Answer:
There is 0.466 KW required to operate this air-conditioning system
Explanation:
<u>Step 1:</u> Data given
Heat transfer rate of the house = Ql = 755 kJ/min
House temperature = Th = 24°C = 24 +273 = 297 Kelvin
Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin
<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.
COPr,c = 1 / ((To/Th) - 1)
COPr,c = 1 /(( 308/297) - 1)
COPr,c = 1/ 0.037
COPr,c = 27
<u>Step 3:</u> The power input cna be given as followed:
Wnet,in = Ql / COPr,max
Wnet, in = 755 / 27
Wnet,in = 27.963 kJ/min
Win = 27.963 * 1 KW/60kJ/min = 0.466 KW
There is 0.466 KW required to operate this air-conditioning system
Answer:
The answer is "
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Explanation:
Please find the correct question in the attachment file.
using formula:
Answer:
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