Refrigeration

RefrigerationApplied Thermodynamics2013N S Seannayake OutlineIntroductionRefrigeration process, COPHeat pump and refrigeratorUnitsThermodynamic cycleTypes of refrigerationRefrigerants

Outline cont Vapor compression refrigerationEffect of compressor output stateEffect of sub cooling Improvements in the simple saturation cycle

IntroductionRefrigeration is the process of removing heat from a space under controlled conditions in order to reduce the temperature in the space. Refrigeration process

Refrigerator and Heat Engine

Units of refrigerationTonne of Refrigeration (TR)J/s or WBTU

1 TR = amount refrigerating effect produced by melting 1tonne of ice at 0oC in 24h.1TR = 3.5kW = 12,000BTUThermodynamic cycle (Reversed Carnot)

Effect of temperatures on COP

Limitations?Types of RefrigerationVapour compression Vapour absorptionGas or air refrigeration RefrigerantsHeat transfer mediumProperties: low boiling temp., high critical temp., high latent heat of vaporization, non corrosive, nonflammable, mixes with oil, leaks easily detectableIdentification by R numbers R11, R12, R134a, R717Environmental impactDepletion of ozone layerGlobal warming Vapour Compression refrigeration

Vapour compression cycle

Schematic representation of vapour compression cycle

Components and ProcessesCompressor :Isentropic compression

Condenser: Constant pressure heat rejection

Processes cont.Expansion Valve : Constant enthalpy pressure reduction

Evaporator: constant pressure heat absorption

COP

T s diagram

This is close to Reversed Carnot CycleProblems of the cycle that gives saturated dry vapor after compression (cycle close to reversed Carnot cycle)This cycle cannot be used in practice for practical limitations, Ex: point 1 which is wet vapour, cannot be successfully compressed in a compressor. So, point 1 is either saturated vapour or superheated. Point 2 is often above Tc and point 3 is very often below Tc.Therefore, there are many refrigeration cycles which have been studied by people aimed at improving the performance and ensuring their practicability. Some of the cycles are:

Practical modificationsCycle with super heated vapour after compression (simple saturated cycle)Cycle with dry saturated vapour after compressionCycle with wet vapour after compressionCycle with super heated vapour before compressionCycle with sub cooling of refrigerant

Procedure for analysisSketch the T s diagram/ p h diagramDetermine enthalpy values from state 1 (compressor inlet) to state 4 (inlet to evaporator coil)Determine work and heat transfer in each componentTo determine capacity use mass flow rateCharts

Property tablesSaturation temperature in(oC)Saturation pressure in bar(p)Specific volumem3/kgSpecific enthalpy in kJ/kgSpecific entropy in kJ/kgKLiquid(vf)Vapour (vg)Liquid(hf)Vapour(hg)Latent(hfg)Liquid(sf)Vapour(sg)-1000.011850.00060010.1951- 51.84142.00193.84- 0.25670.8628-95 0.018640.0006046.6231- 47.56144.22191.78-0.23230.8442-900.028430.0006084.4206-43.28146.46189.74-0.20860.8273Simple saturated cycle on p-h and T-s coordinates

Saturated liquid (f)Saturated vapor (g)Example 1For a simple vapour compression cycle with following data, determine COP and refrigerant mass flow rate.Refrigerant: R12, Capacity: 2TREvaporator temp: -12oC, Condenser temp: 40oC Specific heat capacity of R12 in the superheated region: 0.762kJ/kgK

Sat. Temperature (K)Pressure (bar)Enthalpy (kJ/kg)Entropy (kJ/kgK)liquidvapourliquidvapour-122.034188.95 347.25 0.9589 1.5651409.5909239.03 368.81 1.1315 1.5459

Solution

Evaporator temperature is - 12oC. The state is saturated vapour.Point 1After compression the refrigerant is super heated. For the determination of entropy at 2, we use following equation.Point 2

The condenser output is saturated liquid.Point 3

Point 4Since the process 3 4 is throttling the enthalpy is the same.

