Question

Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is to be heated by electric resistance heaters. If the house is to be maintained at 25 0C, determine the reversible work input for this process and the irreversibility.

Answer 1

Answer:

a) $\dot W = 0.978\,kW$, b) $I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}$

$COP_{HP} = 14.198$

The reversible work input is:

$\dot W = \frac{\dot Q_{H}}{COP_{HP}}$

$\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$\dot W = 0.978\,kW$

b) The irreversibility is given by the difference between real work and ideal work inputs:

$I = \dot W_{real} - \dot W_{ideal}$

$I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$