1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Applying Bernouilli to calculate pressures

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data

    flow rate: 0.057 m^3/s
    cross sectional area of all pipes: 1.864*10^-2 m²
    frictional losses inlet pipe: 1.83m
    frictional losses outlet pipe: 3.66m

    pressure at inlet
    pressure at outlet
    necessary pumping power

    2. Relevant equations
    Bernouilli: http://imgur.com/EEptkNX where h_A is the energy added by the pump and h_r and h_l are the losses
    pumping power= q*rho*g*h

    3. The attempt at a solution
    first of all, i am not sure what location they actually mean exactly by pump inlet and outlet (is it in the tank or really at the pump?)

    inlet: I assumed p1 to be 0, v1=v2 and z1=0 and z2=3.05m which results in a p2 of 47.87kPa (i assumed h_A to be equal to 0, not sure if this is correct)
    outlet: using the previous value as p1 now, assuming v1=v2 and z1=0 and z2=12.2m i found a p2 of 107.71 kPa
    power: 0.057*1000*9.81*15.25 = 8.53kW

    Apparentely, all my answers seem to be wrong. I'm fairly sure that im making a critical mistake for the pressure calculations, but I really don't understand what could be wrong in my power calculation.

    Thanks in advance :)
  2. jcsd
  3. Mar 14, 2015 #2
    The pressure at the top of the water in the lower tank is atmospheric (zero gauge), and the velocity at the top of the water in the lower tank is zero. Between the lower tank and the pump inlet, there is a vertical distance, there is a equivalent vertical distance to allow for the frictional pressure loss, and there is a water velocity in the pipe. You need to use these to get the pressure at the very inlet of the pump (which will be less than zero gauge).

    What value do you get for this pressure at the pump inlet (either in absolute or gauge pressure)?

  4. Mar 14, 2015 #3
    Thank you for your reply!


    0+0+0-1.83 = p2/(1000*9.81)+3.05+3.0579^2/(2*9.81) (i calculated the speed by dividing the flow by the cross sectional area: 0.057/(1.864*10^-2)

    i find p2= -52548.18 Pa

    I wonder, if this is correct, if I have to substitute this value in absolute value or not for p1 in the calculation of the outlet pressure...

    Since when i use the negative value further i find from

    -51548.18/(1000*9.81)+ 0 - 3.66 = p2/(1000*9.81) + 12.2

    that p2= -208134.78 Pa which is also negative and i feel like it should be a positive value in the discharge pipe...
    Last edited: Mar 14, 2015
  5. Mar 14, 2015 #4
    This looks OK so far (I didn't check the arithmetic). This is the gauge pressure at the inlet.
    If you are trying to get the pressure at the outlet of the pump, the signs on the 12.2 and the 3.66 are wrong. You should be working your way down from the surface of the water in the upper tank, which is likewise at 1 atm.


    Last edited: Mar 14, 2015
  6. Mar 14, 2015 #5
    Allright, when changing the signs of both the 3.66 and 12.2, i find an outlet pressure of 31229.22 Pa.
    Do you have any idea what I am doing wrong with calculating the necessary power? Do i have to work with only the highest head of 12.2m and not with the sum of both maybe?


    P.S.: sorry for cross posting, won't happen again.
  7. Mar 14, 2015 #6
    OK. Using the pump elevation at the datum for potential energy:

    0 + 0 + 12.2 = p/(1000*9.81) + v2/2g + 0 -3.66

  8. Mar 14, 2015 #7
    Ah, i forgot that the velocity wasnt the same on both sides apparentely, so i find 150991.10 Pa using your equation.
    I also think i know what I'm doing wrong for the power... I think I have to add both losses to the the total head, is that correct?

    so P= (3.05+12.2+3.66+1.83)*0.057*1000*9.81=11597.19 watt
    Last edited: Mar 14, 2015
  9. Mar 14, 2015 #8
    The pump power is the pressure increase across the pump times the volumetric flow rate.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted