Thermodynamics Final answer is giving, show me how to get to it. If you get prop

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Thermodynamics

Final answer is giving, show me how to get to it.

If you get property values from a table, give the name of the table.

An air conditioner must maintain a computer laboratory at 18°C when the outside temperature is 30°C. The thermal load that must be removed from the room consists of heat transfers entering through the walls and roof of the laboratory at a rate of75,000 kJ/h and from the computers, lighting, and occupants at a rate of 15,000 kJ/h. (a) Determine the maximum theoretical coefficient of performance and the corresponding minimum theoretical power required by the air conditioner, in kW (Ans: COPR, max 24.25: Wmin 1.03 kWS) (b) If the actual power required by the air conditioner for this duty is 8.3 kW, determine the coefficient of performance. CAns: COPR 3.01) (c) If the given temperature and thermal load are observed for a total of 100 hours and electricity costs are 13 cents per kW-h, determine the cost, in over that period for each of cases (a) and (b). ns: case (a), cost S 13.4; case (b), cost S 1079)

Explanation / Answer

total heat to be removed. QC= 75000+15000= 90000 Kj/h=25 Kj/sec

Coefficient of performance= TC/(TH-TC)= (18+273)/(30+273-18-273)= 24.25

COP= QC/W= 25/W

24.25= 25/W

W= 25/24.25=1.03 Kj/sec= 1.03 KW

when W= 8.3 KW

COP=QC/W= 25/8.3= 3.01

cost is 13 cents per Kw-h, total cost of power = 1.03 *13*100 cents =1.03*13*100*0.01S= 13.9$

for the second case, total cost of power= 8.3*13*100*0.01$=107.9$

24.25= W/25

W=25*24.25 Kj/sec

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