Solutions to Final Exam Download FULL Solution Manual for Thermodynamics for Engineers 1st Edition by Kroos at https://getbooksolutions.com/download/solution-manual-forthermodynamics-for-engineers-1st-edition-by-kroos A D W = N ⋅ m/s = ( kg ⋅ m/s )m/s = kg ⋅ m /s3 C Sum forces in the vertical direction (be sure and use Pa and not kPa): ∑ F = x= PA − W − Kx = ( 600 000 × π × 042 − 100 × 9.81) = 0.254 m or 25.4 cm 8000 If absolute pressure is used, the atmospheric pressure acting on the top of the cylinder must be included A B Since the temperature is below the boiling point (120.2°C from Table C-2) of water at 0.2 MPa, refer to Table C-1 and use hf = 461 kJ/kg D The specific volume at 160°C is vg = 0.307 m3/kg (Table C-1) From Table C-3 we search at 800°C and observe at 1.6 MPa that v = 0.309 m3/kg So, P2 = 1.59 MPa (No careful interpolation is needed.) C The volume is assumed to be constant (it doesn’t blow up like a balloon!) The ideal gas law is used: P1V1 = mRT1 and P2 V = mRT2 so P1 / P2 = T1 / T2 Then, using absolute temperatures and pressures (assume atmosphere pressure of 100 kPa since it is not specified), 340 = 273 ∴ P = 539 kPa or 439 kPa gage P2 433 B If the pressure is constant, the work is mP(v2 – v1) The result is W = mP ( v − v1 ) = 8× 800 × ( 2608 − 2404) = 131 kJ kN m Units: kg ⋅ ⋅ = kN ⋅ m = kJ m2 kg A First, find the height H of the piston above the cylinder bottom: m = ρ AH 0.1 = 400 × (π × 0.082 ) H ∴ H = 1.688 m 0.287 × 473 The temperature when the piston hits the stops in this constant-pressure process is T = T V2 = 473× 1.188A = 333 K or 59.9°C 1V 1.688A 1 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 10 A For this constant pressure process, the heat transfer is Q = mC p T = × 1.0 × ( 400 − 20) = 760 kJ 11 B The temperature at the final state must be known It is T = T P2 = 673× 100 = 168 K So 400 1P Q = mC v T = × 717 × ( 673 − 168) = 724 kJ 12 C For an isothermal process Q = W = mRT ln V / V1 so the heat transfer is Q = mRT ln V2 = × 287 × 473× ln = −94.1 kJ V 13 D The heat from the copper enters the water: mc C p , c Tc = mw C p , w Tw 10 × 0.39 × (200 − T2 ) = 50 × 4.18× (T2 − 10) ∴ T2 = 13.5°C 14 A Only A does not have a fluid flowing in and/or out 15 C The density is found from Table C-3 using ρ = 1/ v : × (π × 142 ) × = 1.89 kg/s m = ρ AV = AV = v 0.06525 16 C The power required by a pump involving a liquid is P2 − P1 W=m ρ =5× 6000 − 20 = 29.9 kW 1000 or 40.1 hp 17 D Equate the expression for the efficiency of a Carnot engine to the efficiency in general: η = 1− TL TH 10 W = out ∴ Qin = Qin = 29.6 kJ/s − 313 / 473 Qout = Qin − W = 29 − 10 = 19.6 kJ/s 18 A The maximum possible efficiency would be ηmax = − TL = − 283 = 0.237 Consequently, TH 2000 W out = 0.237 Qin ∴ Wout, max = 371 + 0.237 × 10 = 10.2 kW The engine is improbable 60 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part If COP = 1, C occurs The condition that W > QL is very possible 19 B If COP > 1, A is violated =QL 20 D The heat transfer from the high-temperature reservoir is Q H efficiency of the engine is thus η = = W QH + W = 20 + = 28 kJ/s The = 0.286 The Carnot efficiency yields 28 η = 286 = 1− 293 ∴ T = 410 K or 137°C H TH Obviously the temperatures must be absolute 21 C The entropy change of the copper