Question

A heat pump is to be used to heat a house in winter and then reversed to cool the house in summer. The interior space is to be maintained at 20oC. Heat transfer through walls and roof is estimated to be χ = 0.525 kJ/sec per degree difference in temperature between inside and outside.
(a) In winter if the outside temperature is 5oC, what is the minimum power required to drive the heat pump?
(b) For the same power requirement, in summer what is the maximum outside temperature for which the reversed heat pump (air conditioner) can maintain an inside temperature of 20oC.

Answer 1

Answer:

$35.07\°C$

Explanation:

To solve this exercise, it is necessary to apply the concepts of Performance Coefficient and work.

For part A, we have given the data on the outside temperature, which is 5°C. In this way the rate of heat loss in the room is given by

$\dot{Q} = \xi \Delta T$

where,

$\xi =$ Heat transfer per second

$\Delta T =$Change in Temperature

We have then,

$\dot{Q} = 0.525(20-5)$

$\dot{Q} = 7.8749kW$

Now we can calculate the coefficient of performance which is given by,

$COP = \frac{T}{\Delta T}$

$COP = \frac{20}{20-5}$

$COP = 19.53$

By definition we know that the coefficient of performance of a pump is given by

$COP = \frac{\dot{Q}}{W}$

where,

$\dot{Q} =$Desired effect

W = Work input

Solving for the work input we have

$W = \frac{\dot{Q}}{COP}$

$W = \frac{7.875}{19.33}$

$W = 0.407kW$

For part B we consider $\tau$ as the maximum temperature outside, therefore, calculating the heat rate we have

$\dot{Q}=0.525*(T-293)Kj/S$

$\dot{Q} = 525*(T-293)W$

Returning to the expression of the coefficient of performance we have to,

$COP = \frac{\dot{Q}}{W}$

$0.407kW = \frac{525*(T-293)}{\frac{293}{T-293}}$

$(T-293)^2 = \frac{403*293}{525}$

$T^2-586T+85849=225$

$T^2-586T+85624=0$

Solving the polynomial you have to

$T= 308K = 35\°C$

Therefore the maxium outside temperature is $35\°C$