the coefficient of performance of the vapour compression refrigeration systems.
‘ Relationship between actual work and ideal work.
‘ To determine the second law efficiency of the vapour compression system.
‘ To analyze, evaluate and optimize the performance of the vapour compression systems.
4.2 VCR Test Rig
‘ Simple Vapour compressor test rig consist of compressor, evaporator, condenser, expansion valve, heater, pressure gauges, temperature measurement equipments, Rota meter, cut-off switches, thermo couples etc.
‘ Here we have picture of the experimental setup on which exergy was analyzed.
Fig (m) VCR Test Rig
4.3 Specification:
‘ Refrigerant Used: R 134a
‘ Refrigeration Rate: 1400W maximum, but varies with respect to evaporative and condensing pressure.
‘ Condensing Temperature: 50??C Maximum
‘ Evaporative temperature: -20 to 10??C variable adjusting by load and liquid
‘ Compressor: Hermetically sealed 314 L
‘ Condenser: Air cooled
‘ Evaporator: Compact once through concentric tube with refrigeration load supplied by separate electrical heating elements
‘ Expansion Valve: Automatic expansion valve with two bypass capillary tube.
4.4 Experiment procedure:
1. Ensure that the operation of plant is clearly understood.
2. Start the unit adjust the evaporator heat input control and to set the evaporating pressure adjust the gas flow rate to give the heat required condenser pressure adjust the gas flow rate to give the heat required condenser pressure and hence saturation temperature.
3. For performance curve start with a small duty, say 250W and increase this in increments of about 250W until the maximum duty is reached.
4. The unit is respond quickly after the load changes and stabilizes within 15 min., although it may take a little longer at light loads. Stability is reached when changes in pressure, temperature, flow, etc. have ceased.
Rotameter
Sub cooling section
Chapter-5
Exergy Analysis of VCR Test Rig
5.1 Exergy Concept on VCR System
Heat transfer from a low-temperature medium to a high temperature one requires special devices called refrigerators. Refrigerators are cyclic devices. The most frequently used refrigeration cyclic is the vapor-compression refrigeration cycle. A basic system is composed of a condenser, an evaporator, a compressor and an expansion valve plus auxiliary and connection components. A simple vapor compressor refrigeration system as shown in the Fig.
Fig (n) VCR system and T-S diagram
From the first law point of view, the measure of performance of the refrigeration cycle of is the coefficient of performance (COP). For a vapor compression cycle, COP is defined as the amount of cooling produced per unit work supplied. The COP of a reversible refrigeration cycle can be expressed as
COP= Qe / Wnet =Te/(To-Te)
Where Qe, Wnet, Te, and To are the evaporator load, net work input, the evaporation temperature of refrigerant in the evaporator, and the surrounding ambient temperature respectively. For given refrigeration and ambient temperatures, a reversible refrigeration cycle will have the highest COP. Although, it is difficult to evaluate the performance of a refrigeration system based only on the COP value; it is appropriate to compare performance of actual refrigeration systems with the reversible ones for the same refrigeration and ambient temperatures. Work consumed by the actual cycle is always greater than that in the corresponding reversible one and the difference is the lost work, also called exergy loss, exergy destruction, or irreversibility. The lost work can be obtained by first calculating the entropy generation. The extropy generation is a measure of magnitude of the irreversibility for a process. Entropy generation for a steady-flow process is expressed as
Sgen=??out (mese) ‘ ??in (misi) – ??i (Qi/Ti) ‘ 0
The first two terms on the right-hand side are the sums of exergy outputs and inputs and the third term is the rate of entropy transferred through the portion of the control surface where the instantaneous absolute temperature is Ti. The lost work can be determined from WL = TO Sgen
The second law of efficiency ??II, also called the exergy efficiency or the effectiveness, can be defined for a refrigeration cycle as the ratio of the minimum work requirement to the actual work input. That is,
W rev W rev
?? II ‘ ‘
W ac
W rev ‘ W L
Where Wrev and Wac are the work inputs to a reversible and an actual refrigeration cycle, respectively. The work input for a reversible refrigeration cycle is determined from
Wrev = Qe ((To/Te)-1)
5.2 Analysis Methodology
A reversible thermodynamic process can be reversed without leaving any trace on the surroundings. That is, the system and the surroundings are returned to their initial states at the end of the reverse process. This is possible only if the net heat and net work process. All real processes are irreversible. Some factors causing irreversibility in a refrigeration cycle include friction and heat transfer across a finite temperature difference in the evaporator, compressor, condenser, and refrigerant lines, sub cooling to ensure pure vapor at compressor inlet, superheating to ensure pure vapor at compressor inlet, pressure drops, and heat gains in refrigerant lines. The actual vapor-compression refrigeration cycle investigated is presented in Figure by temperature entropy (T-s) diagram.
