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Physics of superchargers/compressors

  1. Dec 12, 2003 #1
    original post was cut off half way through, so I'm re-posting:

    What I'm really trying to understand has to do with superchargers, and a phenomenon I (and others) observe at the track every season: more boost in colder weather.

    The common explanation for this is that air is denser in the cold weather, resulting in more boost. Another variation is that the (adiabatic) compressibility is higher for the colder air, which sort of reduces to the same argument.

    Now my thermo is a little rusty, but if it's that simple, then I would expect that this effect can be shown with a simple adiabatic compression model on an ideal gas (approximating air as an ideal gas). Is there a simple expression showing that if you do some amount of work W adiabatically (lets say a piston/cylinder configuration) that the final P reached will be higher when the initial air temperature is lower (or initial air density is higher, assuming density is proportional to P/T)?

    I can't find such an expression in any of my thermo books. And when I try to derive an expression, both initial density and initial temperature eventually cancel out of the equation, and P_final becomes only a function of W and the specific heat. Since the process is adiabatic, I've been using [tex]PV^{\gamma} [/tex] = constant and taking dT = W/Mc (M = density * volume, c = specific heat, ignoring temperature dependance of c).

    Another explanation is that the above effect isn't a "density thing", but rather a result of better efficiency in the colder weather.

    I'm hardly an expert on compressor physics. If anyone can shed some light on this, I'd really appreciate it. Sorry for the long post.
     
    Last edited: Dec 12, 2003
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  3. Dec 12, 2003 #2

    chroot

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    Why don't you think that's the correct explanation?

    - Warren
     
  4. Dec 12, 2003 #3

    turin

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    Why is this restricted to superchargers?

    Have you ever heard of a cold air intake?

    Usually, the lower the temp. the higher the density (this is true for the air in normal temp. ranges, not sure about gasoline vapor, though, but I imagine so). This is the basic principle of the supercharger: to give the system more of one of the reactants, more air. But the supercharger, as I understand it, doesn't give the system more fuel, so you run the risk of a lean mixture if the temp. goes too low.

    But I'm sure there is some factor contributed from the fact that you've lowered the temperature of the exhaust reservoir, which should theoretically improve effeciency, and thus give you more bang for your buck.
     
  5. Dec 12, 2003 #4

    chroot

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    Re: Re: Physics of superchargers/compressors

    There is an oxygen sensor in the exhaust stream that tells the engine computer to add more fuel via the fuel injectors to compensate for the extra oxygen added by a supercharger. Unless you exceed the fuel flow rate of your injectors, you'll be fine -- and most cars can handle at least a few pounds of boost with no fuel system modifications. Of course, if your car originally has weeny injectors, you'll have to upgrade them with higher-flow units to avoid damaging the engine.

    - Warren
     
  6. Dec 12, 2003 #5
    Sorry, for some reason my original post was cut off in the middle. Here it is in its entirety:
    ---------------------------

    What I'm really trying to understand has to do with superchargers, and a phenomenon I (and others) observe at the track every season: more boost in colder weather.

    The common explanation for this is that air is denser in the cold weather, resulting in more boost. Another variation is that the (adiabatic) compressibility is higher for the colder air, which sort of reduces to the same argument.

    Now my thermo is a little rusty, but if it's that simple, then I would expect that this effect can be shown with a simple adiabatic compression model on an ideal gas (approximating air as an ideal gas). Is there a simple expression showing that if you do some amount of work W adiabatically (lets say a piston/cylinder configuration) that the final P reached will be higher when the initial air temperature is lower (or initial air density is higher, assuming density is proportional to P/T)?

    I can't find such an expression in any of my thermo books. And when I try to derive an expression, both initial density and initial temperature eventually cancel out of the equation, and P_final becomes only a function of W and the specific heat. Since the process is adiabatic, I've been using [tex]PV^{\gamma} [/tex] = constant and taking dT = W/Mc (M = density * volume, c = specific heat, ignoring temperature dependance of c).

