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Supercharger power requirement

  1. Apr 14, 2010 #1
    My apologies if this has been covered before: I didn't find it in a quick search, so here goes.

    What is the equation for the power required to drive a supercharger? For example, how much power is required to supply 700 cfm of STP air at one atmosphere of boost? Assume 100% efficiency for the moment.

    I'm designing a drive system and want to get in the right ball park for the torque required of the drive for various systems.

  2. jcsd
  3. Apr 14, 2010 #2
    My personal experience is that Engine Builders may give you a more accurate answer than Physicists to this type of question.

    What is a Supercharger? Perhaps an air compressor that actually compresses a fuel/air mix.

    I ride a Hayabusa motorcycle, many of my buddies have turbocharged theirs, in some cases supercharged.

    I would think power required to drive the charger varies with RPM. Type of fuel is also a consideration. Oxygenated fuels change things.

    My suggestion is to seek someone who has had success with your specific application and use their "real world" numbers.
  4. Apr 14, 2010 #3
    Thanks, but engine builders don't usually concern themselves with the power used by the process, just the net output.

    Yes, the power required varies with the rpm as well as the pressure increase and volume of air. The efficiency of the process also impacts the power required because of energy lost to heating the air and heat rejection through the body of the device.

    I'm just after an idea of the power requirement so that I can size the internals of the drive accordingly.

  5. Apr 14, 2010 #4
    The usual answer is none. Most superchargers/turbochargers are driven by engine exhaust. Think of the engine as purely a low-grade heat source. The turbocharger takes cool air in, pressurizes it, heating it in the process, then the engine heats it further (and adds more molecules).* The air then passes back through the turbocharger drive turbine and out the exhaust. You can add multiple stages and intercoolers to make the whole system more efficient, but it still comes back to using the already compressed exhaust to drive the supercharger.

    Now if you want to know how strong the shaft of the turbocharger has to be, you can work out the torque to drive the compression from adiabatic compression tables. I'd do it for you, but the number will be totally useless. The primary stress on the drive shaft is going to be the torque from an unbalanced system as the engine in the middle changes speed. If this is a stationary engine with a fixed speed, you still have startup and shutdown loads. In any case, what matters is the inertial mass of the turbine. That depends on the design parameters of the turbine, including diameter, blade materials, and RPM.

    The usual worst case design load is to assume that the turbine is operating at full speed, and full boost, and the engine seizes. The pressure the front-end is driving into starts increasing rapidly, and the power from the drive disk doesn't go away immediately. Vibration modes of the turbine as a whole now come into play.

    I don't know anyone who does this sort of work without a design package that can simulate the system. Of course, the only people I know--including me for a couple of years--who do this sort of thing are designing aircraft engines or combined cycle power plants. In either of those cases, blades are run within 50 degrees C of melting for efficiency reasons...

    * In general, hydrocarbons plus oxygen from air yields water and CO2. (Ignoring CO and NOx among other contaminants.) You get two molecules of water for each oxygen atom consumed, but only one atom of CO2 per oxygen molecule. If you assume your fuel is pure octane (it isn't) you get 2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O. Going from 27 to 34 molecules, but only using about 20% of the air results in about a 5 to 7% increase in molecules, or in pressure at constant volume and temperature, which you don't have. ;-)
  6. Apr 14, 2010 #5
    You've described the typical turbo-supercharger, which is of course exhaust driven. The other types are the positive displacement type (Roots, Lysholm, etc) and centrifugal impeller type (Paxton); both are shaft driven. I'm looking at the automotive sized applications, small potatoes compared to what you've been doing.

    Compressor efficiency numbers for each type are fairly easy to find but actual shaft torque numbers are hard to come by other than wild guesses for the Top Fuel blowers. For now, I'm considering only the load at full boost and rpm. Typically, blow-off valves are used to prevent the pressure spike when shifting, and in the event of an engine seizure, the owner will likely be more concerned with other things. :biggrin:

    For now, an example would help a lot so I can see which variables to consider and how each affects the power required, if you have one handy.
  7. Apr 15, 2010 #6
    How about a link to those compression tables?
  8. Apr 15, 2010 #7
    Typically, blow-off valves are used to prevent the pressure spike when shifting, and in the event of an engine seizure, the owner will likely be more concerned with other things.

    I guess automotive turbines tend to be 90 degree designs, where the blades are an integral part of a disk which has the same diameter. It is when you get into turbines with separate blades slotted into a disk that turbine failure moves to the top of the worry list. I assure you, lots of flying metal takes first priority. In the (US) Army this is stated as "Incoming fire has right of way." I've never been near a (large) turbine that failed, thank God, but I remember seeing pictures of a co-generation heat exchanger that got hit. Actually you could just see parts of the heat exchanger around the edges...and smears of copper from the fins of the heat exchanger smeared on the exhaust shroud behind it.

