Last Updated on March 13, 2023 by Noaman Adenwala

In this article we will be discussing about refrigeration and air conditioning system, its properties, function, and different parameters in detail.

Introduction:

• Intensive properties do not depend on the size of the system e.g. pressure and temperature etc.
• Extensive properties depend on the size of the system e,g, volume, internal energy, enthalpy, etc.
• Entropy increases in an irreversible process due to two reason which is due to heat addition and due to friction i.e. conversion of mechanical forms of energy into heat.
• Throttling process is an irreversible adiabatic process. It is used to reduce the pressure of the fluid by introducing a restriction to the flow.
• Since enthalpy is a function of temperature only, the temperature also remains constant during the throttling process.

Basic Concepts:

• Expansion of liquid with flashing : This process is accompanied by an increase in Entropy along with a drop in pressure due to which volume increases and a part of the liquid is vaporized, thus cooling the liquid.

• Reversible adiabatic expansion of gas : This method is used with permanent gases such as air.

• Irreversible adiabatic expansion of a real gas : Real gases produce a substantial decrease in temperature under certain condition, which corresponds to initial low temperature and high pressure of the gas.

• Thermoelectric cooling : Cooling is produced at one junction of two dissimilar metals if a current is passed through them. The phenomenon is called the Peltier effect.

• Adiabatic Demagnetization : Cooling could be produced by the adiabatic demagnetization of a paramagnetic salt.

Properties of air:

Cp = 1.005 kJ.kg.K

R = 0.287 kJ/kg.K

M = 28.966

Cv = 0.718 kJ/kg.K

γ = 1.4

Refrigerating Machine / Heat Pump:

• It contains evaporator, compressor, condenser, expander
• The process followed in the cycle is as follows

Heat Q₀ is absorbed in the evaporator by the evaporation of a liquid refrigerant at a low-pressure P₀ and corresponding low saturation temperature T₀. The evaporated refrigerant vapor is compressed to a high-pressure P_k in the compressor consuming work W. Heat Q_k is rejected from the condenser to the surrounding

The heat rejected to the surroundings equals the heat absorbed from the cold body or refrigeration produced plus the work done or mechanical energy consumed. A reversible heat engine may be converted into a refrigerating machine if it runs in the reversed direction. There is no difference in the cycle of operations between a refrigerating machine and a heat pump. The same machine can be utilized either.

• To absorb heat from the cold body at temperature T₀ and reject it to the surroundings at temperature T_k ≥ T_a ( i.e. refrigerator)
• To absorb heat from the surroundings at temperature T₀ ≤ T_a and rejects it to the hot body at temperature T_k (i.e. heat pump)

The main difference between two is in their operating temperature. A refrigerating machine is operated between the ambient temperature T_a and a low-temperature T₀. A heat pump is operated between the ambient temperature T₀ and a high-temperature T_k.

Difference in their functions:

In a refrigerating machine, the heat exchanger that absorbs heat is connected to the conditional space. A refrigerating machine that is used for cooling in summer is to be used as a heat pump for heating in winter, it will be necessary either,

• To rotate the machine by 180⁰ to interchange the positions of the two heat exchangers between the space and surroundings.
• To exchange the functions of the two heat exchangers by the operation of valves e.g. a four-way valve in a window type air conditioner.

Coefficient of performance (COP):

COP = Energy Ratio = Useful heat / work

(COP)_net = Q₀ / W = Q₀ / (Q_k – Q₀)

(COP)_pump = Q_k / W = Q_k / (Q_k – Q₀)

(COP)_pump = 1 + (COP)_net

If the refrigerator/ heat pump cycle is reversed, it becomes a heat engine.

The thermal efficiency of the heat engine:

η_th = W/ Q_k = (Q_k – Q₀)/Q_k

(COP)_pump = Q_k /(Q_k – Q₀) = 1/ η_th

(COP)_ref = (1/η_th)-1

For vapor compression system ξ is of the order of 3 for air conditioning application. For air cycle refrigeration system ξ ≈ 1 and for vapor absorption systems, it is well below unity. Steam ejector machines have still lower values. For the purpose of heating, it is far more economical to use a heat pump rather than an electric resistance heater. The value of ξ for air conditioning applications can range between 3 and 5.

Reverse Carnot cycle:

A reversible Carnot cycle has the maximum COP

Heat absorbed from cold body Q₀ = T₀ ΔS

Heat rejected to hot body Q_k = T_k ΔS

Work done W = Q_k – Q₀ = (T_k – T₀)ΔS

COP for cooling = Q₀ / W = T₀ / (T_k-T₀)

COP for heating = Q_k / W = T_k / (T_k-T₀)

Carnot COP depends on the operating temperature T_k and T₀ only. It does not depend on the working substance used. The lowest possible refrigeration temperature is T₀ = 0 of which ξ = 0. The highest possible refrigeration temperature is T0 = Tk i.e. when the refrigeration temperature is equal to the temperature of surroundings of which ξ= ∞

• COP for cooling varies between 0 and ∞
• COP for heating varies between 1 and ∞

To obtain maximum possible COP in any application

• The cold body temperature T₀ should be as high as possible
• The hot body temperature T_k should be as low as possible.

The selection of temperature T_k depends on the surrounding medium used for heat rejection viz. air, water, and ground.

Water as a cooling medium is preferable to air as it affords a lower value of T_k because of following reasons:

• It is available at a temperature lower than that of air. Its temperature approaches the wet bulb temperature of the surrounding air. This is the limiting temperature to which heated water can be cooled in a cooling tower or a spray pond.
• The specific heat of water is about four times that of air.
• Water has a higher heat transfer coefficient than air mainly because of its high thermal conductivity.

Thus the use of water as a cooling medium results in a lower T_k, higher COP and lower power consumption in a refrigerating plant. Temperature approaching zero have been obtained using adiabatic demagnetization on a limited scale in laboratories.

• Refrigerating efficiency :

η_c = (COP)_c / (COP)_c, carnot

• Heating efficiency:

η_c = (COP)_h / (COP)_h, carnot