    Another explanation is that the above effect isn't a "density thing", but rather a result of better efficiency in the colder weather.

    I'm hardly an expert on compressor physics. If anyone can shed some light on this, I'd really appreciate it. Sorry for the long post.
     
    Last edited: Dec 12, 2003
  7. Dec 12, 2003 #6
    Hmm...the term 'supercharger' is something that was simply called a compressor in turbines in my degree. After some Googling, I think this is what you are talking about (correct me if I am wrong).

    In which case...your point of reference should be the standard Joule-Brayton cycle.

    A definition of subscripts that will be used follows:
    • Subscript 1 = Thermodynamic properties of working fluid before compressor
    • Subscript 2 = Thermodynamic properties of working fluid after compressor
    • Subscript 3 = Thermodynamic properties of working fluid after heat input
    • Subscript 4 = Thermodynamic properties of working fluid after turbine

    Some keys:
    [tex]\eta[/tex] - denotes efficiency
    Cp - denotes heat capacity at constant pressure
    Qh or Ql - denotes thermal energy in (high/low temp. respectively)
    Tx - denotes total temperature at position x.
    Px - denotes total pressure at position x.

    From the First Law:
    [tex]\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{C_p(T_4 - T_1)}{C_p(T_3 - T_2)} = 1 - \frac{T1(\frac{T_4}{T_1} - 1)}{T_2(\frac{T_3}{T_2} - 1)}[/tex]

    Also, P3 = P2 (isobaric heat input) and P4 = P1 (same end and initial states assumed).

    Therefore:
    [tex]\frac{P_3}{P_4} = \frac{P_2}{P_1}[/tex]

    [tex]\frac{P_2}{P_1} = {(\frac{T_2}{T_1})}^{\frac{k}{k-1}} = \frac{P_3}{P_4} = {(\frac{T_3}{T_4})}^{\frac{k}{k-1}}[/tex]

    [tex]{(\frac{T_2}{T_1})}^{\frac{k}{k-1}} = {(\frac{T_3}{T_4})}^{\frac{k}{k-1}}[/tex]

    Thus:
    [tex]{(\frac{T_2}{T_1})} = {(\frac{T_3}{T_4})}[/tex]

    And rearranging:
    [tex]{(\frac{T_3}{T_2})} = {(\frac{T_4}{T_1})}[/tex]

    [tex]{\frac{T_3}{T_2} - 1} = {\frac{T_4}{T_1} - 1}[/tex]


    Finally:
    [tex]\eta = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{(\frac{P_2}{P_1})^\frac{k-1}{k}}[/tex]

    Note the efficiency increases as the pressure ratio across the compressor increases. That is why a supercharger is good for efficiency. Hope this is the 'simple' expression you wanted :wink:
     
  8. Dec 12, 2003 #7
    Thanks for trying Tyro, but not exactly what I'm looking for. Maybe we can reduce my question to this simple problem:

    Suppose we have a cylinder of air (at atmospheric pressure), and we quickly press down on the air with a piston, applying some force F. We do this fast enough, so the air temperature quickly rises and no heat escape (i.e., an adiabatic process). Suppose we have a small instrument inside the cylinder that measures the pressure of the compressed air inside without affecting it. My question is:

    Will we see a difference in the measured (compressed) air pressure if we conduct this experiment in cooler (denser) air versus hot (less dense) air? Is there an equation which predicts how much this difference will be? I came up with a similar result to yours:

    [tex]\frac{P_2}{P_1} = \left[ \frac{T_2}{T_1} \right]^{\frac{\gamma}{\gamma-1}} [/tex],

    where [tex]P_{1(2)}[/tex] is the initial (compressed) air pressure, and [tex]T_{1(2)}[/tex] are the corresponding temperatures. What I'm really concerned about is [tex] P_2 [/tex].


