# Output pressure of centrifugal pump

I'm trying to get my bearings (a somewhat heavy-handed expression) on pump design and I'm finding in the various books I've procured on the subject that there is very little consideration given to the output pressure developed by a pump.

If there isn't much detail to this question, I apologize; I'm still in the far-below-basic level of understanding. Obviously, head is a key parameter of pump design, but my concerns are with output pressure. The reason being, I'm considering for example rocket engine design where the key parameters of pump requirements are chamber pressure and flow rate. I understand that flow rate varies with head by the characteristic Q-H curves.

So, how does pump head vary with or affect the output pressure developed by a pump? Are the two concepts related, or are they independent of one another? If they are related, how would I go about figuring out the required head parameters from the pressure and flow requirements?

## 1 Answer

Head and pressure are effectively the same thing.

It's impossible to look at a pump and say "this pump will provide 10 bar of pressure." One of my favorite aspects of fluids has always been how interconnected an entire fluid system is, and how it stabilizes itself on its own. So the pressure that's actually supplied depends on the rest of the system: the losses in the pipe, the inlet and outlet conditions of the fluid to the system, the fluid itself, and how much power you're supplying to the pump.

Hopefully you've come across the Bernoulli equation somewhere in your reading.

$$P+{\rho}gz+{\frac12}{\rho}v^2=constant$$

(Remember that this can only be used for incompressible flows, but since we're talking about pumps and probably only common liquids at relatively low speeds, the incompressible assumption is fine. It only becomes an issue when $Mach>0.3$.)

In the form above, we see that the terms together will output a constant in units of pressure. However, we can rewrite to be in the form

$$\frac{P}{\rho g}+z+\frac{v^2}{2g}=constant$$

In this case, the constant has units of length (more specifically, units of length of a particular fluid, e.g. mmHg), and it's referred to as the head. But this term is fundamentally no different from the constant we get in the first equation, it's just expressed differently. $H=P_{total}/{\rho g}$, but because this is an imcompressible flow, the density is constant, and we know that $g$ won't change significantly unless we're moving the fluid over a massive height differential.

• Ah. It seems it's like any electrical system. The power delivered to the load depends on the load itself. So, if I'm understanding correctly, if the load and output network were known in terms of pressure (calibrated orifices and what not), it would be trivial to figure out the head requirement on the pump. – ecfedele May 12 '15 at 21:46
• Pretty much. I wouldn't entirely call it "trivial," but once you have the losses identified and quantified (usually in terms of flow velocity), it's certainly not hard to find the head requirement. And for what it's worth, there are a lot of comparisons to electricity in fluids and heat transfer. – Trevor Archibald May 12 '15 at 22:07