    As for charts and tools that may help you, try this site: http://www.couplingcorp.com/ They mostly provide couplings between (electric) motors and large compressors for refrigeration. But the tools do support internal combustion power sources.
  9. Apr 16, 2010 #8
    As you stated 100% efficiency, the compression is isentropic. You can basically lift the equations from something like a brayton cycle for work and power requirements.

    I can't give you the exact equations as I don't have my thermo book. They are very simple but there are quite a few that all look the same.

    Depending on whether you are compressing air or fuel air mix your polytropic contant will change.

    Air - 1.4
    Fuel air mix - 1.3
  10. Apr 16, 2010 #9
    Thanks, guys, it looks like I'll need to drag out my old TD books and have a go! I was hoping for the quick answer but I'll get more out of it this way.
  11. Apr 16, 2010 #10
    I've done a number of R&D compressor projects for race cars. currently I favor the Rotrex unit. I also have a variety of calculations to calculate the Hp requirement to power the charger. There is a lot of information needed to get fairly accurate numbers, if you've ever blueprinted an engine you'll have a fair idea of what is needed.

    this is of course in reference to extracting the most per the application. I'd suggest a book by Kenne Bell, good starting point and enough formula to get what you'll need. also do double check the math, some of it is not accurate. the thermal formulas from what I recall as being a bit suspect, but it's a good book.

    a mojority of mechanics are really techs with no engineering background or desire, it's R&R and next ticket. the 'old school mechanics are fading away as the need for a more engineered approach is too costly and not covered by the MFG underwriters. Then you'll have the group of 'bolt' on mech/techs where it's a shotgun approach to see what works, but once again the engineering behind it is not there.

    There is only a hand full of the crazy mad scientist types calculating every last envelope then fabricating it to life. It's fun, but it does not pay the bills.

    Back on point, the concern shouldn't be over the HP needed to drive the compressor but the efficiency of the unit to the power gains you're looking for. the type of drive unit can be a clutch vs direct pulley. twin setup, turbo/super etc.. then you'll have proper AFR management, cooling system, brakes. etc...

    you can bolt on a blower and increase the power easily, the trick is making reliable and safe.
  12. Apr 16, 2010 #11
    Thanks for the tip, I went to Kenne Bell's site and found what I needed. 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.
    Last edited: Apr 16, 2010
  13. Apr 17, 2010 #12

    jack action

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    I can't believe nobody gave you an answer faster.

    Power is work divided by time. The work done on a gas is the pressure times the the volume displaced. Hence work divided by time turns out to be the pressure times the volumetric flow:

    p = P * (V/t)

    Where, in your case, the pressure differential is the boost pressure and the volumetric flow is your cfm.

    The previous equation is for SI units. Plug the correct conversion factors for hp, psi and cfm and you you get:

    p = 0.00436 * P * (V/t)

    This is the power you actually need to compress the gas. Then you have to factor in the efficiency of your compressor.

    Here are some typical efficiencies for compressor used on car engine:

    http://auto.howstuffworks.com/supercharger4.htm": 69%
    centrifugal (reversible): 55%
    http://www.google.ca/imgres?imgurl=...age_result&resnum=6&ct=image&ved=0CCEQ9QEwBQ": 50%
    http://auto.howstuffworks.com/supercharger2.htm": 55%
    http://auto.howstuffworks.com/supercharger3.htm": 72%
    vane: 62%
    http://en.wikipedia.org/wiki/G-Lader" [Broken]: 68%
    http://imagineauto.wordpress.com/2007/09/25/pressure-wave-superchargers/": 84%
    Last edited by a moderator: May 4, 2017
  14. Apr 17, 2010 #13


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    Given dry air at 14.7 psia, 70 F inlet and 29.4 psia outlet, 100% isentropic efficiency, the theoretical discharge temp is 186 F with a power of 36.1 hp required.
    Last edited: Apr 18, 2010
  15. Apr 18, 2010 #14
    From experience it's a lot hotter than that, a quick run thru the calcs I have show a discharge 259.7*F, that makes the air density ratio .68, so to maintain that 2.99PR you'd need to compress at 3.44PR or 35.86 psi to achieve the 29.46 psi outlet pressure.

