# Typical power consumption of large compressor

This one should be easy to answer by looking at the thermodynamics, but I find it not to be. Say we have a natural gas stream, ~700 N∙m3/h, that we want to compress from 16 bar to 250 bar. What will the power consumption of the compressor be?

The thing that stymies me is that the power uptake is not path independent, but mostly depends on how many cooling steps you have in between. Or so it seems to me. You don't need to do the math for me, I just provided the numbers as an example. I would appreciate a detailed explanation of how to arrive at an approximate real world value.

• How do your typical real world results compare with Pressure_in x volume _per_time x ln*(Pout/Pin) ? | You mention coolling steps. Do you subtract recovered heat energy from work done when calculating power required? (ie you may not want or use the heat but you ARE "producing" it). | [* lower case l may bot be clear - that's "ln" aka LN as in natural logarithm]. – Russell McMahon Mar 30 '15 at 7:07
• I arrive at 323kW from your calculation but as of now I don't have real world values to compare to. What you gave looks like isothermal compression? – mart Mar 30 '15 at 7:25
• @RussellMcMahon's point is that the compression stage(s) must be considered to be adiabatic (no significant heat flow into or out of the gas), even if you subsequently remove the heat on the output side. Cooling increases the density of the gas, but the mass flow rate remains the same. – Dave Tweed Mar 30 '15 at 12:23
• This calculator gives 114 kW. Here's a calculator with more knobs and dials than I can handle. Looks good :-). And this page from Petroskills looks very useful. – Russell McMahon Mar 30 '15 at 13:20

The work is surely path dependent; you can estimate a fast passable value in two way. Isothermal, as in the case when several stages with intercoolers are used

(SI Units) KWiso = 2.78*10^-4 * (VolFlowRate) * P1 * LN( P2/P1 )

And Adiabatic with costant ratio of specific heats (k)

(SI Units) KWad = 2.78*10^-4 * k /( k-1 ) * (VolFlowRate) * P1 * [(P2 / P1)^(k/(k-1)) -1]

Dividing this ideal one for an Efficiency value that is 75-87 % for centrifugal and 80-91 % for Axial. You can also improve the value using polytropic (n) stage efficiency

PolyEff = [(k-1)/k] / [(n-1)/n]

If you want a more accurate interstage cooling calculation consider every stage as a different compressor and get a big balance.

Reference: Perry 's Chemicals Engineer Handbook Section 10-45 • Can you explain the 2.78x$10^{-4}$, where it cxomes from? I don't understand the part about polytropic stage efficiency, can you elaborate? Finally, this is the kind of answer I wanted, thanks! – mart Mar 31 '15 at 5:35
• Oh sorry, the book use (VolFlowRate) of m3/h so the 1/3600 for the seconds is exactly 2.78*10^-4 =). Polytropic stage efficency, which is often considered as the small stage efficency, or the idraulic efficency, consider the continuum variation of the pressure ratio on the specific polytropic path; in fact you remove the k costant hyp. Maybe this picture should be usefull. – LinearLambda Mar 31 '15 at 8:42

Maybe I didn't get the problem it sounds a little bit simple, but as I see you would have to calculate using a PVT correlation (Peng-Robinson for exemple) the power your transformation requires in a ideal system (ΔS=0, the power required for this transformation is path independent) and then you can divide the value you got by the efficiency of your compressor, this will change with your equipment, but 0,8 is a good value, having more cooling stages will give you more efficiency, with less colling stages you will have a lower value for efficiency.

You can use the following equations with a PVT correlation to calculate the power required by the transformation:

dU=CvdT+[T*(∂P/∂T)−P]dV

dS=(Cv/T)dT+(∂P/∂T)dV

dH=dU+d(PV)

Cv=Cp-R