# Why it is desirable to cool a compressor to improve its performance?

Could anybody explain why we should cool a compressor in order to improve its performance with a $p-v$ diagram? I know it should have something to do with the efficiency.

Another related question is that in a centrifugal compressor the working fluid (air) should be considered as incompressible or compressible fluid?

• Why don't you do a model calculations to quantify this. Suppose you are going to compress a gas 9x in volume. In case 1, you do it with one compressor, and, in case 2, you do it with two compressors, each with a compression ratio of 3x, and intercooling between the compressors. Determine how much work you do in each case, and how much heat you need to remove to cool the gas down to its initial temperature. Feb 13, 2018 at 23:47
• If air isn't a compressible fluid, how do you expect to compress it? Feb 21, 2018 at 3:11

In answer to your main question: 'Why it is desirable to cool a compressor to improve its performance?', I shall show the $Pv$ diagram and theoretical/mathematical formulation, and then explain the practical implications of not cooling maximally.

$\underline{Theoretically:}\quad$ In the $Pv$ diagram below, the work done on the air in a partly cooled compressor, is represented by the route a-1-2-b-a (Ref). The bigger the area enclosed in this curve, the more the work done on the gas. The equation for work done on the gas during compression is:

$$W = \int_{V_1}^{V_2}PdV\tag1$$ The mathematical formulation is outlined here. For isentropic compression:

$Pv^n=c\tag2$

where $n$ depends on amount of cooling, and $c$ is a constant, and can be determined from the Universal Gas Law: $$Pv = nRT\tag3$$ (see here for meaning of n, R and T). Continuing from equation $1$, using the 'Effect of Cooling During the Compression Process' formulation here: $$W = {{nR(T_2-T_1)}\over{n-1}}\tag4$$ $$W = {{nRT_1}\over{n-1}}\bigg[\Big({{P_2}\over{P_1}}\Big)^{({n-1})/{n}}-1\bigg]\tag5$$

A reduction in the exponent $n$ decreases the work done on the gas.

For maximum cooling, the process is isothermal, $n = 1$, and the work integral yields:

$$W = RTln\bigg({{P_2}\over{P_1}}\bigg)\tag6$$

With isothermal compression, the work done on the air is at its minimum given the start and end parameters. $\underline {Practically:}\quad$ this means:

For a dynamic displacement compressor (such as centrifugal and axial compressors), the compression functions approximately according to the theoretical: in the isentropic case, the air not cooled to the maximum possible will increase in temperature and occupy a relatively larger volume compared to the cooled case as pressure increases from $P_1$ to $P_2$.

For a positive displacement compressor (such as reciprocating, rotary screw, rotary vane compressors, which traps a fixed amount of air then compresses it), the end volume will be the same regardless of cooling, but without cooling, the gas pressure and temperature will be higher for the same volume throughout the compression stroke $(1-2)$ on the $Pv$ diagram). In this case, the figure can be interpreted as "if an amount of air were compressed to the same pressure, then if there were no cooling, it would occupy more volume at position $2$ on the diagram".

Without cooling, the gas heats up and therefore expands (see the laws that make up the Combined Gas Law), applying a higher back pressure and thus force on the compressor cylinder (or rotor, screw etc): meaning more work for the same amount of compressed air delivered say to a compressed reservoir downstream at a given temperature. For an isothermal compression, heat is removed so rapidly that the gas will not heat up as it is compressed, i.e. the gas will comply more with the compressing force, the back force on the compressor cylinder will be less, and less work will be done for the same amount of air delivered downstream at a given temperature.

Free energy? Not quite, the process benefits from the fact that there is usually low cost 'cool' water (say at least $10^{\circ}$C below the compressor operating temperature) available in most inhabited parts of the earth. All one has to do is make it exchange heat with the compressor chamber.

Pertaining to your second question, the working fluid is indeed considered as compressible, or the $Pv$ diagram and its analysis would look a lot different.

cooling the compressor coils reduces the cold sink temperature of the thermodynamic cycle that the system operates on. this increases the efficiency by increasing the difference between the temperature of the heat source and that of the cold sink.

in any compressor handling air as the working fluid, that air must be considered compressible- or else how could it be compressed?

• Thanks, @niels nielsen. What is your mean by "compressor coils"?
– jsxs
Feb 13, 2018 at 8:47
• connected to the output of the compressor pump you will find a length of thin-walled metal tubing which typically has a fan nearby to blow air over it, so as to carry away the heat from the substance inside. to save space, this tubing is usually bent into a coiled-up or serpentine shape right near the fan blades. these are the compressor coils. Feb 13, 2018 at 17:32

When you compress air it heats up. This increased temperature raises it's pressure, meaning that you have to work harder to compress it. And even if you don't get rid of the heat (to the extent possible) while compressing, the hot gas will cool off naturally once it gets into the holding tank.

So why not cool it as soon as possible to reduce its pressure and make it easier (and more efficient) to compress?

(And if you're about to complain that you don't want to cool the compressed air, because that reduces pressure, and pressure is what you're after in the first place, recall that, as I said above, it will cool eventually -- when it gets into the holding tank -- anyway. If you don't cool it as rapidly as possible then the pressure in the tank will fluctuate wildly as temperature goes up and down.)

• This answer conflates different reasons for compressing air. If the "pressure is what you're after" (e.g., to operate a pneumatic tool), then the compressor system (including its automatic control, and its regulator) will deliver constant pressure to the tool, and the tool will not care about the temperature. But sometimes, the pressure is not what your're after. Sometimes your goal is to squeeze more molecules of air into a fixed volume with a given pressure limit (e.g., filling a scuba tank), in which case, cooling the air is a vitally important part of the process. Mar 12, 2018 at 19:38
• @jameslarge - You didn't read the last paragraph, did you? Mar 12, 2018 at 20:05
• I did read it. That's where I got the quote, "pressure is what you're after." Mar 12, 2018 at 20:39