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A carnot refrigerator absorbs heat form a space at 15 C at a rate of 16,000 KJ/h and rejects heat to a reservior at 36 C. Determine the COP of the refrigerator, the power input, in KW, and the rate of heat rejected to high temperature reservoir, in KJ/h.

Respuesta :

fabianb4235

Answer:

Cop=13.72

Power input=0.324Kw

heat rejected=17150.4Kj/h

Explanation:

Cop:

first you use the equation to calculate the cop in a carnoT cycle

Cop=1/(T1/T2-1)

Remember to use the absolute temperature in Kelvin, adding 273.15

T2=15+273.15=288.15

T1=36+273.15=309.15

Cop=1/(309.15/288.15-1)=13.72

Power input

Q1=heat absorbed from the space

Power input=W

Cop= Q1/W

W=Q1/Cop

Q1=16000kJ/h=4.44Kw

W=4.44Kw/13.72=0.324Kw

heat rejected

taking into account the principle of energy conservation, everything that enters the system must leave, therefore the rejected heat is the sum of the power and the absorbed heat

Q2=0.324Kw+4.44Kw=4.764Kw=17150.4KJ/h

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