1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.

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Roderick Bates
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1 AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature Solution We have h + v = h 0 + v 0 But at the rest state v 0 = 0, so h 0 h = v c P T 0 T = v For air c P = 0045 J/kg/K, so T 0 = T + v c P T 0 = 300 K m/s 0045 J/kg/K T 0 = K 50 A small utility gasoline engine of 50 cm 3 runs at 500 RPM with a compression ratio of 8: The inlet state is 75 kpa, 7 C, and the combustion adds 500 kj/kg to the charge This engine runs a heat pump using R-40a with a high pressure of 4 MPa and an evaporator operating at 0 C Find the rate of heating the heat pump can deliver Assume the gasoline engine is modeled well by an ideal Otto cycle with calorically imperfect air as the working fluid Solution I m going to be a little sloppy about interpolation in this problem You can fix it if you like! We ll be close For the gasoline engine, we have P G = 75 kpa,

2 T G = = 90 K We can also take V G = 50 cm 3 = m 3 Now for the isentropic compression from to, we have s G = s G Now s G s G = 0 = s o T G s o T G R ln P P s G s G = 0 = s o T G s o T G R ln T V T V ln 0 = s o T G 6835 kj 087 kg K kj kg K 8 T 90 K We iterate a few times on T T = 660 K is pretty close When T = 660 K, u = 4834 kj/kg Then u 3G = u G + q 3 = = 98 kj kg At this internal energy, we re pretty close to T 3G = 300 K Let s say we are at this point, so u 3G = kj/kg Thus we really added kj/kg Now s o T 3G = 9489 kj/kg/k Now s 4G s 3G = 0 = s o T 4G s o T 3G R ln P 4 P 3 s 4G s 3G = 0 = s o T 4G s o T 3G R ln T 4 V T 3 V 0 = s o T 4G 9489 kj 087 kj ln kg K kg K 8 T K We iterate and find that T 4G 50 K At this temperature u 4G = 889 kj/kg Now w G = u 3G u4g u G u G = = kj kg Now The power is m = P GV G = 75 kpa00005 m3 = kg RT G 087 kj/kg/k90 K Ẇ = mw G RPM/60 = kg80543 kj/kg500/60 = 454 kw Now for the refrigerator, we have at the compressor entrance x = and T = 0 C The saturation tables for R-40a give P = 7987 kpa, h = 79 kj/kg and s = 0368 kj/kg/k We have s = s = 0368 kj/kg/k We have P = 4000 kpa Interpolating the superheated tables gives h = 339 kj kg We then take x 3 = 0 with P 3 = P = 4000 kpa We interpolated the saturation tables to find h 3 = 7786 kj kg

3 The coefficient of performance for the heat pump is β = q H = h h = w c h h = 3396 So if we put in 454 KW of work we can acquire Q h = β Ẇ = kw = 6 kw 3 40 A fixed mass m of calorically perfect ideal gas with ratio of specific heats k = 7/5 undergoes a thermodynamic cycle At the beginning of the cycle, the gas is known to have pressure P and volume V The cycle is composed of the following steps: : Polytropic compression to V = V /5 via the path PV = P V 3: Isobaric heat addition to V 3 = V 3 4: Isothermal expansion to V 4 = V 4 : Isochoric return to state a Sketch the process on P V and T s diagrams b Find the net work, net heat transfer, and thermal efficiency Solution First from the ideal gas law, we have For the polytropic compression we have T = P V P V = P V Since V = V /5, we have V P 5 = P V P = 5V From the ideal gas law, we get T = P V = 5P V /5 = 5P V Now let s get the work going from to W = PdV

4 W = W = P V P V V dv V V W = P V 5 V V W = 4P V Now let s get the heat transfer going from to The first law tells us U U = Q W, so Q = U U + W Q = mc v T T + W 5P V Q = mc v P V 4P V Q = c v R 4P V 4P V cv Q = 4P V R Now c v /R = c v /c P c v = /c P /c v = /k, so Q = 4P V k Q = 4P V k k Q = 4P V 7/5 7/5 Q = 6P V Now P 3 = P = 5P And V 3 = V = V /5 So W 3 = 3 T 3 = P 3V 3 = 5P V /5 = 0P V 3 PdV = P dv = P V 3 V = 5P 5 V 5 V = 5P V For the heat transfer, we have Q 3 = U 3 U + W 3 Q 3 = mc v T 3 T + W 3 0P V Q 3 = mc v 5P V + 5P V Q 3 = c v R 5P V + 5P V Q 3 = 5P V k + Q 3 = 5P V 7/5 +

5 Q 3 = 35 P V Now for 3 to 4, we have T 4 = T 3 = 0P V / and V 4 = V We get The work is 3W 4 = 4 3 PdV = 4 The heat transfer is For 4 to, we have 3 P 4 = T 4 V 4 = 0P V / V = 0P T 3 dv = T 3 ln V 4 = V V 3 0P V 3Q 4 = U 4 U }{{} W 4 = 0P V ln 5 =0 4W = 4 PdV = 0 Therefore P V 4Q = U U 4 = mc v T T 4 = mc v 0P V ln V V 5 = 0P V ln 5 = c v R 9P V = 9 k P V The net work is 4Q = 9 7/5 P V = 45 P V W net = W + W W W = 4P V + 5P V + 0P V ln The heat transfer that we pay for is W net = P V + 0 ln 5 Q in = Q + Q Q 4, Q in = 6P V + 35/P V + 0P V ln 5 Q in = P V 47/ + 0 ln 5 So the thermal efficiency is η = W net Q in η = + 0 ln = 0345 Appropriate diagrams are sketched in Fig The shapes of the P V diagram is relatively obvious The T s diagram is less so We have shown in class that the slope of an isobar is s = T = T, P c P kc v

6 P T V s Figure : Sketches of P V and T s diagrams and the slope of an isochore is s = T v c v Since k >, the slope of the isochore is greater than the isobar Now consider the slope for a general polytropic process for which Pv n = C Differentiate the polytropic equation: nv n Pdv + v n dp = 0 npdv + vdp = 0 Now differentiate the ideal gas law Pv = RT: Pdv + vdp = RdT Eliminate vdp: Now consider the Gibbs law: Pdv npdv = RdT npdv = RdT Pdv = R n dt Tds = du + Pdv Eliminate du = c v dt and Pdv = R/ ndt: Tds = c v dt + R n dt Tds = c v + c P c v n cv nc v + c P c v Tds = n cp nc v Tds = dt n dt dt

7 k n dt Tds = n c v s = n T polytropic k n c v So when n = and k = 7/5, we get s = 5 T n= 3 c v This is steeper than the isochore, which has s = T v c v When k = 7/5, it is also steeper than the isobar which has s = 5 T P 7 c v

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,