You have carried out steady state tests for three condenser water leaving temperatures (approximately constant refrigerant condensing temperatures (saturated vapour)) and a range of evaporator water leaving temperatures (constant refrigerant evaporating temperatures (saturated vapour)). You are expected to:
Task 1 - Plot the performance characteristics (performance maps) of the refrigeration system as functions of the condenser water leaving and evaporator water leaving temperatures. These characteristics should include power consumption, cooling capacity and COP.
Task 2 – Calculate and plot other characteristics (efficiencies) that you may feel are relevant to your discussion and the performance evaluation of the system.
Task 3 – Calculate and Plot the Carnot COP and compare it with the actual COP of the system discussing the main reasons for the difference in the two values.
Task 4 - Discuss the performance characteristics of the system and their implications with respect to system design and optimal operation.
Task 1 - Plotting Performance Characteristics
Majorly restriction has been injected by the Standard thermodynamics on the thermodynamic measures and all this is grounded on reversible assumptions, which will mean either a zero rate of operation or infinite system size. The extension of thermodynamic analysis in order to include the finite time constraints was facilitated by the finite time thermodynamic, which derives a more accurate restriction on the performance. This method of finite time thermodynamic has been used majorly in several thermodynamic systems. Previously the heat engines were used extensively, with maximum power efficiency production expression that is derived from two heat reservoirs at temperatures and under the limitation of Newtonian heat transfer which is given by expression;
?CA = 1-
The invention of reciprocating chillers lead to analysis and its characteristics was compared to experimental data. The heat leakage and friction on the performance in line with engines were totally addressed and the features were described using either power degradation or power coefficient of performance coordinates.
Numerous objective functions have been proposed. Where combination of both maximization of power with maximization of losses (entropy production)
Extensive merit has been observed through ecological coefficient of performance which is defined as the ratio of the cooling load to the rate of availability loss (or entropy generation rate).
The present study of the finite-time analysis of refrigeration or heat pump systems, has an inclusion of friction, solid, and fluid, with an aim of increasing the importance of source of dissipation.
The characteristics are shown in graph of the cooling rate (r) versus coefficient of performance (ω) coordinates, from this observation a comparison is drawn to the characteristics of real refrigeration or heat pump devices, which later results to a proposal an estimation of temperatures of the working fluid of the hot and cold side.
The presentation of the finite heat transfer rate is the sole source of heat losses [the dashes curved]
- Open circuit in which both (r) and (ω) vanishes in the limit of slow operation
- Short circuit in which both (r) and (ω) vanishes but the limit of fast operation
- Maximum both (ω)
Task 1
Task 2
The calculation of the thermal power of a hydraulic system is based on the mass flow rate, the heat capacity of the medium, and the temperature difference before and after heat dissipation.
