IIAR

By Jerry Von Dohlen, Newark Refrigerated Warehouse

Two systems are generally available to remove moisture from the air in a low temperature refrigeration process. The defrost costs and refrigeration loads of each system are compared in this analysis. The first alternative is a frosted fan-coil evaporator. The assumptions are that the coil is
operated at –20° F and condensing takes place at 90° F. For this comparison, a Vilter 16-cylinder R-22 compressor is being used. The second alternative is a liquid desiccant system.

Beginning with frosted coils, the first demand on the refrigeration system is the creation of the frost on the coil.

Assuming that the air entering the refrigerated space is approximately 50° F, then one pound of water entering with that air is also 50° F. The first cooling effort reduces the vapor temperature and then the phase change from vapor to liquid and the latent heat of condensation is 1076 BTUs per pound at –15° F. The water will now be converted to frost requiring 144 BTUs per pound. That frost must now be reduced in temperature to the coil temperature of –15° F which requires an additional 23 BTUs. Therefore, the amount of heat which must be removed from 50° F water vapor, in the process of being converted to –15° F frost, is 1253 BTUs.

How much energy in the form of electric power is consumed in removing those 1253 BTUs from the refrigerated space? The compressor is operating at about 2 HP per ton and other electric power consumers (evaporators, condenser, pumps, etc.) represent approximately an additional 33% for a total HP per ton of about 2.7. This implies the equivalent of approximately 0.57 kilowatt of electric power to transfer one kilowatt of heat from the refrigerated space. Therefore, for the above calculation (1253 BTUs divided by 3413 BTUs or 0.367 kilowatt hours of heat), it will require 0.57 x 0.367 kilowatt hours or 0.21 kilowatt hours of electric power to be used by refrigeration system to convert the one pound of water to –15° F frost. Obviously, that one pound of frost must now be removed from the coil.

Most systems use hot gas defrost; hot gas is supplied to the coil through a piping system. The hot gas must raise the frost temperature from –15° F to 32° F which requires 24 BTUs. The frost is then converted through the application of the latent heat of frost to water at 144 BTUs per pound. According to Stoecker, that process is 20% efficient, at best. Therefore, an additional 4 x 144 BTUs must be applied. This heat is used to heat pipes, the evaporator itself, maintain the water temperature as it drains out of the room, etc. Therefore, an additional 576 BTUs must be applied to the coil. Coils are generally defrosted at around 42° F and the water must be raised from 32° to 42° F which requires 10 BTUs. Therefore, defrosting the frost requires about 754 BTUs. Unfortunately, both sublimation (conversion from frost to vapor) and evaporation (water to vapor) occur in the process of defrosting a coil. According to the Ned Hoeckler paper presented at the 1994 IIAR Annual Meeting, approximately 16% of the frost sublimes and 14% of the frost evaporates into the surrounding air space, so heat for those loads must also be supplied. Obviously, this vapor is returned to the room, so approximately 30% of 754 BTUs (or 227 BTUs) must be supplied to this process. Therefore, a total of 981 BTUs of defrost heat must be supplied to the coil. At this point, only .7 pounds of water vapor has been removed.

How much did the 981 BTUs cost? Compressors are relatively efficient heat pumps; in this case moving 3.31 BTUs for each BTU of input energy. Some argue that the hot gas is free but this author believes that this assumption is incorrect because the vapor condensed into liquid is returned to the low side of the system at defrost pressure and therefore propagates flash gas which is an additional load on the compressor. This parasitic load can be significantly higher in defrost piping designs which rely on a simple back pressure regulator instead of a float drainer to remove the condensed defrost hot gas from the coil. Unnecessarily long defrost cycles plus the addition of unnecessary defrost cycles can add significantly to the parasitic load on the refrigeration system.

