SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamc Sxth Edton
CONTENT SUBSECTION PROB NO. Concept-Study Gude Problem 134-141 Steady Sngle Flow Devce 142-149 Steady Irreverble Procee 150-155 Tranent Procee 156-157 Reverble Shaft Work, Bernoull 158-163 Devce Effcency 164-170 Revew Problem 171-172 Th problem et compared to the ffth edton chapter 9 et and the current SI unt problem. New 5th SI New 5th SI New 5th SI 134 new 6 147 new 33 160 105 82 135 new 8 148 new 35 161 106 84 136 new 9 149 97 39 162 new 94 137 new 14 150 94 47 163 107 90 138 new 17 151 95 52 164 109 96 139 new 18 152 108 54 165 110 101 140 new 19 153 new 53 166 111 107 141 new 20 154 98 56 167 117 111 142 91mod 22 155 99 60 168 115-143 29 24 156 101 69 169 114 114 144 96 30 157 100 73 170 118-145 new 31 158 103 77 171 102 131 146 93 37 159 104 79 172 112 133
Concept Problem 9.134E A compreor receve R-134a at 20 F, 30 pa wth an ext of 200 pa, x = 1. What can you ay about the proce? Properte for R-134a are found n Table F.10 Inlet tate: = 0.4157 Btu/lbm R Ext tate: e = 0.4080 Btu/lbm R Steady tate ngle flow: e = + e dq T + gen Snce decreae lghtly and the generaton term can only be potve, t mut be that the heat tranfer negatve (out) o the ntegral gve a contrbuton that maller than - gen.
9.135E A large condener n a team power plant dump 15 000 Btu/ at 115 F wth an ambent at 77 F. What the entropy generaton rate? Th proce tranfer heat over a fnte temperature dfference between the water nde the condener and the outde ambent (coolng water from the ea, lake or rver or atmopherc ar) C.V. The wall that eparate the nde 115 F water from the ambent at 77 F. Entropy Eq. 9.1 for teady tate operaton: Condenng water Sea water ds dt = 0 = Q.. T + Ṡ gen = Q T Q. 115 T + Ṡ gen 77 cb 115 F 77 F Ṡ gen = [ 15 000 536.7 15 000 115 + 459.7 ] Btu R = 1.85 Btu R
9.136E Ar at 150 pa, 540 R throttled to 75 pa. What the pecfc entropy generaton? C.V. Throttle, ngle flow, teady tate. We neglect knetc and potental energe and there are no heat tranfer and haft work term. Energy Eq. 6.13: h = h e T = T e (deal ga) Entropy Eq. 9.9: e = + e dq T + gen = + gen Change n Eq.8.24: e = e dt C p T R ln P e P = R ln P e P gen = e = 53.34 778 ln 75 150 = 0.0475 Btu lbm R
9.137E A pump ha a 2 kw motor. How much lqud water at 60 F can I pump to 35 pa from 14.7 pa? Incompreble flow (lqud water) and we aume reverble. Then the haftwork from Eq.9.18 w = v dp = v P = 0.016 ft 3 /lbm (35 14.7) pa = 46.77 lbf-ft/lbm = -0.06 Btu/lbm Ẇ = 2 kw = 1.896 Btu/ ṁ = Ẇ -w = 1.896 0.06 = 31.6 lbm/
9.138E A team turbne nlet at 200 pa, 900 F. The ext at 40 pa. What the lowet poble ext temperature? Whch effcency doe that correpond to? We would expect the lowet poble ext temperature when the maxmum amount of work taken out. Th happen n a reverble proce o f we aume t adabatc th become an entropc proce. Ext: 40 pa, = n = 1.8055 Btu/lbm R T = 483.7 F The effcency from Eq.9.27 meaure the turbne relatve to an entropc turbne, o the effcency wll be 100%. 9.139E A team turbne nlet at 200 pa, 900 F. The ext at 40 pa. What the hghet poble ext temperature? Whch effcency doe that correpond to? The hghet poble ext temperature would be f we dd not get any work out,.e. the turbne broke down. Now we have a throttle proce wth contant h aumng we do not have a gnfcant ext velocty. Ext: 40 pa, h = h n = 1477.04 Btu/lbm T = 889 F Effcency: η = w w = 0 P T h = C e e v Remark: Snce proce rreverble there no area under curve n T- dagram that correpond to a q, nor there any area n the P-v dagram correpondng to a haft work.
