The air-conditioner (AC) in the sketch inputs 10 hp while the cooling load is 10 tons. The heat rejected is nearest:
A) 43 kJ/s
B) 39 kJ/s
C) 32 kJ/s
D) 22 kJ/s
A) 43 kJ/s
Heat transferred by the condenser goes for a useful purpose or it is discarded to the atmosphere with a cooling tower or to a river or large lake. For this problem, it is
Q C = m ( h1 ? h4 )
= 12 × [ 251? ( 251+ 0.9× 2358)] = ?25 500 kJ/s or ?25.5 MJ/s
We can check our calculations using Eq. 4.45: Is 43.6 = 18. 1 + 25. 5? 43.6 = 43.6 ? Yes! Or, we could have used
Eq. 4.45 to find
Q C = Q B ? W T = 43.6 ? 18. 1 = 25.5 MJ/s
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