# Supercharger power requirement

jack action
Gold Member
Hi Jack,

If you multiply your answer (24.6 hp) times the ratio of specific heats you do indeed get 34.4 hp (the answer I gave).

Let's consider a cylinder with a piston that's compressing air. We put a closed control volume around the air in the cylinder and work is done on the air, compressing it. How does this process compare to an open control volume in which you have air entering at some mass flow rate and then exiting the control volume at higher pressure and temperature due to work done on it?

The first law for the closed control volume gives:
dU = W

The first law for the open control volume gives:
dU = W + dH
where dU = 0 since there is no air stored inside the control volume. Thus:
dH = W

You calculated dU = W. I calculated dH = W. Which is correct? For a compressor, you have air in and air out (enthalpy in and enthalpy out). If you only do an analysis on the closed control volume of air being compressed, you don't get the entire picture. There is additional work being done by the compressor to force the air out of the cylinder at the higher pressure. The work done by a compressor is dH = W. The correct answer looks at the open control volume with a mass flow in, mass flow out and work added.

http://www.grc.nasa.gov/WWW/K-12/airplane/compth.html
http://www.codecogs.com/reference/engineering/thermodynamics/air_compressors.php
You are correct. I've been looking deeper to see where I was wrong. I knew I saw what I demonstrated somewhere (I told you I've been doing this from memory) and found http://en.wikipedia.org/wiki/Adiabatic_process" [Broken] which gives the same answer as me. But they say it's the work done by the system.

Then I found http://en.wikipedia.org/wiki/Enthalpy#Open_systems", which essentially say the same thing you say, which is you have to add the shaft work. So dW = dH in a adiabatic process.

And then, from http://en.wikipedia.org/wiki/Isentropic_process#Derivation_of_the_isentropic_relations", dH = Vdp in an isentropic process and by re-doing the whole derivation I did earlier it gives:

$$W= \frac{k}{k-1}P_1 V_1 \left( PR^{\frac{k-1}{k}} - 1 \right)$$

Hence, the same one you gave, i.e. with Cp. So my initial equation (and hopefully the final one!) becomes:

$$p=0.00436 \frac{PR^{0.286}-1}{0.286 \left( PR-1 \right)}\frac{\Delta P \ CFM}{\eta_s}$$

Where the units are psi, cfm and hp.

Thanks, that's why I like this forum.

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jack action
Gold Member
I understand the efficiency islands of the flow maps, however that is only in reference to the rpm at which your plotted to it. no two engines will be the same due to the Ve of them and the specifications. The power needed to operate the compressor is not included directly in any of the maps, should be simple enough when testing them for flow based on amp load to drive the unit, however the pulley ratio the end user utilizes would need to be factored to get the final per app number. there again is the next point, each and every flow map does not include the data for what size pulley was used as it was most likely a direct drive test. change the available options for pulley size and the power/rpm points move about to get to where you want, this in turn effects the Hp required.

I don't think you understand the efficiency islands of the flow maps. Look carefully at a http://www.turbobygarrett.com/turbobygarrett/tech_center/turbo_tech103.html" [Broken]: for a given mass flow rate and a given pressure ratio you have a given compressor rpm ("speed lines"). It doesn't matter what type of engine it is connected to. That is what we are doing when matching a compressor to an engine: we put the appropriate pulley ratio to match the compressor rpm to the one of the engine such that the compressor gives the pressure ratio we want with the mass flow rate we want at the engine rpm we want.

But no matter what will be the engine rpm you select, the compressor power will still be the same, if the PR and mass flow rate of the compressor are the same. And, at those specified values, we will be stuck with an isentropic efficiency which results on the compressor design.

if torque varies so does power.
No, the power is constant. If the torque varies, so does the rpm (One goes up, the other goes down, just like an engine that produces 300lb.ft at 5000rpm is connected to the driven wheels at 833 rpm with 1800lb.ft of torque, which is 286hp for both).

yep, thing is you'll never get 100%. even with the 85% of some of the new units it's still only for a fraction of the overall rpm range. it may be a 10hp gain but that's if you can plot the map exactly that same for each choice, you can't. each type of compressor operates differently and although the lysholm blower is not as efficient as some centrifugal or vice versa the increase in CFM per rpm point is significantly different so the less efficient positive displacement blower will deliver more CFM sooner and comparatively then the 'loss' of Hp isn't the same.
Again, I don't think you get the isentropic efficiency concept and what a compressor does. Compressor don't have to be used with an engine. Actually, the first use for blower was for http://www.cementsilos.com/Applications/blower.htm" into silos (still used today). We can still measure the power we need to actuate the pump even if the blower is not connected to an engine. All we need to know is the mass flow rate, the PR and the efficiency of the compressor. And this will be constant for a given compressor rpm and for a given compressor design.

Efficiency doesn't tell you how much boost or how much CFM you can produce, but how much power you need to make that boost and that CFM.

not all losses are heat, the centrifugal at low rpm is not increasing air temp as that's a function of pressure, it simply not compressing the same volume due to low impeller speed so at low rpm the amount of power needed to operate the compressor drops. whereas the roots and lysholm require more however they also compressor more air, as the rpms increase the positive (fixed) displacement falls to the centrifugal in efficiency.
If the centrifugal isn't increasing pressure at low rpm, it also doesn't require power. But the roots and lysholm do increase pressure even at low rpm, so they do require power. That is the concept we're describing here: Compressing power varies with pressure boost and mass flow rate!

