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Level 3 Air Conditioning Inspections for Buildings

Understand the efficiency of air conditioning systems through concepts like Cooling Load, COP, COSP, and SEER. Learn how factors like temperature difference and refrigerant type affect the performance. Discover tips on optimizing COP through evaporator, condenser, compressor efficiency, refrigerant amount, and type selection.

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Level 3 Air Conditioning Inspections for Buildings

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  1. . Level 3 Air Conditioning Inspections for Buildings 14. Efficiency of Air Conditioning Systems (Day 3) PRESENTED BY Anthony Balaam aircon@stroma.com

  2. Cooling Load The Refrigeration Load • Cooling load on a refrigeration system determines:- • 1. The plant size • 2. and the power consumption • This does not effect the Coefficient of Performance (COP). • No refrigeration load = no COP or efficiency The smaller the load, the lower the electricity power consumption.

  3. Coefficient of System Performance (COSP) Efficiency of a Refrigeration Cycle • Thecoefficient of system performance (COSP) is expressed as the capacity (kW)divided by the power input (kW) required to produce the cooling:- COSP = • This takes into account all fans, motors and controls associated with the system. • The Seasonal Energy Efficiency Ratio (SEER) is now commonly used as it takes into account the seasonal performance.

  4. Coefficent of Performance Efficiency of a Refrigeration Cycle • Thecoefficient of performance (COPc) is expressed as the useful coolingduty (kW)divided by the power input (kW) required to produce the cooling: • The maximum efficiency, ‘Carnot cycle coefficient of performance’, is based upon a theoretical thermodynamic cycle on which the actual cycle is based: COPc = Te/(Tc-Te) Where:- Tc is the condensing temperature (K) Te is the evaporating temperature (K) ‘c’ denotes a cooling COP This does not take into account of all fans, motors and controls associated with the system, compressor only.

  5. Performance Efficiency of a Refrigeration Cycle An actual COP would be typically 50% less than the theoretical COP due to:- • 1. Deviations from the theoretical cycle • 2. Inefficiencies within the practical cycle like:- • A. Pressure losses • B. Heat transfers Example: A COP of 3.7 means… The refrigeration unit will produce 3.7kW ofcooling per kW of electrical consumption by thecompressormotor. Question: Why is the efficiency greater than 1? Surely this is against the laws of thermodynamics…?

  6. Performance Efficiency of a Refrigeration Cycle Answer:-Cooling duty is in kW of heat exchange energy, whereas the power input is electricalorwork done energy. If:power input was expressed as heat energy (i.e. considering the energy used to produce the electricity), then the ratio would be less than 1(or… 100%). COP provides useful information about the running costs of the refrigeration system with respect to the cooling duty.

  7. Answer 3.7kW Heat Rejected Atmosphere COP = Cooling Output (kW) / Power Input (kW) COP = 3.7kW / 1kW = 3.7 1kW Electrical Power 70% Losses 3.7kW Heat Absorbed Occupied Zone 2.5 kW Fuel

  8. Performance Efficiency of a Refrigeration Cycle • COP is mostdependant on temperature difference (or temperature lift) between the condensing and evaporating temperatures. • Smaller the difference, greater the COP • Bigger the difference, lower the COP • You will get around 2-4% increase in performance for:- • 1. 1K increase in evaporating temperature • 2. 1K decrease in condensing temperature • COP is also affected by:- • Refrigerant type • Equipment used • Controls and maintenance

  9. COP Factors 1. Raising the Evaporating Temperature • To achieve this use an evaporator with a higher basic rating is used (usually a larger evaporator). • The evaporator is defrosted ifnecessary. • Also ensure the evaporator is cleanandfreeof blockage.

  10. COP Factors 2. Lowering the Condensing Temperature • To achieve this - Use a condenserwith ahigh basic rating (usually a larger condenser). • Allow the condensing temperaturetofloat down with the ambient temperature. • Average UK temp is around10˚C, this should be used to advantage, as opposed to holding a condensing temperature artificially high. • Can save in excess of around 25% energy. • Use water instead of air as condensermedium. • Also - Ensure condensers do not become blocked, or flow of cooling air or water becomes impeded in any other way.