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
C = 4.185 kJ/Kgk
Evaporation temperature = 50C
= 300C – 50C = 25 K
Pth = 0.0032*4.185 kJ/kg/K *25 K *1000
= 334.8 kJ
Evaporation temperature = 60C
= 300C – 60C = 24 K
Pth = 0.0032*4.185 kJ/kg/K *24 K *1000
Task 2 - Calculation and Plotting of Efficiencies
= 321.41 kJ
Evaporation temperature = 70C
= 300C – 70C = 23 K
Pth = 0.0032*4.185 kJ/kg/K *23 K *1000
= 308.016 kJ
Evaporation temperature = 80C
= 300C – 80C = 22 K
Pth = 0.0032*4.185 kJ/kg/K *22 K *1000
= 294.624 kJ
Evaporation temperature = 90C
= 300C – 90C = 21 K
Pth = 0.0032*4.185 kJ/kg/K *21 K *1000
= 281.232 kJ
Evaporation temperature = 100C
= 300C – 100C = 20 K
Pth = 0.0032*4.185 kJ/kg/K *20 K *1000
= 267.84 kJ
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
C = 4.185 kJ/Kgk
Evaporation temperature = 50C
= 350C – 50C = 30 K
Pth = 0.0032*4.185 kJ/kg/K *30 K *1000
= 401.76 kJ
Evaporation temperature = 60C
= 350C – 60C = 29 K
Pth = 0.0032*4.185 kJ/kg/K *29 K *1000
= 388.37 kJ
Evaporation temperature = 70C
= 350C – 70C = 28 K
Pth = 0.0032*4.185 kJ/kg/K *28 K *1000
= 374.98 kJ
Evaporation temperature = 80C
= 350C – 80C = 27 K
Pth = 0.0032*4.185 kJ/kg/K *27 K *1000
= 361.58 kJ
Evaporation temperature = 90C
= 350C – 90C = 26 K
Pth = 0.0032*4.185 kJ/kg/K *26 K *1000
= 348.19 kJ
Evaporation temperature = 100C
= 350C – 100C = 25 K
Pth = 0.0032*4.185 kJ/kg/K *25 K *1000
= 334.8 kJ
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
Evaporation temperature = 50C
= 400C – 50C = 35 K
Pth = 0.0032*4.185 kJ/kg/K *35 K *1000
= 468.72 kJ
Evaporation temperature = 60C
= 400C – 60C = 34 K
Pth = 0.0032*4.185 kJ/kg/K *34 K *1000
= 455.328 kJ
Evaporation temperature = 70C
= 400C – 70C = 33 K
Pth = 0.0032*4.185 kJ/kg/K *33 K *1000
= 441.94 kJ
Evaporation temperature = 80C
= 400C – 80C = 32 K
Pth = 0.0032*4.185 kJ/kg/K *32 K *1000
= 428.54 kJ
Evaporation temperature = 90C
= 400C – 90C = 31 K
Pth = 0.0032*4.185 kJ/kg/K *31 K *1000
= 415.152 kJ
Evaporation temperature = 100C
= 400C – 100C = 30 K
Pth = 0.0032*4.185 kJ/kg/K *30 K *1000
= 401.76 kJ
The Energy Efficiency Ratio - EER - is the cooling efficiency of unitary air-conditioning and heat pump system.
The efficiency is always determined at a single rate of condition specified by an appropriate equipment standard and is defined as the ratio of net cooling capacity - or heat removed to the total input rate of electric power applied in Watts.
Task 3 - Comparing and Discussing Actual vs Carnot COP
EER =
where
EER = energy efficient ratio
Ec = net cooling capacity = 25KW
Pa = applied power (Watts)
Energy efficiency ratio for 1st condenser of 300C
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
C = 4.185 kJ/Kgk
Evaporation temperature = 50C
= 300C – 50C = 25 K
Pth = 0.0032*4.185 kJ/kg/K *25 K *1000
= 334.8 kJ
Applied power per second
Power =
= 334.8 KW
EER =
= 0.0747
Evaporation temperature = 60C
= 300C – 60C = 24 K
Pth = 0.0032*4.185 kJ/kg/K *24 K *1000
= 321.41 kJ
Applied power per second
Power =
= 321.41 KW
EER =
= 0.0778
Evaporation temperature = 70C
= 300C – 70C = 23 K
Pth = 0.0032*4.185 kJ/kg/K *23 K *1000
= 308.016 kJ
Applied power per second
Power =
= 308.016 KW
EER =
= 0.0812
Evaporation temperature = 80C
= 300C – 80C = 22 K
Pth = 0.0032*4.185 kJ/kg/K *22 K *1000
