A refrigeration system requires 1.3 kW of power for a refrigeration rate of 3 kJ/s and a heat rejection at 60 °C.

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A refrigeration system requires 1.3 kW of power for a refrigeration rate of 3 kJ/s and a heat rejection at 60 °C. Calculate the coefficient of performance and lowest temperature the system can possibly operate at, a) 4.30, 232.18 K b) 2.30, 232.18 K c) 2.30, 352.18 K d) 3.30, 252.18 K

The Correct Answer and Explanation is:

To solve this problem, we’ll first need to understand the coefficient of performance (COP) for a refrigeration system and the lowest temperature the system can possibly operate at.

1. Coefficient of Performance (COP)

The COP of a refrigeration system is the ratio of the refrigeration effect (cooling rate) to the work input required by the system. Mathematically, it can be expressed as:COP=QLWCOP = \frac{Q_L}{W}COP=WQL​​

Where:

  • QLQ_LQL​ is the refrigeration rate or the amount of heat removed from the cold reservoir (in kJ/s).
  • WWW is the power input required to operate the system (in kW).

From the problem, we are given:

  • QL=3 kJ/sQ_L = 3 \, \text{kJ/s}QL​=3kJ/s
  • W=1.3 kWW = 1.3 \, \text{kW}W=1.3kW

Now, calculate the COP:COP=31.3=2.30COP = \frac{3}{1.3} = 2.30COP=1.33​=2.30

So, the COP is 2.30.

2. Lowest Temperature of Operation

To calculate the lowest temperature the system can operate at, we can use the thermodynamic relationship between the COP, the temperatures of the cold and hot reservoirs. For a refrigeration cycle, the COP is related to the temperatures by:COP=TLTH−TLCOP = \frac{T_L}{T_H – T_L}COP=TH​−TL​TL​​

Where:

  • TLT_LTL​ is the temperature of the cold reservoir (in Kelvin).
  • THT_HTH​ is the temperature of the hot reservoir (in Kelvin).

We are given that the heat rejection temperature is 60°C. Converting this to Kelvin:TH=60+273.15=333.15 KT_H = 60 + 273.15 = 333.15 \, \text{K}TH​=60+273.15=333.15K

Now, rearranging the COP formula to solve for TLT_LTL​:COP=TLTH−TL⇒2.30=TL333.15−TLCOP = \frac{T_L}{T_H – T_L} \quad \Rightarrow \quad 2.30 = \frac{T_L}{333.15 – T_L}COP=TH​−TL​TL​​⇒2.30=333.15−TL​TL​​

Solving for TLT_LTL​:2.30(333.15−TL)=TL2.30 (333.15 – T_L) = T_L2.30(333.15−TL​)=TL​2.30×333.15−2.30TL=TL2.30 \times 333.15 – 2.30 T_L = T_L2.30×333.15−2.30TL​=TL​766.245−2.30TL=TL766.245 – 2.30 T_L = T_L766.245−2.30TL​=TL​766.245=3.30TL766.245 = 3.30 T_L766.245=3.30TL​TL=766.2453.30=232.18 KT_L = \frac{766.245}{3.30} = 232.18 \, \text{K}TL​=3.30766.245​=232.18K

Thus, the lowest temperature TLT_LTL​ is 232.18 K.

3. Answer

Based on these calculations:

  • The COP is 2.30.
  • The lowest temperature the system can operate at is 232.18 K.

Thus, the correct answer is (b) 2.30, 232.18 K.

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