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 (r)

- 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.

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

C = 4.185 kJ/Kgk

Evaporation temperature = 5^{0}C

= 30^{0}C – 5^{0}C = 25 K

P_{th} = 0.0032*4.185 kJ/kg/K *25 K *1000

= 334.8 kJ

Evaporation temperature = 6^{0}C

= 30^{0}C – 6^{0}C = 24 K

P_{th} = 0.0032*4.185 kJ/kg/K *24 K *1000

## Task 2 - Calculation and Plotting of Efficiencies

= 321.41 kJ

Evaporation temperature = 7^{0}C

= 30^{0}C – 7^{0}C = 23 K

P_{th} = 0.0032*4.185 kJ/kg/K *23 K *1000

= 308.016 kJ

Evaporation temperature = 8^{0}C

= 30^{0}C – 8^{0}C = 22 K

P_{th} = 0.0032*4.185 kJ/kg/K *22 K *1000

= 294.624 kJ

Evaporation temperature = 9^{0}C

= 30^{0}C – 9^{0}C = 21 K

P_{th} = 0.0032*4.185 kJ/kg/K *21 K *1000

= 281.232 kJ

Evaporation temperature = 10^{0}C

= 30^{0}C – 10^{0}C = 20 K

P_{th} = 0.0032*4.185 kJ/kg/K *20 K *1000

= 267.84 kJ

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

C = 4.185 kJ/Kgk

Evaporation temperature = 5^{0}C

= 35^{0}C – 5^{0}C = 30 K

P_{th} = 0.0032*4.185 kJ/kg/K *30 K *1000

= 401.76 kJ

Evaporation temperature = 6^{0}C

= 35^{0}C – 6^{0}C = 29 K

P_{th} = 0.0032*4.185 kJ/kg/K *29 K *1000

= 388.37 kJ

Evaporation temperature = 7^{0}C

= 35^{0}C – 7^{0}C = 28 K

P_{th} = 0.0032*4.185 kJ/kg/K *28 K *1000

= 374.98 kJ

Evaporation temperature = 8^{0}C

= 35^{0}C – 8^{0}C = 27 K

P_{th} = 0.0032*4.185 kJ/kg/K *27 K *1000

= 361.58 kJ

Evaporation temperature = 9^{0}C

= 35^{0}C – 9^{0}C = 26 K

P_{th} = 0.0032*4.185 kJ/kg/K *26 K *1000

= 348.19 kJ

Evaporation temperature = 10^{0}C

= 35^{0}C – 10^{0}C = 25 K

P_{th} = 0.0032*4.185 kJ/kg/K *25 K *1000

= 334.8 kJ

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

Evaporation temperature = 5^{0}C

= 40^{0}C – 5^{0}C = 35 K

P_{th} = 0.0032*4.185 kJ/kg/K *35 K *1000

= 468.72 kJ

Evaporation temperature = 6^{0}C

= 40^{0}C – 6^{0}C = 34 K

P_{th} = 0.0032*4.185 kJ/kg/K *34 K *1000

= 455.328 kJ

Evaporation temperature = 7^{0}C

= 40^{0}C – 7^{0}C = 33 K

P_{th} = 0.0032*4.185 kJ/kg/K *33 K *1000

= 441.94 kJ

Evaporation temperature = 8^{0}C

= 40^{0}C – 8^{0}C = 32 K

P_{th} = 0.0032*4.185 kJ/kg/K *32 K *1000

= 428.54 kJ

Evaporation temperature = 9^{0}C

= 40^{0}C – 9^{0}C = 31 K

P_{th} = 0.0032*4.185 kJ/kg/K *31 K *1000

= 415.152 kJ

Evaporation temperature = 10^{0}C

= 40^{0}C – 10^{0}C = 30 K

P_{th} = 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

E_{c} = net cooling capacity = 25KW

P_{a} = applied power (Watts)

Energy efficiency ratio for 1^{st} condenser of 30^{0}C

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

C = 4.185 kJ/Kgk

Evaporation temperature = 5^{0}C

= 30^{0}C – 5^{0}C = 25 K

P_{th} = 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 = 6^{0}C

= 30^{0}C – 6^{0}C = 24 K

P_{th} = 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 = 7^{0}C

= 30^{0}C – 7^{0}C = 23 K

P_{th} = 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 = 8^{0}C

= 30^{0}C – 8^{0}C = 22 K

P_{th} = 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 = 9^{0}C

= 30^{0}C – 9^{0}C = 21 K

P_{th} = 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 = 10^{0}C

= 30^{0}C – 10^{0}C = 20 K

P_{th} = 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 2^{nd} condenser of 35^{0}C

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

C = 4.185 kJ/Kgk

Evaporation temperature = 5^{0}C

= 35^{0}C – 5^{0}C = 30 K

P_{th} = 0.0032*4.185 kJ/kg/K *30 K *1000

= 401.76 kJ

Applied power per second

Power =

= 401.76 KW

EER =

= 0.0622

= 35^{0}C – 6^{0}C = 29 K

P_{th} = 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 = 7^{0}C

= 35^{0}C – 7^{0}C = 28 K

P_{th} = 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 = 8^{0}C

= 35^{0}C – 8^{0}C = 27 K

P_{th} = 0.0032*4.185 kJ/kg/K *27 K *1000

= 361.58 kJ

Applied power per second

Power =

= 361.58 KW

EER =

= 0.0691

= 35^{0}C – 9^{0}C = 26 K

P_{th} = 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 = 10^{0}C

= 35^{0}C – 10^{0}C = 25 K

P_{th} = 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 3^{rd} condenser of 40^{0}C

P_{th} = M*C**1000

M = mass flow rate = 0.0032 m^{3}/s

Evaporation temperature = 5^{0}C

= 40^{0}C – 5^{0}C = 35 K

P_{th} = 0.0032*4.185 kJ/kg/K *35 K *1000

= 468.72 kJ

Applied power per second

Power =

= 468.72 KW

EER =

= 0.0533

= 40^{0}C – 6^{0}C = 34 K

P_{th} = 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 = 7^{0}C

= 40^{0}C – 7^{0}C = 33 K

P_{th} = 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 = 8^{0}C

= 40^{0}C – 8^{0}C = 32 K

P_{th} = 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 = 9^{0}C

= 40^{0}C – 9^{0}C = 31 K

P_{th} = 0.0032*4.185 kJ/kg/K *31 K *1000

= 415.152 kJ

Applied power per second

Power =

= 415.152 KW

EER =

= 0.0602

30^{0} 35^{0} 40^{0}

= 40^{0}C – 10^{0}C = 30 K

P_{th} = 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] = [5^{0}C to 10^{0}C] = 278 K to 283 K

Tc: condensing temperature [K] = 30^{0} = 303K, 35^{0}C = 308K and 45^{0}C = 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 T_{c} (^{0}C)

Note that the mass flow rate = 0.0032 m^{3}/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 (-120^{0}C to -90^{0}C).

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

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