Sixty-seven point six refrigeration tons (67.6) is the transfer of 811,200 BTUs. The energy of the motor itself (138 HP) is also transferred to the refrigerant vapor: 138 HP x 0.745 (to convert to kilowatts) x 3413 (to convert to BTUs) or 350,890 BTUs for a total of 1,162,090 heat infused into the gas by a 138 HP compressor. 350,890 BTUs of energy in the form of electric power transfers 1,162,000 BTUs or 3.31 BTUs of heat is contained in the refrigerant vapor for each BTU applied to the electric motor. Therefore, 296 BTUs (981 BTUs divided by 3.31) of electric energy is required to produce the 981 BTUs of defrost heat. At 3413 BTUs per kilowatt hour (296/3413), 0.09 kilowatt hours are required to produce the defrost heat. Therefore, the cost of defrost heat is approximately 40% of the refrigeration cost of creating the frost. Remember, the system is operating at one kilowatt of energy transfer for each 0.57 kilowatts of electric energy and the defrost heat of 983 BTUs must be removed from the space. Therefore, on a per hour time basis, 559 BTUs (981 BTUs x 0.57) of energy must be applied at 3413 BTUs per kilowatt hour or 0.16 kilowatt hours must be used to remove the heat which defrosted the coil.

In total, approximately 0.46 kilowatt hours are required to remove 0.70 of one pound of water, remembering that .30 of the pound was re-introduced as water vapor into the room, or approximately 0.66 (0.46/0.7) kilowatt hours are required to remove one pound of moisture from air. If a kilowatt hour costs ten cents, the cost to remove one pound of water from 50° F air is about $.066.

A second way of removing the moisture is with a brine/ liquid desiccant system. Generally, liquid desiccant systems require about a 20 percent premium (energy) to remove vapor from the air and about 1.65:1 premium to remove the moisture from the brine solution. The water is never converted to frost so the latent heat of condensation of 1100 BTUs (at –20° F) plus approximately 35 BTUs to reduce the vapor from 50° F to –20° F, or 1143 BTUs, is required by the system. With 20 percent inefficiency (heat of absorption and higher temperature of regenerated brine), a total of 1362 BTUs is required to remove one pound of water vapor using a liquid desiccant. At 0.57 kilowatts of electric usage per kilowatt of heat transferred, it requires 776 BTUs of electric power at 3413 BTUs per kilowatt or approximately 0.23 kilowatts would be required to remove vapor from the air and reduce it to –20° F. At 10 cents per kilowatt hour that is approximately .23 cents per pound.

At this point in the process, the brine contains the additional 1lb of water and 1362 additional BTUs. This pound must be vaporized by a heating process with 60% efficiency. Therefore, regeneration (returning the brine to its original concentration) will require 1683 BTUs (1020 x 1.65 at 150° F). If natural gas is used, then 1683 BTUs costs about $.017 with natural gas costing $1 per therm (100,000 BTUs). The refrigeration system must remove the 1362 BTUs from the brine at a cost of .23 kwh (.57 X 1284 / 3413 BTU) at $.10 per kwh or $.023 on a per hour time basis.

The refrigeration system will experience the entire $.066 (.66 kwh) for the coil system but only $.023 (.23 kwh) in the desiccant system. The remaining $.017 is purchased natural gas. If waste heat is available from any source including compressor oil cooler, engine block heat, process heat, etc., the cost advantage of a desiccant system goes from about 40% to 65%. Sufficient heat is available from a screw compressor oil cooler. If the system is properly designed, the regeneration cost would be zero.

If typical warehouse infiltration loads in a building are 15%, and 33% of that is latent, that 5% will be accomplished at about a 40% cost advantage with a liquid desiccant system when paying for regeneration heat. So about 2% of the cost disadvantage of a desiccant system versus a traditional fan coil air cooler is recovered.

If heat from the screw compressor oil cooler is used for regeneration, the cost advantage of 65% translates to a system cost efficiency improvement of over 3%. The assumption was that each evaporator was only defrosted when needed and exactly the correct amount of heat applied to it. The adverse effect of frost on the coils has been ignored. In practice, the coil system will be less efficient than this analysis assumes. Several conclusions emerge:

1. Frosted coils are very inefficient for moisture removal.
2. Liquid desiccants below freezing are much more efficient.
3. If regeneration heat is available from a waste heat source, such as a screw compressor oil cooler, liquid desiccant systems significantly outperform those with frosted coils.
4. Systems with frosted coils would benefit from a liquid desiccant removing the latent load to whatever extent possible, usually on the docks.
5. Using a liquid desiccant system as a primary refrigeration technique imposes an energy premium because of the additional fluid and heat exchange (refrigerant to brine) but the higher latent efficiency will at least partially offset that disadvantage.