# Coefficient of Performance Formula

Every appliance that people buy has some efficiency and based on the efficiency people prefer to buy it. One of the most important factors in determining the efficiency of air conditioners, refrigerators, and heat pumps is the coefficient of performance. When we go to the store to buy an air conditioner, refrigerator, or heat pump, the first thing we do is look at the COP value. Let us take a closer look at the Coefficient of Performance (COP) formula.

### Coefficient of Performance (COP)

The Coefficient of Performance (COP) is defined by the ratio of heat dissipation and electrical power intake. It is the parameter by which the air conditioning unit’s performance is evaluated. It shows how effective heating and cooling systems are, following points show the characteristics of COP,

- It’s the ratio of compressor power input to refrigeration output.
- When the coefficient of performance (COP) rises, it means that the cooling output rises in proportion to the power input.
- Increased COP equates to greater efficiency.
- It means that as COP rises, so does efficiency, and as COP falls, so does efficiency.
- It refers to the ratio of evaporator cooling output to the air conditioner or refrigerator work input.
- When the COP is higher, the power consumption is higher, and the running costs are higher.

### Formula of Coefficient of Performance (COP)

The Coefficient of Performance is calculated as follows,

COP = Output / Input

**Coefficient of Performance for heating is,**

K = Q_{H}/ W_{in}Where,

K = Coefficient of Performance,

Q

_{H}= The heat pump’s output,W

_{in}= The work required by the considered system.

**Coefficient of Performance for cooling is,**

K = Q_{C}/ W_{in}Where,

K = Coefficient of Performance,

Q

_{C}= the refrigerator’s heat had dissipated,W

_{in}= The work required by the considered system.

If the Work input is zero then the Coefficient of Performance is equal to infinity.

**COP of Refrigerator is,**

K_{R}= Q_{2}/ (Q_{1 }– Q_{2})For reversible Refrigerator,

Q

_{1}/Q_{2}= T_{1}/T_{2}

K_{R}= T_{2}/ (T_{1 }– T_{2})

**COP of Heat pump,**

K_{H}= Q_{1}/ (Q_{1 }– Q_{2})For reversible Heat pump,

Q

_{1}/Q_{2}= T_{1}/T_{2}

K_{H}= T_{1}/ (T_{1 }– T_{2})

### Sample Questions

**Question 1: Define the Coefficient of Performance (COP).**

**Answer**:

The Coefficient of Performance (COP) is defined by the ratio heat dissipation and electrical power intake. More the COP of any system, better will be the efficiency and vice-versa. However, increased COP also means costlier system as it consumes more power.

**Question 2: How Can I Raise the Coefficient of Performance of Heat Pumps?**

**Answer**:

Heat pumps Coefficients of Performance (COP) can be improved by reducing temperature differences, enhancing insulation, utilizing proper flooring, and, when available, using a ground-source heat pump system.

**Question 3: A Carnot cycle machine works between 402 and 365 K. Calculate the COP of the refrigerator.**

**Solution:**

Given: T

_{1}= 402 K, T_{2}= 365 KSince,

K

_{R}= T_{2}/ (T_{1 }– T_{2})∴ K

_{R}= 365 / (402 – 365)∴ K

_{R}= 365 / 37∴

K_{R}= 9.86

**Question 4: A pump operating on cycle pump heat from a reservoir at 300 K equivalent to 27 KJ to a reservoir at 600 K equivalent to 128 KJ. Calculate COP?**

**Solution:**

Given: Q

_{1}= 128 KJ, T_{1}= 600 K, Q_{2}= 27 KJ, T_{2}= 300 KSince,

K

_{H}= Q_{1}/ (Q_{1 }– Q_{2})∴ K

_{H}= 128 / (128 – 27)∴ K

_{H}= 128 / 101∴

K_{H}= 1.26

**Question 5: The temperature in a residential food freezer is kept at -25 degrees Celsius. The temperature outside is 43 degrees Celsius. If 2.75 kJ/s of heat seeps into the freezer on a constant basis. What is the smallest amount of energy required to constantly pump this heat out?**

**Solution:**

Given: T

_{2 }= -25°C = -25 + 273 = 248 K, T_{1}= 43°C = 43 + 273 = 316 K, Q_{2}= 2.75 kJ/sSince,

For reversible Refrigerator,

Q

_{1}/Q_{2}= T_{1}/T_{2}∴ Q

_{1}= (Q_{2}× T_{1}) / T_{2}∴ Q

_{1}= (2.75 × 316) / 248∴ Q

_{1}= 869 / 248∴ Q

_{1}= 3.50 kJ/sW = Q

_{1}– Q_{2}∴ W = 3.50 – 2.75

∴

W = 0.75 kJ/s

**Question 6: A refrigerator’s food compartment is kept at 4 degrees Celsius by eliminating heat at a rate of 360 kJ/min. If the refrigerator’s needed power input is 2 kW. Find the refrigerator’s COP.**

**Solution:**

Given: T

_{2}= 4°C = 4 + 273 = 277 K, Q_{2}= 360 kJ/min = 360/60 = 6 kW, W = 2 kWSince,

K

_{R}= Q_{2}/W∴ K

_{R}= 6/2∴

K_{R}= 3