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.