9.140E A team turbne nlet at 200 pa, 900 F. The ext at 40 pa, 600 F. What the entropc effcency? from table F.7.2 Inlet: h n = 1477.04 Btu/lbm, n = 1.8055 Btu/lbm R Ext: h ex = 1333.43 Btu/lbm, ex = 1.8621 Btu/lbm R Ideal Ext: 40 pa, = n = 1.8055 Btu/lbm R h = 1277.0 Btu/lbm w ac = h n - h ex = 1477.04 1333.43 = 143.61 Btu/lbm w = h n - h = 1477.04 1277.0 = 200 Btu/lbm η = w ac w = 143.61 200 = 0.718 P e e ac T 200 pa 40 pa e ac e v
9.141E The ext velocty of a nozzle 1500 ft/. If η nozzle = 0.88 what the deal ext velocty? The nozzle effcency gven by Eq. 9.30 and nce we have the actual ext velocty we get V 2 e = V2 ac /η nozzle V e = V ac / η nozzle = 1500 / 0.88 = 1599 ft/
Steady Sngle Flow Devce 9.142E Steam enter a turbne at 450 lbf/n.2, 900 F, expand n a reverble adabatc proce and exhaut at 130 F. Change n knetc and potental energe between the nlet and the ext of the turbne are mall. The power output of the turbne 800 Btu/. What the ma flow rate of team through the turbne? C.V. Turbne, Steady ngle nlet and ext flow. Adabatc: Q. = 0. Contnuty Eq.6.11: ṁ = ṁ e = ṁ, Energy Eq.6.12: ṁh = ṁh e + Ẇ T, Entropy Eq.9.8: ṁ + 0/ = ṁ e ( Reverble Ṡ gen = 0 ) Explanaton for the work term n Sect. 9.3, Eq.9.18 P 1 2 v T 1 2 Inlet tate: Table F.7.2 h = 1468.3 btu/lbm, = 1.7113 btu/lbm R Ext tate: e = 1.7113 Btu/lbm R, T e = 130 F aturated x e = (1.7113 0.1817)/1.7292 = 0.8846, h e = 97.97 + x e 1019.78 = 1000 Btu/lbm w = h - h e = 1468.3 1000 = 468.31 Btu/lbm. m = Ẇ / w = 800 / 468.3 = 1.708 lbm/
9.143E In a heat pump that ue R-134a a the workng flud, the R-134a enter the compreor at 30 lbf/n. 2, 20 F at a rate of 0.1 lbm/. In the compreor the R- 134a compreed n an adabatc proce to 150 lbf/n. 2. Calculate the power nput requred to the compreor, aumng the proce to be reverble. C.V. Compreor, Steady ngle nlet and ext flow. Adabatc: Q. = 0. Contnuty Eq.6.11: ṁ 1 = ṁ 2 = ṁ, Energy Eq.6.12: ṁh 1 = ṁh 2 + Ẇ C, Entropy Eq.9.8: ṁ 1 + 0/ = ṁ 2 ( Reverble Ṡ gen = 0 ) Inlet tate: Table F.10.2 h 1 = 169.82 Btu/lbm, 1 = 0.4157 Btu/lbm R Ext tate: P 2 = 150 pa & 2 h 2 = 184.46 Btu/lbm Ẇ c = ṁw c = ṁ(h 1 - h 2 ) = 0.1 (169.82-184.46) = -1.46 btu/ Explanaton for the work term n Sect. 9.3 Eq.9.18 P 2 1 v T 2 1
9.144E A dffuer a teady-tate, teady-flow devce n whch a flud flowng at hgh velocty decelerated uch that the preure ncreae n the proce. Ar at 18 lbf/n. 2, 90 F enter a dffuer wth velocty 600 ft/ and ext wth a velocty of 60 ft/. Aumng the proce reverble and adabatc what are the ext preure and temperature of the ar? C.V. Dffuer, Steady ngle nlet and ext flow, no work or heat tranfer. Energy Eq.: h + V 2 /2g c = h e + V 2 e /2g c, => h e - h = C Po (T e - T ) Entropy Eq.: + dq/t + gen = + 0 + 0 = e (Reverble, adabatc) Energy equaton then gve (converon 1 Btu/lbm = 35 037 ft 2 / 2 from A.1): C Po (T e -T ) = 0.24(T e - 549.7) = 6002-60 2 2 25 037 T e = 579.3 R k P e = P (T e /T ) k-1 = 18 579.3 549.7 3.5 = 21.6 lbf/n 2 P e T e Inlet Ext v H V Low P, A Low V H P, A
9.145E The ext nozzle n a jet engne receve ar at 2100 R, 20 pa wth neglble knetc energy. The ext preure 10 pa and the proce reverble and adabatc. Ue contant heat capacty at 77 F to fnd the ext velocty. C.V. Nozzle, Steady ngle nlet and ext flow, no work or heat tranfer. Energy Eq.6.13: h = h e + V 2 e /2 ( Z = Z e ) Entropy Eq.9.8: e = + dq/t + gen = + 0 + 0 Ue contant pecfc heat from Table F.4, C Po = 0.24 The entropc proce ( e = ) gve Eq.8.32 Btu lbm R, k = 1.4 k-1 => T e = T ( P e /P ) k = 2100 (10/20) 0.2857 = 1722.7 R The energy equaton become (converon 1 Btu/lbm = 25 037 ft 2 / 2 n A.1) V 2 e /2 = h - h e C P ( T - T e ) V e = 2 C P ( T - T e ) = 2 0.24(2100-1722.7) 25 037 = 2129 ft/ P T H P Low P e v e Low V H V
9.146E Ar at 1 atm, 60 F compreed to 4 atm, after whch t expanded through a nozzle back to the atmophere. The compreor and the nozzle are both reverble and adabatc and knetc energy n/out of the compreor can be neglected. Fnd the compreor work and t ext temperature and fnd the nozzle ext velocty. 1 -W 2 1 T 2 1 P2 P 1 Separate control volume around compreor and nozzle. For deal compreor we have nlet : 1 and ext : 2 Adabatc : q = 0. Reverble: gen = 0 Energy Eq.6.13: h 1 + 0 = w C + h 2 ; Entropy Eq.9.8: 1 + 0/T + 0 = 2 - w C = h 2 - h 1, 2 = 1 The contant from Eq. 8.25 gve k-1 T 2 = T 1 (P 2 /P 1 ) k = (459.7 + 60) (4/1) 0.2857 = 772 R -w C = h 2 - h 1 = C P (T 2 - T 1 ) = 0.24 (772 519.7) = 60.55 Btu/lbm The deal nozzle then expand back down to tate 1 (contant ) o energy equaton gve: 1 2 V2 = h 2 - h 1 = -w C = 60.55 Btu/lbm V = 2 60.55 25 037 = 1741 ft/ Remember converon 1 Btu/lbm = 25 037 ft 2 / 2 from Table A.1.
