3
$\begingroup$

The Net Positive Suction Head of a pump is defined as 

NPSH=(pi-pv)/(g0*rho)

  pi  : pressure at pump inlet [Pa]
  pv  : vapore pressure of a medium (propellant) [Pa]
  rho : density of a medium (propellant) [kg/m^3]
  g0  : gravitational constant [m/s^2]

Why is there g0 in this equation?

It makes complete sense here on earth, but for a system in space shouldn't NPSH better be calculated with the acceleration of the vehicle?

| improve this question | |
$\endgroup$
  • 2
    $\begingroup$ Dimensional analysis tells me that there is an error in your formula. The units of g0 do not match those of rho as you have described it. $\endgroup$ – Erik Mar 17 '15 at 16:31
  • $\begingroup$ You are right Erik! I corrected it :) $\endgroup$ – cl10k Mar 17 '15 at 16:40
3
$\begingroup$

NPSH is a quantity derived from water pumps. On earth tells you how high you can put the pump above the water line before the medium starts cavitating.

Expressing the cavitation potential of a space-born turbopump in meters is an anachronism. Dividing by g0 is just a way of expressing a pressure as a length. It has nothing to do with the actual acceleration at the place.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! :) Your answer supports what I already had assumed. (It would be nice if the standard literature on propulsion systems like Sutton or Humble would express that fact more clearly) $\endgroup$ – cl10k Mar 17 '15 at 19:49
1
$\begingroup$

NPSH is measured in meters (length). It tells you how high a pump (assuming no friction losses) on Earth can be put above a fluid before you start damaging it due to cavitation. If you were putting the pump on the moon, the height would be different. You could use this on a rocket under static acceleration, but that would be a bit silly.

It is similar to describing acceleration in g's -- an arbitrary scaling factor. All that matters in your pump design (generally) is that the number is greater than 0.

NPSH is different than NPSP. NPSP removes scaling and expresses the value in pressure (mass/(length*time^2)). The pressures involved might increase or decrease due to the system's acceleration, but the value is still measured in pressure.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ We use it in rocket turbopumps. It's just an anachronistic way of expressing a pressure. $\endgroup$ – Rikki-Tikki-Tavi Mar 17 '15 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.