature will also have an eEEcct on the partial presaun: ol the air. which can he found by the following relation: _ P5: = Flaws! T23 {1131" llea{¢ll.35 ”W415i ”31} {1139 MPa Finally, in order to find the final containment pressure. we apply Dalton's law of partial pressures: Pa = Fe. + PM: P; = {1139 MPa 4- [3.459% MPa = 111.5933 Page '2 of 2 pages ...
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