9.147E An expander receve 1 lbm/ ar at 300 pa, 540 R wth an ext tate of 60 pa, 540 R. Aume the proce reverble and othermal. Fnd the rate of heat tranfer and work neglectng knetc and potental energy change. C.V. Expander, ngle teady flow. Energy Eq.: ṁh + Q. = ṁh e + Ẇ Entropy Eq.: ṁ + Q. /T + ṁ gen = ṁ e Proce: T contant and gen = 0 Ideal ga and othermal gve a change n entropy by Eq. 8.24, o we can olve for the heat tranfer Q. = Tṁ( e ) = ṁrt ln P e P = - 1 540 53.34 778 60 ln 300 = 59.6 Btu/ From the energy equaton we get Ẇ = ṁ(h h e ) + Q. = Q. = 59.6 Btu/ P T e Q e e v W exp
9.148E A flow of 4 lbm/ aturated vapor R-22 at 100 pa heated at contant preure to 140 F. The heat uppled by a heat pump that receve heat from the ambent at 540 R and work nput, hown n Fg. P9.35. Aume everythng reverble and fnd the rate of work nput. C.V. Heat exchanger Contnuty Eq.: ṁ 1 = ṁ 2 ; Energy Eq.: ṁ 1 h 1 + Q. H = ṁ 1 h 2 Table F.9.2: h 1 = 109.01 Btu/lbm, 1 = 0.2179 Btu/lbm R h 2 = 125.08 Btu/lbm, 2 = 0.2469 Btu/lbm R 1 2 Q H HP T L Q L Notce we can fnd Q. H but the temperature T H not contant makng t dffcult to evaluate the COP of the heat pump. W C.V. Total etup and aume everythng reverble and teady tate. Energy Eq.: ṁ 1 h + Q. 1 L + Ẇ = ṁ 1 h 2 Entropy Eq.: ṁ 1 1 + Q. L /T L + 0 = ṁ 1 2 (T L contant, gen = 0) Q. L = ṁ 1 T L [ 2-1 ] = 4 540 [0.2469 0.2179] = 62.64 Btu/ Ẇ = ṁ 1 [h 2 - h 1 ] - Q. L = 4 (125.08 109.01) 62.64 = 1.64 Btu/
9.149E One technque for operatng a team turbne n part-load power output to throttle the team to a lower preure before t enter the turbne, a hown n Fg. P9.39. The teamlne condton are 200 lbf/n. 2, 600 F, and the turbne exhaut preure fxed at 1 lbf/n. 2. Aumng the expanon nde the turbne to be reverble and adabatc, determne a. The full-load pecfc work output of the turbne b. The preure the team mut be throttled to for 80% of full-load output c. Show both procee n a T dagram. a) C.V. Turbne full-load, reverble. 3a = 1 = 1.6767 Btu/lbm R = 0.132 66 + x 3a 1.8453 x 3a = 0.8367 h 3a = 69.74 + 0.8367 1036.0 = 936.6 Btu/lbm w = h 1 - h 3a = 1322.1-936.6 = 385.5 Btu/lbm b) w = 0.80 385.5 = 308.4 = 1322.1 - h 3b h 3b = 1013.7 Btu/lbm 1013.7 = 69.74 + x 3b 1036.0 x 3b = 0.9112 3b = 0.13266 + 0.9112 1.8453 = 1.8140 Btu/lbm R 2b = 3b = 1.8140 h 2b = h 1 = 1322.1 P 2 = 56.6 lbf/n 2 T 2 = 579 F T 1= 2a 2b h = C 1 2 3 3a 3b W T
Steady Irreverble Procee 9.150E Analye the team turbne decrbed n Problem 6.161. I t poble? C.V. Turbne. Steady flow and adabatc. Contnuty Eq.6.9: ṁ 1 = ṁ 2 + ṁ 3 ; Energy Eq.6.10: ṁ 1 h 1 = ṁ 2 h 2 + ṁ 3 h 3 + Ẇ Entropy Eq.9.7: ṁ 1 1 + Ṡ gen = ṁ 2 2 + ṁ 3 3 1 2 3 W T State from Table F.7.2: 1 = 1.6398 Btu/lbm R, 2 = 1.6516 Btu/lbm R, 3 = f + x fg = 0.283 + 0.95 1.5089 = 1.71 Btu/lbm R Ṡ gen = 40 1.6516 + 160 1.713 200 1.6398 = 12.2 Btu/ R Snce t potve => poble. Notce the entropy ncreang through turbne: 1 < 2 < 3
9.151E Two flowtream of water, one at 100 lbf/n. 2, aturated vapor, and the other at 100 lbf/n. 2, 1000 F, mx adabatcally n a teady flow proce to produce a ngle flow out at 100 lbf/n.2, 600 F. Fnd the total entropy generaton for th proce. Contnuty Eq.6.9: ṁ 3 = ṁ 1 + ṁ 2, Energy Eq.6.10: ṁ 3 h 3 = ṁ 1 h 1 + ṁ 2 h 2 State properte from Table F.7.2 h 1 = 1187.8, h 2 = 1532.1, h 3 = 1329.3 all n Btu/lbm 1 = 1.6034, 2 = 1.9204, 3 = 1.7582 all n Btu/lbm R => ṁ 1 /ṁ 3 = (h 3 h 2 ) / (h 1 h 2 ) = 0.589 Entropy Eq.9.7: ṁ 3 3 = ṁ 1 1 + ṁ 2 2 + Ṡ gen => Ṡ gen /ṁ 3 = 3 (ṁ 1 /ṁ 3 ) 1 (ṁ 2 /ṁ 3 ) 2 = 1.7582-0.589 1.6034-0.411 1.9204 = 0.0245 Btu lbm R 1 2 T 100 pa Mxng 3 chamber 2 3 1
