# Pump power and fluid viscosity

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;

• Are you using the correct formula? See neutrium.net/equipment/pump-power-calculation Jun 3 '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 Jun 3 '18 at 20:59
• If you now say R=0 then are you assuming the pump is 100% efficient? Jun 3 '18 at 21:09
• 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 4 '18 at 1:46
• @SolarMike I have never said that! I said R=0 in YOUR example. Jun 4 '18 at 21:26

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: 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)