  11. COP Factors Compressor Efficiency • Varies withtype and manufacturer. • The most efficient compressor for an application, should be selected. • This depends on:- • 1. Size of cooling load. • 2. Refrigerant used. • 3. Temperatureof the application. • 4. The average temperatureof thecooling medium (air/ water)

  12. COP Factors Amount of Refrigerant • This has a significant effect on the temperature lift. • Too muchortoo little,reduces efficiency. • Systems that leakrefrigerant consume more power than necessary. • Costs UK refrigerant plant owners an extra 11% per year. • Over charged systems have more refrigerant tolose in the event of a leak which is environmentally detrimental.

  13. COP Factors Refrigerant Type • Variation of this can effect the COP by up to around 10%. • Hardwareneeds to be optimised to the refrigerant for benefit. • The most efficient refrigerant for an application depends upon:- • 1. Compressor type • 2. Temperatureof theapplication • 3. Average temperatureof thecooling medium (air/ water).

  14. COP Factors Superheat of the Suction Vapour • This needs to be as low as possible. • Warmer vapourreduces the capacityof thecompressor; • Also, it does not reduce itspower input. • On direct expansion systems it is achieved by:- • 1. Correctlycontrolling the expansion device; • 2. Insulatingthesuction line.

  15. COP Factors Amount of Sub-cooling • This should be as high as possible. • By Increases the capacityof thesystem. • But does not increase its power input. • The Liquid line should not:- • 1. Should not beinsulated. • Should not pass through any hot areas (like kitchens, direct sunlight).

  16. Comparison

  17. NDHCVCG Non-Domestic Heating, Cooling and Ventilation Compliance Guide (Second Tier Ref: ADL2B) For Cooling:- Energy Efficiency Ratio (EER) • For Chillers; EER is defined as the ratio of cooling energy delivered, divided by the energy input to the cooling plant. • For Packaged Air Conditioners; EER is defined as the ratio of energy removed from air within conditioned space, divided by the effective energy input to the unit

  18. Cooling Seasonal Energy Efficiency Ratio (SEER)

  19. SEER Profile COP100 COP75 A B C D Heating COP50 COP25 0 Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec EER25 EER50 Cooling EER75 EER100

  20. Cooling Energy Efficincy Ratio (EER)

  21. Cooling Cooling Controls:-

  22. Cooling Determining The SEER - no part load data

  23. Cooling Determining The SEER – by part load data

  24. Reference Material “Heating, Ventilation, Air Conditioning and Refrigeration”, CIBSE Guide B, Chartered Institute of Building Services Engineers, 2005 “CIBSE KS13: Refrigeration”, CIBSE Knowledge Series, Chartered Institute of Building Services Engineers, 2008 “ASHRAE Handbook: Fundamentals”, American Society of Heating, Refrigeration and Air Conditioning Engineers, 2001 “BS EN 378: Specification for Refrigeration Systems and Heat Pumps; Part 1: 2000: Basic Requirements, Definitions, Classification and Selection Criteria; Part 2: 2000: Design, Construction, Testing, Marking, and Documentation; Part 3: 2000: Installation Site and Personal Protection; Part 4: 2000: Operation, Maintenance, Repair and Recovery”, London: British Standard Institution, 2000 “Non-Domestic Heating, Cooling, and Ventilation Compliance Guide”, Department For communities and local Government Building Regulations Approved Document L2B”, Department For communities and local Government

  25. LEVEL 3 Air ConditioningENERGY ASSESSORS TRAINING ANY QUESTIONS OR FEEDBACK ON ANY SLIDE

  26. STROMA Certification Ltd – Contacts Web Links www.stroma.com/certification Contacts:- STROMA Certification Ltd. 4 Pioneer Way, Castleford, WF10 5QU 0845 621 11 11 training@stroma.com

  27. End of this section

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