EGR3030M Machines and Energy Conversion: Refrigeration Cycle Report

This report presents a detailed analysis of refrigeration and heat pump cycles, focusing on the application of thermodynamic principles. The assignment involves calculating the coefficient of performance (CoP) for various refrigerants, including Ammonia (R717), R134a, R245fa, and RF1234yf, using P-h diagrams. The report examines the impact of refrigerant changes on CoP and performance. Furthermore, it explores a modified refrigeration/heat pump system designed for food processing, calculating CoP values for refrigeration heat input, heat rejected for steam generation, and heat rejected to the atmosphere. The report concludes with a discussion on potential design improvements for enhanced system efficiency.

STUDENT
NAME:
TUTOR NAME: Prof R A Bickerton
PROGRAMME: B Eng / M Eng Mechanical Engineering MODULE CODE:
EGR3030M
MODULE
TITLE: Machines and Energy Conversion
SUBJECT: CALCULATIONS OF REFRIGERATION/HEAT PUMP CYCLES
COURSEWORK
TITLE:
ENGINEERING REPORT ON THE USE OF THERMODYNAMIC PRINCIPLES
TO CALCULATE CONDITIONS THROUGHOUT VAPOUR POWER CYCLES
COURSEWORK
WEIGHTING (%):
40% OF 25% FOR
COURSEWORK
(SEE LECTURE
ZER0)
Issue Date:
04/07/18
Due Date:
6/8/18
Feedback Date:
PERFORMANCE CRITERIA:
TARGETED LEARNING OUTCOMES
Technical report writing.
Mechanical & thermodynamic concepts of refrigerants and vapour power cycles
Application of thermodynamic principles
Use of vapour P-h diagrams.
Important Information – Please Read Before Completing Your Work
All students should submit their work by the date specified using the procedures specified in the Student Handbook.
An assessment that has been handed in after this deadline will be marked initially as if it had been handed in on
time, but the Board of Examiners will normally apply a lateness penalty.
Your attention is drawn to the Section on Academic Misconduct in the Student’s Handbook.
All work will be considered as individual unless collaboration is specifically requested, in which case this should be
explicitly acknowledged by the student within their submitted material.
Any queries that you may have on the requirements of this assessment should be e-mailed to . No queries will
be answered after respective submission dates.
You must ensure you retain a copy of your completed work prior to submission.
Page 1 of 8

COURSEWORK BRIEF:
EGR3030M
Assignment 4 re-sit Refrigeration
a) Draw the diagram for a simple refrigeration circuit identifying the 4 major constituents [4
marks]
The chemical compositions of refrigerant gasses have changed over the
years from Ammonia (R717) to fluorocarbon blends such as R134a, R
245fa and RF1234yf etc.
b) Using the P-h diagrams for the 4 refrigerants given below calculate the coefficient of
performance (CoP) when these are used as a refigerant between the following
temperatures.
Cold box -20°C outside temperature for the condenser 30°C
Assume a 5°C temperature (delta T) for both heat exhangers, evapourator
and condenser
[40 marks]
Comment on the differences in CoP and on the performance effects of
changing from NH4 to flurocarbons.
[1 mark]
c) A company wishes to use a refrigerator/ heat pump for food processing where an initial high
temperature heat exhanger is set to produce steam at 100⁰C. For this the condenser pressure is
set to give 105⁰C at heat exchager input and this pressure is then maintained in the normal air
condenser. The processed food is then to be chilled to -20°C
Assume a 5°C temperature (delta T) for all heat exhangers, steam
generator ( ie 105⁰C pressure setting) evapourator and condenser
Draw a modified refrigertion diagram shown this new configuration [5 marks]
Cold box -20°C, outside temperature for the condenser 30°C, steam
generator (105⁰C)
Using the P-h diagrams for the 4 refrigerants given below calculate the
coefficient of performance (CoP) for:-
i) The refigeration heat input
ii) The heat rejected for steam generation
iii) The remaining heat rejected by the condenser to atmosphere
[40 marks]
Pepare a short engineering report on your findings and explain how you
could improve design of the refigerator/heatpump system
[10 marks]
Page 2 of 8