Fig (o) Actual Temperature-Entropy Diagram
The lines a-b-c-d-a shows reversible refrigeration cycle and the lines 1-2-3-4-1 shows are the T ‘s diagram for the actual refrigeration cycle in Fig. 2. The compression process for the actual cycle is irreversible. Compressor takes the refrigerant from the evaporator and compresses to the condenser through the line 1-2. The line 1-2s represents the isentropic compression process. In the ideal case, the refrigerant is assumed to leave the condenser as saturated liquid at state 3′ at the compressor exit pressure. In the actual case, it is unavoidable to have some pressure drop in the condenser so that is exits the condenser at state 3, and then enters the expansion value. It leaves the expansion valve at state 4, and enters the evaporator. The cycle is completed as the refrigerant exits from the evaporator at state 1 instead of state 1’ because of the pressure drop in the evaporator.
Entropy generation and lost work can be calculated for an irreversible-adiabatic compression process can be obtained from:
I in Compressor = TO (s2-s1)
This lost work may be represented by the area b’-b-m-n-b on T-s diagram. Heat is rejected from the refrigerant in the condenser to the surrounding cooling fluid. The lost work in the condenser can be obtained from
I in condenser = qcon + To (s3-s2)
Where, qcon is the amount of heat rejected from the refrigerant as it flows through the condenser, and it can be calculated from
qcon = h2 ‘ h 3 = h2 ‘ h 3”
In that case, according to the Eq. (8) and Fig. 2, area 2-3′-3′-p-r-s-b’-2 represents the lost work due to the heat transfer and circulating fluid friction in the condenser. The lost work produced in the expansion value is represented by the area t-s-r-u-t, and it can be obtained from:
I in expansion valve = TO (s4 ‘ s 3)
The friction due to the flow of refrigerant in the evaporator and the heat transfer from the refrigerated space at a temperature of Tc are the sources for lost work, and it can be calculated from
I in evaporator ‘ To (s1 ‘ s 4) ‘ To (Qb/Tc)
Ws is the isentropic compressor work, which is obtained from
Ws = h2s ‘ h 1
Then, the actual work requirement of the refrigeration cycle becomes:
Wac = Ws + I in compressor + I in condenser + I in expansion valve + I in evaporator
5.3 Assumptions
‘ Steady state, steady flow operation.
‘ Negligible pressure drops except in the evaporator and condenser.
‘ Adiabatic compressor and expansion device.
‘ Saturated states at the condenser and evaporator outlets.
‘ Chemical, kinetic and potential energy and exergy of the components are omitted
‘ Mass flow rate is taken as a unity(1 kg/s).
‘ Isenthalpic expansion of refrigerants in expansion valve
‘ Heat transfers from/to the compressor and expansion valve are neglected
‘ Pressure losses in pipelines are neglected.
‘ Concept of sub cooling and super heating is not considered.
5.4 Observation Table
Series Test No:1 Test no:2 Test No:3
Condenser Pressure(psi) 250 135 189
Evaporator Pressure(psi) 28 8 17
Compressor Delivery/ Condenser inlet (??C) 67 61.9 64
Liquid leaving condenser(??C) 45.5 29.6 37
Evaporator inlet (??C) -7.1 -16.7 -12
Compressor suction/
Evaporator outlet 15.1 -12.7 2
Table A: Observation Table
5.5 Analysis on Test Rig
Sample Calculation:
The required readings are taken with the usual test rig of the vapour compression refrigeration system using R-134a as the refrigerant.
