1

I used to calculate the power needed by a pump using Bernoulli equation,

$ W =(p_2-p_1) + {1 \over 2}\rho(u_2^2-u_1^2) + g\rho(h_2-h_1) + R $

where $\rho$ is fluid density (and assuming it is constant), subscript 1 is inlet, 2 is outlet, $p$ is pressure, $u$ is speed, g is acceleration of gravity, $h$ is and $R$ are pressure drops (see EDIT2).

In this formula fluid properties are represented by $\rho$. What will happen if I change the flowing fluid? e.g. water for mineral oil.

With water, I need more power than to flow the same amount of mineral oil, due to its lower density.

How this assumption could be right/complete!? Since does not appear the property of viscosity of the fluid anywhere and that's just the resistance of a fluid to the deformation/movement?

EDIT1: fixed the Bernoulli formula; add $\rho$ assumption;
EDIT2: clarify $ R$ term: since we are talking about a pump, losses are lumped, not distributed, we are not talking about losses in a pipe, thus they are not Reynolds dependants. Thus $R$ does not depend on fluid properties but relies only on inlet/outlet pump geometry;

mattia.b89
  • 335
  • 3
  • 15
  • Are you using the correct formula? See https://neutrium.net/equipment/pump-power-calculation/ – Solar Mike Jun 03 '18 at 18:15
  • @SolarMike I have corrected the OP formula, now it's correct! anyway the equation you point out it's just a simplification of the Bernoulli equation, under the hypothesis: 1. p1=p2, u1=u2 and R=0 – mattia.b89 Jun 03 '18 at 20:59
  • If you now say R=0 then are you assuming the pump is 100% efficient? – Solar Mike Jun 03 '18 at 21:09
  • @SolarMike I have never said that! I said R=0 in YOUR example. – mattia.b89 Jun 04 '18 at 21:26
  • So 3 on here seem to point out you have it wrong so.... – Solar Mike Jun 05 '18 at 04:03
  • Your equation does not have units of power. What do you mean by "the same amount": volume flow rate or mass flow rate? – Bob Jun 04 '18 at 01:46
  • Yeah, it has the pressure unit, which you can also see it as work over volume and again as power over flow rate. Pa=N/m^2=(N*m)/(m^2*m)=J/m^3=(J*s)/(m^3*s)=W/(m^3/s). Second, because density is constant, there is no need to underline it out, since they are the "same" – mattia.b89 Jun 04 '18 at 21:36
  • Actually, it would be power "per unit volume flow rate." You mentioned calculating power, not normalized power/V_dot, with the Bernoulli equation. Which again raises the question: what do mean by 'same amount' because V_dot is not the same for a constant mass flow rates of both water and mineral oil, unless there are temperature games to add to the word game. It seems you already received an answer, but it wasn't to your liking. If this needs further discussion, please clarify your question. – Bob Jun 05 '18 at 11:38

2 Answers2

2

Strictly speaking, the energy content of the moving fluid is a function of pressure/head and velocity, so that vscostiy does not show up in your equation is no wonder.

The viscosity mostly changes the behavior in the pipe system, you will have more friction losses, so for the same flow rate you need a higher head.

Also, with higher viscostiy the efficiency of your pump falls. This nomogram, cribbed from a planning handbook from KSB & based on the methodology of the Hydraulics Insititue, shows how you can find the correct efficiency + correction factors for other values:

enter image description here

So if you know the values for water for flow rate $Q_w$, head $H_w$ and efficency $\eta_w$, you find the correction values $k_{Q}$, $k_{H}$ and $k_{\eta}$ from the nomogram. This relation holds true from Q between 0.8 - 1.2 time Q at best efficiency point. So do the conversion at BEP and one operating point higher and one lower than BEP and you should be able to draw a new pumping curve.

As for power, I believe your formula is wrong because of the way it accounts for efficiency. Power is given by:

$$P=\rho g H Q\frac{1}{\eta}$$

obviously using the corrected values!

Source: KSB, Auslegung von Kreiselpumpen (apparently no longer available online)

mart
  • 4,711
  • 1
  • 15
  • 28
1

Bernoulli equation is derived assuming the viscosity of fluid is zero (inviscid fluids). Thus, this equation is not reliable when viscous forces are comparable to the inertial forces in the Naiver-Stokes equations, which is the case in the flow of mineral oil for example.

Mohammad Amin Kazemi
  • 11
  • 1