    Also superchargers are not 100% isentropic efficient as they require a gear ratio to increase the impeller speed thru mechanical means, they're not direct drive like a turbo which utilizes the exhaust gases, also the engine rpm is not in a steady state work load.
  16. Apr 18, 2010 #15
    It's not quite that simple. If your lucky enough sometimes the manufacturer will provide you with the power consumption data for the flow charts, however that's still not the most accurate. the efficiency of the compressor also needs to factored in to the power consumption as that is the volume of air the compressor can compress for that given area on the PR flow. those values change as the rpm changes. then you have the thermal effects to calculate as well.

    CFM is a known value from the flow charts but is still only half the eq. the CFM of the engine at each delta rpm needs to be calculated and worked into the numbers as well, for that you need extensive data on the engine.

    I'll post what I can when I get the opportunity, with the race season getting into swing again I'm getting busy with engine development and testing.
  17. Apr 18, 2010 #16
    I stayed on the conservative side for my numbers but thanks for confirming that I'm in the ballpark, guys.

    Madhatter, it looks like the PR got bumped a bit in the mix; the original intent was only 2, not 3 but interesting info nonetheless.

    I've been building race cars and engines (GT1, Trans Am, etc; road course mostly) for the last couple of decades but have only played with turbos a bit. I'm doing a couple of twin turbo projects, one for me and the other for a customer, so have been studying up. I've done my share of R&D and dyno testing on NA engines, so am quite familiar with the whys and wherefores before adding boost. It's my busy season too but I'll watch for any tips that you want to share.

    Thanks to all who replied! Don't stop yet!:smile:
    Last edited: Apr 18, 2010
  18. Apr 18, 2010 #17

    jack action

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    When I saw the equation you got from Kenne Bell' site, I thought it was enough of a check, but after reading the last comments, I've double check what I wrote and I was wrong. The equation I gave you is valid only for a constant pressure compression (Although it takes a lot more power than an isentropic one, so you are on the safe side).

    When I do the calculations, I obtain the same numbers than Q_Goest ... except for the power! (Can anyone check this?) The power that I found is based on:

    [tex]p=0.00436 \left(\frac{PR^{0.286}-1}{0.4}\right)\frac{P_o \ CFM}{\eta_s}[/tex]

    Which gives me 24,6 hp with a 100% isentropic efficiency ([tex]\eta_s[/tex]) and an inlet pressure ([tex]P_o[/tex]) of 14,7 psia (CFM is the volumetric flow at the inlet). All my estimated efficiencies for the different compressors would still be good, though.

    @ madhatter106:

    It is normal that the actual temperature is higher than the isentropic one: That's what isentropic efficiency means. If you have an actual temp of 259.7°F that's because you have an isentropic efficiency of 61.2%:

    [tex]\eta_s = \frac{T_s - T_o}{T_{act}- T_o}[/tex]

    (Of course, the temp are either in Rankine or Kelvin in this equation)

    Do not mix the isentropic efficiency with the mechanical efficiency. If you have a compressor with some gearing (centrifugal, roots, screw, etc), you will have a noticeable mechanical efficiency (probably around 90% - 95%) that you will have to multiply with the isentropic efficiency.

    Furthermore, nobody cares about the engine when looking for shaft power of a compressor. Yes, the isentropic efficiency will vary depending on inlet and outlet conditions of the compressor, but once you know the actual volumetric flow (actually the mass flow is usually better) the actual Pressure Ratio and the air temperature, you have everything you need to determine shaft power. In the problem posted here, the PR is a given,the CFM (@ STP) is a given, so all that needs to be estimated is the isentropic efficiency. If you have the http://www.turbobygarrett.com/turbobygarrett/tech_center/turbo_tech103.html" [Broken], then you are in business for more precision, but what is needed here is a ball park figure.
    Last edited by a moderator: May 4, 2017
  19. Apr 19, 2010 #18
    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.

    I did check my old thermodynamics texts to make sure I was close but I'm not too worried about exact numbers yet. I'm going to play with my engine simulator to see what the difference might be between a typical turbo exhaust with backpressure vs an optimized exhaust (belt-driven supercharger).

    I use Turbocalc to speed up my compressor map reading to get a reasonable compressor match.
    Last edited: Apr 19, 2010
  20. Apr 19, 2010 #19

    jack action

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    So I confirm my equation is good. If I want to relate my equation to yours, I have to do this transformation:

    [tex]p=0.00436 \frac{PR^{0.286}-1}{0.4 \left( PR-1 \right)}\frac{\Delta P \ CFM}{\eta_s}[/tex]


    [tex]PR-1 = \frac{\Delta P}{P_o}[/tex]

    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.

    That makes sense.

