# Introduction and Basic Concepts of Refrigeration and Air Conditioning, PDF

## Introduction and Basic Concepts of Refrigeration and Air Conditioning:

#### First, let’s understand a few things.

• 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 it is isoenthalpic process h_1 = h_2.
• Since enthalpy is a function of temperature only, the temperature also remains constant during the throttling process.

#### Production of low temperature:

• 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 a certain initial low temperature and high pressure of the gas.

• Thermoelectric cooling:

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

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 involved 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 Pk in the compressor consuming work W.

Heat Qk 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 run 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 Tk ≥ Ta ( i.e. refrigerator)
• To absorb heat from the surroundings at temperature T0 ≤ Ta and rejects it to the hot body at temperature Tk (i.e. heat pump)

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

#### Another essential difference is in their useful function.

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 / (Qk – Q0)

(COP)pump = Qk / W = Qk /

(Qk – 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:

ηth = W/ Qk = (Qk – Q0)/Qk

(COP)pump = Qk /(Qk – Q0) = 1/ ηth

(COP)ref = (1/ηth)-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 machined 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 Qk = Tk ΔS

Work done W = Qk – Q0 = (Tk – T0)ΔS

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

COP for heating = Qk / W = Tk / (Tk-T0)

Carnot COP depends on the operating temperature Tk 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 Tk should be as low as possible.

The selection of temperature Tk 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 Tk 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 Tk, higher COP and lower power consumption in a refrigerating plant.

• Refrigerating efficiency :

ηc = (COP)c / (COP)c, carnot

• Heating efficiency:

ηc = (COP)h / (COP)h, carnot

Temperature approaching zero have been obtained adiabatic demagnetization on a limited scale in laboratories.

Amrit Kumar

Amrit Kumar is the founder of Learn Mechanical, an Advisor at The Mechanical Engineering- a content-based website in Mechanical Engineering based in Delhi. He has 5+ years of teaching experience in the Core Mechanical Field.