= 294.624 kJ
Applied power per second
Power =
= 294.624 KW
EER =
= 0.0849
Evaporation temperature = 90C
= 300C – 90C = 21 K
Pth = 0.0032*4.185 kJ/kg/K *21 K *1000
= 281.232 kJ
Applied power per second
Power
= 281.232 KW
EER =
= 0.0889
Evaporation temperature = 100C
= 300C – 100C = 20 K
Pth = 0.0032*4.185 kJ/kg/K *20 K *1000
= 267.84 kJ
Applied power per second
Power =
= 267.84 KW
EER =
= 0.0933
Energy efficiency ratio for 2nd condenser of 350C
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
C = 4.185 kJ/Kgk
Evaporation temperature = 50C
= 350C – 50C = 30 K
Pth = 0.0032*4.185 kJ/kg/K *30 K *1000
= 401.76 kJ
Applied power per second
Power =
= 401.76 KW
EER =
= 0.0622
= 350C – 60C = 29 K
Pth = 0.0032*4.185 kJ/kg/K *29 K *1000
= 388.37 kJ
Applied power per second
Power =
= 388.37 KW
EER =
= 0.0644
Evaporation temperature = 70C
= 350C – 70C = 28 K
Pth = 0.0032*4.185 kJ/kg/K *28 K *1000
= 374.98 kJ
Applied power per second
Power =
= 374.98 KW
EER =
= 0.0667
Evaporation temperature = 80C
= 350C – 80C = 27 K
Pth = 0.0032*4.185 kJ/kg/K *27 K *1000
= 361.58 kJ
Applied power per second
Power =
= 361.58 KW
EER =
= 0.0691
= 350C – 90C = 26 K
Pth = 0.0032*4.185 kJ/kg/K *26 K *1000
= 348.19 kJ
Task 4 - Implications for System Design and Optimal Operation
Applied power per second
Power =
= 348.19 KW
EER =
= 0.0718
Evaporation temperature = 100C
= 350C – 100C = 25 K
Pth = 0.0032*4.185 kJ/kg/K *25 K *1000
= 334.8 kJ
Applied power per second
Power =
= 334.8 KW
EER =
= 0.0747
Energy efficiency ratio for 3rd condenser of 400C
Pth = M*C**1000
M = mass flow rate = 0.0032 m3/s
Evaporation temperature = 50C
= 400C – 50C = 35 K
Pth = 0.0032*4.185 kJ/kg/K *35 K *1000
= 468.72 kJ
Applied power per second
Power =
= 468.72 KW
EER =
= 0.0533
= 400C – 60C = 34 K
Pth = 0.0032*4.185 kJ/kg/K *34 K *1000
= 455.328 kJ
Applied power per second
Power =
= 455.328 KW
EER =
= 0.0549
Evaporation temperature = 70C
= 400C – 70C = 33 K
Pth = 0.0032*4.185 kJ/kg/K *33 K *1000
= 441.94 kJ
Applied power per second
Power =
= 441.94 KW
EER =
= 0.0566
Evaporation temperature = 80C
= 400C – 80C = 32 K
Pth = 0.0032*4.185 kJ/kg/K *32 K *1000
= 428.54 kJ
Applied power per second
Power =
= 428.54 KW
EER =
= 0.0583
Evaporation temperature = 90C
= 400C – 90C = 31 K
Pth = 0.0032*4.185 kJ/kg/K *31 K *1000
= 415.152 kJ
Applied power per second
Power =
= 415.152 KW
EER =
= 0.0602
300 350 400
= 400C – 100C = 30 K
Pth = 0.0032*4.185 kJ/kg/K *30 K *1000
= 401.76 kJ
Applied power per second
Power =
= 401.76KW
EER =
= 0.0622
Task 3
Calculation of Carnot COP for the three types of condensers
COP = Qo/Pe = To/(Tc-To),
where:
COP: coefficient of performance [-]
Qo: cooling capacity [kW] = 25 kW
Pe: driving power for the compressor [kW]
To: evaporating temperature [K] = [50C to 100C] = 278 K to 283 K
Tc: condensing temperature [K] = 300 = 303K, 350C = 308K and 450C = 318 Kfor the three condensers
Note that; The theoretical COP needs to be multiplied by a factor of 0.35 in order to give a realistic value
At 278K
COP =
= 11.12
Actual COP = 11.12 * 0.35 = 3.892
At 279K
COP =
= 11.625
Actual COP = 11.625 * 0.35 = 4.069
At 280K
COP =
=
= 12.174
Actual COP = 12.174 * 0.35 = 4.261
At 281K
COP =
= 12.773
Actual COP = 12.773 * 0.35 = 4.471
Objective Functions and Ecological Coefficient of Performance