Inlet temperature of the evaporator = 265.9 K
Out let temperature of the compressor = 340K
Out temperature of the condenser = 318.5 K
Inlet temperature of the compressor = 288.1 K
Temperature of the refrigerant space = 273 K
Ambient temperature = 312 K
h1 = 326.627 Kj/kg
h2 = 428.096 Kj/kg
h3 = 264.439 Kj/kg
h4 = 323.344 Kj/kg
s1 = 1.49341 Kj/kg K
s2 = 1.69789 Kj/kg K
s3 = 1.2156 Kj/kg K
s4 = 1.3091 Kj/kg K
‘ Carnot Coefficient of Performance = T evp / (T cond -T evp) = 4.264
‘ Refrigerating effect = Q ref =h1 ‘ h 4 = 3.283 Kj/Kg
‘ Actual work input = h2-h1=101.46 Kj/kg
‘ Carnot work = Refrigerating work/ Carnot COP =0.769 Kj/Kg
‘ Exergetic Efficiency = Carnot Work / Actual Work = 0.7586
‘ Heat released by the condenser = Qr = h2 ‘h 3 = 163.65 Kj/Kg
Analysis of evaporator
‘ Loss of Exergy in Evaporator = Tref(S1-S4) ‘ Q ref = 54.22 Kj/Kg
‘ % loss of exergy in the evaporator = 20.70 %
Analysis of compressor
‘ Loss of exergy in compressor = T0(S2-S1) = 63.79 Kj/kg
‘ % loss of exergy in the compressor = 24.36 %
Analysis of condenser
‘ Loss of exery in the condenser = Qr ‘ T 0 (S2 ‘ S 3 ) = 13.18 Kj/kg
‘ % Loss of exery in the condenser = 5.03 %
Analysis of throttling process
‘ Loss of exergy in throttling = T0 (S4 ‘ S 3) = 29.17 kj/kg
‘ % Loss of exergy in throttling = 5.014 %
Chapter-6
Softwares and Refrigerant Charts
6.1 C program
Exergy loss in each component, the coefficient of performance and the second law efficiency is calculated by an iterative scheme.
For making all calculations easy and quick one equation solver software developed in C program and its basic fundamentals.
Program contains inputs of all temperatures enthalpy and entropy and from that we can calculate each and every value which is required
C program, input and output of a reading given as below:
Program:
#include<stdio.h>
#include<conio.h>
void main()
{
float ws,h1,h2,lec,s1,s2,t0,qr,h3,h4,lecon,s3,s4,lexpan,qref,levap,wac,carnot,tevp,tcond,carnotwrk,exeff,pcomp,pcon,pevap,pexpan;
clrscr();
printf("n Ambient temperature To(K)=");
scanf("%f",&t0);
printf("nInlet Evaporator temperature Tevp(K)=");
scanf("%f",&tevp);
printf("nOutlet condenser temperature Tcon(K)=");
scanf("%f",&tcond);
printf("nEnthalpy at point1 h1(Kj/Kg)=");
scanf("%f",&h1);
printf("nEnthalpy at point2 h2(Kj/Kg) =");
scanf("%f",&h2);
printf("nEnthalpy at point3 h3(Kj/Kg) =");
scanf("%f",&h3);
printf("nEnthalpy at point3 h4(Kj/Kg) =");
scanf("%f",&h4);
printf("nEntropy at point1 S1(Kj/KgK) =");
scanf("%f",&s1);
printf("nEntropy at point2 S2(Kj/KgK) =");
scanf("%f",&s2);
printf("nEntropy at point3 S3(Kj/KgK)=");
scanf("%f",&s3);
printf("nEntropy at point4 S4(Kj/KgK)=");
scanf("%f",&s4);
ws=h2-h1;
printf("nWork in compressor Ws (Kj/Kg)=%f",ws);
lec=t0*(s2-s1);
printf("nnLoss of exergy in compressor(Kj/Kg)=%f",lec);
qr=h2-h3;
lecon=qr-(t0*(s2-s3));
printf("nLoss of exergy in condenser (Kj/Kg)=%f",lecon);
lexpan=t0*(s4-s3);
printf("nLoss of exergy in expansion (Kj/Kg)=%f",lexpan);
qref=h1-h4;
levap=t0*(s1-s4)-qref;
printf("nLoss of exergy in evaporator(Kj/Kg)=%f",levap);
wac=ws+lec+lecon+lexpan+levap;
printf("nActual Work Of Refrigeration Cycle=Total exergy analysis (Kj/Kg)=%f",wac);