Again, at which rpm a particular compressor does its compression is irrelevant. If a centrifugal compressor delivers 700 cfm of air at 10 psi of boost @ 50 000 rpm with an efficiency of 70%, it will take the exact same amount of power that a lysholm compressor that delivers 700 cfm of air at 10 psi of boost @ 3 000 rpm does, if its efficiency is also 70%. Nobody cares about the engine it is connected to, they will still take the exact same amount of power (which, in this case, is 35,9 hp). The compressor rpm and efficiency are only design dependent. And if you can find another compressor that will do the same CFM and the same pressure boost but with an efficiency of 80%, well you've just gain 4,5 hp on your engine output.

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Replacing the data you just gave us, it would mean that the efficiencies are: 59.0% @ 5.8 psi, 66.4% @ 8.8 psi and 65.8% @ 11.8 psi.
The measured test data are as follows:

5.8 psi, 592 cfm: 23 hp
8.8 psi, 585 cfm: 29 hp
11.8 psi, 568 cfm: 36 hp

From that, I assumed that the compressor efficiency was 70%.

If the trend of improving efficiency indicated by the higher boost numbers continues, 14.7 psi should be about .0035*cfm*psi, resulting in 36 hp for 700 cfm.

It occurs to me that I'm still missing part of the equation, that of the ratio of specific heat (and very likely more than just that) but this seems to work well enough for now. I appreciate the discussion!

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jack action
Gold Member
The measured test data are as follows:

5.8 psi, 592 cfm: 23 hp
8.8 psi, 585 cfm: 29 hp
11.8 psi, 568 cfm: 36 hp

From that, I assumed that the compressor efficiency was 70%.

If the trend of improving efficiency indicated by the higher boost numbers continues, 14.7 psi should be about .0035*cfm*psi, resulting in 36 hp for 700 cfm.

It occurs to me that I'm still missing part of the equation, that of the ratio of specific heat (and very likely more than just that) but this seems to work well enough for now. I appreciate the discussion!
If I put those in my last equation, solving for $$\eta_s$$:

5.8 psi, 592 cfm, 23 hp: 57.6%
8.8 psi, 585 cfm, 29 hp: 64.9%
11.8 psi, 568 cfm, 36 hp: 64,9%

These numbers seem pretty realistic. I think even if the boost increases, you won't get more than 65%. It probably will stabilized at that level. It will probably even drop at a PR too high. Furthermore, you estimate that at a higher boost (14.7 psi) and a higher flow rate (700 cfm) you would still take 36 hp? (that would mean an 95.5% eff.) The power has to increase with those values, for sure.

So, if we assume an efficiency of 65% in my last equation:

14.7 psi, 700 cfm, 65%: 53 hp

That will be the expected power from the compressor shaft of that particular compressor.

About those numbers you mentioned earlier:

The equation I posted dropped out of the charts showing the hp required for various levels of boost. I chose to use .004 as my constant but the actual numbers were .00467 @ 5.8 psi boost, .00394 @ 8.8 psi and .00379 @ 11.8 psi boost.
They don't related to the last one you've posted:

0.00467 * 5.8 * 592 = 16 hp (82.6% eff)
0.00394 * 8.8 * 585 = 20 hp (92.7% eff)
0.00379 * 11.8 * 568 = 25 hp (91.9% eff)

Is this for another compressor? The efficiencies look pretty high with the correct equation.

Jack Action, thank you for the explanations. I have a neurological disorder that makes it difficult for me to write or type my thoughts out efficiently or clearly, there's a time delay of sorts making it frustrating. I appreciate your time on this.

I see the isentropic efficiency, but the centrifugal compressor is dynamic and not the same as the positive or fixed compressor, and also in the environment of the engine bay it becomes a polytropic process. there is a variation to the density of the air due to the fluctuations in heat.

normalizing the efficiency for the density and volume between compressors may be the same for that point but as you point out it will vary on the rpm. however since the gas is not static in density and the thermal transfer of heat the density and volume changes for the same given rpm point, and as this fluctuates along with the density of the air being compressed so does the power requirement for it, so the maps are only relevant for the air temp stated on them.

exhaust turbo compressors tend to have larger efficiency islands then the centrifugal compressors. It's possible to move from one to another effecting the efficiency and thus the power requirement for a given CFM of air. I've seen this on the dyno, thermal transfer becomes a real problem.

My apologies, as usual I'm skipping steps as I type. Since I only have the three test samples, I'm not too worried about exact numbers but I should be a little more careful when I post!

So, if we assume an efficiency of 65% in my last equation:

14.7 psi, 700 cfm, 65%: 53 hp

That will be the expected power from the compressor shaft of that particular compressor.
Which meshes well with my understanding.
It works out to about (0.004 hp/cfm*psi boost)/compressor efficiency, so the answer to my original question is 41.2 hp for 700 cfm at 14.7 psi and 100% efficiency. That's about 58.8 hp for a compressor that is about 70% efficient.
Adjusting my constant as mentioned earlier results in the lower 36 hp calculated requirement.

The 36 hp is a "pure" number, before the drive and compressor efficiencies are accounted for. Given a drive efficiency of 96% for each of two steps (pulley/belt, internal gears) and a sweet spot in the compressor map of 75%, the required hp would be 52.1 hp. Using your number of 65% with my equation results in 55.4 hp.

For me, that's close enough. With what has been covered so far, I can plot the trends which is what I was after.

0.00467 * 5.8 * 592 = 16 hp (82.6% eff)
0.00394 * 8.8 * 585 = 20 hp (92.7% eff)
0.00379 * 11.8 * 568 = 25 hp (91.9% eff)

Is this for another compressor? The efficiencies look pretty high with the correct equation.
Same compressor but I assumed an efficiency of 70% and took that out so I could compare other methods of compression and drives. Putting that back in yields 23 hp, 29 hp, and 36 hp respectively. Hope that made sense; my shortcuts don't always to others!

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