Answer:
A) 0.03382 kg/s
B) 7.0372 Kw
C) 4.3982
D) 0.7396 kw
Explanation:
Given data:
Evaporator at 60 C
Space temperature = 25 C
power consumed by compressor = 1.6 kw
T1( evaporator temperature ) = 12°C
attached below is the detailed solution
Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.
Answer:
I don't know ☺️☺️☺️❌‼️
Explanation:
I don't understand this question