Pressure enthapy R717 Ammonia NH4
Pressure enthalpy RF1234YF
Page 3 of 8

HAND IN OF REPORT AND ANY APPENDICES IE SPREADSHEET CALCULATIONS, SKETCHES
AND GRAPHS ETC NEEDS TO BE BY TURNITIN AS PER SCHOOL PROCEEDURES.
Page 4 of 8

OUTLINE OF MARKING SCHEME
a) Refrigeration diagram 4 marks
b) CoP calculations for refrigerants 40 marks
Comments of change from ammonia 1 mark
c) Refrigerator/heat pump diagram 5 marks
CoP calculations for refrigerator /heat pump 40
marks
Engineering report 10
marks
total 100 marks
As usual I am hoping that you will demonstrate engineering
professionalism and strive for the highest marks using your initiative
deriving your calculations for these combined vapour power cycles.
Page 5 of 8

MARKING CRITERIA:
COURSEWORK WILL BE MARKED ACCORDING TO THE FOLLOWING UNIVERSITY CRITERIA.
90-100%: a range of marks consistent with a first where the work is exceptional in all areas;
80-89%: a range of marks consistent with a first where the work is exceptional in most
areas.
70-79%: a range of marks consistent with a first. Work which shows excellent content,
organisation and presentation, reasoning and originality; evidence of independent reading
and thinking and a clear and authoritative grasp of theoretical positions; ability to sustain an
argument, to think analytically and/or critically and to synthesise material effectively.
60-69%: a range of marks consistent with an upper second. Well-organised and lucid
coverage of the main points in an answer; intelligent interpretation and confident use of
evidence, examples and references; clear evidence of critical judgement in selecting,
ordering and analysing content; demonstrates some ability to synthesise material and to
construct responses, which reveal insight and may offer some originality.
50-59%: a range of marks consistent with lower second; shows a grasp of the main issues
and uses relevant materials in a generally business-like approach, restricted evidence of
additional reading; possible unevenness in structure of answers and failure to understand
the more subtle points: some critical analysis and a modest degree of insight should be
present.
40-49%: a range of marks which is consistent with third class; demonstrates limited
understanding with no enrichment of the basic course material presented in classes;
superficial lines of argument and muddled presentation; little or no attempt to relate issues to
a broader framework; lower end of the range equates to a minimum or threshold pass.
35-39%: achieves many of the learning outcomes required for a mark of 40% but falls short
in one or more areas.
30-34%: a fail; may achieve some learning outcomes but falls short in most areas; shows
considerable lack of understanding of basic course material and little evidence of research.
0-29%: a fail; basic factual errors of considerable magnitude showing little understanding of
basic course material; falls substantially short of the learning outcomes for compensation.
BEGIN YOUR WORK ON THE FOLLOWING PAGE IF YOU ARE WORD PROCESSING YOUR
COURSEWORK
Page 6 of 8

SOLUTION
CALCULATIONS OF THE REFRIGERATION/HEAT PUMP CYCLES
Refrigeration is reversed heat engine, where in the heat will be transferring from the body of
the low temperature (Sink) to the body of the higher temperature (Source). However still as
it is against to the laws of thermodynamics, it is required to have certain amount of external
work to let the heat flow from the body of lower temperature to the body of the higher
temperature. If during the process the net effect is cooling the cold body then it is said to be
working as refrigerator and if the net effect of the process is to heat the surroundings, then it
is called as heat pump. The performance of Heat engine is measured in efficiency and the
performance of the refrigerator of Heat pump will be measured in COP.
REFRIGERATION
1)The Fig-1 shown below is a simple diagram of refrigeration circuit. The main constituents
of the refrigeration cycle are compressor, condenser, expansion valve and evaporator.
Evaporator is the component that works on to cool the surroundings by taking in heat from
the surroundings. Condenser will work to release heat from the fluid compressed in the
compressor. Expansion valve will work to reduce the pressure of the compressed gas and this
inturn will work to let the gas take in heat from the surroundings.
Figure 1 Pressure and Enthalpy plot
Figure 2 Refrigeration circuit
2) The given data conditions:
As per the diagram given above the refrigeration circuit is in operation and four different
refrigerants are proposed to be employed in the refrigeration cycle as part of the performance
analysis. The relative analysis of the performance of the cycle on four different refrigerants is
evaluated and COP (coefficient of performance) for each cycle is determined and analyzed.
The four refrigerants being employed in the cycle.
Page 7 of 8