pcomp=lec/wac*100;
printf("nnnloss of exergy in compressor in percent=%f",pcomp);
pcon=lecon/wac*100;
printf("nloss of exergy in condenser in percent=%f",pcon);
pexpan=lexpan/wac*100;
printf("nloss of exergy in expansion in percent=%f",pexpan);
pevap=levap/wac*100;
printf("nloss of exergy in evaporatorin percent=%f",pevap);
printf("nnnRefrigeratin effect(Kj/Kg)=%f",qref);
printf("nHeat released by the condenser (Kj/Kg)=%f",qr);
carnot=tevp/(tcond-tevp);
printf("nCarnot Cofficient of Performance=%f",carnot);
carnotwrk=qref/carnot;
printf("nCarnot Work(Kj/Kg)=%f",carnotwrk);
exeff=carnotwrk/ws*100;
printf("nExergetic Efficiency=%f",exeff);
getch();
}
Input:
Ambient temperature To(K) = 312
Inlet Evaporator temperature Tevp(K) = 258
Outlet condenser temperature Tcond (K) = 318.5
Enthalpy at point 1 h1 (Kj/kg) = 326.62
Enthalpy at point 1 h2 (Kj/kg) = 428.096
Enthalpy at point 1 h3 (Kj/kg) = 264.439
Enthalpy at point 1 h4 (Kj/kg) = 323.344
Entropy at point 1 s1 (Kj/kg K)= 1.4934
Entropy at point 1 s2 (Kj/kg K)= 1.6978
Entropy at point 1 s3 (Kj/kg K)= 1.2156
Entropy at point 1 s4 (Kj/kg K)= 1.3091
Output:
Work in compressor Ws (Kj/Kg) =101.476013
Loss of exergy in compressor (Kj/Kg) = 63.800900
Loss of exergy in condenser (Kj/Kg) = 13.182524
Loss of exergy in expansion (Kj/Kg) = 26.332781
Loss of exergy in evaporator (Kj/Kg) = 57.064808
Actual work of Refrigeration Cycle = Total exergy analysis (Kj/Kg) = 261.857025
Loss of exergy in compressor in percent = 24.3647
Loss of exergy in condenser in percent = 5.03425
Loss of exergy in expansion in percent = 10.05616
Loss of exergy in evaporator in percent = 21.79235
Refrigeration effect (Kj/Kg) = 3.276001
Heat released by the condenser (Kj/Kg) = 163.6570
Carnot coefficient of performance = 4.2644
Carnot Work (Kj/kg) = 0.768210
Exergetic Efficiency = 0.75703
6.2 Refrigerant and Charts
R 134a: 1,1,1,2-Tetrafluoroethane, R-134a, Forane 134a, Genetron 134a, Florasol 134a, Suva 134a or HFC-134a, is a haloalkane refrigerant with thermodynamic properties similar to R-12 (dichlorodifluoromethane) but with less ozone depletion potential. It has the formula CH2FCF3 and a boiling point of ‘26.3 ??C (‘15.34 ??F) at atmospheric pressure. R-134a cylinders are colored light blue.
Refrigeration Chart of R134a is used for finding out enthalpy and entropy of given pressure and given temperature in refrigeration system.
Fig (p) : R134a Enthalpy- Entropy Chart
Fig (q) P-H diagram of R 134a
Fig (r) T-S diagram of R134a
Chapter-7
Results and Discussion
Results and Discussions
‘ A vapour compression cycle using R-134a as the refrigerant is considered for the present analysis. The cold room and ambient air temperatures are assumed to be 0?? C and 39??C.
‘ The exit temperature of the evaporator is varied between – 15?? C and 5?? C while the condensing temperature is taken as constant.
‘ Here we have three different readings of all temperature and based on those readings can generate more comparative analysis graphs.
‘ Table shows the different values which are gained from VCR test rig, Refrigeration charts.
‘ Also it includes the value find out from using software C program.