9.152E A mxng chamber receve 10 lbm/mn ammona a aturated lqud at 0 F from one lne and ammona at 100 F, 40 lbf/n. 2 from another lne through a valve. The chamber alo receve 340 Btu/mn energy a heat tranferred from a 100-F reervor. Th hould produce aturated ammona vapor at 0 F n the ext lne. What the ma flow rate at tate 2 and what the total entropy generaton n the proce? CV: Mxng chamber out to reervor Contnuty Eq.6.9: ṁ 1 + ṁ 2 = ṁ 3 Energy Eq.6.10: ṁ 1 h 1 + ṁ 2 h 2 + Q. = ṁ 3 h 3 Entropy Eq.9.7: ṁ 1 1 + ṁ 2 2 + Q. /T re + Ṡ gen = ṁ 3 3 2 1 3 MIXING CHAMBER Q. 2 P 1 3 v From Table F.8.1: From Table F.8.2: From Table F.8.1: From the energy equaton: h 1 = 42.6 Btu/lbm, 1 = 0.0967 Btu/lbm R h 2 = 664.33 Btu/lbm, 2 = 1.4074 Btu/lbm R h 3 = 610.92 Btu/lbm, 3 = 1.3331 Btu/lbm R ṁ 2 = ṁ 1(h 1 - h 3 ) + Q. 10(42.6-610.92) + 340 h 3 - h = 2 610.92-664.33 = 100.1 lbm/mn Ṡ gen = ṁ 3 3 - ṁ 1 1 - ṁ 2 2 - Q. /T re ṁ = 110.1 lbm/mn 3 = 110.1 1.3331-10 0.0967-100.1 1.4074-340 559.67 = 4.37 Btu R mn
9.153E A condener n a power plant receve 10 lbm/ team at 130 F, qualty 90% and reject the heat to coolng water wth an average temperature of 62 F. Fnd the power gven to the coolng water n th contant preure proce and the total rate of enropy generaton when condener ext aturated lqud. C.V. Condener. Steady tate wth no haft work term. Energy Eq.6.12: ṁ h + Q. = ṁh e Entropy Eq.9.8: ṁ + Q. /T + Ṡ gen = ṁ e Properte are from Table F.7.1 h = 98.0 + 0.9 1019.8 = 1015.8 Btu/lbm, h e = 98.0 Btu/lbm = 0.1817 + 0.9 1.7292 = 1.7380 Btu/lbm R, e = 0.1817 Btu/lbm R Q. out = Q. = ṁ (h h e ) = 10(1015.8 98.0) = 9178 btu/ Ṡ gen = ṁ ( e ) + Q. out/t = 10(0.1817 1.738) + 9178/(459.7 + 62) = 15.563 + 17.592 = 2.03 Btu/-R
9.154E Ar at 540 F, 60 lbf/n. 2 wth a volume flow 40 ft 3 / run through an adabatc turbne wth exhaut preure of 15 lbf/n. 2. Neglect knetc energe and ue contant pecfc heat. Fnd the lowet and hghet poble ext temperature. For each cae fnd alo the rate of work and the rate of entropy generaton. T = 540 F = 1000 R v = RT /P = 53.34 1000/(60 144) = 6.174 ft 3 / lbm ṁ = V /v = 40/6.174 = 6.479 lbm/ a. lowet ext T, th mut be reverble for maxmum work out. k-1 T e = T (P e /P ) k = 1000 (15/60) 0.286 = 673 R w = 0.24 (1000 673) = 78.48 Btu/lbm ; Ẇ = ṁw = 508.5 Btu/ Ṡ gen = 0 b. Hghet ext T, for no work out. T e = T = 1000 R. W= 0 Ṡ gen = ṁ (e ) = - ṁr ln (P e / P ) = - 6.479 53.34 778 ln (15/60) = 0.616 Btu/ R
9.155E A upply of 10 lbm/ ammona at 80 lbf/n. 2, 80 F needed. Two ource are avalable one aturated lqud at 80 F and the other at 80 lbf/n. 2, 260 F. Flow from the two ource are fed through valve to an nulated mxng chamber, whch then produce the dered output tate. Fnd the two ource ma flow rate and the total rate of entropy generaton by th etup. C.V. mxng chamber + valve. Steady, no heat tranfer, no work. Contnuty Eq.6.9: ṁ 1 + ṁ 2 = ṁ 3 ; Energy Eq.6.10: ṁ 1 h 1 + ṁ 2 h 2 = ṁ 3 h 3 Entropy Eq.9.7: ṁ 1 1 + ṁ 2 2 + Ṡ gen = ṁ 3 3 T 2 1 MIXING CHAMBER 3 1 3 2 State 1: Table F.8.1 h 1 = 131.68 Btu/lbm, 1 = 0.2741 Btu/lbm R State 2: Table F.8.2 h 2 = 748.5 Btu/lbm, 2 = 1.4604 Btu/lbm R State 3: Table F.8.2 h 3 = 645.63 Btu/lbm, 3 = 1.2956 Btu/lbm R A all tate are known the energy equaton etablhe the rato of ma flow rate and the entropy equaton provde the entropy generaton. ṁ 1 h 1 + (ṁ 3 - ṁ 1 )h 2 = ṁ 3 h 3 => ṁ 1 = ṁ h 3 - h 2 3 h 1 - h = 10-102.87 2-616.82 = 1.668 lbm/ ṁ 2 = ṁ 3 - ṁ 1 = 8.332 lbm/ Ṡ gen = ṁ 3 3 - ṁ 1 1 - ṁ 2 2 = 10 1.2956 1.668 0.2741 8.332 1.46 = 0.331 Btu/ R
