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Valve Exhaust Port Calculation

  1. Oct 27, 2016 #1
    Hi,

    I am following this particular article from machinedesign.com about properly sizing a pneumatic system - http://machinedesign.com/archive/right-sizing-pneumatic-motion-systems

    In the section Valve Exhaust Port, he does a calculation for the valve loss coefficient and arrives at 1.833 cfm. I have tried to follow his numbers but am not coming up with the same value. I am reaching out to the board to see if you can help me understand where the author came up with this value.

    I used the equation he specified in the beginning -
    Q = 22.48*Cv (((P1-P2)*P1)/(T*G))^1/2

    and used the following values
    Q = 9.94 cfm
    P1 = 79.7 psi
    P2 = 14.7 psi
    T = 273 + 70 = 343
    G = 1 (SG of Air)

    Please let me know where I am going wrong in this equation.

    Thanks
     
  2. jcsd
  3. Oct 27, 2016 #2

    JBA

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    How did you come up with the P1 = 79.7 value?

    In the article he states that P2 = 14.7 and ΔP = 5 psi so then P1 = 14.7 + 5 = 19.7 psi

    Additionally, I did not see any temperature given but 70°F is a STP value so that is OK; but, Fahrenheit should be converted to Rankin not Kelvin so T = 70 + 459.67 = 529.67°R

    I have done the calculation but I think these changes will get you the answer you are looking for.
     
  4. Oct 28, 2016 #3
    JBA,

    Thanks for responding. I took your values came up with the value the author cites in the article - 1.185.

    **Please note to anyone reading this threat that I made a typo in my original post - 1.833 cfm should be 1.185 cfm.**

    Q = 22.48*Cv (((P1-P2)*P1)/(T*G))^1/2

    9.94 = 22.48 * (Cv) * (((19.7-14.7)*14.7)/(530*1))^1/2

    Cv = 1.187 (close to 1.185 as stated in article)

    Thanks for the clarification with the values.
     
  5. Oct 28, 2016 #4
    I have a follow-up question. In the delay time section, the author cites the following equation:

    Td = (Pe*g*Ve*ks)/(Qm * Vs^2)

    where Td = time delay (seconds)
    Pe = change in pressure (79.7 - 34.91 psi)
    g = .075 lb/ft^3
    Ve = exhaust volume - 5.07 in^3
    ks = 1.4 for air
    qm = flow rate - this I'm not sure what value he's using
    vs = 1127 fps, volume of sound

    Please let me know if someone can explain how the author is calculating .054 sec from this equation. I can't seem to identify what value I am misinterpreting.

    Thanks,
     
  6. Oct 28, 2016 #5

    JBA

    User Avatar

    From the use of Q=11.75 scfm in the four calculations just above this one it might be assumed that he is using that value for q in this calculation. The symbol change from Q to q may be just a careless mistake when copying the equation into the article but we don't know if that is true. I suggest that you try using that value for this calculation in see if the resulting answer matches his.


    Just as quick note for your general information on units. The correct designation for vs is "velocity of sound"
     
  7. Oct 31, 2016 #6
    Hi JBA,

    I used 11.75 as my Qm value but did not receive the correct answer. I thought, perhaps, my units were incorrect so I converted all inch measurements to feet and all minutes to seconds. I don't see how the pounds value, from value Pe, which is either PSI or PSF cancels out. All the other values are either a volume or a velocity and don't have an expression of pounds to offset the Pe value. Please let me know what I am missing.

    Thank you for all of your input thus far. You have been a wonderful help!
     
  8. Oct 31, 2016 #7

    JBA

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    I have finally worked through the units issue and sorted out what the units are for qm and how the equation can end up with sec as it finished units. See the next post for all of the details because I cannot paste everything I need using the editing function on this earlier post.
     
    Last edited: Oct 31, 2016
  9. Oct 31, 2016 #8

    JBA

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    First your value for g is an error g is the symbol for the acceleration of gravity = 386.4 ips2 (inches / sec^2 ) as he uses it in the Physics Section:
    "Assuming a linear ramp (constant acceleration), peak velocity is twice this or 62.5 ips. Divide by time again to get acceleration, 651 ips2 or 1.685 g (651 ips2/386.4 ips2 = 1.685 g). Thus, the required force is 1.685 X 10 lb (due to acceleration) + 10 lb (due to gravity) = 26.85-lb force. If there are other forces, such as guide friction, they would be added here."