    But putting your value of 0.004 with PR = 2 would mean an efficiency of 59.7%.
    Last edited: Apr 19, 2010
  21. Apr 19, 2010 #20


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    Hi Jack,
    I wonder where you got your equation for power from. I went back and checked my calculator (a computer program set up for reciprocating compressors) and found I left ring frictional loads in. When I remove frictional loads I get 34.43 hp (down slightly from 36.1 hp). That program uses actual values for enthalpy since the first law can be reduced to:
    All I need to do is pull in properties for enthalpy given an isentropic pressure rise and power drops out. No need to look up other properties such as R or cp. It's a much cleaner way of determining power, but for most people that don't have a computerized database, it obviously won't work.

    Just to check it, I tried by hand using a more conventional equation:

    [tex]\dot{W} = \frac{\dot{m}C_pT_1}{\eta_c} \left[\left(\frac{P_2}{P_1}\right)^{(\frac{\gamma-1}{\gamma})}-1\right][/tex]

    The value I get using this equation is 34.47 hp, so it comes out to essentially the same number. Anyway, thought you might want to check your equation.
  22. Apr 19, 2010 #21

    jack action

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    The difference between your equation and mine is that you use the value of [tex]C_p[/tex] and I use the value of [tex]C_v[/tex]. I can't explain really the physical - "thermodynamic" - reason why I'm right (and my thermo book is not close so I do everything from memory), but here I go on how I find my equation and how it relates to yours:

    Assume a process where:


    The work is found by:

    [tex]W=\int PdV = \int^{V_2}_{V_1} \frac{constant}{V^k}dV = constant \frac{V^{1-k}}{1-k}\right]^{V_2}_{V_1}[/tex]


    [tex]constant = P_1 V_1^k[/tex]

    [tex]V_2 = V_1\left(\frac{P_1}{P_2}\right)^{\frac{1}{k}}=V_1 PR^{- \frac{1}{k}}[/tex]


    [tex]W= \frac{P_1 V_1^k}{1-k} \left( \left( V_1 PR^{- \frac{1}{k}} \right)^{1-k} - V_1^{1-k} \right)[/tex]

    [tex]W= - \frac{P_1 V_1}{k-1}\left( PR^{\frac{k-1}{k}} - 1 \right)[/tex]

    Replace [tex]V_1[/tex] with the volumetric flow rate, [tex]k[/tex] with [tex]\gamma[/tex] and you got my equation. The minus sign only states negative work is compression, positive is an expansion.

    The fun thing with this equation is that not only it can represents an isentropic process but if [tex]k[/tex] = [tex]\infty[/tex], you have an isochoric process, if [tex]k[/tex] = 1, then it is an isothermal process and if [tex]k[/tex] = 0, it will be an isobaric process.

    But how does it relates to your equation? Well:

    [tex]P_1 V_1 = mRT_1[/tex]


    [tex]W= - \frac{ mRT_1}{\gamma-1}\left( PR^{\frac{\gamma-1}{\gamma}} - 1 \right)[/tex]


    [tex]C_v = \frac{R}{\gamma-1}[/tex]


    [tex]W= - m C_v T_1 \left( PR^{\frac{\gamma-1}{\gamma}} - 1 \right)[/tex]

    If I was wrong and you were right, that would mean that my equation would be multiply by a factor [tex]\gamma[/tex] since:

    [tex]C_p = \frac{\gamma R}{\gamma-1}[/tex]

    And if I put the numbers given by mender in this new equation, then the efficiencies would be all multiplied by 1.4 giving us something in the order of 85% which is pretty high.
  23. Apr 19, 2010 #22
    good info.

    Here's a quick and simple x-cell sheet I threw togther a few yrs back and cleaned up for some clarity today. it's just a basic pressure to hp est. you'll need to know your Ve of the engine your working on for close numbers. If you have head flow data you'll be close, or if your using a good sim that has all that input then in the dyno sim run you should be able to pull Ve, temp & pressures from the data.

    2 good sims I've used, dynomation-5 and Engine Analyzer Prov3.5.

    btw, the spreed sheet is just the bell eq. lay'd out for easier input.

    As to the OP and topic about the Hp needed to operate the compressor, I wonder how accurate it is without actually having the unit tested. We can calculate the force needed to compress the air, but chargers are not direct shaft from input, they are gear multiplied to get the impeller up to the speeds needed, each one is different. The Rotrex I spec now is a 'traction' drive, it multiplies the input rev by 7.5 for a max impeller rpm of 90k. In order to figure the actual Hp needed I would have to work from the impeller back to the input shaft and increase the torque by the ratio, neglecting frictional and heat losses. and to avoid impeller damage the units are always run with a filter which restricts the input and can severely hinder it if not sized correctly. I've seen losses of 10psi from poor filter choices. the intake tube dia and route also need to be calculated.

    then you also have belt load and friction, crank to SC pulley input ratio. the belt slip from improper tension will show up on the bar gauge as the bar jumps around. * FYI* DO NOT use a fixed tensioner, a dynamic spring one allows for the inertia of the SC to re-sync with the engine at shift points.

