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Power Eng 2nd Class study materials
Power Eng 2nd Class study resources

Power Eng 2nd Class: Refrigeration, Air Conditioning, and Compressed Air

Power Eng 2nd Class · Work through your exam, one topic at a time.

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Refrigeration, Air Conditioning, and Compressed Air

This domain covers the vapour-compression cycle, psychrometrics, cooling towers and compressed-air systems. Calculations are usually COP, tonnage or cooling-tower approach, and the qualitative items test whether you can spot the operating fault costing the most energy.

Vapour compression

  • Components: compressor, condenser, expansion device, evaporator. The compressor adds work and raises pressure and temperature; the expansion valve drops pressure without work.
  • COP = refrigeration effect / compressor work. Real machines reach about 40-60% of the Carnot COP between the same evaporating and condensing temperatures.
  • One ton of refrigeration = 3.517 kW = 12,000 Btu/h.
  • Subcooling improves COP; evaporator superheat of roughly 5-8 K protects the compressor from liquid return.
  • Refrigerants: ammonia (R-717) is efficient but toxic and flammable at higher concentrations; HFCs such as R-134a and R-410A have no ozone-depletion potential but high global-warming potential. Never mix refrigerants or use the wrong lubricant; mineral oil and POE are not interchangeable.

Air conditioning and cooling towers

  • Psychrometrics: dry bulb, wet bulb, dew point, relative humidity, enthalpy. Coil duty follows the air mass flow times enthalpy difference.
  • Tower range = hot water in minus cold water out. Tower approach = cold water out minus wet bulb. Approach cannot be zero, and the wet bulb sets the practical limit.

Compressed air

  • Pressure dew point must sit below the coldest point in the distribution system. Refrigerated dryers reach about 3 C; desiccant dryers reach -40 C or lower.
  • Leaks typically waste 20-30% of output; several small leaks can equal one compressor's capacity.
  • Compressor power falls when discharge pressure is reduced; lowering header pressure by 100 kPa saves several percent of energy.

Worked example

A chiller delivers 350 kW of cooling with 100 kW of compressor input: COP = 350/100 = 3.5. Carnot between an evaporating temperature of 2 C (275 K) and condensing at 40 C (313 K) is 275/(313 - 275) = 7.2, so the plant runs at 3.5/7.2 = 49% of ideal. A fouled condenser lifting condensing temperature 4 K raises compressor power by roughly 8-12%; cleaning it is the cheapest efficiency gain available.

Traps

  • Calling the difference between cold water and wet bulb the range instead of the approach.
  • Charging refrigerant by feel rather than by weighing in the specified charge.
  • Assuming any dryer delivers instrument-quality air regardless of its dew point specification.
  • Chasing low suction pressure while ignoring high condensing pressure, which costs more power.