At 282K
COP =
= 13.429
Actual COP = 13.429 * 0.35 = 4.7
At 283K
COP =
= 14.15
Actual COP = 14.15 * 0.35 = 4.953
At 278K
COP =
= 9.267
Actual COP = 9.267 * 0.35 = 3.2435
At 279K
COP =
= 9.621
Actual COP = 9.621 * 0.35 = 3.3674
At 280K
COP =
= 10.00
Actual COP = 10.00 * 0.35 =
At 281K
COP =
= 10.407
Actual COP = 10.407 * 0.35 = 3.642
At 282K
COP =
= 10.846
Actual COP = 10.846 * 0.35 = 3.796
At 283K
COP =
= 11.32
Actual COP = 11.32 * 0.35 = 3.962
At 278K
COP =
= 6.95
Actual COP = 6.95* 0.35 = 2.433
At 279K
COP =
= 7.154
Actual COP = 7.154 * 0.35 = 2.504
At 280K
COP =
= 7.368
Actual COP = 7.368 * 0.35 = 2.579
At 281K
COP =
= 7.595
Actual COP = 7.595 * 0.35 = 2.658
At 282K
COP =
=
= 7.833
Actual COP = 7.833 * 0.35 = 2.742
At 283K
COP =
= 8.086
Actual COP = 8.086 * 0.35 = 2.8301
Graphical presentation of COP against Tc (0C)
Note that the mass flow rate = 0.0032 m3/s
Task 4
Discuss the performance characteristics of the system and their implications with respect to system design and optimal operation.
A number of performance characteristics were measured, the first was the time required to bring the system down from the room temperature to operating temperature, since the system could not be able to absorb the the requisite heat load until it was at operating temperature.
Adjusting this results to take into aacount the thermal capacitance of a large system, the cool down time will approach about 70 minutes, and this will turn out to be an obverse concern, but in order to address this concern a special system would be set, having a throttle designed with an aim of enabling a high flow rate during the cool down period, then it should be switched over to continuously enable operation setting, and this will lead to a cool down period which improves to less than 20 minutes when using the throttling scheme.
Also, the heat load – temperature behaviour of the system considering the heat heat load as a function of average evaporator temperature, that is cold head, it will be observed that at around 173K the unit is able to absorb approximately 120 Watts.
The maximum load that the system is able to sustain at a stable temperature is found to be about 140 Watts at around 183 K, and any power above this level will result to a temperature runaway, with its sensitivity that change to heat load will be approximately 0.5 K/Watts.
The coefficient of the performance(COP) is evaluated as a function of the the average evaporator surface temperature, and it is seen to appear to operate consistently at about 16% of the Carnot coefficient of performance(COP) over the range of about 153K to about 183K (-1200C to -900C).
Note that the higher EER the more efficient the system is.
References
K.C., M. G. (2000). Cool Thermodynamics: The Engineering and Physics of Predictive, Diagnostic and Optimization Methods for Cooling Systems. Cambridge International Science Publishing.
James B. Rishel, Thomas H. Durkin and Benny L. Kincaid. HVAC Pump Handbook - McGraw-Hill Handbooks, second edition (2006). ISBN: 0-07-145784-4.
D. Eastop and A. McConkey, Applied Thermodynamics for Engineering Technologists - Prentice Hall, fifth edition (1996). ISBN: 0582091934
Applied Thermal Engineering: Design, Processes, Equipment and Economics – ELSEVIER (2009). ISSN: 1359-4311
