Elementary Principles 273 - 7.50 100 L H 2 O( v ), 25o C ....

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7-28 7.50 100 .) , L H O( 25 C m( k g ) 2 o v1 v 400 , L H O( 25 C k g ) 2 o L1 l m[ k g HO (] =m m v2 2 v1 e v ) + k g m L2 2 L1 e l ) + Q=2915 kJ U V W = Assume not all the liquid is vaporized. Eq. at kg H O vaporized. 2 TPm ff e ,. Initial conditions: Table B.5 kJ kg ⇒= ± . U L 1 104 8 , ± . V L 1 1003 = Lkg P = 0 0317 . b a r T 25 C, sat'd ± . U v 1 2409 9 = kJ kg , ± , V L 1 43 400 = m v 1 5 43400 2 304 10 == × 1.00 l l kg kg bg b g . , m LI 4 00 1 3988 .. l .003 l kg kg b g Energy balance: Δ UQ mUT evf eLf =⇒ × + + × 2 304 10 2 304 10 2409 9 5 5 . ± . ± di d i did i −= 3988 104 8 2915 .( . ) kJ 2 304 10 3333 5 . ± . ± ×+ + = E mU T e vf d i −× m UU e vL 3333 2 304 10 5 . ± . ± ±± (1) VVV m V Tm V T m VV Lv e L fe L f e += × + F H G G I K J J +− = A A tan . ± . ± . ± . ± k kg liters kg L 2 304 10 500 2304 10 5 5 2 12 3333 2 304 10 0 5 5 bg b g d i d i −⇒ = = fT UT f vf Lf . ± . ± ± . ± Procedure: Assume Table 8.5 TU U V V f T fv L v L f ⇒⇒ ± , ± , ± , ± Find T f such that f = 0 TUUV V f TP fvL v L ±±± ± . . . . . . . . . . 2014 25938
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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