Question

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a low-temperature heat source; during the summer, it acts as a high-temperature heat sink. The heat-transfer rate through the walls and roof of the house is 0.75 kJ s—l for each C of temperature difference between the inside and outside of the house, summer and winter. The heat-pump motor is rated at 1.5 kW. Determine the minimum outside temperature for which the house can be maintained at 200C during the winter and the maximum outside temperature for which the house can be maintained at 250C during the summer.

Answer 1

Answer:

$T_{C}$ = -4.2°C

$T_{H}$ = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

|$\frac{Q_{C} }{Q_{H} }$| = $\frac{T_{C} }{T_{H} }$

Similarly,

|$Q_{H}$| = |$Q_{C}$| + |$W_{S}$|

During winter, the value of |$T_{H}$| = 20°C = 273.15 + 20 = 293.15 K and |$W_{S}$| = 1.5 kW. Therefore,

|$Q_{H}$| = 0.75($T_{H}$ -  $T_{C}$)

Similarly,

|$\frac{W_{S} }{Q_{H} }$| = 1 - $\frac{T_{C} }{T_{H} }$

1.5/0.75*(293.15-$T_{C}$) = 1 - ($T_{C}$/293.15

Further simplification,

$T_{C}$ = -4.2°C

During summer, $T_{C}$ = 25°C = 273.15+25 = 298.15 K, and |$W_{S}$| = 1.5 kW. Therefore,

|$Q_{C}$| = 0.75($T_{H}$ -  $T_{C}$)

Similarly,

|$\frac{W_{S} }{Q_{C} }$| = $\frac{T_{H} }{T_{C} }$ - 1

1.5/0.75*($T_{H}$ - 298.15) = ($T_{H}$/298.15

Further simplification,

$T_{H}$ = 49.4°C