Comparing Work Required At Compressing Gas & Pumping Liquid

The work neded from any fluid mover is the difference between the finish conditions and the start conditions, multiplied by the flowrate.

• In the case of a gas, the conditions can be evaluated by comparing the change in enthalpy. The pressure and temperature of the gas before and after the compressor can be compared on a suitable chart, or looked up in tables. Multiply by the mass flowrate.
• In the case of a liquid, the work required is the change in pressure multiplied by the volumetric flowrate.

In either case, the electrical input is the calculated figure divided by the efficiency of the equipment.

The electrical load for an absorption chiller is VERY low- the big energy "hog" is the thermal input required to re-generate the liquid desiccant.

Absorption chillers "cool" the water much like cooling towers- evaporating water sprayed on the "chilled water" coils which extracts heat from the chilled water to vaporize the water which is under a very high vacuum (so the water will "boil"- evaporate) and the gaseous water vapor is absorbed by the concentrated desiccant being sprayed in a coarse mist near the water spray on the tubes. The "cool" concentrated desiccant becomes slightly warmer saturated desiccant, which is then heated enough so that the water in it "boils" again, and then is condensed by some of the cooling water. The cooling water also cools the heated desiccant (now concentrated again) so it can do its thing all over again.

Using "English" terms- a high efficiency electric chiller can provide cooling at about 0.6 kW per ton- meaning that the cooling water needs to remove about 0.6 x 3412 PLUS 12,000 BTU's (or 14,500 BTUs) every hour per ton provided. A high-efficiency absorber uses about 9,400 BTUs of thermal input (and about 0.05 kW of power) for every ton, so the cooling water needs to remove 9,400 + 1,700 +12,000 BTUs (or 23,100 BTUs for every ton. That also means that the cooling pump will work about twice as hard, and the tower will have about twice the load (tower fan) to operate- which will use more electricity. NET power for an electric machine is about 0.75 kW per ton and for an absorption chiller is about 0.30 kW per ton- plus the thermal input.

Obviously, IF the thermal input energy (high grade waste heat) is FREE, the absorber is definitely the way to go- AND its "refrigerant"- water- has a zero GHG impact- always better than ANY refrigerant.

IF the thermal is coming from a "purchased" source, it is worth about 2.75 kW per ton so- the overall energy load is about 3.05 kW per ton- about 4 times more than an electric machine.

Hope this helps your decision process.

This is an interesting question!

Leaving aside the details of refrigeration, and looking at compression of gases and liquids, for one-off compression in a closed cylinder, gas compression (isothermal, adiabatic or somewhere in between) needs more work, obvious because as liquid is practically non-compressible the piston movement → 0. But in continuous compression the liquid case uses more power.

For one-off compression it is non-flow work and

W = 1/(k - 1)*P₁*V₁*((P₂/P₁)^(k-1)/k - 1) It's convenient to define n = (k - 1)/k.

Isothermal compression k = 1, n = 0. Can show that W = (P₁*V₁)*ln(P₂/P₁)

Polytropic compression (somewhere between isothermal and adiabatic) usual figures are k = 1.3, n = 0.23.

Liquid compression k → ∞, n → 1 and W → 0 as 1/(k -1) → 0.

For continuous compression, flow work = k*non-flow work.

W = k/(k - 1)*P₁*V₁*((P₂/P₁)^(k-1)/k - 1) = (P₁*V₁)/n*((P₂/P₁)ⁿ - 1)

Isothermal W = (P₁*V₁)*ln(P₂/P₁) as for non-flow (as k = 1)

Liquid, k/(k - 1) → 1 as k → ∞, so W = P₁*V₁*(P₂/P₁ - 1) = V₁*(P₂ - P₁) = V₁*ΔP as expected for liquid flow. You can check power is greater than for gas compression.

This was discussed a while back under thread below.

http://cr4.globalspec.com/thread/99106/Compressor-Work-in-Brayton-Cycle