Tranent Procee 9.156E An old abandoned altmne, 3.5 106 ft 3 n volume, contan ar at 520 R, 14.7 lbf/n. 2. The mne ued for energy torage o the local power plant pump t up to 310 lbf/n. 2 ung outde ar at 520 R, 14.7 lbf/n. 2. Aume the pump deal and the proce adabatc. Fnd the fnal ma and temperature of the ar and the requred pump work. Overnght, the ar n the mne cool down to 720 R. Fnd the fnal preure and heat tranfer. C.V. The mne volume and the pump Contnuty Eq.6.15: m 2 - m 1 = m n Energy Eq.6.16: Entropy Eq.9.12: m 2 u 2 - m 1 u 1 = 1 Q 2-1 W 2 + m n h n m 2 2 - m 1 1 = dq/t + 1 S 2 gen + m n n Proce: Adabatc 1Q 2 = 0, Proce deal 1S 2 gen = 0, 1 = n m 2 2 = m 1 1 + m n n = (m 1 + m n ) 1 = m 2 1 2 = 1 Contant Eq.8.28 o T2 = o T + R ln(p e / P ) Table F.4 o T2 = 1.63074 + 53.34 778 310 ln ( 14.7 ) = 1.83976 Btu/lbm R T 2 = 1221 R, u 2 = 213.13 Btu/lbm Now we have the tate and can get the mae m 1 = P 1 V 1 /RT 1 = 14.7 3.5 106 144 53.34 520 m 2 = P 2 V 2 /RT 2 = 310 3.5 106 144 53.34 1221 m n = m 2 - m 1 = 2.1319 10 6 lbm = 2.671 10 5 lbm = 2.4 10 6 kg 1W 2 = m n h n + m 1 u 1 - m 2 u 2 = 2.1319 10 6 124.38 + 2.671 10 5 88.73-2.4 10 6 213.13 = -2.226 10 8 Btu = -pump work W pump = 2.23 10 8 Btu 2W 3 = 0/, P 3 = P 2 T 3 /T 2 = 310 720/1221 = 182.8 lbf/n 2 2Q 3 = m 2 (u 3 - u 2 ) = 2.4 10 6 (123.17-213.13)= -2.16 10 8 Btu
9.157E Ar from a lne at 1800 lbf/n. 2, 60 F, flow nto a 20-ft 3 rgd tank that ntally contaned ar at ambent condton, 14.7 lbf/n. 2, 60 F. The proce occur rapdly and eentally adabatc. The valve cloed when the preure nde reache ome value, P 2. The tank eventually cool to room temperature, at whch tme the preure nde 750 lbf/n. 2. What the preure P 2? What the net entropy change for the overall proce? CV: Tank. Ma flow n, o th tranent. Fnd the ma frt m 1 = P 1 V/RT 1 = 14.7 144 20 53.34 520 = 1.526 lbm Fll to P 2, then cool to T 3 = 60 F, P 3 = 750 pa m 3 = m 2 = P 3 V/RT 3 = Cont. Eq.: 750 144 20 53.34 520 = 77.875 lbm m = m 2 - m 1 = 77.875-1.526 = 76.349 lbm T v = C 1800 pa 2 1 lne 3 750 pa 14.7 pa Conder the overall proce from 1 to 3 Energy Eq.: Q CV + m h = m 2 u 3 - m 1 u 1 = m 2 h 3 - m 1 h 1 - (P 3 - P 1 )V But, nce T = T 3 = T 1, m h = m 2 h 3 - m 1 h 1 Q CV = -(P 3 -P 1 )V = -(750-14.7) 20 144/778 = -2722 Btu S NET = m 3 3 - m 1 1 - m - Q CV /T 0 = m 3 ( 3 - ) - m 1 ( 1 - ) - Q CV /T 0 = 77.875 0-53.34 778 ln 750 1800-1.526 0-53.34 778 ln 14.7 1800 + 2722/520 = 9.406 Btu/R The fllng proce from 1 to 2 ( T 1 = T ) 1-2 heat tranfer = 0 o 1t law: m h = m 2 u 2 - m 1 u 1 T 2 = m C P0 T = m 2 C V0 T 2 - m 1 C V0 T 1 76.349 0.24 + 1.526 0.171 77.875 0.171 520 = 725.7 R P 2 = m 2 RT 2 /V = 77.875 53.34 725.7 / (144 20) = 1047 lbf/n 2
Reverble Shaft Work, Bernoull 9.158E Lqud water at ambent condton, 14.7 lbf/n. 2, 75 F, enter a pump at the rate of 1 lbm/. Power nput to the pump 3 Btu/. Aumng the pump proce to be reverble, determne the pump ext preure and temperature. C.V. Pump. Steady ngle nlet and ext flow wth no heat tranfer. Energy Eq.6.13: w = h h e = Ẇ/ṁ = -3/1 = - 3.0 btu/lbm Ung alo ncompreble meda we can ue Eq.9.18 w P = vdp v (P e P ) = 0.01606 ft/lbm(p e 14.7 pa) from whch we can olve for the ext preure 3 0.01606(P e - 14.7) 144 778 P e = 1023.9 lbf/n 2 e Pump -Ẇ -Ẇ = 3 Btu/, P = 14.7 pa. T = 75 F m = 1 lbm/ Energy Eq.: h e = h w P = 43.09 + 3 = 46.09 Btu/lbm Ue Table F.7.3 at 1000 pa => T e = 75.3 F
9.159E A freman on a ladder 80 ft above ground hould be able to pray water an addtonal 30 ft up wth the hoe nozzle of ext dameter 1 n. Aume a water pump on the ground and a reverble flow (hoe, nozzle ncluded) and fnd the mnmum requred power. C.V.: pump + hoe + water column, total heght dfference 35 m. Here V velocty, not volume. Contnuty Eq.6.3, 6.11: Energy Eq.6.12: Proce: ṁ n = ṁ ex = (ρav) nozzle ṁ(-w p ) + ṁ(h + V 2 /2 + gz) n = ṁ(h + V 2 /2 + gz) ex h n h ex, V n V ex = 0, z ex - z n = 110 ft, ρ = 1/v 1/v f -w p = g(z ex - z n ) = 32.174 (110-0)/25 037 = 0.141 Btu/lbm Recall the converon 1 Btu/lbm = 25 037 ft 2 / 2 from Table A.1. The velocty n the ext nozzle uch that t can re 30 ft. Make that column a C.V. for whch Bernoull Eq.9.17 : gz noz + 1 2 V2 noz = gz ex + 0 V noz = 2g(z ex - z noz ) = 2 32.174 30 = 43.94 ft/ 110 ft 30 ft Aume: v = v F,70F = 0.01605 ft 3 /lbm ṁ = π v f D 2 2 Vnoz = ( π/4) (1 2 /144) 43.94 / 0.01605 = 14.92 lbm/. W pump = ṁw p = 14.92 0.141 (3600/2544) = 3 hp