Exergy anlaysis of VCR test rig and it’s all data are given as below:
Reading 1
Evaporator temp. 1 Evaporator temp. 2 Evaporator temp. 3 Evaporator temp. 4 Evaporator temp. 5
To(K) 312 312 312 312 312
Tcomp(K) 337 334.9 334.9 334.9 334.9
Tevp(K) 258 263 268 273 278
Tcond(K) 302.6 302.6 302.6 302.6 302.6
h1(Kj/Kg) 326.627 330.77 334.8576 338.894 342.964
h2(Kj/Kg) 427.52 427.52 427.52 427.52 427.52
h3(Kj/Kg) 251.73 251.73 251.73 251.73 251.73
h4(Kj/Kg) 324.91 324.91 324.91 324.91 324.91
s1(Kj/Kg K) 1.49341 1.49811 1.50338 1.50866 1.51376
s2(Kj/Kg K) 1.699 1.699 1.699 1.699 1.699
s3(Kj/Kg K) 1.1755 1.1755 1.1755 1.1755 1.1755
s4(Kj/Kg K) 1.33089 1.33089 1.33089 1.33089 1.33089
Table B: Inputs of Reading 1
Evaporator temp. 1 Evaporator temp. 2 Evaporator temp. 3 Evaporator temp. 4 Evaporator temp. 5
Actual work(Kj/Kg) 100.893 96.75 92.6624 88.626 84.556
Exergy Loss in Compressor(Kj/Kg) 64.14408 62.67768 61.03344 59.38608 57.79488
Exergy Loss in Condenser(Kj/Kg) 12.458 12.458 12.458 12.458 12.458
Exergy Loss in Expansion Valve(Kj/Kg) 48.48168 48.48168 48.48168 48.48168 48.48168
Exergy Loss in Evaporator(Kj/Kg) (Kj/Kg) 48.98924 46.31264 43.86928 41.48024 39.00144
Toatal Exergy Loss Wac 274.966 266.68 258.5048 250.432 242.292
% Exergy loss in Compressor 23.328 23.50295 23.61018 23.71346 23.8534
% Exrgy Loss in Condenser 4.530742 4.671516 4.819253 4.974604 5.14173
% Exergy Loss in Expansion Valve 17.63188 18.17972 18.75465 19.35922 20.00961
% Exergy Loss in Evaporator 17.81647 17.36637 16.97039 16.56347 16.09687
Refrigeration Effect Qe(Kj/Kg) 1.717 5.86 9.9476 13.984 18.054
Heat Released by Condenser Qr(Kj/Kg) 175.79 175.79 175.79 175.79 175.79
Carnot COP 4.961538 6.641414 7.745665 9.222973 11.30081
Carnot work(Kj/Kg) 0.346062 0.882342 1.28428 1.516214 1.597584
Exergetic Efficiency 0.342999 0.911982 1.385977 1.7108 1.88938
Table C: Output of Reading 1
Effect of Evaporator Temperatures:
Fig (s): Exergy Loss In Each Component Vs Evaprator Temperature
The graph plotted above between the exergy losses in various components Vs evaporator temperature reveals the fallowing points.
‘ Greater the exergy losses takes place in the compressor and less in other components.
‘ The exergy loss in the evaporator decreases with increase of evaporator temperature.
‘ The exergy losses in the condenser remain constant with increase of temperature.
‘ The trend of the exergy loss in the evaporator decreasing with the evaporator temperature can be explained by the fact that the average temperature difference b/n the evaporator and the refrigerated space decrease with increasing evaporator temperature.
‘ The higher the temperature differences the high the exergy loss in evaporator.
‘ Exergy loss in the compressor decreases with decrease in evaporator temperature.
Fig (t): Second Law Efficiency Vs Evaporator Temperature
The graph plotted above between the total exergy loss Vs Evaporator temperature reveals the fallowing points
‘ Total exergy loss decreases with increase of evaporator temperature can be explained by the fact that decrease of exergy loss in evaporator is more compared with increase of exergy loss in other components with increase in evaporator temperature.
‘ Therefore total exergy loss of the system decreases with increase of evaporator temperature.
Fig (u) Second Law efficiency Vs Evaporator Temperature
The graph plotted above between the second law efficiency Vs evaporator temperature reveals the fallowing points.
‘ The second law efficiency increases with increase of temperature.
‘ As the minimum exergy intake to perform to perform the given task decreases, the second law efficiency increases.
Fig (v) COP Vs Evaporator Temperature
The graph plotted above between COP Vs evaporator temperature reveals the fallowing points.
‘ COP increases with increase of evaporator temperature because the refrigerating effect is increases and work required is decreases.