Mass flow rate required

Effect of compressor output state on the refrigeration parametersExample 2(Saturated vapour after compression)A vapour compression refrigerator works between the pressure limits of 60bar and 25bar. The working fluid is dry saturated after compression and there is no sub cooling in the condenser. Determine (1) Refrigerating effect, (2) COP (3) mass flow rate for 1TR (4) Compressor powerPressure (bar)Sat. Temperature (K)Enthalpy (kJ/kg)Entropy (kJ/kgK)liquidvapourliquidvapour2526156.32322.580.2261.246460295151.96293.290.5541.0332

SolutionPoint 1 lies between saturated liquid and saturated vapour points on the 25bar constant pressure line as shown in the diagrams above. Since we do not know exact position, we have to relate this state to state point 2, which is known.Point 1

To find out x1 we use entropy values at points 1 and 2.

Since 1 2 is isentropic,

Therefore, Point 2State of point 2 is saturated vapour at 60bar pressure. Therefore, from the table we directly get the enthalpy.

State of point 3 is saturated liquid at 60bar pressure. Therefore, enthalpy can be obtained directly from the table.Point 3

Since, 3 4 is throttling, the enthalpy remains same.Point 4

Mass flow rate 1TR

Since 1TR = 210kJ/minRefrigerating effect

COP

Compressor power

Example 3(Wet vapour after compression)Determine the refrigerating effect, COP, mass flow for 1TR and compressor power of the refrigeration plant in example 2 above, when the refrigerant after compression is 70% dry.Pressure (bar)Sat. Temperature (K)Enthalpy (kJ/kg)Entropy (kJ/kgK)liquidvapourliquidvapour2526156.32322.580.2261.246460295151.96293.290.5541.0332Solution

70% dryPoint 2Since dryness fraction at point 2 is known, we start form here

Point 1First find out the dryness fraction at point 1

Point 3State of point 3 is saturated liquid.

Point 4Since 3 4 is throttling process the enthalpy remains same.

Refrigerating effect

COPMass flow rate 1TR

Compressor power

Example 4(Super heated vapour after compression)Determine the refrigerating effect, COP, mass flow for 1TR and compressor power of the refrigeration plant in example 2 above, when the refrigerant after compression is superheated.Pressure (bar)Sat. Temperature (K)Enthalpy (kJ/kg)Entropy (kJ/kgK)liquidvapourliquidvapour2526156.32322.580.2261.246460295151.96293.290.5541.0332Solution

Point 1Saturated vapor

Point 2Super heated vapor, enthalpy is written as follows;

To find out T2 , use entropy at 1 and 2

Point 3Saturated liquid at 60bar

Point 4Since the process 3 4 is throttling

Refrigerating Effect

COP

Mass flow for 1TRSince 1TR = 210kJ/min

Compressor power

The effect of state after compression State of vapor after compressionRefrigerating effect (kJ/kg)COPMass flow for 1TR (kg/min)Comp. power (W)Wet vapor 77.4273.62.712972Saturated vapor 114.7 4.31.831813Super heated vapor170.623.91.231884The effect of sub cooling the liquid on refrigeration parametersThe effect of sub cooling the liquid Example 4Evaporator temperature: - 10oCCondenser temperature: 40oCSub cooled temperature: 30oCRefrigerant: R12Specific heat capacity of liquid R12: 1.03kJ/kgKSpecific heat capacity of superheated vapour R12 = 0.762kJ/kgKCalculate COPTemp(oC)Sat. pressure (bar)Sp. volume (vapour)m3/kgEnthalpy( kJ/kg)Entropy (kJ/kgK)LiquidVapourLiquidVapour-102.18930.07731190.78348.170.56981.5639409.59090.01834239.03368.811.13151.5459

SolutionPoint 1Point 1 is dry saturated.

Point 2Because 1 2 is isentropic,

Since point 2 is super heated,

Point 3Point 3 is sub cooled liquid.

