1.

2 – 3.

4 – 6.

7 – 8.

9 – 10.

11 – 12.

13 – 15.

16 – 18.

19. A Carnot engine operates between 300°C and 40 °C. What is the efficiency of the engine?

20. A heat engine takes in 1000 J of energy at 1000 K and exhausts 700 J at 400 K. What is the actual efficiency of this heat engine?

22. What is the theoretical maximum efficiency of the heat engine in number 20?

23. The valve is now opened and saturated vapor flows from A to B until the pressure in B Consider two tanks, A and B, connected by a valve as shown in fig.

Each has a volume of 200 L and tank A has R-12 at 25°C, 10 % liquid and 90% vapor by volume, while tank B is evacuated has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at

ηc=(T H−T L )/T H=1−T L /T H

∴ηc=1−( 40+273)K(300+273 )K

=. 45×100=45 %

25 °C throughout the process. How much has the quality changed in tank A during the process?

Given: P= 651.6 KPa νg= 0.02685 m3/Kg

νf = 0.763*10-3 m3/Kg

24.

25.

26.

27 – 29. and

40 – 47.

48 – 50.

51 – 54.

51 – 57.

58 – 60.

61 – 63.

64 – 67.

68 – 73.

74 – 78.

78 – 82.

83 – 87.

87 – 91.

92 – 96.

98 – 101

102 – 106.

107 - 110

111 - 114

115 – 116

116 -120.

126 - 127. A heat engine performs 200 J of work in each cycle and has an efficiency of 30%. For each cycle, how much thermal energy is

absorbed and expelled?

Solution:a. Efficiency = e = 0.3 = 1 - Qc/Qh = W / Qh.Therefore, Qh = 200 J / 0.3 = 666.7 J.b. Qc = Qh - W = 466.7 J.

128. A heat engine operates between two reservoirs at 20EC and 300EC. What is the maximum efficiency possible for this engine?

Solution:emax = 1 - (20+273.15) / (300+273.15) = 0.488

129 – 130.. A heat engine operates in a Carnot cycle between 80EC and 350EC. It absorbs 2.0 x 104 J of thermal energy per cycle from the hot reservoir. The duration of each cycle is 1.0 s.

What is the maximum power output of this engine? How much thermal energy does it expel in each cycle?

Solution:a. Qh = 2.0x104 Je = 1 - (80+273.15)/(350+273.15) = .4333therefore, W = 0.4333 x 2.0x104 J = 8665 J per cycletherefore, Power = 8.665 kJ/cycle x 1 cycle/s = 8.665 kWb. Qc = Qh - W = 2.0x104 J - 8665 J = 11,334 J

131. The efficiency of a 1000 MW nuclear power plant is 33%; that is, 2000 MW of heat isrejected to the environment for every 1000 MW of electrical energy produced. If a river of flow rate 106 kg/s were used to transport the excess thermal energy away, what would be the average temperature increase of the river?

Solution:An energy balance on a segment of the river yields:2000 MW = 2000 MJ/s = 2x106 kJ/s= flow (kg/s) x heat capacity (kJ/kg/EC)x )T(EC)= 106 kg/s x 4.186 kcal/kg x )T(EC)Therefore, )T = 2 x 106 / (106 x 4.186) = 0.478 EC

132 - 133. A gasoline engine has a compression ratio of 6 and uses a gas for which (= 1.4.

What is the efficiency of the engine if it operates in an idealized Otto cycle? If the actual efficiency is 15%, what fraction of the fuel is wasted as a result of

friction and unavoidable heat losses? (Assume complete combustion of the air-fuel mixture.)

Solution:a. emax = 1 - 1/(V1/V2)(-1 = 1 - 1/(6)0.4 = 0.512b. eactual = 0.15, thus if Qh = 1, then 0.15 is used for work and0.85 is rejected or wasted. Compare this to the maximumefficiency situation where 0.512 is used and 0.488 is wasted:the extra waste caused by friction and other unavoidablelosses is 0.85 - 0.488 = 0.362. Thus, the extra wastage is36.2%.

134 - 135. A refrigerator has a coefficient of performance equal to 5. If the refrigerator absorbs 120 J of thermal energy from a cold reservoir in each cycle, find

the work done in each cycle and the thermal energy expelled to the hot reservoir.

Solution:a. COP = Qc / W, therefore W = Qc / COP = 120/5 = 24 J.b. W = Qh - Qc, therefore, Qh = 24 + 120 = 144 J.