‘ This can be explained by the fact that with increase of evaporator temperature, the refrigerating effect of the system increases and the work required is decreases, causing increase in COP.
Fig (w) Comparison of Exergy Losses In Different Components
The graph plotted above shows the comparison of exergy loss in each component of the system. It reveals the fallowing points.
‘ Under the design conditions a comparison of exergy loss during each process are shown.
‘ The lost of work in the compressor appears to be much higher than that in the evaporator.
‘ This can be explained by the fact that the refrigerant under goes almost an isothermal heat addition process during phase change in the evaporator with a relatively small temperature difference b/n the evaporator and cold space.
‘ In condenser only part of the heat rejection takes place during phase change process larger temperature difference between the condenser and outside air.
Reading 2
Evaporator
Temp 1 Evaporator
Temp 2 Evaporator
Temp 3 Evaporator
Temp 4 Evaporator
Temp 5
To(K) 312 312 312 312 312
Tcomp(K) 334.9 334.9 334.9 334.9 334.9
Tevp(K) 258 263 268 273 278
Tcond(K) 302.6 302.6 302.6 302.6 302.6
h1(Kj/Kg) 326.627 330.77 334.8576 338.894 342.964
h2(Kj/Kg) 427.02 427.02 427.02 427.02 427.02
h3(Kj/Kg) 240.908 240.908 240.908 240.908 240.908
h4(Kj/Kg) 325.25 325.25 325.25 325.25 325.25
s1(Kj/Kg K) 1.49341 1.49811 1.50338 1.50866 1.51376
s2(Kj/Kg K) 1.7009 1.7009 1.7009 1.7009 1.7009
s3(Kj/Kg K) 1.14136 1.14136 1.14136 1.14136 1.14136
s4(Kj/Kg K) 1.2946 1.2946 1.2946 1.2946 1.2946
Table D Inputs of Reading 2
Evaporator
Temp 1 Evaporator
Temp 2 Evaporator
Temp 3 Evaporator
Temp 4 Evaporator
Temp 5
Actual work(Kj/Kg) 100.393 96.25 92.1624 88.126 84.056
Exergy Loss in Compressor(Kj/Kg) 64.73688 63.27048 61.62624 59.97888 58.38768
Exergy Loss in Condenser(Kj/Kg) 11.53552 11.53552 11.53552 11.53552 11.53552
Exergy Loss in Expansion Valve(Kj/Kg) 47.81088 47.81088 47.81088 47.81088 47.81088
Exergy Loss in Evaporator(Kj/Kg) 60.65172 57.97512 55.53176 53.14272 50.66392
Total Exergy Loss Wac 285.128 276.842 268.6668 260.594 252.454
% Exergy loss in Compressor 22.7045 22.85436 22.9378 23.01622 23.12805
% Exrgy Loss in Condenser 4.045734 4.166824 4.293616 4.426625 4.569355
% Exergy Loss in Expansion Valve 16.76822 17.2701 17.7956 18.34688 18.93845
% Exergy Loss in Evaporator 21.27175 20.94159 20.66938 20.39292 20.06857
Refrigeration Effect Qe(Kj/Kg) 1.377 5.52 9.6076 13.644 17.714
Heat Released by Condenser Qr(Kj/Kg) 186.112 186.112 186.112 186.112 186.112
Carnot COP 5.784753 6.641414 7.745665 9.222973 11.30081
Carnot work(Kj/Kg) 0.23804 0.831148 1.240384 1.479349 1.567498
Exergetic Efficiency 0.237108 0.863531 1.345868 1.678675 1.864826
Table E: Outputs of Reading 2
Effect of Evaporative Temperature:
Fig (x) Exergy Losses in Each Components Vs Evaporative temperature
Fig (y) Total Exergy Loss Vs Evaporative Temperature
Fig (z) COP Vs Evaporative Tempreature
Fig (aa) Second Law of Efficiency Vs Evaporative Temperature
Fig (bb) Comparison Of Exergy Losses In Different Components
Reading 3
Evaporator
Temp 1 Evaporator
Temp 2 Evaporator
Temp 3 Evaporator
Temp 4 Evaporator
Temp 5
To(K) 312 312 312 312 312
Tcomp(K) 340 340 340 340 340
Tevp(K) 258 263 268 273 278
Tcond(K) 318.5 318.5 318.5 318.5 318.5
h1 326.627 330.77 334.8576 338.894 342.964
h2 428.096 428.096 428.096 428.096 428.096
h3 264.439 264.439 264.439 264.439 264.439
h4 323.344 323.344 323.344 323.344 323.344
