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Charra [1.4K]
3 years ago
5

An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De

termine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.
Engineering
1 answer:
Paraphin [41]3 years ago
5 0

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.

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Answer:

t=2.025 inches

Explanation:

Given that

P = 400 Psi

Yield stress ,σ = 80 ksi

Diameter ,d= 45 ft

We know that

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The required thickness given as

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\dfrac {400\times 540\times 3}{4\times 80\times 1000}=t

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