Heat Pump

It is a device that transfers heat from a colder area to a hotter one using external energy, typically in the form of work. Unlike traditional heating or cooling systems, a heat pump can both heat and cool a space. It operates on the principles of thermodynamics, similar to a refrigerator but with a reversed purpose.Heat pumps are widely used in air conditioning systems (heat pump AC units) and heating systems. They are practical examples of how energy transfer and refrigeration cycles work.

1.0Definition of  Heat Pump

• It is a device that transfers heat from a low-temperature source to a high-temperature sink. It operates much like a refrigerator but with a reverse intent. While a refrigerator removes heat from the food compartment and rejects it outside, a heat pump extracts heat from the environment (like air, water, or ground) and delivers it into a living space.

2.0Types of Heat Pumps

3.0Block Diagram of Heat Pump

4.0System Components of Heat Pump

A heat pump generally consists of:

• Evaporator: Absorbs heat from the environment.
• Compressor: Compresses the refrigerant, increasing its pressure and temperature.
• Condenser: Releases heat to the indoor environment.
• Expansion Valve:Lowers the refrigerant’s pressure to initiate the cycle again.

5.0Working of Heat Pump

• It works by taking heat from a cooler place—like the air outside or the ground—and moving it into a warmer area, such as inside your home. It does this by circulating a special fluid called refrigerant through different parts of the system. As the refrigerant moves, it picks up heat in one place and releases it in another, repeating this cycle over and over to keep your home comfortable.

1.Evaporator (Heat Absorption)

• Location: Outside the house (in heating mode)
• Function: The refrigerant, in low-pressure liquid form, flows through the evaporator coil.
• It absorbs heat from the outside environment, even if the air feels cold.
• The refrigerant evaporates into a low-pressure gas.

2.Compressor (Pressure and Temperature Increase)

• The low-pressure vapor enters the compressor.
• It compresses the vapor, increasing both its pressure and temperature.
• Now, the refrigerant becomes a high-pressure, high-temperature gas.

3.Condenser (Heat Release)

• Location: Inside the house (in heating mode)
• The heated gas moves through the condenser coil.
• It transfers heat to the indoor air through a heat exchange process.
• As it loses heat, the refrigerant condenses into a high-pressure liquid.

4.Expansion Valve (Pressure Reduction)

• The liquid refrigerant at high pressure moves through the expansion valve.
• This valve rapidly reduces the pressure, causing the temperature to drop.
• The refrigerant returns to a low-pressure, cold liquid state and re-enters the evaporator.

Cycle Repeats Continuously

• This process repeats for as long as the heat pump remains in operation.
• In cooling mode, the cycle reverses: the evaporator becomes the indoor coil, and the condenser becomes the outdoor coil.

6.0The Coefficient of Performance (COP)

$(COP{)}_{HP}=\frac{Q_H}{W}$

Carnot Ideal Case

$(COP)=\frac{T_H}{T_H-T_C}$

Apply First Law of Thermodynamics 

$\begin{array}{rl}& Q_H=Q_C+W\\ & Q_H=\text{ Heat deliver to hot reservoir(output) }\\ & Q_C=\text{ Heat absorbed from cold reservoir(Input) }\\ & W=\text{ Work done on the system }\end{array}$

COP of the Heat Pump

$\begin{array}{rl}& (COP{)}_{HP}=\frac{\text{ Useful Heat Delivered }}{\text{ Work input }}=\frac{Q_H}{W}\\ & W=Q_H-Q_C\\ & (COP{)}_{HP}=\frac{Q_H}{Q_H-Q_C}\end{array}$

Carnot Heat Pump(Ideal Case)

$\begin{array}{rl}& \frac{Q_C}{Q_H}=\frac{T_C}{T_H}\Rightarrow \frac{Q_H}{Q_C}=\frac{T_H}{T_C}\\ & (COP{)}_{HP}=\frac{Q_H}{Q_H-Q_C}\end{array}$

Dividing numerator and denominator by $Q_c$

$\begin{array}{rl}& (COP{)}_{HP}=\frac{\frac{Q_H}{Q_C}}{\frac{Q_H}{Q_C}-1}=\frac{\frac{T_H}{T_C}}{\frac{T_H}{T_C}-1}\\ & (COP{)}_{HP}=\frac{T_H}{T_H-T_C}\\ & T_H=\text{ Temperature of hot reservoir(Kelvin) }\\ & T_C=\text{ Temperature of cold reservoir(Kelvin) }\end{array}$

Note:Real systems will have lower COP due to irreversibilities

7.0Factors Affecting Heat Pump Performance

• Outdoor Temperature – Colder air reduces heating efficiency (especially air-source).
• Indoor Thermostat Setting – Extreme settings increase energy use.
• Heat Pump Type – Air-source, ground-source, and mini-splits vary in efficiency.
• System Sizing – Oversized/undersized units lower efficiency and comfort.
• Ductwork Quality – Leaks and poor insulation waste energy.
• Maintenance – Dirty filters/coils reduce airflow and efficiency.
• Defrost Cycles – Ice buildup triggers energy-consuming defrost modes.\
• Refrigerant Level – Incorrect charge lowers performance.
• Climate & Humidity – Extreme weather and poor humidity control affect performance.

8.0Advantages and Disadvantage of Heat Pumps

9.0Comparison of an Air Conditioner And Heat Pump

Illustration-1.A heat pump heats a house to 22°C using ambient air at -3°C as the source. The house loses heat at a rate of 4000 kJ/h per °C difference between inside and outside.

Find:

a) The total heat loss per hour

b) The minimum power input required by an ideal (Carnot) heat pump to maintain the temperature.

Solution:

$\begin{array}{rl}& \text{ (a) Temperature difference }=22-(-3)=25^{\circ }\mathrm{C}\Rightarrow Q_L=4000\times 25=100000\mathrm{kJ}/\mathrm{h}\\ & \text{ (b) }\begin{array}{rl}{T}_H& =22+273=295K\\ T_C& =-3+273=270K\end{array}\end{array}$

Carnot COP,

$CO{P}_{Carnot,HP}=\frac{T_H}{T_H-T_C}=\frac{295}{25}=11.8$

Power Input Required,

$W=\frac{Q_H}{COP}=\frac{100000}{11.8}\approx 8474.6\mathrm{kJ}/\mathrm{h}\approx 2.35\mathrm{kW}$

Illustration-2A Carnot heat pump operates between -10°C and 35°C. Calculate the COP.

Solution:

$\begin{array}{rl}& T_C=-10+273=263\mathrm{K}\\ & T_H=35+273=308\mathrm{K}\\ & {}^{COP}{}_{\text{Carnot },HP}=\frac{T_H}{T_H-T_C}=\frac{308}{308-263}=\frac{308}{45}=6.84\end{array}$