ANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is

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1 Problm 4.47 Fgur P4.47 provds stady stat opratng data for a pump drawng watr from a rsrvor and dlvrng t at a prssur of 3 bar to a storag tank prchd 5 m abov th rsrvor. Th powr nput to th pump s 0.5 kw. Th watr tmpratur rmans narly constant at 5 o C, thr s no sgnfcant chang n kntc nrgy from nlt to xt, and hat transfr btwn th pump and ts surroundngs s nglgbl. Dtrmn th mass flow rat of watr, n /s. Lt g = 9.8 m/s. KNOWN: Opratng and stat data ar provdd for a pump and accompanyng ppng at stady stat. FIND: Dtrmn th mass flow rat of watr. SCHEMATIC AND GIVEN DATA: T = 5 o C p = bar Q cv = 0 W cv = 0.5 kw ENGINEERING MODEL: () Th control volum shown on th sktch s at stady stat. () For th control volum, Q cv = 0, and Δk = 0. (3) Th watr tmpratur rmans narly constant from nlt to xt. (4) For th watr, th approxmaton of Eq. 3.3 appls. (5) g = 9.8 m/s. ANALYSIS: Th mass rat balanc for th on-nlt, on-xt control volum at stady stat s m = m m. And, th nrgy rat balanc (Eq. 4.0a) rducs as follows: Wth Eq = Q cv W cv + m [(h h ) + (V V ) + g(z z )] () (h h) {hf(t) + vf(t)[p psat(t)]} {hf(t) + vf(t)[p psat(t)]} Or, wth T = T: hf(t = hf(t), so (h h) = vf(t) [p p] () Collctng rsults and solvng for th mass flow rat m = W cv [v f (T) (p p)+g(z z )] Insrtng valus and convrtng unts, wth vf(5 o C) =.0009 x 0-3 m 3 / from Tabl A-: m = /s ( 0.5 kw) kw 03 N m (.0009 x 0 3 ) m3 ( 3)bar 05 N/m +(9.8 m bar s )( 5 m) N m/s =.49 /s

2 Problm 4.56

3 Problm 4.59 An opn fdwatr hatr oprats at stady stat wth lqud watr ntrng nlt at 0 bar, 50 o C. A sparat stram of stam ntrs nlt at 0 bar and 00 o C wth a mass flow rat of 6 /s. Saturatd lqud at 0 bar xts th fdwatr hatr at xt 3. Ignorng hat transfr wth th surroundngs and nglctng kntc and potntal nrgy ffcts, dtrmn th mass flow rat, n /s, at nlt. KNOWN: Lqud watr at gvn prssur and tmpratur and stam at gvn prssur, tmpratur and mass flow rat ntr a fdwatr hatr. Saturatd lqud xts th fdwatr hatr at gvn prssur. FIND: Dtrmn th mass flow rat at nlt. SCHEMATIC AND GIVEN DATA: p = 0 bar T = 00 o C m = 6 /s m 60 /s p = 0 bar T = 50 o C Fdwatr Hatr 3 Saturatd lqud p 3 = 0 bar ENGINEERING MODEL:. Th control volum shown on th accompanyng fgur s at stady stat.. Hat transfr and kntc and potntal nrgy ffcts can b nglctd. 3. = 0 snc a fdwatr hatr has no work assocatd wth t. 4. For th lqud watr ntrng at, h hf(t) appls W cv ANALYSIS: Th stady-stat mass rat balanc gvs m = m Th stady-stat nrgy balanc gvs m m m3 0 = Q cv W cv + m (h + ½ V + gz) m (h + ½ V + gz)

4 Problm 4.59 (Contnud) Th nrgy balanc smplfs to Substtutng for m 3 0 = 0 = m m h + from th mass rat balanc h m h m h m 3 h3 0 = m h + m h ( m + m )h3 Solvng for m m = m (h h 3 ) (h 3 h ) At nlt, th watr s comprssd lqud. From Tabl A-, h hf = /. At nlt, th stam s suprhatd. From Tabl A-4, h = 87.9 /. At xt 3, th watr s saturatd lqud. From Tabl A-3, h3 = hf3 = 76.8 /. Substtutng valus ylds m = 6 s 87.9 ( ) = /s. Th T-v dagram for th thr stat ponts s shown blow. T ( o C) bar 3 50 v (m 3 /)

5 Problm 4.65 A horzontal constant-damtr pp wth a buld-up of dbrs s shown n Fg. P4.65. Ar modld as an dal gas ntrs at 30 K, 900 kpa, wth a vlocty of 30 m/s and xts at 305 K. Assumng stady stat and nglctng stray hat transfr, dtrmn for th ar xtng th pp (a) th vlocty, n m/s. (b) th prssur, n kpa. KNOWN: Ar flows at stady stat through a constant-ara pp scton wth a buld-up of dbrs. Data ar known at th nlt and xt. FIND: For th ar xtng th pp scton, dtrmn th vlocty and th prssur. SCHEMATIC AND GIVEN DATA: Ar T = 30 K p = 900 kpa V = 30 m/s Dbrs buld-up T = 305 K D = D ENGINEERING MODEL:. Th control volum shown on th schmatc s at stady stat.. Ar s modld as an dal gas. 3. For th control volum, W cv 0, Q cv 0, and p = 0. ANALYSIS: (a) Th nrgy rat balanc for th on-nlt, on-xt control volum at stady stat smplfs to 0 = Q W m [(h h) + ½ (V V ) + g(z z)] cv cv 0 = (h h) + ½ (V V ) Solvng for xt vlocty gvs V V ( h h ) From Tabl A-, h = 30.9 / and h = 305. /. Solvng for xt vlocty V m s 305. m 000 N m s N V = 76. m/s

6 Problm 4.65 (Contnud) (b) From th mass rat balanc and dal gas quaton of stat AV AV m m v v AV p AV p RT RT Snc th nlt and xt damtrs ar th sam, th aras cancl as wll as th gas constant. Solvng for xt prssur ylds p V T p V T p (900 kpa) 30 m/s 76. m/s p = 46. kpa 305 K 30 K