Question

A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and the heat transfer from the building through its roof and walls is 3 × 106 kJ. For this duty heat pumps are under consideration that would operate between the dwelling and
a. the ground at 15°C.
b. a pond at 10°C.
c. the outside air at 5°C.

Answer 1

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

$(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }$

$W=Q_{1} \frac{T_{1}-T_{2} }{T_{1} }$

a. the ground at 15°C.

$T_{1}$=20°C = 273 K + 20 = 293 K

$T_{2}$=15°C = 273 K + 15 = 288 K

$Q_{1}=3x10^{6} kJ$

$W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}$

$W = 0.051195x10^{6} kJ$

W = 51,194.54 kJ

b. a pond at 10°C.

$T_{2}$=10°C = 273 K + 10 = 283 K

$W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}$

$W = 0.102390x10^{6} kJ$

W = 102,390 kJ

c. the outside air at 5°C.

$T_{2}$=5°C = 273 K + 5 = 278 K

$W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}$

$W = 0.153585x10^{6} kJ$

W = 153,585 kJ

Hope this helps!