s1 1.49341 1.49811 1.50338 1.50866 1.51376
s2 1.69789 1.69789 1.69789 1.69789 1.69789
s3 1.2156 1.2156 1.2156 1.2156 1.2156
s4 1.309102 1.309102 1.309102 1.309102 1.309102
Table F: Inputs of Reading 3
Evaporator
Temp 1 Evaporator
Temp 2 Evaporator
Temp 3 Evaporator
Temp 4 Evaporator
Temp 5
Actual work(Kj/Kg) 101.469 97.326 93.2384 89.202 85.132
Exergy Loss in Compressor(Kj/Kg) 63.79776 62.33136 60.68712 59.03976 57.44856
Exergy Loss in Condenser(Kj/Kg) 13.18252 13.18252 13.18252 13.18252 13.18252
Exergy Loss in Expansion Valve(Kj/Kg) 29.17262 29.17262 29.17262 29.17262 29.17262
Exergy Loss in Evaporator(Kj/Kg) (Kj/Kg) 54.2211 51.5445 49.10114 46.7121 44.2333
Toatal Exergy Loss Wac 261.843 253.557 245.3818 237.309 229.169
% Exergy loss in Compressor 24.36489 24.58278 24.73171 24.87885 25.06821
% Exrgy Loss inCondenser 5.034513 5.199036 5.372248 5.555002 5.752314
% Exergy Loss in Expansion Valve 11.14127 11.50535 11.88867 12.2931 12.72974
% Exergy Loss in Evaporator 20.70748 20.32856 20.0101 19.68408 19.30161
Refregeration Effect Qe(Kj/Kg) 3.283 7.426 11.5136 15.55 19.62
Heat Released by Condenser Qr(Kj/Kg) 163.657 163.657 163.657 163.657 163.657
Carnot COP 4.264463 4.738739 5.306931 6 6.864198
Carnot work(Kj/Kg) 0.769851 1.567084 2.16954 2.591667 2.858309
Exergetic Efficiency 0.758705 1.610139 2.326874 2.905391 3.357503
Table G: Outputs of Reading 3
Fig(cc) Exergy Losses in Each components Vs Evaporator Temperature
Fig (dd) Total Exergy Losses Vs Evaporator Temperature
Fig (ee) COP Vs Evaporator Temperature
Fig (ff) Second Law Efficicncy Vs Evaporator Temperature
Fig (gg) Comparison of Exergy Losses in Different Components
Chapter-8
Conclusion and Future Work
8.1 Conclusion
A computational model based on exergy analysis is presented for the investigation of the effects of the evaporating temperature on the second law efficiency and COP of the vapor compression refrigeration cycle.
‘ It is found that the evaporating have strong effects on the exergy losses in the evaporator, second law efficiency and COP of the cycle but little effects on the other components of exergy losses.
‘ The second law efficiency,COP increases with increase of evaporator temperature.
‘ Total exergy losses decreases with increases of evaporator temperature.
‘ Compressor and Evaporative Exergy losses decrease with increase in temperature.
‘ Maximum Loss found in the compressor, for effective refrigeration it must be serviced periodically.
‘ In this work, It can be concluded that the exergy losses are due to irreversibility of the process leading to the reduction of the useful effects of the process and the exergy analysis of vapour compression refrigeration system can locate the inefficient areas and point out the areas with great potential for the improvement.
‘ Thus it is obvious that, the purpose of exergy analysis is to improve analysis of system by introducing ways of concurrently suggesting improvements to the analyzed system.
8.2 Future Enhancement
‘ Exergy analysis by changing in evaporative temperature has evaluated in this experiment but by changing the condenser pressure and temperature we also can analyze the system.
‘ Here we are neglecting the effect of super cooling and sub cooling but with consideration of that we can analyze this system.
‘ Also with change of mass flow rate exergy analysis also will be possible.
‘ Exergy is a wide area in which numbers of assumptions are taken, including all those assumptions we can gain all exergy losses of the system.