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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.

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User Hayk
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1 Answer

3 votes

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Step-by-step explanation:


(COP)_(HP) =(Desired-effectx)/(Work-done)= (Q_(1) )/(W) \\\\(COP)_(HP) =(COP)_(Ideal)\\\\(Q1)/(W) =(T_(1) )/(T_(1) -T_(2) )


W=Q_(1) (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 (293 K-288 K)/(293 K)=3x10^(6) kJ (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 (293 K-283 K)/(293 K)=3x10^(6) kJ (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 (293 K-278 K)/(293 K)=3x10^(6) kJ (15 K)/(293 K)=3x10^(6) kJ x 0.051195}


W = 0.153585x10^(6) kJ

W = 153,585 kJ

Hope this helps!

answered
User Leeroy Brun
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7.6k points