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# Rated Power and Actual Operational Expenditure of Centrifugal Pump

## Rated Power and Actual Operational Expenditure of Centrifugal Pump

(OP)
Dear All,

I am new in engineering world and currently I am struggling to solve one issue:
To calculate minimum needed motor capacity, I need to count :
1. Theoritical value of the pump (hydraulic power); (e.g : 1,96kW)
2. Rated power (hydraulic power / pump efficiency); (e.g : 1.96kW/ 0.55 = 3.59kW)
3. Actual needed power (rated power / motor efficiency), (e.g : 3.59kW / 0.9 = 3.95kW)
resulted maximum power needed by pump (3.95 kW)
And then I need to search motor capacity that is higher than 3.95kW (e.g : 5.5kW)

My Question is :
1. Is my calculation above correct?
2. To Calculate OPEX (I didn't use inverter or any other variable speed controller); Should I calculate using pump capacity (5.5kW); or actual needed power by pump (3.95kW); if pump run with same condition (same flowrate, head) with the one I use for theoritical value?
3. If no 2's answers is actual needed power by pump (3.95kW); why can I calculate OPEX from this value even though I didn't use variable speed controller?

Thank you very much for your support :)

Sincerely,

Felicia

### RE: Rated Power and Actual Operational Expenditure of Centrifugal Pump

1. Is my calculation above correct?
Yes - the pump and efficiency looks a bit low, but the calculation is correct.

2. To Calculate OPEX (I didn't use inverter or any other variable speed controller); Should I calculate using pump capacity (5.5kW); or actual needed power by pump (3.95kW); if pump run with same condition (same flowrate, head) with the one I use for theoritical value?
The motor capacity listed is normally the MAXIMUM the motor can supply as a SHAFT power. The motor will only consume the electricity it needs to meet the shaft power demand, hence first work out your flow and head required, work out the shaft power and then add motor efficieny to work out electrical power required.

3. If no 2's answers is actual needed power by pump (3.95kW); why can I calculate OPEX from this value even though I didn't use variable speed controller?
Because that is the actual electrical consumption of the motor.

Think of this like a car if you like. You want to travel at 70 mph which needs a certain amount of power from the engine. However you can only buy a car which is capable of 90 MPH flat out. However when you drive at 70 MPH you will use a less fuel than driving the same car at 90.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Rated Power and Actual Operational Expenditure of Centrifugal Pump

In engineering the mechanical engineer is required to give a quick ballpark estimate of equipment loads to the electrical engineer so they can start their design simultaneous with all other trades. Also you have to indicate who provides starter and strarter type or VFD (provided by mechanical contractor, equipment vendor or electrical contractor)

Quick estimate for pump bhp = (GPM x Ft Water Gauge Total Pressure)/(3960 x Eff)

Quick estimate for fan bhp = (CFM x Inches water gauge total pressure)/(6362 x Eff)

In both cases initial estimated Eff = 0.65

Later on in the design final selection of the equipment is made to finalize motor hp. Refer to pump manufacturer's pump selection programs that would also print out the pump curve and show intersection with system curve.

For pumps you need to calculate required total head and NPSH available then select type and rpm using pump curve plotted against system head. You have to make sure the motor hp is conservatively chosen so the pump does not overload at any possible location on its system curve. You have to make sure available NPSH is more than the required minimum NPSH. You have to make sure you have included all pressure drops through piping, fitting, equipment, strainers etc plus if open system include static head and the larger pipe/fitting pressure drop for open piping which accounts for more pipe corrosion (see carrier Handbook chapter on piping). Then you have to show on detail drawings how pump is to be installed. The engineering off ice would have stock details based on if pump is on the basement floor or on upper floors. Seismic design require flexible connection at pump inlet & outlet.

### RE: Rated Power and Actual Operational Expenditure of Centrifugal Pump

(OP)
Dear LittleInch,

Well noted, thank you very much for the explanation.
So if the actual flowrate is way lower than designed one, amount of electricity consumed will be lower as well without using variable speed controller?
For Example :
Designed :
Flowrate : 40 m3
Theoritical value of the pump (hydraulic power); (e.g : 1,96kW)
Rated power (hydraulic power / pump efficiency); (e.g : 1.96kW/ 0.55 = 3.59kW)
Actual needed power (rated power / motor efficiency), (e.g : 3.59kW / 0.9 = 3.95kW)
resulted maximum power needed by pump (3.95 kW)

Selected motor capacity is 5.5 kW and variable speed controller is not used,

1.If actual flowrate is 40m3 and head 18m, amount of electricity that is needed to be paid to electricity will be 3.95kW?
2. If actual flowrate is 30m3 and head 10m, amount of electricity that is needed to be paid to electricity vendor will be lower than 3.95kW?
3. If that so, then variable speed controller is actually not needed ?

Thank you very much for your attention.

Dear Liliput1,

Noted, thank you for the information :)

### RE: Rated Power and Actual Operational Expenditure of Centrifugal Pump

1 - Correct
2- partly correct. For the same pump at the same speed the pump will deliver 30m3/hr almost certainly at a higher head than it does at 40m3/hr - how much you need to see the pump curve and also find out it's new efficiency as this also varies with flow.

So yes, it should use less electricity, but how much less is not possible to say without a new calculation and new figures ( head, efficiency). To get to 10m head you need to add a restriction / control valve in the line to drop the outlet head from say 20m to your 10m that is required for the flow. This could happen next to the pump or at the end of the pipe, but either way there will need to be a pressure drop somewhere to force the flow to be reduced to 30m3/hr

3) If you understand the answer to 2 then you will be able to see that a pressure drop = "lost" energy. A VFD will reduce the flow and head as speed falls so that the pump is putting out closer to the head for the required flow. But in many cases a vFD might not make much of a difference to the energy used as the VFD itself uses some electricity.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

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