Optimized suction pressure
(1)
A majority of storage coolers and storage
freezers have single speed evaporator fans. In the following 2 newsletters I
will show you the opportunity to optimize suction pressure for these coolers and
freezers.
It is common knowledge in the
industry that raising the suction pressure improves efficiency of the compressor
(BHP/TR). Typical improvement might be 1% to
2% increase in saturated suction pressure per degree F. If all space
temperatures are satisfied and evaporator coils are in a low load mode, the
suction pressure usually increases to the maximum until some limiting
temperatures are approached; thus, the efficiency of the compressors is
improving.
However, in order to operate the entire
refrigeration plant efficiently, we have to pay attention to evaporator
fans.
There are 4 steps to optimized
suction pressure.
Step 1. Compressor
efficiency.
Example. Freezer temperature is 0 °F; single
stage screw compressor with economizer Frick RWB - II 134E;
condensing temperature is 75°F at 125.8
psig head pressure; saturated suction temperatures are – 5 °F, –10 °F, –15 °F, –
20 °F. Compressor efficiency at suction temperature of –10 °F
Ecomp = 221.1 BHP / 163.9 TR = 1.349 BHP/TR
Table 1. Compressor efficiency at
different suction temperatures
Suction temperature,
°F |
Compressor efficiency,
BHP/TR |
–
5 |
1.237 |
–
10 |
1.349 |
–
15 |
1.469 |
–
20 |
1.599 |
If we increase the suction temperature,
our compressors are using less energy (BHP)
per unit of refrigeration (TR) and their efficiency is improving.
Step 2. System
efficiency.
To estimate efficiency of the system
(compressors + evaporators) we have to add efficiency of compressors and
efficiency of evaporators.
We have compressor efficiencies in
the Table 1.
Efficiency of evaporators can be
estimated as follows.
Example. Evaporator coil fan power is 15 HP;
capacity of the coil is 20 TR at temperature difference (TD) of 10
°F.
Evaporator coil efficiency at TD of 10 °F
(suction temperature –10 °F)
Eevap (10) = 15 HP/ 20 TR = 0.75
HP/TR
Capacity of the coil is proportional to
TD. At TD of 5 °F capacity of our coil is 10 TR. Evaporator coil efficiency at
TD of 5 °F (suction temperature is –
5 °F)
Eevap (5) = 15 HP/ 10 TR = 1.5
BHP/TR.
Table 2. System efficiency at different
suction temperatures.
Suction
temperature, °F |
Compressor efficiency,
BHP/TR |
Evaporator efficiency,
BHP/TR |
System
efficiency, BHP/TR |
– 5 |
1.237 |
1.500 |
2.737 |
– 10 |
1.349 |
0.750 |
2.099 |
– 15 |
1.469 |
0.500 |
1.969 |
– 20 |
1.599 |
0.375 |
1.974 |
From this table we can see that
efficiencies of the system are better (less
BHP per TR) at suction temperatures –15 °F
and – 20 °F, than efficiencies at suction temperatures –5 °F and –10 °F. This
means that the highest suction temperature (pressure) is not the most efficient
for this refrigeration plant.