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  • Pump pressure question

    Discussion in 'The main mechanical design forum' started by loveit, Apr 22, 2010.

    1. loveit

      loveit New Member

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      Guys,

      I am a structural engineer but at work I am doing some minor mechanical work for the mech engineer while he is away on leave. I need clarification on the following problem and how to solve it.

      I need to deliver a certain capacity in m3/s water. However the pipeline diameter measured on site is found to be 95% of what was originally designed. How much additional pump pressure is required to
      deliver still the same capacity. How can you achieve this?

      I don't know the real figures of the orignal flow rate or pipeline but I guess if someone can show me with a sample calculation I can then apply this with the real figures.

      Cheers,

      Loveit
       
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    3. ConnectUTS

      ConnectUTS Active Member

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      I assume you are talking about a change in internal diameter and not outside diameter. You the critical diameter is the inside diameter.

      A 5% loss of inside diameter will result in a proximately a 10% loss in flow area. To maintain the same flow rate you velocity will increase by the same amount. For continuity volume flow rate is a function of area x flow rate.

      Since the work is a function of velocity square this means that amount of energy will go up by approximately 20%.

      Niel Leon
       
    4. npanchal_007

      npanchal_007 Member

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      pls mention following Detail

      required head ,capacity
       
    5. maniacal_engineer

      maniacal_engineer Well-Known Member

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      I think ConnectUTS is right. All other things being equal the pressure drop will increase by 20%. Power is pressure times volume flow (with corrected units to get customary power units). So for the same flow it will be about 20% more power (actually 22.7%). However pressure drop is calculated in terms of fL/D (friction factor, length, diameter) and q (dynamic pressure) If we assume f is constant then L/D also changes by 5%.

      I would call it a 25% increase just to be conservative and call it good
       
    6. rushinpatel

      rushinpatel Active Member

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      please provide me some data,

      head of pump(hd)=mm (height of at which your fluid will be reached by pump)
      n=rpm of your pump,
      D=required discharge pressure(n/mm^2)
      f=friction factor in pipe,
      v=angular velocity or velocity(m/s),

      so you provide available parameters which are maintain above
       
    7. MAHADEV DABARIA

      MAHADEV DABARIA New Member

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      the pressure is the most important factor in the pump, the lifting of the fluid from one datum head to another datum
      head,by help of the pressure. because pressure =forcexarea, and another factor is the high pressure above the atmospheric, the water will convert in to the steam & this steam produce the cavitaion on the wall of the pipe & casing of pump.
      the rise of the fluid is depend on the pressure= density x sp.gravity x height of the liquid.

      so pressure is proportional to the height or of liquid.
       
    8. maniacal_engineer

      maniacal_engineer Well-Known Member

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      oh yeah "how can this be achieved?"
      pressure is proportional to velocity squared, so for the same RPM of the pump (same number of poles in an AC motor) the diameter of the impeller needs to be about 12% bigger. Fluid velocity is proportional to impeller tip speed.

      I think it is interesting that some people responding want all the data, and others look at the basic equations, what is in the numerator/denominator and their exponents, and do the ratios.
       
    9. vsk_design

      vsk_design New Member

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      We required flow rate of water & we should calculate the height to lift the water & add frictional losses to the head then we will get exact pressure required.
       
    10. Karl A Petersen

      Karl A Petersen Member

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      Most of this is fine in the stable range of flow, so if the flow is smooth, laminar, at present and you are going to increase the velocity. but if you are near the limits of the design which you indicate is the case and obtain turbulent flow, all bets are off. If you pull a vacuum when crossing intersecting pipes or rounding corners or changing diameters, then any of these things will greatly increase the pressure drop through the line and thus increase the power required. Also, if your pump is running at max and starts to cavitate, then you lose everything. This is the kind of thing that might work when cold and then magically drops out when it warms up.

      Since this was posted long ago, the thing has probably been modified and is running, so how did it go?
       
    11. prieto01

      prieto01 New Member

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      Foundry with High Pressure Moulding Lines at here
       

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