9.160E Saturated R-134a at 10 F pumped/compreed to a preure of 150 lbf/n. 2 at the rate of 1.0 lbm/ n a reverble adabatc teady flow proce. Calculate the power requred and the ext temperature for the two cae of nlet tate of the R- 134a: a) qualty of 100 %. b) qualty of 0 %. C.V.: Pump/Compreor, ṁ = 1 lbm/, R-134a a) State 1: Table F.10.1, x 1 = 1.0 Saturated vapor, P 1 = P g = 26.79 pa, h 1 = h g = 168.06 Btu/lbm, 1 = g = 0.414 Btu/lbm R Aume Compreor entropc, 2 = 1 = 0.414 Btu/lbm R h 2 = 183.5 Btu/lbm, T 2 = 116 F 1 t Law Eq.6.13: q c + h 1 = h 2 + w c ; q c = 0 w c = h 1 - h 2 = 168.05 183.5 = - 15.5 Btu/lbm; => Ẇ C = ṁw C = -15.5 Btu/ = 21.9 hp b) State 1: T 1 = 10 F, x 1 = 0 Saturated lqud. Th a pump. P 1 = 26.79 pa, h 1 = h f = 79.02 Btu/lbm, v 1 = v f = 0.01202 ft 3 /lbm 1 t Law Eq.6.13: q p + h 1 = h 2 + w p ; q p = 0 Aume Pump entropc and the lqud ncompreble, Eq.9.18: w p = - v dp = -v 1 (P 2 - P 1 ) = -0.01202 (150 26.79) 144 = -213.3 lbf-ft/lbm = - 0.274 Btu/lbm h 2 = h 1 - w p = 79.02 - ( - 0.274) = 187.3 Btu/lbm, Aume State 2 approxmately a aturated lqud => T 2 10.9 F Ẇ P = ṁw P = 1 (- 0.274) = -0.27 Btu/ = -0.39 hp P 2b 1b 2a 1a v T 2b 1b 2a 1a
9.161E A mall pump take n water at 70 F, 14.7 lbf/n. 2 and pump t to 250 lbf/n. 2 at a flow rate of 200 lbm/mn. Fnd the requred pump power nput. C.V. Pump. Aume reverble pump and ncompreble flow. Th lead to the work n Eq.9.18 w p = - vdp = -v (P e - P ) = -0.01605(250-14.7) 144 778 = -0.7 Btu/lbm Ẇ p n = ṁ(-w p ) = 200 60 (0.7) = 2.33 Btu/ = 3.3 hp
9.162E An expanon n a ga turbne can be approxmated wth a polytropc proce wth exponent n = 1.25. The nlet ar at 2100 R, 120 pa and the ext preure 18 pa wth a ma flow rate of 2 lbm/. Fnd the turbne heat tranfer and power output. C.V. Steady tate devce, ngle nlet and ngle ext flow. Energy Eq.6.13: h + q = h e + w Neglect knetc, potental energe Entropy Eq.9.8: Proce Eq.8.37: + dq/t + gen = e n-1 n T e = T (P e / P ) 0.25 1.25 = 2100 (18/120) = 1436.9 R o the ext enthalpy from Table F.5, h = 532.6 Btu/lbm h e = 343.0 + 36.9 40 (353.5 343.0) = 352.7 Btu/lbm The proce lead to Eq.9.19 for the work term. W = ṁw = -ṁnr n-1 (T e - T 1.25 53.34 ) = -2 (1436.9-2100) 0.25 778 = 454.6 Btu/ Energy equaton gve. Q = ṁq = ṁ(h e - h ) + Ẇ = 2(352.7 532.6) + 454.6 = -359.8 + 454.6 = 94.8 Btu/ P e n = 1 n = 1.25 v T n = k = 1.4 n = 1 e n = 1.25 Notce: dp < 0 o dw > 0 d > 0 o dq > 0 Notce th proce ha ome heat tranfer n durng expanon whch unuual. The typcal proce would have n = 1.5 wth a heat lo.
9.163E Helum ga enter a teady-flow expander at 120 lbf/n. 2, 500 F, and ext at 18 lbf/n. 2. The ma flow rate 0.4 lbm/, and the expanon proce can be condered a a reverble polytropc proce wth exponent, n = 1.3. Calculate the power output of the expander. Q e CV: expander, reverble polytropc proce. From Eq.8.37: W exp T e = T P e P n-1 0.3 n = 960 18 120 1.3 = 619.6 R Work evaluated from Eq.9.19 nr w = - vdp = - n - 1 (T e - T ) = - = +731.8 Btu/lbm. 3600 W = ṁw = 0.4 731.8 2544 = 414 hp Table F.4: R = 386 lbf-ft/lbm-r 1.3 386 (619.6-960) 0.3 778 P n = 1 e n = 1.3 v T n = k = 1.667 n = 1 e n = 1.3
Devce Effcency 9.164E A compreor ued to brng aturated water vapor at 103 lbf/n. 2 up to 2000 lbf/n. 2, where the actual ext temperature 1200 F. Fnd the entropc compreor effcency and the entropy generaton. C.V. Compreor. Aume adabatc and neglect knetc energe. Energy Eq.6.13: w = h 1 - h 2 Entropy Eq.9.8: 2 = 1 + gen We have two dfferent cae, the deal and the actual compreor. State: 1: F.7.1 h 1 = 1188.36 Btu/lbm, 1 = 1.601 Btu/lbm R 2ac: F.7.2 h 2,AC = 1598.6 Btu/lbm, 2,AC = 1.6398 Btu/lbm R 2: F.7.2 (P, = 1 ) h 2, = 1535.1 Btu/lbm IDEAL: -w c, = h 2, - h 1 = 346.7 Btu/lbm ACTUAL: -w C,AC = h 2,AC - h 1 = 410.2 Btu/lbm Defnton Eq.9.28: η c = w c, /w c,ac = 0.845 ~ 85% Entropy Eq.9.8: gen = 2 ac - 1 = 1.6398-1.601 = 0.0388 Btu/lbm R
