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Operating Point of a Pump

  1. Jul 19, 2017 #1
    1. The problem statement, all variables and given/known data

    A centrifugal pump has the following pressure – capacity characteristics:

    Capture.PNG

    It is planned to use this for a process having the following system characteristic:

    Capture2.PNG

    (a) Determine the operating point for this pump with this system.

    (b) The actual flowrate required by the process is 50 m3 h–1. If the overall efficiency of the pump at this flowrate is 70%, determine the power consumed when the liquid being pumped has a density of 1200 kg m–3.
    2. Relevant equations
    For Part a I think I need to plot the data given onto a graph and find where the lines intersect.

    Part b: Po = qv ρ g Hp

    where Hp is the system characteristic

    Hp = (Ht2 - Ht1) + hf + hm
    3. The attempt at a solution
    (a)
    Capture3.PNG

    After graphing the data given in excel the lines intersect at a flowrate of 100 m3h-1 and a Pressure of 465 KPa

    (b)
    I haven't figured out how to calculate the system characteristic as I'm not sure where to find the initial total head and final total head. I must have misread or overlooked something in my notes on how to calculate it. I'll be reading through my notes again. Plaese let me know if I've got anything wrong that would be much appreciated thanks.
     
  2. jcsd
  3. Jul 20, 2017 #2

    CWatters

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    Part a) looks correct.

    For part b)... What is the pressure required to push 50 m3/h through the system? This will be the pressure the pump has to deliver eg the pressure or head across the pump.
     
  4. Jul 24, 2017 #3
    Hi CWatters, thanks for the reply. The pressure required according to the tables would be 400KPa. So I've got density, flow rate and pressure now and looking through some notes and other resources I think I was missing some equations.

    b)
    y = ρg =1200 x 9.81 = 11772

    where y is specific weight.

    h = (P2 - P1)/y = 400x103/11772 = 33.98m

    Po = qv ρ g Hp = 50 x 1200 x 9.81 x 33.98 = 20000628W

    Actual Power output = 70% of Po = 0.7 x 20000628 = 14000439.6 W

    I hope this is right, I'm a bit out of practice with these questions on pumps.
     
  5. Jul 24, 2017 #4
    Hi i used this equation P=Q x dP/3600 and i came out with 3.89kW @ 70% eff. Sounds about right for size of pipework and distance pumped.
     
  6. Oct 25, 2017 #5
    Hi I am struggling on this question myself and seem to be having the same problem in that I don't have an equation for finding Hp using the data given, jayohtwo can I ask where you found h = (P2 - P1)/y ?
    Thanks
     
  7. Oct 25, 2017 #6

    CWatters

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    I was thinking along same lines as mechengstu123 but got different numbers..

    Power = Pressure * Flow Rate

    50m3h–1 = 13.9 L/S

    From the table the pressure at 50 m3 h–1 = 400kPa

    So the Hydraulic power required is..

    13.9 * 400 = 5,560W

    The efficiency is only 70% so the motor power required is..

    5560/0.70 = 7,942W
     
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