Answer 5

5-4 A lump of copper having a mass of 10(kg) at a temperature of 540(℃) is dropped into a well-insulated bucket containing 100(kg) of water at a temperature 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 total entropy change resulting from the process.
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.
Solution a. Heated by the saturated steam is at 0.7 MPa 1). The entropy change of water: Δ S1 = 10(S65.6 – S 15.6) =10(0.900-0.233) =6.677 kJ/K
2). From heat balance, the steam needed is: ms = 10(65.6-15.6) 4.186/(Hv - HL) = 500 (4.186) / (2762 – 697) = 1.014 kg The entropy change of saturated steam at 0.7 MPa: Δ S2 = 1.014 (SL – Sv) = 1.014 (1.992 - 6.705)= 4.784kJ/K 3).Δ Stotal =Δ S2+Δ S1 =- 4.784+ 6.677= 1.893 kJ/K
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. 93 – 21 = 180 – t’2 , t’2 = 108℃
parallel countercurrent 1) 21℃(t1) 93℃ (t2) 2) 93℃ (t2) 21℃ (t1) 180℃(t’1) 108℃(t’2) 180℃(t’1) 108℃(t’2) Change in entropy also is the same, no matter whether it is parallel or countercurrent flow in the exchanger.
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 heating air are to be used. Assume that heat losses from the exchanger are negligible. a. Calculate the entropy changes of the heated air and the heating air for both parallel and countercurrent flow in the exchanger. b. What is the total entropy change in each case? c. Repeat parts a and b for countercurrent flow if the heating air enters at 108(℃).
5-6 Calculate the entropy changes for the heated water and the heating material, and the total entropy changes, when 10(kg) of water is heated from 15.6 to 65.6(℃):
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.
5-3. Please indicate which cycle is the Carnot refrigeration cycle from Fig. 5-3. Answer: A D C B A
5-4. What is the Carnot’s theorem? Answer: The Carnot’s theorem is that the efficiency of converting heat into work in a reversible heat engine must depend not upon the medium employed to run the engine, but only upon the temperature levels θ1 and θ2.
a. Change in entropy of heating air(one mole): ΔS1 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(381.15/453.15) = -5.035 J/K Change in entropy of heated air(one mole): ΔS2 = ∫dQ/T = ∫Cp(dT/T) = 7/2(8.314)ln(366.15/294.15) = 6.371 J/K
b. Total: ΔS = 6.371 – 5.035 = 1.336 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
5-1. What is the restriction on the direction of energy transformation? Answer: The second law, which can be described by many general statements, is the restriction on the direction of energy transformation.
Change in entropy of heating 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): 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
5-2. What are the two statements of the second law of thermodynamics? Answer: 1. No apparatus can operate in such a way that its only effect (in system and surroundings) is to convert heat absorbed by a system completely into work. 2. Any process which consists solely(单独) in the transfer of heat from one temperature to a higher one is impossible.
Solution From heat balance: 0.095(10)(540-t2) = 1(100)(t2 – 21), t2 = 25.88℃ Change in entropy of copper: by Eq.(5-28), ΔS =∫dQ/T =∫CpdT/T = 0.095(10)ln(298.15/813.15) = - 0.9507 kJ/K