    Now, [tex] T_2 [/tex] needs to be solved for. Now, because it's an adiabatic process, the work done by the applied force, [tex]W = \Delta U = M c \Delta T [/tex], where [tex]M [/tex] is the mass of the air (= constant throughout compression process), and [tex] c [/tex] is the specific heat. Solving, we have:

    [tex] T_2 = T_1 + \frac {W}{\rho_1 V_1 c} [/tex]

    where [tex] \rho_1 [/tex] is the initial air density and [tex] V_1 [/tex] is the initial volume of air. Putting this into the first equation, we have:

    [tex]\frac{P_2}{P_1} = \left[ \frac{T_1 + \frac {W}{\rho_1 V_1 c}}{T_1} \right]^{\frac{\gamma}{\gamma-1}} = \left[ 1 + \frac {W}{\rho_1 V_1 c T_1}\right]^{\frac{\gamma}{\gamma-1}}[/tex].

    Here is where I get stuck. I need to show that the rhs will be larger in colder air. Now if W is fixed, we can expect [tex]\Delta T [/tex] will be lower in the colder air (because [tex] \rho_1 [/tex] is larger, i.e., there's more thermal mass in the cylinder). This is going in the wrong direction, although its balanced by the [tex] T_1 [/tex] in the denominator. Anyway, lets just say the force applied to the piston is fixed, as this is more realistic. Now, in colder temperatures, will the air compress more, giving the desired larger values of [tex] W, T2, \rho_2 [/tex] and hence [tex] P_2 [/tex]? If that's the answer, how can this be shown mathmatically?

    See my dilemma?
     
    Last edited: Dec 13, 2003
  9. Dec 13, 2003 #8

    Bystander

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    "More boost." You might look at T effects on transport properties --- viscosities of gases decrease with decrease in T. You're talking about turbulent flow, not my department, and were you to hit the aeros with this question they might be able to tell you how much power is required to move a given quantity of air through the blower as a function of T.
     
  10. Dec 13, 2003 #9
    I am not quite sure what you mean by "difference". Are you talking about fractional or absolute differences?

    Your starting point should be the work input into the container and its initial total temperature.

    [tex]Work done = {\int}{P}{\cdot}{dV}[/tex]

    If you set up the experiment so that the work done is the same, then the final total temperature rise will be the same. Otherwise, you have to evaluate the integral to find the work done, and then determine which one is higher.

    Also, I am unsure if you realise this, so I will point it out anyway:

    Your expression for work is OK, if the process is adiabatic (as you point out). But remember the mass is constant. So what exactly was the point of substituting M with density & volume? Are you keeping your starting mass or volume the same in both the cold/hot cases? If it is the latter, I can understand your substitution, but if it is the former, it is pointless.

    More clarification needed...
     
  11. Dec 13, 2003 #10
    Tyro,

    This expression:

    [tex]\frac{P_2}{P_1} = \left[ 1 + \frac {W}{\rho_1 V_1 c T_1}\right]^{\frac{\gamma}{\gamma-1}}[/tex],

    is used because I'm trying to understand how P2/P1 is affected by the initial air temperature. [tex]V1[/tex] will always be the same (= volume of cylinder), while [tex]\rho_1 [/tex] will be larger when the air is cold (i.e., there will be a greater mass of air in the cold case). However this effect is balanced by the [tex]T_1 [/tex] term.

    This brings us to [tex]W[/tex]. Let's assume a force F on the piston (which is fixed). The piston compresses an amount [tex]\Delta x[/tex], so that [tex]W = F\Delta x [/tex]. Now will the air compress more (i.e., will [tex]\Delta x[/tex] be larger), for colder, denser air?

    Another way to calculate W is with

    [tex]W = {\int}{P}{dV}[/tex].

