DIFFERENCE BETWEEN A HEAT ENGINE, REFRIGERATOR AND HEAT PUMP

Heat Engine: In a heat engine, as shown in Fig. 18.4(a), the heat supplied to the engine is converted into useful work. If Q₂ is the heat supplied to the engine and Q₁ is the heat rejected from the engine, then the net work done by the engine is given by,

 

W_e =Q₂ −Q₁

Efficiency of heat engine,

Refrigerator: A refrigerator is an equipment used to remove the heat continuously from the space (sink). It maintains the temperature below atmospheric temperature and reject the heat to the atmosphere (source) by increasing the temperature potential of the heat to be rejected with the help of mechanical work input (compressor).

The refrigeration system is shown in Fig. 18.4(b). In a refrigerator system Q₁ quantity of heat is removed from the source where its temperature T₁ is maintained below atmospheric temperature T₂. COP of refrigerator is,

Heat Pump: It is an equipment used to supply the heat continuously to the space. It maintains the temperature above atmospheric temperature by absorbing the heat from the atmosphere and increasing its temperature potential with mechanical work input (compressor). The heat pump system is shown in Fig. 18.4(c). In a heat pump system, Q₂ quantity of heat is supplied to the room where the temperature T₂ is maintained above atmospheric temperature T₁. The heat Q₁ is obtained from the atmosphere.

Figure 18.4 (a) Heat engine, T₁> T>_atm, (b) Refridgerator T₁ < T_atm, (c) Heat pump

COP or Energy performance ratio of heat pump is,

Example 18.1

Find the COP of a refrigeration system if the work input is 10 kJ/kg and refrigeration effect produced is 180 kJ/kg of refrigerant flowing.

Solution

 

w_ref =100 kJ/kg, q=180 kJ/kg

Example 18.2

A machine working on a Carnot cycle operates between 310 K and 270 K. Calculate the COP when it is operated as:

1. a refrigerator, and
2. a heat pump.

Solution

Given: T₁ = 270 K, T₂ = 310 K

1. Refrigerator 
2. Heat pump: 

Example 18.3

A Carnot refrigeration cycle absorbs heat at 268 K and rejects it at 298 K.

1. Calculate the COP of this refrigeration cycle.
2. If the cycle is absorbing 1150 kJ/min at 268 K, find the rate of work input required.
3. How many kJ/s will be delivered if it works as a heat pump at 298 K and absorbs 1150 kJ/min at 268 K.

Solution

Given: T₁ = 268 K, T₂ = 298 K, Q₁ = 1150 kJ/min

1. Carnot refrigeration cycle: Work required, 
2. Heat pump:(COP)_HP =1 + (COP)_ref =1 + 8.933= 9.933