(i) Refrigerant 1 Ammonia (R717)
(ii) Refrigerant 2 R134a
(iii) Refrigerant 3 R245fa
(iv) Refrigerant 4 RF1234yf
The given operational conditions of the refrigeration cycle are,
 (cold box is mentioned to be operating at -20 degree Celsius)
Temperature of the evaporator = -25 degree Celsius
 (Outside temperature for condenser is 30 degree Celsius)
Temperature of the condenser = 35 degree Celsius
Coefficient of performance of the refrigerator is given by,
COP of refrigerator equation = QL/W
Where QL is the refrigerant load
W is the work consumed to let the refrigerator function.
The problem is solved using P-H and T-H plots for each of the refrigerant and there by
computing the COP for each of their functioning.
For Ammonia:
COP calculations for Ammonia
Figure 3 Temperature Vs entropy
Page 8 of 8

Figure 4 COP Calculation for Ammonia
Page 9 of 8

Figure 5 COP Calculations for R134a
Figure 6 COP calculations for R245fa
Page 10 of 8

Figure 7 COP calculation for R1234yf
For all above figures consider hd is the point where the isenthalpic expansion will start, ha is
the point where the isenthalpic expansion will end. hb is the enthalpy before isentropic
compression and hc is the enthalpy after isentropic compression process.
Comparative COP values:
S.No Refrigerant details COP
1 Ammonia 3.237
2 R134a 3.1125
3 R245fa 2.89
4 R1234YF 1
Comment
Hence from the above observations it is evident that the cop of the refrigeration cycles has
come down with the change over from ammonia to flouro carbon blends. Hence the obvious
performance effect is the fact that it will consume more energy for than the previous versions
for the sake of same amount of refrigeration effect.
PART C
As per the new strategic plan of the organization, if the company is wishing to make up
refrigerator/heat pump for food processing where an initial high temperature heat exchanger
is set to produce steam at 100 degree Celsius. For this the condenser pressure is set to give
105 degree Celsius at heat exchanger input and this pressure is then maintained in the normal
air condenser. The processed food is then to be chilled to -20 degree Celsius.
Page 11 of 8

Then the new operating conditions of the circuit will be,
The set higher temperature of the heatl pump = 100 degree celsius
Condenser pressure is set to give temperature output of 105 degree celsius
The processed food is chilled at -20 degree celsius.
As there is 5 degree celsius temperature difference
Condenser actual temperature is 105 degree celsisus
Evaporator actual temperature is -25 degree celsius
In this environmental conditions, the COP calculatios are as follows,
Figure 8 Modified PH and TS diagram of the refrigeration system
(i) For Ammonia
hd=780KJ/kg
ha=780KJ/kg
hb=1410KJ/kg
hc=2128KJ/kg
For ammonia:Refrigeration heat input = 630kj/kg
Heat rejected for steam generation = 626kj/kg
Heat rejected to the atmosphere = 720kj/kg
(ii) For R134a
Hd=400KJ/kg
Hc=490KJ/kg
Ha=360KJ/kg
Hb=385KJ/kg
R134a refrigeration heat input = 25kj/kg
Heat rejected for steam generation = 90kj/kg
Heat rejected for atmosphere = 40kJ/kg
(iii) For R245a
Hd=355KJ/kg
Hc=490KJ/kg
Ha=355KJ/kg
Page 12 of 8