is mCplnT2/T1 and that of the water is Q/T Hence, T S =mC net ln cu p ,cu + T1 Q mcu C p ,cu T T =mC Twater ln cu p,cu + T1 Twater = 10 × 0.39ln 293 +10 × 0.39 × (100 − 20) = 0.123 kJ/K 373 293 The copper loses entropy and the water gains entropy Make sure the heat transfer to the water is positive 22 C The maximum work occurs with an isentropic process: s1 = s2 = 7.168 kJ/kg·K The enthalpy at the turbine exit and the work are found as follows (use Tables C-3 and C-2): s2 = s1 = 168 = 6491+ x ( 7.502) ∴ x2 = 0.869 ∴ h2 = 192 + 0.869 × 2393 = 2272 kJ/kg and wT = h = 3658 − 2272 = 1386 kJ/kg The copper loses entropy and the water gains entropy Make sure the heat transfer to the water is positive 23 A The minimum work occurs with an isentropic process for which T2 = T1 ( P2 / P1)k −1/ k 2587 = 300× = 477 K ∴ wC = C p (T2 − T1) = 1.0 × ( 477 − 300) = 177 kJ/kg 24 D The maximum possible turbine work occurs if the entropy is constant, as in Problem 22 which is 1386 kJ/kg The actual work is wT = h1 − h2 = 3658 − 2585 = 1073 kJ/kg The efficiency is then η= wT w = 1073 = 0.774 or 77.4% 1386 T,s 25 D The heat that leaves the condenser enters the water: mw C p , w Tw = ms ( h1 − h2 ) ( 2609.7 − 251) mw × 4.18× 20 = × ∴ mw = 113 kg/s © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 26 B Heat transfer across a large temperature difference, which occurs in the combustion process, is highly irreversible The losses in A, C¸ and D are relatively small 27 B The combustion process is not reversible 28 A The rate at which heat is added to the boiler is Q B = m( h3 − h2 ) = × ( 3434 − 251) = 6366 kJ/s 29 C The turbine power output is WT = m( h3 − h4 ) = × ( 3434 − 2609 7) = 1648 kW 30 D The rate of heat transfer from the condenser is QC = m( h4 − h1) = × ( 2610 − 251) = 4718 kJ/s 31 B The required pump horsepower follows: 5000 − 20 P WP =m ρ =2× 1000 = 9.96 kW or 13.35 hp k−1 = T1 32 C First, find the temperature at state 2: T2 = 293× 80.4 = 673 K Then V1 V2 q -3 = C v (T3 − T2 ) = 717 × (1473 − 673) = 574 kJ/kg 33 A The efficiency of the Otto cycle is η = 1− k−1 = 1− r 0.4 = 0.435 or 43.5% 34 A The work output can be found using the efficiency and the heat input: w net = ηq = − 1q 2-3 in rk−1 =1 − × 574 = 250 kJ/kg 80 4 © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 35 D The heat input is provided by the combustor: qin = C p (T3 − T2 ) = 1.0 × (1073 − 513) = 560 kJ/kg 36 B The pressure ratio for the assumed isentropic process is P P1 T = 513 3.5 k /( k−1) = T1 = 7.1 293 37 A The back-work ratio is found to be BWR = wcomp w = C p (T2 − T1) = 240 − 20 = 0.44 or 44% −T) C (T T p 800 − 300 38 C The efficiency is the net work divided by the heat input: η= wT − wcomp = C p (T3 − T4 ) − C p (T2 − T1) qin = 800 − 300 − ( 240 − 20) C p (T3 − T2 ) = 0.5 800 − 240 39 D The vapor pressure, using Pg = 5.628 kPa from Table C-1, is Pv = Pgφ = 5.628× 0.9 = 5.065 kPa Then we have ω = 0.622 P v Pa 5.065 = 0.622 × = 0.0332 kgw /kga 100 − 5.065 40 C Locate state at 10°C and 60% humidity Move at constant ω (it is assumed that no moisture is added If it were, the amount of water would be needed) to the right until T2 = 25°C There the chart is read to provide the humidity as φ = 24% © 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part