Point 4Since 3 4 is throttling

Refrigerating effect

For sub cooled cycleFor simple saturation cycle;

For sub cooled cycle;

Compressor work:

Coefficient of performance:For sub cooled cycle;

For Simple saturation cycle;

Mass flow rate needed for 1TR capacity:1TR = 210kJ/minFor simple saturation cycle;

Compressor volume capacity:Compressor volume capacity is determined in terms of the volume of vapour handled in unit time.Vol. of vapour compressed per unit time = mass flow x specific volume

For both sub cooled and simple saturation cycles

For sub cooled cycle;For simple saturation cycle

Power of compressor for 1TR:

For sub cooled cycle;

For simple saturation cycle;

Effect of sub coolingParameter Unit of measurementSimple saturation cycleSub cooled cycle Refrigerating effect per kg of refrigerant kJ/kg109.14119.44Compressor work per kg of refrigerantkJ/kg25.9725.97Mass flow rate per 1TRKg/min1.9241.76COP4.214.6Compressor volume capacitym3/min0.1487 0.1361 Power for 1TRkW0.8330.762Improvements to simple saturation cycleFlash ChamberPre coolerSub cooling condenser outputBy liquid refrigerantBy vapor refrigerant

Flash Chamber

Used to separate vapour and liquid and vapour is directed to the compressorNo improvement to COP and reduce mass flow through evaporatorConsidering thermal equilibrium of flash chamber

Refrigerating Effect, QE

This is same as for simple saturation cycleRefrigerating effect and COP are same as that of a simple saturation cycleThe effect flash chamber is only the reduction of mass flow through the evaporator and hence the reduction in size of evaporator

Compressor work, WC

This is same as for simple saturation cycleAccumulator or pre-coolerNeed for accumulatorThe liquid refrigerant passing through the evaporator is not completely evaporated to give dry saturated vapour. If the compressor is supplied with liquid, with vapour refrigerant, the compressor has to do an additional work of evaporating and raising the temperature of the liquid. It also upset the normal functioning of the compressor, which has been made to compress dry vapour. The accumulator ensures that only the dry vapour is fed into the compressorAccumulator or pre-cooler

Ensures no liquid enters the compressorNo change in refrigerating effect, compressor work and COP

Considering thermal equilibrium for the accumulator1144m2m1m1m2

Refrigerating effect, QE

This is same as for simple saturation cycleCompressor work, WC

Refrigerating effect , compressor power and COP are same as that of a simple saturation cycleThe accumulator is used to prevent he liquid refrigerant flowing into the compressor

This is same as for simple saturation cycleSub cooling by liquidThe refrigerant leaving the condenser is at a higher temperature than the liquid refrigerant leaving the expansion valve. The liquid refrigerant leaving the condenser is sub cooled by passing through a heat exchanger which is supplied with liquid refrigerant from the expansion valve.Sub cooling by liquid

No change in COPm1= mass of refrigerant leaving the evaporatorm2= mass of refrigerant leaving through the condenserm3= mass of refrigerant supplied to the heat exchangerConsidering thermal equilibrium of the heat exchanger

Refrigerating effect QE

This is same as for simple saturation cycleCompressor work WC

Coefficient of PerformanceThis is same as for simple saturation cycleThis is same as for simple saturation cycleMass flow rate through evaporator for 1TR

This is less than that of simple saturation cycle

Sub cooling by liquidSince the COP and the compressor power are the same as those of simple saturation cycle, this arrangement of sub cooling the liquid refrigerant is of no advantage.Sub cooling by vapourThe refrigerant leaving the condenser is at a higher temperature than the liquid refrigerant leaving the evaporatorThe liquid refrigerant leaving the condenser is sub cooled by passing through a heat exchanger which is supplied with dry saturated vapour from the evaporator.

Sub cooling by vapour

COP, refrigerating increased; power for 1TR reduced

Considering thermal equilibrium of the heat exchanger

Refrigerating effect QE

This is higher than that of simple saturation cycle

Compressor work WC

This is higher than that of simple saturation cycleCoefficient of Performance

This is less than that of simple saturation cycleMass flow rate through evaporator for 1TR

This is less than that of simple saturation cycleSub cooling by vapourSub cooling the liquid by vapour refrigerant reduces the COP.However, this is used in practice despite some loss of COPMass flow rate per 1TR is reduced, and this is an advantage