    General info for those curios,
    Until WOT or wide open throttle the Pr of an NA engine is less than 1, even at WOT it's not 1 as the restriction and Ve losses prevent this, hence the 'vacuum' or low pressure in the manifold. adding in a compressor this needs to be accounted for as pressure and temps rise faster due to the air 'stacking' faster than it's used, this of course is applicable to roots and turbos. with SC's it's not as noticeable due to a linear rpm input.

    All-in all the Hp needed is not a major consideration you'll always get more than what you started with esp with the technology today with the new turbos and SC's. as long as you make sure the engine and compressor are sized right to each other and that the rpm spread is broad enough to avoid on/off of the power. that requires having a very detailed 'blueprint' of the engine, most good tuners do this. and also camdoctor files or the equivalent.

    So based on the above the internal ratio in this case 1:7.5 would be that for every 1n applied on the impeller it would need 7.5n input to do that work. being that the SC sits in the engine and will raise to a temp of 160*F on average that needs to be accounted for in the eq as well. I use temp sensors to get finer resolution for the fuel requirements. the air density in engine applications is constantly changing, this makes it harder to say exactly how much Hp at a given rpm it will need or make.

    It's one of those things of where to draw the line, how far to go with the equations, interplay etc... the 'cascade' effect is very noticeable in this field and I find that is what makes it fun, looking at the larger picture and working with the changes to 'engineer' the most optimum result then constantly 'tweaking' it to get it one step further.... down the rabbit hole we go...

    Attached Files:

  24. Apr 19, 2010 #23

    jack action

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    This is exactly how we find the efficiency of a compressor. We test the unit, measure the mass flow rate, the pressure and the temperature at both the inlet and the oulet. Then we calculate the theorical work and outlet temperature and we find the efficiency like I wrote in a previous post. That's why the equation work as well to determine the power needed (as long as somebody has tested a similar unit before).

    Just like the power that goes from the engine to the wheel, compressor power is constant. No matter how you modify the rpm of the input shaft. Only the torque will vary accordingly [power = (torque X RPM) / 5252].

    Again, this is all part of the mechanical efficiency and it is fairly easy to estimate.

    Restrictions will lower your CFM, hence it will lower your compressor power. If you know your actual CFM (we don't care if it could be more if unrestricted), you know the compressor power. Again fairly easy to estimate.

    Air is 'stacking up', no matter the type of compressor you are using; that's what a compressor do. The only way you can increase the pressure in the intake system is if the compressor is 'pushing' more air than the engine is 'pulling'. Take the volume/rev of a blower times its rpm vs the volume/rev of the engine times its rpm and the first one is alway larger than the other. Increase the pulley ratio, you increase the rpm of the blower vs the one of the engine, increasing the amount of air 'pushed' by the blower, increasing the air 'stacking up', hence you increase the intake air pressure. The difference with a supercharger is that at low rpm it is extremely inefficient at 'pushing air'. Contrary to positive displacement compressor, it uses the mass flow rate energy (½[tex]\rho[/tex]v²) to increase the pressure.

    It is extremely important. First, as the OP mentioned, just to know how to size your input components, it is important.

    But, performance wise, you want to achieve the 100% isentropic efficiency. Just take the example on this thread. It requires 24,6 hp from your engine to compress the air with 100 % efficiency. Doing it with a blower (55% eff) means 44.7 hp. Doing it with a screw compressor (72% eff) means 34.2 hp. That is a 10 hp gain just by selecting the proper compressor.

    In addition, these losses are in form of heat, which means a higher outlet temperature. High temperature for the intake air is never good news as it reduces the mass flow rate of the engine (more hp loss) and raises detonation problems (which means lower CR, hence more hp loss).
  25. Apr 19, 2010 #24
    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.

    if torque varies so does power.

    I didn't say it was impossible but that it needed to be factored in.

    Agreed, and I didn't go into detail as you did. I'm very aware of the cfm/rpm and the mass density charge as it's very critical to fuel mass requirements and the time in which it can be delivered into the time sequence available. many, many factors that would take a book to fill, and there are many books for that thankfully.

    Yes it's a factor but from my experience not a major one as the power required is going to be what it is after all the other factors are accounted for and plotted.

    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.

    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.
  26. Apr 19, 2010 #25


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    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.

    See also:
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