study problems ch8 - CHAPTER 8 STUDY PROBLEMS ENTROPY The...

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CHAPTER 8 STUDY PROBLEMS ENTROPY The inequality of Clausius The property entropy Entropy changes in processes Entropy generation Entropy changes for solids and liquids Entropy changes for an ideal gas The reversible polytropic process FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN 73
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8.1 A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at T H and T L . Use the second law in the form of Eq.8.1 for the whole heat engine to prove that a. If the heat engine is ideal then η = 1 – T L T H b. If the heat engine is non-ideal then η η Carnot HE Solution: C.V. Heat engine out to the reservoirs at T H and T L . The heat engine whether ideal or not does not store any energy or entropy so the energy equation is Energy Eq.: 0 = Q H – Q L – W Clausius : dQ T = Q H T H Q L T L 0 Q L Q H T L T H Substitute this into the energy equation and solve for the work W = Q H – Q L Q H [ 1 – T L T H ] Case a . Ideal, the equal sign applies and the work is W = Q H [ 1 – T L T H ] = Q H η and the efficiency is equal to the Carnot cycle efficiency. Case b . Non-ideal, the inequality applies and the efficiency η = W/Q H becomes η [ 1 – T L T H ] = η Carnot HE And we see that the efficiency is smaller than the Carnot efficiency. Remark: After writing the entropy equation as Eq.8.11 we can express how much smaller the efficiency is due to S gen as Entropy Eq.8.11: 0 = dQ T + δ S gen = Q H T H Q L T L + S gen ; S gen 0 0 = Q H T H Q L T L + S gen Q L = Q H T L T H + T L S gen Notice that Q L is larger than the ideal case by an amount proportional to S gen W = Q H – Q L T L S gen = Q H [ 1 – T L T H ] T L S gen = Q H η η = [ 1 – T L T H ] T L S gen Q H 74
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8.2 Change in s from the steam tables Water at 100 o C is heated to 150 o C at constant P. What are changes in u and s when the starting state is at a. 10 000 kPa b. 500 kPa c. 100 kPa d. 10 kPa Solution: a. From Table B.1.4 compressed liquid states at 10 000 kPa, we get u 2 – u 1 = 627.39 – 416.09 = 211.3 kJ/kg s 2 – s 1 = 1.8304 – 1.2992 = 0.5312 kJ/kg K b. From Table B.1.4 compressed liquid states at 500 kPa, we get u 2 – u 1 = 631.66 – 418.8 = 212.9 kJ/kg s 2 – s 1 = 1.8415 – 1.3065 = 0.535 kJ/kg K c. From Table B.1.3 superheated vapor states at 100 kPa, we get u 2 – u 1 = (2582.75 – 2506.06) 50 150 – 99.62 = 76.11 kJ/kg s 2 – s 1 = (7.6133 – 7.3593) 50 150 – 99.62 = 0.252 kJ/kg K d. From Table B.1.3 superheated vapor states at 10 kPa, we get u 2 – u 1 = 2587.86 – 2515.5 = 72.1 kJ/kg s 2 – s 1 = 8.6881 – 8.4479 = 0.2402 kJ/kg K s T 500 100 10 v P 150 C 100 C 10 000 kPa 150 100 10 100 500 10000 The changes in u and s are larger for the liquid than for the gas. Within the liquid phase the influence of P is very modest. Water vapor at 100 kPa is close to an ideal gas (Z = Pv g /RT sat = 0.985). In the vapor phase changing P (fixed T) from 100 kPa to 10 kPa changes s, but u is nearly constant. The influence of P becomes much stronger as the state approaches the dense fluid region near the critical point.
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study problems ch8 - CHAPTER 8 STUDY PROBLEMS ENTROPY The...

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