Homework # 7

5.22
a) First find the total heat rejected which equals 145 + 8 = 153 GJ/h. The power output = QH - QL = 280 - 153 = 127 GJ/hr = 35.3 MW b) The thermal efficiency = W/QH = 127 GJ/hr/280 GJ/hr = .454

5.58
Find the COP = QL/W = 8000 kJ/hr/1 kW = 2.22 Watch units. Then QH = QL + W = (8000 kJ/hr) + (1 kW*3600kJ/hr) = 11,6000 kJ/h

5.63
The heat pump needs to cover for the 60,000 minus the 4000 already provided so it needs to provide a QH of 56,000 kJ/hr. Then W = QH/COP = 56000 kJ/h/2.5 6.22 kW.

5.85
The highest efficiency a heat engine can have is 1 - (TL/TH) = 1 - 298/823 = .638 which is a Carnot heat engine. So the maximum power is W = h*QH = 0.638*12000 kJ/min = 756.6 kJ/min = 12.8 kW.

5.101
The highest COP a refrigerator can have is 1/(TH/TL)-1 = 1/(295/268)-1 = 9.9. The inventor's refrigerator is lower so it is possible.

5.110
a) The highest thermal efficiency (Carnot) is 1 - (TH/TL) = 1 - 300/1173 = 0.744. So the maximum power = h*QH = .744*800 kJ/min = 595.2 kJ/min which is also the power input to the refrigerator. The rate of heat removal for the refrigerator is a maximum if a Carnot refrigerator is used so the COP = 1/((TH/TL) - 1) = 8.37. So the rate of heat removal becomes (COP)(W) = 8.7*595.2 kJ/min = 4982 kJ/min
b) To find total rate of heat rejection you add the QL from the heat egnine and the QH from the refrigerator. QL = QH - W = 800 - 595.2 = 204.8 kJ/min. QH = QL + W = 4982 + 595.2 = 5577.2 kJ/min so the total = 5782 kJ/min.