9.165E A mall ar turbne wth an entropc effcency of 80% hould produce 120 Btu/lbm of work. The nlet temperature 1800 R and t exhaut to the atmophere. Fnd the requred nlet preure and the exhaut temperature. C.V. Turbne actual energy Eq.6.13: w = h - h e,ac = 120 Table F.5: h = 449.794 Btu/lbm h e,ac = h 120 = 329.794 Btu/lbm, T e = 1349 R C.V. Ideal turbne, Eq.9.27 and energy Eq.6.13: w = w/η = 120/0.8 = 150 = h - h e, h e, = 299.794 Btu/lbm From Table F.5: T e, = 1232.7 R, o Te = 1.84217 Btu/lbm R Entropy Eq.9.8: = e, adabatc and reverble To relate the entropy to the preure ue Eq.8.28 nverted and tandard entropy from Table F.5: P e /P = exp[ ( o Te o T )/R ] = exp[(1.84217-1.94209) 778 53.34 ] = 0.2328 P = P e / 0.2328 = 14.7/0.2328 = 63.14 pa If contant heat capacty wa ued T e = T - w/c p = 1800-120/0.24 = 1300 R T e, = T - w /C p = 1800-150/0.24 = 1175 R The contant relaton Eq.8.32 P e /P = (T e /T ) k/(k-1) P = 14.7 (1800/1175) 3.5 = 65.4 pa P T P e, ac e, = C v e, P e e, ac
9.166E Ar enter an nulated compreor at ambent condton, 14.7 lbf/n. 2, 70 F, at the rate of 0.1 lbm/ and ext at 400 F. The entropc effcency of the compreor 70%. What the ext preure? How much power requred to drve the compreor? C.V. Compreor: P 1, T 1, T e (real), η COMP known, aume contant C P0 Energy Eq.6.13 for real: -w = C P0 (T e - T ) = 0.24(400-70) = 79.2 Btu/lbm Ideal -w = -w η = 79.2 0.7 = 55.4 Btu/lbm Energy Eq.6.13 for deal: 55.4 = C P0 (T e - T ) = 0.24(T e - 530), T e = 761 R Contant entropy for deal a n Eq.8.32: k P e = P (T e /T ) k-1 = 14.7(761/530) 3.5 = 52.1 lbf/n 2 -Ẇ REAL = ṁ(-w) = 0.1 79.2 3600/2544 = 11.2 hp P T Pe e, e, e, ac e, ac = C v P
9.167E A watercooled ar compreor take ar n at 70 F, 14 lbf/n. 2 and compree t to 80 lbf/n. 2. The othermal effcency 80% and the actual compreor ha the ame heat tranfer a the deal one. Fnd the pecfc compreor work and the ext temperature. Ideal othermal compreor ext 80 pa, 70 F Reverble proce: dq = T d => q = T( e ) q = T( e ) = T[ o Te o T1 R ln(p e / P )] = - RT ln (P e / P ) = - (460 + 70) 53.34 778 A ame temperature for the deal compreor w = q = -63.3 Btu/lbm => 80 ln 14 = - 63.3 Btu/lbm h e = h w ac = w /η = - 79.2 Btu/lbm, q ac = q Now for the actual compreor energy equaton become q ac + h = h e ac + w ac h e ac - h = q ac - w ac = - 63.3 (-79.2) = 15.9 Btu/lbm C p (T e ac - T ) T e ac = T + 15.9/0.24 = 136 F
9.168E A nozzle requred to produce a teady tream of R-134a at 790 ft/ at ambent condton, 15 lbf/n. 2, 70 F. The entropc effcency may be aumed to be 90%. What preure and temperature are requred n the lne uptream of the nozzle? C.V. Nozzle, teady flow and no heat tranfer. Actual nozzle energy Eq.: h 1 = h 2 + V 2 2/2 State 2 actual: Table F.10.2 h 2 = 180.975 Btu/lbm h 1 = h 2 + V 2 2/2 = 180.975 + Recall 1 Btu/lbm = 25 037 ft 2 / 2 from Table A.1. Ideal nozzle ext: h 2 = h 1 - KE = 193.44 - State 2: (P 2, h 2 ) T 2 = 63.16 F, 790 2 = 193.44 Btu/lbm 2 25 037 790 2 /0.9 = 179.59 Btu/lbm 2 25 037 2 = 0.4481 Btu/lbm R Entropy Eq. deal nozzle: 1 = 2 State 1: (h 1, 1 = 2 ) Double nterpolaton or ue oftware. For 40 pa: gven h 1 then = 0.4544 Btu/lbm R, For 60 pa: gven h 1 then = 0.4469 Btu/lbm R, T = 134.47 F T = 138.13 F Now a lnear nterpolaton to get P and T for proper 0.4481 0.4544 P 1 = 40 + 20 0.4469 0.4544 = 56.8 pa T 0.4481 0.4544 T 1 = 134.47 + (138.13 134.47) 0.4469 0.4544 = 137.5 F 1 2 2 h 1 1
