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Rating of Refrigeration Systems
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Unit of Refrigeration

The standard unit of refrigeration is ton refrigeration or simply ton denoted by TR. It is equivalent to the rate of heat transfer needed to produce (or that results in the freezing) of 1 short ton (2000 lbs; 907kg) of pure ice at 32°F/0°C from water at 32°F/0°C in one day, i.e., 24 hours.

As a practical matter, additional refrigeration is required to take water at room temperature and turn it into ice.

One ton of refrigeration is 3.517 kW. This is derived as follows:

The latent heat of ice (also the heat of fusion) = 333.55 kJ/kg = 144 Btu/lb

One short ton = 2000 lb

Heat extracted = 2000 x 144/24 hr = 288000 Btu/24 hr = 12000 Btu/hr = 200 Btu/min If Btu ton unit is expressed into SI system, it is found to be 210 kJ/min.

1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP

This unit of refrigeration is currently in use in the USA, the UK and India.

A much less common definition is: 1 tonne of refrigeration is the rate of heat removal required to freeze a metric ton (1000 kg) of water at 0°C in 24 hours. Based on the heat of fusion being 333.55 kJ/kg, 1 tonne of refrigeration = 13,898 kJ/h = 3.861 kW. Thus, 1 (metric) tonne of refrigeration is 10% larger than 1 ton of refrigeration.

In many countries, the standard MKS unit of kcal/hr is used. In the MKS it can be seen that: 1 TR 12000 Btu/hr = 12000/3.968 = 3024.2 kcal/hr ~ 50 kcal/min (approx).

Most residential air conditioning units range in capacity from about 1 to 5 tons of refrigeration or 3.5 kW ~ 17.5 kW, or 12,000 Btu/h ~ 60,000 Btu/h. Large industrial chiller systems range up to 800 tons of refrigeration (2.8 MW or 9.6 million Btu/h).

ILLUSTRATIVE PROBLEMS

Example 1

A refrigeration system produces 40 kg/hr of ice at 0°C from water at 25°C. Find the refrigeration effect per hour and TR. If it consumes 1 kW of energy to produce the ice, find the COP. Take latent heat of solidification of water at 0°C as 335 kJ/kg and specific heat of water 4.19 kJ/kg°C.

Solution

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Example 2

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Solution

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ENERGY EFFICIENCY RATIO (EER), AND BEE STAR RATING

Energy Efficient Rating/Ratio (EER) is a labelling scheme implemented by the Bureau of Energy Efficiency (BEE). Under this scheme, appliances (mainly refrigerators and air conditioners) were rated according to their energy consumption and energy saving.

The EER is the ratio of output cooling energy (in BTU) to electrical input energy (in Watt-hour).

EER = output cooling energy (in BTU) / electrical input energy (in Watt-hour)

The units are therefore BTU/W/hr.

To convert EER to COP, we need to accommodate for the units used. We convert the BTU energy and the electrical input energy to a common energy unit, namely Joule. One BTU equals 1055 J. One Whr equals 3600 Ws or 3600 J. So:

COP = output cooling energy / electrical input energy = EER (Btu) / (Whr) x (1055 J/Btu)/(3600 J/Whr) = EER x 0.293

Alternatively, EER = COP x 3.41

The BEE rating is at 5-point scale where 1-star shows the worst in terms of energy saving and 5-star is best. In India, we don’t have appliance less than 3-star these days. If you will look the label closely you will see Energy Efficiency Ratio (EER) which is defined as cooling capacity/power input. The higher the EER, more energy efficient the appliance.

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One important thing to be noted in the image above is that EER is given by W/W (which incidentally is also the same as COP), whereas we have defined EER by Btu/W/hr.

So the EER in the above image is 2.9 x 3.41 = 9.889 Btu/W/hr.

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In another example, the units per year is mentioned, which is kWh per year, which describes the power consumption of the unit throughtout the whole year. This rating is used for appliances which are usually kept on for the whole year, such as refrigerators.

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