# Centrifugal Pump curve Performance

1. Nov 26, 2015

### peter010

Dear,

I have a confusion related to the attached image which represents the curve performance of centrifugal pumps. My point is, why is the consumed power, on the same pump curve, is increased as the flow increases?!!! since the logic for me is the opposite at which the consumed power should increase as the pressure drop increases !

From another perspective, I know that the consumed current is always in a proportional relationship with the torque/ load, so how its come when the pump has less load\pressure-drop and thus higher flow rate to consume more current by its driving motor?

<The attache drawing was taken from: http://blog.craneengineering.net/how-to-read-a-centrifugal-pump-curve

Regards.

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• ###### Centrifugal_Pump_Curve_1-3.jpg
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2. Nov 26, 2015

### stockzahn

The power consumed by a pump is

P = Q⋅h⋅ρ⋅g

Q ... volume flow
ρ ... density of the fluid
g ... gravitational acceleration

That means, if you double the volume flow and the pressure head is not halved, the power consumption is higher at higher flow (if the efficiency η = 100 %).

Imagine, theoretically, the pump curves are horizontal. If you increase the volume flow with the same head, the power must increase. If the progress of the curve depending on the geometry of the pump is "flat" the additional volume flow outweighs the decrease in pressure head.

3. Nov 26, 2015

### peter010

Hello,

@stockzahn I appreciate your participation :)

Is this example apply over when we double the head but keep the flow constant ? will it then consume the same power as if we double the flow at constant head?

As per the pump curve performance it does not ! since when we decrease the flow, this means higher head, but the consumed power will be less !!!

Last edited: Nov 26, 2015
4. Nov 26, 2015

### stockzahn

Yes, assuming the efficiency is constantly 100 % over the entire range of the pump field. Whereas the first example corrisponds better to piston pumps and the second one to centrifugal pumps.

I'm not sure, if I understand that correctly, but the shape of the curves depend on the efficiency of the geometry of the pump. If a vertical pipe is installed at the pump outlet, longer than the maximal pressure head of the pump, there would be no volume flow (the efficiency is zero) and all the electrical power would heat up the fluid. So zero flow, maximal head and the necessary electrical power is lower compared to a state, where an actual volume flow is achieved.

5. Nov 26, 2015

### peter010

Thank you @stockzahn ,

May I have the following inquiries:

"If a vertical pipe is installed at the pump outlet, longer than the maximal pressure head of the pump, there would be no volume flow (the efficiency is zero) and all the electrical power would heat up the fluid. So zero flow, maximal head and the necessary electrical power is lower compared to a state, where an actual volume flow is achieved."

-> if there is no flow because of the maximal head, then according to the law: P = Q⋅h⋅ρ⋅g, P would equal Zero since v=0. Is this logical, because I'am wondering about the power used to move the fluid almost until the pipe outlet.

-> From another perspective, i'm wondering regarding the case when the valve after the pump is totally close, then v=0, so the consumed power would be zero as well?! even though the impeller is trying to rotate and to pump a flow.

6. Nov 26, 2015

### Staff: Mentor

Since the power is proportional to BOTH the flow and head, if one is going up and the other going down, I depends on how much they are changing.

7. Nov 27, 2015

### peter010

8. Nov 28, 2015

### stockzahn

Of course the fluid won't be heated directly, but accelerated/pressurized, while small flow systems are generated, where all the energy is transformed to heat due to friction/turbulence. So generally the kinetic/mechanic energy of the fluid is zero and all the electrical power/work done by the pump is stored in a temperature increase of the fluid. For this point the electrical power necessary is smaller than at the pump's design point → although the generated heat is higher, the power is lower.

As written above, the consumed power is not zero, it just has different values along the pump curve. The benefit (PFl = Q⋅h⋅ρ⋅g) at Q=0 and Δh=0 would be zero, but all the consumed electrical power is converted into heat, which is not the form of energy a pump should achieve, therefore the efficiency is zero:

η=PFl/Pel

- The necessary electrical power along the pump curve is not constant.

- Depending on the type of pump (geometry, pumping concept) the power consumption increases with increasing head or increasing volume flow (piston pumps generally consume more electrical power, if the head increases (when the valve at the outlet is closed) / hydrodynamic pumps tend to consume more electrical energy, if the mass flows are increased).

-The efficiency gives you the fraction of the consumed energy, which is converted into the wanted form (PFl / benefit) and which is converted into an unwanted form (heat). If the pressure head is maximal (wheter by a closed valve or a vertical pipe) all the consumed power is converted into heat - but not directly, because the blades of the pump are not heated up, but by generating small flow systems (which create the necessary pressure profile to preserve the head / the pressure gradient before the closed valve), where the kinetic energy of the fluid after accelerated by the impeller is converted into pressure and back to kinetic energy, which is converted into heat due to (viscous) friction. These transformations between the electrical energy and the heat are normally neglected in linguistic usage and it is said, that all the power is converted into heat.