Answer to Question #239553 in Molecular Physics | Thermodynamics for zzz

Question

A heat pump heats a house in the winter and then reverses to

cool it in the summer. The interior

temperature should be 20°C in the winter and 25°C in the summer. Heat transfer th

rough the

walls and ceilings is

estimated to be 2400 kJ per hour per degree temperature difference

between the inside and outside. (a) If t

he winter outside temperature is 0°C, what is the

minimum power required to drive the heat pump? (b) For the same power as in part (a), what

is the maximum outside summer temperature for which the house can be maintained at 25°C?

Show the schematic diagram

s of the systems.

Answer 1

Inside temperature

$T_{iw}= 20 \;°C$ in winter

$T_{is} = 25 \;°C$ in summer

Heat transfer through the walls ans ceilings Q = 2400 kJ

$Q^* = \frac{2400 \;kJ}{3600 \;sec \cdot °C} = 0.667 \; \frac{kW}{°C}$

(a) In winter

$Q = Q^*\times ΔT = 0.667 \times (293-273) \\ Q= 13.333 \;kW$

For ideal hezt pump

$\frac{Q_i}{Q_i-Q_0} = \frac{T_i}{T_i-T_0} = COP \\ \frac{Q_i}{P} = \frac{T_i}{T_i-T_0 } \\ \frac{13.333}{P} = \frac{293}{293-273} \\ P= 0.910 \;kW$

The minimum power required to drive the heat pump is 0.910 \;kW

(b) in summer

$Q=Q^* \times ΔT = 0.667 (T_0-298) \\ COP = \frac{Q_i}{P} = \frac{T_i}{T_0-T_i} \\ \frac{0.667(T_{max} -298)}{0.91} = \frac{298}{T_{max} -298} \\ T_{max} -298 = 20.168 \\ T_{max} = 318.168 \;K \\ T_{max} = 45.168 \;°C$