    Using [tex]PV^{\gamma} [/tex]=constant, I get

    [tex]W = \frac{P_1V^{\gamma}_1}{(1-\gamma)}\left[V^{1-\gamma}_2 - V^{1-\gamma}_1\right].
    [/tex]

    But the rhs depends on [tex]V_2[/tex], which we don't know...so I'm kind of stuck here.

    At this point, it looks like the problem is reduced to knowing how [tex]V_2[/tex] or [tex]\Delta x[/tex] behaves when we go from warmer, less dense air to colder, denser air (and keeping F constant).
     
  12. Dec 13, 2003 #11
    The only factor which determines how much the piston is going to compress is the maximum force you can exert on the piston. If the piston is ideal (perfectly sealed, etc.) then as you reduce the volume of the working fluid, you will approach infinite pressures.

    The initial pressure, [tex]P_1[/tex], I am assuming to be atmospheric (the container is initially in equilibrium with its surroundings).

    The end pressure, [tex]P_2[/tex], I am assuming to be the same as well. Why? Because I take it the maximum force you can exert in the hot/cold case will be the same.

    Therefore, since [tex]{PV^\gamma} = constant[/tex]:

    [tex]{V_2} = {V_1}{\cdot}({\frac{P_1}{P_2}})^{\frac{1}{\gamma}}[/tex]

    As I said, [tex]P_2[/tex] and [tex]P_1[/tex] is the same for the hot/cold cases. [tex]V_1[/tex] is the same for both cases too. Thus [tex]V_2[/tex] is the same in both cases. The piston moves equally in both cases.
     
  13. Dec 13, 2003 #12
    I agree Tyro! It became very obvious to me right after I wrote the last response, that the volume change will be:

    [tex]\Delta V = V_1 \left[ \left[ \frac{P_{atm} A}{F} \right]^{\frac{1}{\gamma}} - 1 \right] [/tex],

    which would be the same if the air was cold or hot...


    Now I need to investigate how/if the "extra pressure" that I'm observing is related to the efficiency, and/or the reduced viscosities at lower temperatures.

    Thanks for your help!
     
  14. Dec 14, 2003 #13
    My pleasure, Tony. Would my first reply with regard to the Joule-Brayton cycle answer your first new question? If you can elaborate a bit more on what else is puzzling you, maybe I can shed some light. It was pretty fun to exercise those cerebral thermo muscles which I have not flexed in ages [b(]
     
  15. Dec 15, 2003 #14
    Cool question.
    Since I drank too much last night I can't really muster a set of equations from my pickled head right now.

    However I though of a few things that might be helpfull.

    First off: What type of compressor are we talking about?
    By looking at some of your equations you are modeling a fixed volume compressor (such as a roots or meshed rotor compressor).
    Are the sysems you are looking at really constant volume systems? Most systems these days are running centrifugal inertial/centrifugal compressors.



    If this is the case, and you are trying to model a centrifugal compressor then things are different. Cenrifugal compressors are very common in modern supercharger systems and used in virtually all turbo systems). A centrifugal supercharger compresses air by rotating radially arranged columns of air around a fixed axis, thus using the mass of the air as the driving force. The pressure at the column outlet could be roughed up by integrating he (density)*(centripital accelleration)*d(lengh) down he lengh of the column (ignoring the inertial and Bernoulli effects a the entrances an exits of the column vane).The density is not constant down the length for a compressible like air but will be offset throughout due to the initial inlet density. This calculation is clearly a funcion of the density which is in turn a function of the temp. whether this is a significant enough effect for explaining the pressure difference you are observing: I don't know.

    If it really is a fixed volume compressor you are looking at I will try to follow up with some calculations on some effects that mighrt be significant.

    Take Care.

