化工热力学 热力学第二定律--第五章习题课

5-7. How do you increase the thermal efficiency of actual heat engines? Answer A significant increase in the efficiencies of actual heat engines can be made by increasing the temperature T1 at which heat is absorbed by the engine.
5-5. What is the thermal efficiency in thermodynamics?
Answer The thermal efficiency refers to the ratio of work obtained to the heat absorbed.
5-6. Please indicate in which condition that the thermal efficiency can approach to 1 or 0? Answer The only conditions under which the efficiency of a heat engine can approach unity are those for which T1 approaches infinity or T2 approaches zero. When T1 approaches T2 the efficiency approaches 0.
108- t3= t3-21.1 t3=64.55℃=337.7K
Change in entropy of heating air(one mole):
ΔS1 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(337.7/ 381.15) = -3.522 J/K
Change in entropy of heated air(one mole):
• PROBLEM
5-4. A lump of copper （铜） having a mass
of 10(kg) at a temperature of 540(℃) is dro
pped into a well-insulated （绝热） bucket
containing 100(kg) of water at a temperatur e of 21.1(℃). If the specific heats of copper and water are 0.095 and 1.000, respectively, calculate the resulting changes in entropy of the water and copper, and calculate the tota l entropy change resulting from the process.
93.3 – 21.1 = 182.2 – t’2 , t’2 = 110℃
parallel
1) 21.1℃(t1) 93.3℃ (t2) (t2)
182.2℃(t’1) 110℃(t’2)
countercurrent
2) 21.1℃ (t1) 93.3℃ 182.2℃(t’1) 110℃(t’2)
Change in entropy also is the same, no matter whether it is parallel or countercurrent flow in the exchanger.
a. Change in entropy of heating air(one mole):
5-1. What is the restriction on the d irection of energy transformation?
Answer The second law, which can be describe d by many general statements, is the restriction on the direction of energy transformation.
Solution From heat balance: 0.095(10)(540-t2) = 1(100)(t2 – 21.1), t2 = 25.98℃
Change in entropy of copper: by Eq.(5-28), ΔS =∫dQ/T =∫CpdT/T
= 0.095(10)ln(299.13/813.15) = - 0.9500 kJ/K
ΔS2 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(337.7/294.25) = 4.008 J/K
Total: ΔS = 4.008-3.522 = 0.486 J/K
5-6. Calculate the entropy changes for the heated water and the heating ma terial, and the total entropy changes, when 10(kg) of water is heated from 15.6 to 65.6(℃):
Change in entropy of heatiBaidu Nhomakorabeag air(one mole):
ΔS1 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(308.98/ 381.15) = -6.111 J/K
Change in entropy of heated air(one mole):
5-5. A heat exchanger is to be constructed in which air (assume Cp = 7/2 R) is to be heated from 21.1 to 93.3(℃) by another stream of air, originally at 182.2(℃). Equal amounts of heated air and heatin g air are to be used. Assume that heat losses fro
a. By saturated steam at 0.7 (MPa)
b. By saturated steam at 0.35(MPa)
c. By superheated steam at 0.35(MPa) and 176.7(℃) of superheat
d. In a perfect countercurrent heat exchanger with 10 (kg) of water at 65.6 (℃)
5-3. Please indicate which cycle is the Ca rnot refrigeration cycle from Fig. 5-3.
Answer:
ADCBA
5-4. What is the Carnot’s theorem? Answer The Carnot’s theorem is that the effic iency of converting heat into work in a reversible heat engine must depend no t upon the medium employed to run th e engine, but only upon the temperatur e levels θ1 and θ2. ( 卡诺热机的热效率与其工作介质无关， 仅与温度水平有关 )
ΔS2 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(366.15/294.15) = 6.371 J/K
Total: ΔS = 6.371 – 6.111 = 0.26 J/K
If heating air entered at 108℃ for parallel flow, then the temperature of heating air in the exit would be the same as the heated steam, assuming it as t3.
ΔS1 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(383.15/455.35) = -5.024 J/K
Change in entropy of heated air(one mol e):
ΔS2 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(366.45/294.25) = 6.385 J/K
2.Any process which consists solel y( 单独 ) in the transfer of heat f rom one temperature to a higher one is impossible. （从某以温度向 另一比较高的温度传热的过程不可 能单独存在）
m the exchanger are negligible （可忽略的） .
a. Calculate the entropy changes of the heated air a nd the heating air for both parallel and countercu rrent flow in the exchanger.
Change in entropy of water: by Eq. (5-28),
ΔS =∫dQ/T =∫Cp(dT/T) = 1(100)ln(299.13/294.25) = 1.645kJ/K
Total change in entropy:
ΔS = 1.645-0.9500= 0.6950 kJ/K
b. What is the total entropy change in each case?
c. Repeat parts a and b for countercurrent flow if t he heating air enters at 108(℃).
Solution Because the amount of heated and heating air are the same, and heat losses are negligible, assuming the Cp is a constant value in this problem, then the difference of temperature for both air must be the same.
5-2. What are the two statements of the second law of thermodynamics? Answer 1. No apparatus can operate in suc h a way that its only effect (in system and surroundings) is to convert heat a bsorbed by a system completely into w ork. （没有一种装置能把系统吸收的热 量全部转换为功）
b. Total: ΔS = 6.385 – 5.024 = 1.361 J/K
c. If heating air entered at 108℃ for
countercurrent flow and the exchanger were infinitely large, then the temperature of heating air in the exit would be 35.8℃, for the difference of the temperature of the heating air is identical with that of the heated air.