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

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

Step-by-step explanation:

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

`COP_(HP) = (T_(H))/(T_(H)-T_(L))`

`COP_(HP) = (298.15\,K)/(298.15\,K - 277.15\,K)`

`COP_(HP) = 14.198`

The reversible work input is:

`\dot W = (\dot Q_(H))/(COP_(HP))`

`\dot W = \left((50000\,(kJ)/(h) )/(14.198) \right)\cdot \left((1)/(3600)\,(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\,(kJ)/(h) \right)\cdot \left((1)/(3600)\,(h)/(s)\right)\cdot \left((1)/(COP_(real)) \right) - 0.978\,kW`