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C. Calculation problems (12 marks each) 6. Generating electricity from a nuclear power plant A thermal heat at a rate of 1.2 GW and at a temperature of 450 C. This heat is used in an engine that is connected to the sea water at T 20°C as low temperature reservoir. The engine has an efficiency of 51%. nuclear power plant generates (A) What is the electrical power output of this power plant? (The electrical enerzy output is equal to the amount of work.) (B) What would be the ideal efficiency of an engine operating between these two temper- atures? How does this relate to the real efficiency? (C) The cold water flows through the power plant at a rate of 5.8. 101 k/s. By how much will the temperature of the cold water rise? (It takes 4.184 kJ to heat one kg of water by one degree Celsius.) 7. A thermodynamic cycle. A gas expands adiabatically from state A to state B, in the to state C, where its process doing 5.9 kJ of work. It is then compressed isothermally volume equals that of state A. During the compression, 4.8 kJ of work is done on the gas. The gas is then heated at constant volume until it returns to state A. (A) Draw this cycle in a pV diagram. (B) What is the net heat added to the gas during one complete cycle? (C) What is the efficiency of this cycle? Hint: First determine how much heat is added in process from C to A

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Inputpower to the heat engine-Pin 1.2GW Temperature at low temperature T 20° C (273+20)K-293K Temperature at high temperature TH450° C 450273 723K Efficiency of heat engine-η-5190-0.51 (A) Efficiency of engine is defined as out 0.51 =-out 1.2 GW Pout 1.2 G W (0.51) 0.612GW Ans (B) Efficiency of heat engine can be defined in terms of temperature as, TL TH 293K 723K. η (0.595x 100)% 59.5% This efficiency is greater than the given (real) efficiency in the question, this is because, some of 0.595 the energy loss due to friction and other dragging force. So, we get the output less than the expected ideal output energy4 kg (C) Given, rate of flow ofmass-R Consider mass of water = m 5.8 × 10 dt aise of temperature of water ΔΤ Specific heat of water-s-4186 J/kg° C The rate of heat gain by water-Pwater Now, by conservation of energy, (msAT) - (SAT)- RsAT dt dt Pout-Pwater 0.612 GW 5.8 × 104 × 4186th K2.52° C Ans AT-out = 0612 × 109 W 5.8 x 104 x 4186

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