    :smile: :smile:
     
    Last edited: Dec 15, 2003
  16. Dec 15, 2003 #15
    Thanks Achy47,

    It is a centrifugal blower. But I think that this effect (more boost when cold) is seen with all types of forced induction set-ups -even turbos. But you may be on to something here. The (oversimplified) model I used was just static adiabatic compression. I did take the initial (uncompressed) volume of air to be a constant. I think that part may be OK, but your statement about the air's own mass being the driving force for compression may be the key. I've been taking the compressing force/work to be independent of the air properties. I'll re-think it now in this new light. Thanks!
     
  17. Dec 15, 2003 #16

    turin

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    TonyG,
    This is obviously not my area as you no doubt saw in my first post. But something doesn't seem right about the overall treatment that I'm seeing. Correct me if I'm wrong. You are trying to figure out some final pressure of some closed thermodynamic process (a full stroke or half cycle in a given cylinder) because you see it to be somehow related to the output at the wheels? Shouldn't the consideration be more focused on the ignition, and how the colder air/fuel mixture reacts differently for the hot vs. cold air conditions. I figured that would be the most important consideration. The variables like the force exerted by the piston and the compression ratio seem to be well controlled and not sensitive to environmental variations. And, according to chroot, apparently the mixture is also well controled.
     
    Last edited: Dec 15, 2003
  18. Dec 15, 2003 #17
    turin,

    We're not talking about fuel delivery, ignition timing, etc., but rather the properties of the compressed air. Specifically, more boost (i.e., peak air pressure in the intake manifold) in cold weather. No matter how I approach the problem, I get the answer that the peak pressure depends on the initial pressure (which is always the same, [tex]P_{atm}[/tex], and the work done by the compressor (blower).

    So the only way I can explain it right now is that the efficiency of the blower is higher in the cold air, therefore it's actually doing more work.

    Of course, the decription above about centrifugal superchargers using the air's own mass to create boost might explain it, but so far I haven't been able to model it correctly (trying to calculate the pressure distribution inside a long, thin column of air rotating about one end, with frequency [tex]\omega[/tex].
     
  19. Dec 15, 2003 #18
    Not really.In modern fuel delivery systems the the entire intake tract including virtually all of the control volume between the atmospheric inlet and the intake valves is a pure air system. Fuel is usually injected at a point very close to the intake valve and thus the portion of the control volume consisting of air fuel mixture is quite miniscule. Furthermore, the injector flow rate is not continuous, rather it is best approximated as having on or off states with a specific frequency and duty cycle.

    Reply to Tonyg

    I think you are right about this bieng common to both types of systems, I believe I have witnessed the same thing before. Since modeling a centrifugal sysem is insanely complex (I think I have read before that some design issues are beyond our scientific undersanding and purely emperical based), I am thinking hat we may be able o find the common thread by examining the more simple roots configuraion? You think?

    If so I am guessing that the the the best way to model this system is to consider two sequential compressors with the second (the reciprocating piston or rotary engine) always bieng a fixed volume compressor (this is really not true but I think we can assume this strictly from its inlet behavior: what happens next could be ignored). Simply stated, we are assuming that the engine itself consumes a fixed volume per intake cycle at a specific frequency resulting in a approximate volume flow rate. The first compressor can be of any type and we want to know why we observe that the pressure meassured between the first and the second is higher when the inlet temperature of the first is lower.

    Remember though that when the first compressor is a fixed volume unit and is driven by a mechanical attatchment to the engine there is a fixed volume ratio between the two. Meaning for a given rpm the volume processed by the first (compressor) is theoretically a linear function of the volume processed by the second (the engine) (since they are fixed volume and both related via a mechanical ratio). For there to be a positive pressure difference across the first it only makes sense that the first should process more theoretical volume per per unit time or rotation via a fixed ratio.........

    Out of time gotta go .... Be back later. I will edit..
    :smile:
     
    Last edited: Dec 16, 2003
  20. Dec 15, 2003 #19

    turin

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    OK, I was very confused with the context of consideration. I didn't know that you had a pressure gauge in the intake manifold. I thought you were concerned with what goes on inside the cylinders.
     
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