An 18,000 Btu/h air conditioner running at full load is used to keep a room at 22°C in an environment at 33°C. The power input to the compressor is 3.4 kW. Determine:

(a) The rate of heat rejected in the condenser, in kJ/h.

(b) The COP (Coefficient of Performance) of the air conditioner.

(c) The rate of cooling, in Btu/h, if the air conditioner operated as a Carnot refrigerator for the same power input.

The rate of heat rejected in the condenser is 15,629.08 kJ/h,the COP of the air conditioner is approximately 5.023 and the rate of cooling for the Carnot refrigerator is approximately 96,136.71 Btu/h.

To solve this problem, we'll use the formulas and equations related to the Coefficient of Performance (COP) and Carnot refrigeration cycle.

Given:

Power input to the compressor (P) = 3.4 kW

Heat capacity of the air conditioner (Q) = 18,000 Btu/h

Room temperature (T2) = 22°C = 295 K

Environment temperature (T1) = 33°C = 306 K

(a) Rate of heat rejected in the condenser:

We can use the formula:

Q = P + Q_condenser

Since the power input is given in kilowatts, we need to convert Q to the same unit:

Q = 18,000 Btu/h * 1.05506 kJ/Btu ≈ 19,029.08 kJ/h

Now we can calculate the heat rejected in the condenser:

Q_condenser = Q - P = 19,029.08 kJ/h - 3.4 kW * 1,000 J/kW ≈ 19,029.08 kJ/h - 3,400 kJ/h ≈ 15,629.08 kJ/h

Therefore, the rate of heat rejected in the condenser is approximately 15,629.08 kJ/h.

(b) Coefficient of Performance (COP) of the air conditioner:

The COP is defined as the ratio of the desired output (cooling) to the required input (power):

COP = Q / P

Using the given values:

COP = 18,000 Btu/h / (3.4 kW * 1,055.06 J/Btu) ≈ 18,000 Btu/h / 3,585.24 J/s ≈ 5.023

Therefore, the COP of the air conditioner is approximately 5.023.

(c) Rate of cooling if the air conditioner operated as a Carnot refrigerator:

The Carnot refrigerator's coefficient of performance can be calculated using the formula:

COP_carnot = T2 / (T1 - T2)

Using the given temperatures:

COP_carnot = 295 K / (306 K - 295 K) ≈ 295 K / 11 K ≈ 26.82

The rate of cooling for the Carnot refrigerator is given by:

Q_carnot = P * COP_carnot

Substituting the given values:

Q_carnot = 3.4 kW * 26.82 ≈ 91.128 kW

Converting Q_carnot to Btu/h:

Q_carnot = 91.128 kW * 1,055.06 J/Btu ≈ 96,136.71 Btu/h

Therefore, the rate of cooling for the Carnot refrigerator is approximately 96,136.71 Btu/h.

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