How To Solve This Thermodynamics Question 1 (274.16 KiB) Viewed 27 times
How To Solve This Thermodynamics Question 2 (44.92 KiB) Viewed 27 times
T/°C 60 501 40 6. In a demo heat pump, heat Qin is taken from a cold reservoir containing 10 litres of a mixture of water and glycol. Heat, Qout, is delivered to a hot reservoir also containing 10 litres of water. (In a practical application, Vin is the amount of heat taken from a suitable reservoir such as the cold outdoor air and Qout is the heat we use to warm up our house (or hot water or something). The compressor supplies the work W net which we pay for through the electricity bill.) 30- 20 10 t/s 0 1000 1200 1400 1600 1800- The graph in Figure 2 shows how the temperature Figure 2. The variation of temperature with in the hot reservoir increases with time t. The time in the hot reservoir. measurement started when the heat pump started. For simplicity, we fit a straight line to our measurements using the least squares method. It gives T = a.t + b where a = dT/dt = 0.0163 °C/s and b = 25.7 °C. This is an approximation that shows that on average, the temperature in the "hot reservoir" increases by 0.0163 °C/s during this time, when the temperature changes from about 26 °C to 52 °C. The power of the compressor is approximately constant at 158 W during the measurement. Calculate the coefficient of performance for the heat pump. The practical coefficient of performance of the heat engine is defined as Qout COP = Wnet where Qout is the heat deposited in the hot reservoir and Wnet is the energy used to drive the compressor. The COP may also be written as a ratio of powers (energy/time);
COP = lout _dQout/dt Wnet dWnet/dt Pout/Pnet where Pout is the heat power delivered to the hot reservoir and Pnet is the electric power that drives the compressor. To calculate Pout we use that dQout dT m. dt dt Pout where m is the mass of the liquid changing temperature, and c is the specific heat capacity of the liquid. Answer: 4,3
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