Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reciprocating Compressors Energy Losses

  1. Mar 14, 2007 #1
    Hi, I was wondering if anybody could tell me what are the losses in a reciprocating compressor, and how would I calculate them. Specifically I'm interented to know how does the input energy (electrical) translate into output energy (pressure), and where would I look for the losses that may occur during that "transaltion"
  2. jcsd
  3. Mar 14, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi fab. There's no easy list of things that cause losses. In fact, the list would be absolutely enormous if you tried to write down every single item that causes losses.

    Note that depending on how you define "loss", you might also consider all the heat that must be rejected during compression. Or you might consider only isentropic inefficiency. Can you define what you mean as "losses"?

    If you're only refering to isentropic efficiency, then just as a starter, I'd suggest breaking down losses into different catagories. Very broadly, there are mechanical losses, and thermodynamic losses. Unfortunately, these losses often overlap, such as for example piston rings which create frictional resistance to the motion of the pistons (mechanical loss) but that friction also creates heat which can be transfered to the gas being compressed (thermodynamic loss).

    In addition to mechanical versus thermodynamic, you might break it down into drive train losses and compression end losses. Drive train for example might cover bearing losses, pully friction, oil pump or water pump power use, etc. These are mostly mechanical losses. Compression end losses might include pressure drop between stages, re-expansion of gas that doesn't get expelled from the compression chamber, gas leakage past valves and dynamic seals, etc... These are a combination of mechanical and thermodynamic losses.

    Perhaps you'd like to take a shot at catagorizing various inefficiencies and others could then add on to the list and take a shot at how those inefficiencies might be determinined analytically.
  4. Mar 15, 2007 #3
    Reciprocating Compressor Losses

    The way I'm looking at losses is the difference between the energy in (electrical) and energy out (Kw % of compressed gasses). I may not be looking at this the right way, but the idea goes along the lines that if xkW/energy is provided to the (gas compression) system, then that energy must go through changes (mechanical/tourqe/etc) and at the end you get to the compression target sought. Now some of that energy (xkW) provided to the system is being lost or maybe not fully used, in any case, energy out (in the form of compressed gas) is the difference between the energy in minus the sum of the (systematic) losses. I also understand that there are many variations and many calculations that may provide a theoretical gas compression system losses, I'm just wondering if there are some "major" losses that are always considered when looking at a system as a whole.
  5. Mar 15, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi fab. I think it's best to take an overall look at what energy is going in and what energy is going out. To do this, let's put a control volume around the entire compressor, including all utilities used. If we apply he first law of thermodynamics and simplify, we find:

    dU = Hin - Hout + Win - Qout
    dU = 0 for a steady state system
    Hin = Enthalpy entering the compressor determined by the state of the gas
    Hout = Enthalpy coming out of the compressor determined by the state of the gas
    Win = Work going in such as electricity or shaft work
    Qout = Heat being rejected

    Generally, the compressor system rejectes heat such that the gas leaving the compressor is at aproximately the same temperature as the gas entering. For example, an air compressor taking ambient temperature air at 70 F and compressing it isentropically to 100 psig will result in the air temperature climbing to 489 F. But before we discharge that gas, a compressor will typically cool it back down to roughly ambient temperature. If 1 lbm/s of air is compressed this way isentropically, that compressor requires 144 hp of power to operate. This 144 hp is the work per unit time entering the control volume in the form of electricity or shaft work assuming no losses to friction or other factors.

    So we put 144 hp in to 1 lbm/s of air at 70 F and 0 psig, and we get 100 psig out at 489 F. But before we allow that air to continue down the pipe, we must cool it back down to 70 F. The heat we remove is the Qout in the above equation. The energy removed per unit time equates to a power which we might want to compare to the 144 hp power input. We might then say that the energy of this system in the form of the gas coming out, minus the energy of the gas going in, plus the power going in, is equal to the heat rejected.

    Qout = Hout - Hin - Win

    We might say this heat represents all that energy that "is being lost or maybe not fully used". If all the work put in, stayed in the gas, the temperature of the gas would be 489 F, but there is heat rejected, so that much energy is 'lost' so to speak. That of course, assumes we have 100% isentropic efficiency. So how much heat must be rejected? How much power do we loose by cooling the air back down?

    Guess what… Heat out equals work in for an ideal gas, and for this particular example, very slightly more heat is rejected than work in! Heat removed is roughly 0.7% more than work in - not much but the point is the heat out is roughly equal to the work in. I know that seems funny, but it's true. The amount of heat rejected is often slightly larger than the work put in. Another way of loooking at it is that the enthalpy of the air exiting the compressor at 100 psig and 70 F is actually less than the enthalpy of the air going into the compressor at 0 psig and 70 F. So we put all this work into compressing the air, but we also take out heat at a rate roughly equal to, or even slightly faster than, the work we put in.

    If the isentropic efficiency is less than 100%, even more energy in the form of heat must be removed when compared to work in. This gets back to my original question, are you interested in this additional loss, or are you more interested in the overall energy balance? As you can see, the energy put into the gas is going to be roughly equal to the energy in the form of heat that must be rejected.
  6. Mar 16, 2007 #5
    electric motor loss (electric to mech conversion)
    motor-compressor coupling loss + gear box loss (if any).
    piston friction loss
    compressed air heat loss
    and etc..
    the question needs answer of big book. if the question is very specific, i can give formulae for calculating.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook