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!

Centrifugal pump cavitation prevention

  1. Feb 27, 2014 #1

    Maylis

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    The 9.75 in. impeller option of the Taco Model 4013 F1 series centrifugal pump (pump curve given by Figure P3-2) is used to pump 200 GPM of water at 25 C from an open reservoir whose surface is above the centerline
    of the pump inlet (Figure P3-1). The piping system from the reservoir to the pump consists of 10.0
    ft of cast iron pipe with an ID of 2.0 in. and an average roughness of 0.02 in. There are several
    minor losses: a sharp-edge inlet (Kc = 0.5), three flanged smooth 90 degree elbows (Ke = 0.3 each)
    and a fully open flanged globe valve (Kv = 6.0).

    Estimate the minimum elevation of the liquid surface above the pump inlet required to operate the pump without cavitation.

    Hint: Recall that the NPSH is the difference between the head at the pump inlet and the vapor
    pressure head (Equation 1).



    Data: Vapor pressure of water at 25 C = 3.169 kPa
    Density of water = 997.0 kg /m3
    Viscosity of water = 8.91 X 10-4 km / m s
    Patm = 101.3 kPa






    2. Relevant equations



    3. The attempt at a solution
    I keep getting a pressure that is negative. using the chart that I attached for the Taco Pump, I get a NPSH of about 1.12 m (3.7 feet) and when I use equation one, the pressure at the inlet of the pump is negative.
     

    Attached Files:

    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2
    From the graph, what is the minimum NPSH required to avoid cavitation? For 200 gpm, what is V2/2g coming into the pump? What does P/ρg have to be at the pump inlet to aviod cavitation?
     
  4. Feb 27, 2014 #3

    Maylis

    User Avatar
    Gold Member

    From looking at the graph, I see the NPSH head is about 4 feet, just slightly under. In my attempt, I just divided Q/A to get V, and the only unknown is P. When I calculated it, I got -5.05 kPa, which doesn't make sense to me.

    I see that there is an NPSH-available and NPSH-required. The equation given is for the available NPSH according to wikipedia
    http://en.wikipedia.org/wiki/NPSH#NPSH_in_a_Pump

    And the graph labels ''required NPSH''. To prevent cavitation, does the available NPSH = required NPSH?
     
    Last edited: Feb 27, 2014
  5. Feb 27, 2014 #4
    Yes. That's my understanding. 108 ft.

    Chet
     
  6. Feb 27, 2014 #5

    Maylis

    User Avatar
    Gold Member

    I see where you came up with 108 ft. Look at the right side of the graph. It says NPSH and gives a separate scale that reads NPSH feet.
     
  7. Feb 28, 2014 #6
    Ah, yes. I missed that. Thanks.

    It looks like you used P2=1 atm in the calculations when you should have been using the pressure at the entrance to the pump. You first need to determine what the pressure at the entrance to the pump has to be so that you match the NPSH. You then have to determine what the height of the water in the reservoir has to be to give you that pressure at the entrance to the pump.

    Chet
     
  8. Feb 28, 2014 #7

    Maylis

    User Avatar
    Gold Member

    I think you have misread. My P2 = -5.05 kPa, not 1 atm. That is what I calculate the pressure at the entrance of the pump to be. Just look at the last line and you will see that I am solving for P2 using the equation given to us for NPSH. My whole problem is that when I use that formula, I get a negative pressure for P2.

    I know once I get P2, all I have to do is use bernoulli to get the height difference.
     
  9. Feb 28, 2014 #8
    I see what you are saying now. The NPSH concept seems a little confusing. It looks like, to calculate the NPSH, one is comparing the stagnation pressure within the entry pipe with the vapor pressure of the water. Apparently what happens is that the cross sectional area inside the pump is much higher than in the entry pipe, so the fluid velocity within the pump is much smaller than in the entry pipe. Therefore, inside the pump, the pressure starts out as being roughly equal to the stagnation pressure prior to the pump. This is what is compared to the vapor pressure to guarantee (by the pump manufacturer) that there is no cavitation inside the pump. However, in your problem, what the calculation seems to be indicating is that there may be cavitation in the pipe leading up to the pump. I don't know whether this matters or not. Still I would not allow P2 to come out this low. If I were designing the system, I would choose P2 no lower than the vapor pressure.

    Chet
     
  10. Feb 28, 2014 #9

    Maylis

    User Avatar
    Gold Member

    The professor is here and we came to the conclusion that the problem is messed up, because the velocity head is overcoming the vapor pressure, which accounts for the negative pressure calculation.
     
    Last edited: Feb 28, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Centrifugal pump cavitation prevention
  1. Centrifugal Pump (Replies: 1)

Loading...