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Air consumption of machine tool

  1. Apr 6, 2010 #1

    Ranger Mike

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    Ok fellow posters..
    I am trying to compare two machine tools for economy of operation.
    machine tool 1 uses 32 Nl/m air at idle (not moving) under power and moving uses 70 Nl/m
    machine tool 2 uses 73.5 Nl/m air at idle (not moving) under power and moving uses 73.5 Nl/m
    under power uses 70 Nl/m
    the recommended air compressor should be 250 to 300 Nl/m
    electricity cost 12 cents per KWH
    the proposed machine tool would be under power and moving 5 out of 8 hours per day and idle the remaining time.
    how do i calculate the cost savings of machine tool 1 over machine tool 2 based on air consumption at idle ?
    any help is most heartily appreciated..
    thank you
     
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  3. Apr 6, 2010 #2

    Q_Goest

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    Machine tool 2 consumption is confusing. Maybe there's a misprint there. Regardless, the difference in the amount of air used per unit time is what you're interested in. The additional air consumption of one tool over the other has to be provided for by the compressor. The size or flow rate of the compressor isn't important, only the power used per normal meter of air is needed. (I'm assuming Nl/m is normal meters of air?) The other issue is that typical air compressors put air into a reservoir and shut down at a high pressure, and start up at a low pressure. So the average pressure is best to use since the power the compressor uses is roughly linear with respect to discharge pressure.

    The power needed for the compressor is
    P = (dhtheoretical) mdot / u
    where u = isentropic efficiency of the compressor. If you don't have, then describe the machine and the inlet and outlet pressure and I'll suggest an aproximate value.
    dhtheoretical = difference in enthalpy for a isentropic compression of air from atmosphere to average discharge pressure
    mdot = the flow rate difference between the two tools. Note that one tool is using more than the other, so calculate the difference as if it were constant and determine the average flow per unit time.

    You now have power per unit time that the compressor must run in order to make up the difference between the air consumption of the two tools. Multiply by the cost of power and you have the cost per unit time.

    If you have questions such as how to determine enthalpy for an isentropic compression, just provide the inlet conditions and outlet pressure and I can provide the enthalpy for the two.

    Hope that helps.
     
  4. Apr 6, 2010 #3

    Ranger Mike

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    Thank you so much..
    (I'm assuming Nl/m is normal meters of air?) yes you are correct..

    machine 2 was typo error should be 73.5 Nl/m

    air pressure required is 6 Bar or 87 psi for both machines.

    if i read the post right..i should calculate the power required to compress air for the difference between the two and then figure cost per kilowatt hour..

    i will study this and hope to be able to do the math and arrive at correct figure
    thanks again

    rm
     
  5. Apr 7, 2010 #4

    Ranger Mike

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    Rule of thumb, i have been told, for compressors under 10HP in size, bank on getting 2-3 CFM of compressed air flow at 90 PSI for each HP of motor size.
    Is is a valid assumption that 1 horsepower = .745 Kwh
    so it appears that it will take .745 Kwh to make 1.5 cfm of compressed air?

    if this true, assuming an 80 gallon air reservoir, and the air compressor starts when pressure drops below 87 psi and shuts off at 150 psi. all i would need to do is calculate the amount of compressor run time per hour to figure cost of electricity..right?
     
  6. Apr 7, 2010 #5

    Q_Goest

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    That's probably a reasonable value for a very low efficiency machine. I figure the isentropic efficiency would have to be as low as around 55% for 2.5 CFM. I would think larger air compressors are upwards of 80% isentropic efficiency.
    The conversion of hp to kw is fine but I don't know where you got the 1.5 CFM when the rule of thumb was 2 - 3 per hp unless you're counting on a higher discharge pressure.

    That's basically correct. I would use (87+150)/2 = 118.5 psi as an average value. To get to 150 psi, a 2 stage machine would be preferred, if not necessary for any larger machine. Smaller machines might get away with doing it in 1 stage because the amount of heat needing to be disipated is small per unit area. The discharge temperature for a single stage would be very high, so I suspect the system is using a 2 stage or possibly a scroll machine. The point being that adding stages increases efficiency.

    I'd suggest using 6 CFM air flow at the average discharge pressure of 118.5 psi per kW of power used. That's for a 2 stage machine with an isentropic efficiency of roughly 75%.
     
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