9.169E Redo Problem 9.159 f the water pump ha an entropc effcency of 85% (hoe, nozzle ncluded). C.V.: pump + hoe + water column, total heght dfference 35 m. Here V velocty, not volume. Contnuty Eq.6.3, 6.11: Energy Eq.6.12: Proce: ṁ n = ṁ ex = (ρav) nozzle ṁ(-w p ) + ṁ(h + V 2 /2 + gz) n = ṁ(h + V 2 /2 + gz) ex h n h ex, V n V ex = 0, z ex - z n = 110 ft, ρ = 1/v 1/v f -w p = g(z ex - z n ) = 32.174 (110-0)/25 037 = 0.141 Btu/lbm Recall the converon 1 Btu/lbm = 25 037 ft 2 / 2 from Table A.1. The velocty n the ext nozzle uch that t can re 30 ft. Make that column a C.V. for whch Bernoull Eq.9.17 : gz noz + 1 2 V2 noz = gz ex + 0 V noz = 2g(z ex - z noz ) = 2 32.174 30 = 43.94 ft/ 110 ft 30 ft Aume: v = v F,70F = 0.01605 ft 3 /lbm ṁ = π v f D 2 2 Vnoz = ( π/4) (1 2 /144) 43.94 / 0.01605 = 14.92 lbm/. W pump = ṁw p /η= 14.92 0.141 (3600/2544)/0.85 = 3.5 hp
9.170E Repeat Problem 9.160 for a pump/compreor entropc effcency of 70%. C.V.: Pump/Compreor, ṁ = 1 lbm/, R-134a a) State 1: Table F.10.1, x 1 = 1.0 Saturated vapor, P 1 = P g = 26.79 pa, h 1 = h g = 168.06 Btu/lbm, 1 = g = 0.414 Btu/lbm R Aume Compreor entropc, 2 = 1 = 0.414 Btu/lbm R h 2 = 183.5 Btu/lbm, T 2 = 116 F 1 t Law Eq.6.13: q c + h 1 = h 2 + w c ; q c = 0 w c = h 1 - h 2 = 168.05 183.5 = - 15.5 Btu/lbm; Now the actual compreor w c, AC = w c /η = - 22.1 = h 1 h 2 AC h 2, AC = 168.06 + 22.1 = 190.2 T 2 = 141.9 F => Ẇ C n = ṁ(-w C ) = 22.1 Btu/ = 31.3 hp b) State 1: T 1 = 10 F, x 1 = 0 Saturated lqud. Th a pump. P 1 = 26.79 pa, h 1 = h f = 79.02 Btu/lbm, v 1 = v f = 0.01202 ft 3 /lbm 1 t Law Eq.6.13: q p + h 1 = h 2 + w p ; q p = 0 Aume Pump entropc and the lqud ncompreble, Eq.9.18: w p = - v dp = -v 1 (P 2 - P 1 ) = -0.01202 (150 26.79) 144 = -213.3 lbf-ft/lbm = - 0.274 Btu/lbm Now the actual pump w c, AC = w c /η = - 0.391 = h 1 h 2 AC h 2 = h 1 - w p = 79.02 - ( - 0.391) = 79.41 Btu/lbm, Aume State 2 approxmately a aturated lqud => T 2 11.2 F Ẇ P n = ṁ(-w P ) = 1 (0.391) = 0.39 Btu/ = 0.55 hp P 2b 1b 2a 1a v T 2b 1b 2a 1a
Revew Problem 9.171E A rgd 35 ft 3 tank contan water ntally at 250 F, wth 50 % lqud and 50% vapor, by volume. A preure-relef valve on the top of the tank et to 150 lbf/n. 2 (the tank preure cannot exceed 150 lbf/n. 2 - water wll be dcharged ntead). Heat now tranferred to the tank from a 400 F heat ource untl the tank contan aturated vapor at 150 lbf/n. 2. Calculate the heat tranfer to the tank and how that th proce doe not volate the econd law. C.V. Tank. v f1 = 0.017 v g1 = 13.8247 m LIQ =V LIQ / vf1 = 0.5 35/0.017 = 1029.4 lbm m VAP =V VAP / vg1 = 0.5 35/13.8247 = 1.266 lbm m = 1030. 67 lbm x = m VAP / (m LIQ + m VAP) = 0.001228 u = u f + x u fg = 218.48 + 0.001228 869.41 = 219.55 = f + x fg = 0.3677 + 0.001228 1.3324 = 0.36934 tate 2: v 2 = v g = 3.2214 u 2 = 1110.31 h 2 = 1193.77 2 = 1.576 m 2 = V/v 2 = 10.865 lbm Q = m 2 u 2 - m 1 u 1 + m e h e+ W = 10.865 1110.31 1030.67 219.55 + 1019.8 1193.77 = 1003187 Btu Ṡ gen = m 2 2 - m 1 1 - m e e - 1 Q 2 / T ource = 10.865 1.576 1030.67 0.36934 + 1019.8 1.57 1003187/860 = 77.2 Btu/ R
9.172E Ar at 1 atm, 60 F compreed to 4 atm, after whch t expanded through a nozzle back to the atmophere. The compreor and the nozzle both have effcency of 90% and knetc energy n/out of the compreor can be neglected. Fnd the actual compreor work and t ext temperature and fnd the actual nozzle ext velocty. 1 -W 3 5 T 2 3 1 4 5 Steady tate eparate control volume around compreor and nozzle. For deal compreor we have nlet : 1 and ext : 2 Adabatc : q = 0. Reverble: gen = 0 Energy Eq.: h 1 + 0 = w C + h 2 ; Entropy Eq.: 1 + 0/T + 0 = 2 Ideal compreor: w c = h 1 - h 2, 2 = 1 The contant from Eq. 8.25 gve k-1 T 2 = T 1 (P 2 /P 1 ) k = (459.7 + 60) (4/1) 0.2857 = 772 R -w C = h 2 - h 1 = C P (T 2 - T 1 ) = 0.24 (772 519.7) = 60.55 Btu/lbm Actual compreor: w c,ac = w c, /η c = -67.3 Btu/lbm = h 1 - h 3 T 3 = T 1 - w c,ac /C P = 519.7 + 67.3/0.24 = 800 R Ideal nozzle: 4 = 3 o ue Eq.8.25 agan k-1 T 4 = T 3 (P 4 /P 3 ) k = 800 (1/4) 0.2857 = 538.4 R V 2 /2 = h 3 - h 4 = C P (T 3 - T 4 ) = 0.24(800-538.4) = 62.78 Btu/lbm V 2 AC /2 = V 2 η NOZ /2 = 62.78 0.9 = 56.5 Btu/lbm V AC = 2 56.5 25 037 = 1682 ft/ Remember converon 1 Btu/lbm = 25 037 ft 2 / 2 from Table A.1.