Basics of Refrigeration

Refrigeration

Literal meaning of refrigeration is the production of cold confinement relative to its surroundings. In this, temperature of the space under consideration is maintained at a temperature lower than the surrounding atmosphere. To achieve this, the mechanical device extracts heat from the space that has to be maintained at a lower temperature and rejects it to the surrounding atmosphere that is at a relatively higher temperature. Since the volume of the space which has to be maintained at a lower temperature is always much lower than the environment, the space under consideration experiences relatively higher change in temperature than the environment where it is rejected.

The precise meaning of the refrigeration is thus the following: Refrigeration is a process of removal of heat from a space where it is unwanted and transferring the same to the surrounding environment where it makes little or no difference. To understand the above definition, let us consider two examples from the daily life.

Unit of Refrigeration and COP

The standard unit of refrigeration is ton refrigeration or simply ton denoted by TR. It is equivalent to the rate of heat transfer needed to produce 1 ton (2000 lbs) of ice at 32 0F from water at 32 0F in one day, i.e., 24 hours. The enthalpy of solidification of water from and at 32 0F in British thermal unit is 144 Btu/lb.

Refrigeration effect is an important term in refrigeration that defines the amount of cooling produced by a system. This cooling is obtained at the expense of some form of energy.

Therefore, it is customary to define a term called coefficient of performance (COP) as the ratio of the refrigeration effect to energy input.

Refrigerator

Refrigeration is the process of maintaining the temperature of a body below that of its surroundings.

The working fluid used for this purpose is called Refrigerant the equipment used is called Refrigerator.

The refrigerant is compressed to high pressure and high temperature in the compressor and it gets cooled in the condenser at constant pressure.

Fig 1 Basic Refrigeration Flow Diagram

The high pressure refrigerant vapour is expanded in the expander and the required quantity of refrigerant evaporates in the evaporator by absorbing heat from the space to be cooled and the cycle is repeated.

Fig 2 Refrigeration cycle Co-efficient of Performance (C.O.P)

Co-efficient of performance

= Refrigeration Effect/ Workdone = Q2 / w

COP = Q2 / Q1- Q2

COP is always greater than Unity.

Heat Pump

Heat pump is a device which maintains temperature of a body greater that of its surroundings.

A refrigeration system produces 40 kg/hr of ice at 0oC from water at 25oC. Find the refrigeration effect per hour and TR. If it consumes 1 kW of energy to produce the ice, find the COP. Take latent heat of solidification of water at 0oC as 335 kJ/kg and specific heat of water

4.19 kJ/kg oC.

Solution

Heat removal rate to form 40 kg of ice at 0oC from water at 25oC

Qc = sensible cooling from 25oC to 0oC + latent heat of solidification of water

= 40 kg/hr X(25 – 0) oCx.19 kJ/kg. oC + 40 kg/hr = 335 kJ/kg

= 4190 kJ/hr + 13400 kJ/hr

= 17590 kJ/hr

Refrigeration effect (QC) = 17590 kJ/hr

We know that 1 TR = 210 kJ/min = 12600 kJ/hr

Therefore, TR equivalent to 17590 kJ/hr = 17590kJ/hr 1.396 12600kJ/hr Refrigeration effect 17590kJ/hr

COP = 4.886

A house hold refrigerator is maintained at a temperature of 2° Every time the door is opened, warm material is placed inside, introducing an average of 420kJ. But making only a small change in the temperature of the refrigerator. The door is opened 20 times a day, and the refrigerator operates at 15%of ideal COP. The cost of work is 32 paise per kWhr. What is the monthly bill for this refrigerator? The atmosphere is at 30°C.

Air Refrigeration Cycles

Introduction

In an air refrigeration cycle, the air is used as refrigerant. In olden days, air was widely used in commercial applications because of its availability at free of cost. Since air does not change its phase i.e. remains throughout the cycle, therefore the heat carrying capacity per kg of air is very small as compared to vapour absorbing systems. The practical unit of refrigeration is expressed in terms of ‘tone of refrigeration’

(briefly written as TR). A tone of refrigeration is defined as the amount of refrigeration effect produced by the uniform melting of one tone (1000 kg) of ice from and at 00C in 24 hours. Since the latent heat of ice is 335 kJ/kg, therefore one tone of refrigeration.

Air enters the compressor of air craft system at 100kpa, 277k and is compressed to 300kpa with an isentropic efficiency of 72%. After being cooled to 328k and air expands is 100kpa and an ηIsen=78% the load is 3 tons and find COP, power, mass flow

Given data:

P1= 100kpa, T3=38k T1= 277k, P4=100kpa P2= 300kpa, ηT=78%

ηc= 72%

Solution:

process 1-2 Isentropic compression T2= (P2/P1) γ-1/γ × T1

T2= (300/100)1.4-1/1.4 T2=379.14k

ηc=(T2-T1)/(T2′-T1)

0.72= (379.14-277)/ (T2′-277) T2=418.86k

Process 3-4 isentropic compression

T3/T4= (P3/P4) γ-1/γ

328/T4= (300/100) γ-1/γ

T4=239.64k

ηt=(T3-T4′)/(T3-T4)

0.78= (328-T4′)/ (328-239.64) T4’= 259.08k

COP= (T1-T4′)/ (T2′-T1)

COP= (277-259.08)/ (418.86-277) =0.17

1 tonne= 3.5kw of heat 3tonne= 3×3.5=10.5kw

Energy balance.

Heat energy absorbed by Ice=Heat rejected by air

= m×Cp× (T1-T4′)

10.5= ma×1.005×(277-259.08)

Mass of air, ma= 0.583Kg/sec.

Power, P=ma×Cpa× (T2′-T1)

= 0.583×1.005× (418.86-277)

= 83.12kw