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Optimized hot gas defrosting (2)


The goal of optimized hot gas defrosting is to minimize the penalties related to frost and defrosting.

There are 2 types of penalties:

  • frost penalties
  • hot gas defrosting penalties

If we operate the frosted evaporator coil at optimum suction pressure at 95% capacity, we have to operate another coil additional 5% of operating time to compensate the losses of capacity. For our coil with 15 HP fans we have to spend additional 0.75 HP/Hr of fan power. Usually, for every 2 HP of fan power, we have to spend 1 HP of compressor and condenser power to remove the heat of evaporator fans. In our example, total penalties of operating coil at 95% capacity are 0.75 + 0.375 = 1.125 HP/Hr.


Efficiency of hot gas defrosting is very low and usually less than 10%. It means that from 100 units of heat provided by hot gas, less than 10 units is used for frost melting. Over 90 units of heat should be removed by the refrigeration plant as a parasitic refrigeration load.


As I mentioned in the previous newsletter, hot gas supply during defrosting usually doubles the amount of gas generated during the cooling mode. It means that after 30 min. of hot gas defrosting, next 1 Hr we have to operate this coil in cooling mode just to remove the heat of defrosting.


Example.
  We can compare hot gas defrosting of 2 identical evaporator coils. First coil will be defrosted every 6 Hrs, length of defrosting is 30 min. Second coil will be defrosted every 24 Hrs.

Capacity of this coil during 5.5 Hrs operations will change from 100% to 98%. Average capacity of first coil is  ( 100 + 98 ) / 2 = 99%  and average frost losses are

0.15 + 0.075 = 0.225 HP/Hrs.

Hot gas defrosting losses can be estimated as following:

( system efficiency ) *  ( coil capacity ) * ( time )

 System efficiency (excluding condenser power) at -20 degF suction temperature is 1.974 HP/TR (October 2005 newsletter). System efficiency including condenser power will be around 2.2 HP/TR. Capacity of the coil at chosen suction temperature is 40 TR. Operating time is 1 Hr. Hot gas defrosting losses would be 2.2 * 40 * 1 = 88 HP.


Capacity of the second coil during 23.5 Hrs operation will change from 100% to 90%. Average capacity of this coil is

( 100 + 90 ) / 2 = 95% and average frost losses are 1.125 HP/Hr.

Coil 1

Coil 2

Total operating time, Hrs

24

24

Hot gas defrosting time, Hrs

0.5 * 4 = 2

0.5

Cooling time, Hrs

24 – 2 = 22

24 - 0.5 = 23.5

Hourly frost losses, HP/Hr

0.225

1.125

Total frost losses, HP

0.225 * 22 = 4.95

1.125 * 23.5 = 26.4

Total hot gas defrosting Losses, HP

88 * 4 = 352

88

Total losses related to the frost, HP

4.95 + 352 = 356.95

88 + 26.4 = 114.4

                                                         

From this example we can see that total losses for the first coil are 3 times greater than for the second coil.


Coil 1 is overdefrosted. I found that overdefrosting very often reduces efficiency of the refrigeration plant. Usually, evaporator coils should be defrosted at around 90% of capacity.


Check the amount of water (melted frost) during defrosting. If you have just a few liters of water for 20 TR coil,  probably this coil is overdefrosted.

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