    Also I was in error in selecting the 11.75 scfm value for this calculation I now think it should be 9.94 scfm as per the following:
    "Multiply by 79.7/14.7 to convert to standard conditions, 9.94-scfm exhaust. In the same manner, calculate the supply flow during motion (without the rod) and get 11.74-scfm supply. These exhaust and supply flows will be used to estimate all other component pressure drops in the circuit."

    Next, the units for qm are Q in scfm (ft^3/min) converted to qm in "lbs/sec" in the manner that the author used in the original Delay Time: Section as follows:
    "Calculate delay time, td, as follows:
    td = ΔPegVeks/qmvs2. Here, ΔPe = 79.70 - 34.91 psia (the piston low side pressure). Density of air is approximately 0.075 lb/ft3 at 528°R. Velocity of sound vs is approximately 1,127 fps at 528°R. Ratio of specific heats ks for air = 1.4."

    With that sorted out this is how the units reduction works out:
    upload_2016-10-31_20-0-30.png
    and then (with the unit cancellations shown by matching colors)
    upload_2016-10-31_20-1-35.png

    I think if you do these conversions and use the new g and Q = 9.94 scfm you will have a good chance of getting the stated time answer.

    Don't feel bad if you didn't find all of this. As I stated earlier, the author did not make it easy and really the only way I found the right path to "qm" is that I worked with gas flow problems for last 20 years of my career and mass flow in lbs/sec is common in these types of calculations; so, when I discovered I needed a "lb" for my units conversion I went looking for the possibility that he had used mass flow rather than volume flow in his calculations, and, finally found it.
    .
     
    Last edited: Oct 31, 2016
  10. Nov 1, 2016 #9
    JBA,

    Thanks for your help with this. I was able to calculate the correct answer and follow your method, with the exception of the 60 sec in the 1/qm value. When we calculate qm, we obtain a value of .0124 lbs/sec. If I multiply this by 60 sec, I get .7455. Taking the inverse of this for my final Td equation results in the correct value of Td per the article. But if I multiplied this value by 60 sec., wouldn't I only have lbs as my qm value. Then I wouldn't have any value in seconds for my final Td answer.

    So for my Td equation, I have the following:

    Td = Pe * g * Ve * k * (1/qm) * (1/vs^2)

    Td = 44.9 psi * 1.6885 g * 5.07 in^3 * 1.4 * (1/.7455 lbs) * (1/13524 s^2/in^2)

    Td = .053 seconds

    I followed your method and put this into excel just as you have and it helped me arrive at the answer (attachment). I'm just a little confused though on the 60 sec calculation and how the seconds still remains for the final Td (time delay) value.
     

    Attached Files:

  11. Nov 1, 2016 #10

    JBA

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    I included the 60 in the seconds conversion was because the original Q was in cfm cubic feet / minute and I was converting it to cubic feet / second in my final conversions. You included a minutes to seconds conversion in your Q conversion before entering it into your final qm equation; and, this resulted in the 60 seconds per minute conversion being duplicated in the calculation. So, your confusion was basically my fault by sticking the 60 value in my otherwise strictly units conversion equation.

    Edit: One main problem with the article is that the author was not careful to define all of his symbol units; but, this is not unusual even in published technical books (in that case, often the units are explained once in some paragraph in chapter 2 and you are expected to remember that when trying some example problem in chapter 10) and it really drove me crazy when it happened in my engineering course textbooks.
     
    Last edited: Nov 1, 2016
  12. Nov 1, 2016 #11
    JBA,

    That makes sense now. Thanks for all your help with this! I believe this could be really beneficial for me to apply at work.
     
  13. Nov 1, 2016 #12

    JBA

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    I commend you for your determination and persistence in working your way through this learning process. Just assisting with the two issues you requested caused me cringe at the thought of going through all of the calculations in the lesson.
     
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