I have a water jet pump pushing water through a 6" diameter open ended pipe 100 ft long, I need to get 12 bar gauge pressure on the pump discharge, what size (m3/hour) pump do I need ?

  • $\begingroup$ Is the only thing creating the back pressure the pressure drop in the pipe? There's no valve or anything? $\endgroup$
    – ericksonla
    Jun 7, 2018 at 15:46
  • $\begingroup$ no, no valve anywhere in the line or pump discharge flange, straight through all the way $\endgroup$
    – Dave Ward
    Jun 8, 2018 at 17:54
  • $\begingroup$ Mark, thanks---I do believe that the bulls eye, .5 m3/second is 30m3 a min which is somewhere around 1,800 m3/hour-- pipe is actually 20 ft of steel then 80 ft of flexi hose and discharges under water jetting a hole in the seabed so sure there some back pressure to add there, which I forgot to add at the begining so ---I'm going to use a 1200 m3 pump with 14 bar --biggest I can find here in Africa, thanks for the fancy formulas $\endgroup$
    – Dave Ward
    Jun 8, 2018 at 18:05

1 Answer 1


Related: System curve for use in determining pump operating point

The equation, copied, is: $$\Delta P =f_D \frac{\rho V^2}{2}\frac{L}{D} = f_D \frac{8\rho Q^2}{\pi^2}\frac{L}{D^5}$$

The friction factor, $f_D$, is based on the moody chart. $Q$ is the flowrate, $\rho$ is the density of the liquid your pumping, $L$ is the length of the pipe, $\Delta P$ is the pressure buildup and $D$ is the diameter. The basis seems to be that you want a system that would build up 12 bar of gauge pressure on the pump discharge, and at the end of 100ft of pipe, you would discharge to jet.

We rearrange as follows:

$$\frac{12\:bar \: \pi^2 (6\:in)^5 }{(8)\frac{1\:kg}{1\:L}\:100ft} = f_DQ^2 = 0.004 \frac{m^6}{s^2}$$

We estimate a relative roughness of 0.001 with relatively smooth cast iron pipes, then the friction factor would be 0.02. This would make the flow rate run at about 0.5 cubic meters per second, with a water jet of 27.5 m/s. The pump would need to put out 12 bar at 0.5 cubic meter per second, which would be difficult to source, but not impossible.

  • $\begingroup$ Love the mixture of units in your final expression - personally I would have converted to either all metric or all imperial... $\endgroup$
    – Solar Mike
    Jun 7, 2018 at 18:44
  • $\begingroup$ The question came in mixed units. It’s a single step calculation to run the numbers in most CAS systems to convert and compute. $\endgroup$
    – Mark
    Jun 7, 2018 at 18:45
  • 2
    $\begingroup$ Perhaps it’s just habit - forced into us by some tough lecturers - if you turned up without steam tables : leave, if your calculator did not work : leave ... with comments like « worse than useless »... :) $\endgroup$
    – Solar Mike
    Jun 7, 2018 at 18:49

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