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Pump VFD

08/01/2015 12:19 PM

If I put VFD on 50HP 460V pump and decrease speed 10%, how much do I save energy in percentage? Is it proportional or not?

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#1

Re: Pump vfd

08/01/2015 12:39 PM

It's not worth the effort.

The cost of the VFD and installation will never be recovered on a pump installation with a 10% speed reduction..

Trimming the impeller will be much cheaper.

Throttling the output 10% will be the easiest way to reduce power consumption.

My advice, DO NOTHING.

Talk to a pump expert.

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#2

Re: Pump vfd

08/01/2015 12:56 PM

I decry it will depend on a lot of things you are not telling us.

The added losses of the VFD could eat up all running speed efficiency at the pump. A decrease in speed could also move the pump out of the most efficient speed itself. Then again if your operation requires frequent stopping and starting the higher efficiency of starting with a VFD will help greatly in the overall process. If the impeller is spinning so fast that cavitation is happening the decrease in speed to below cavitation levels will be a significant non-linear improvement. Pumping a compressible fluid will change all of the equations all together.

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#3

Re: Pump vfd

08/03/2015 10:29 AM

I agree with redfred

I like to add that your process requirements should dictate your pump requirements and speed.

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#4

Re: Pump vfd

08/03/2015 10:38 AM

By pump affinity laws, 1.00 - 0.93 = 1.00 - 0.729 = 0.271 = 27.1% savings. However, as others have noted, there may be a decrease in efficiency, so the savings probably won't be quite that high.

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#5

Re: Pump VFD

08/03/2015 12:25 PM

Dear Mr ZORANDODO,

We have studied in our college days that for Pumps (or Fans) the EQUATION IS AS FOLLOWS:

1. Capacity "C" is Proportional to the SPEED - N

2. Pressure "P" is Proportional to the SQUARE of the SPEED - N^2

3. Power "POWER" is Proportinal to the CUBE of SPEED - N^3

You have stated that the speed is reduced by 10% i.e equipment is running now at 0.9 N

So "C" = 0.9 N, Pressure "P" (0.9 x N )^2 = 0.81 x P, and POWER = (0.9 x N )^3 = 0.729 x ORIGINAL POWER at Speed N

Power Saved = (Power consumed at Speed N - Power consumed at Speed 0.9) = (1-0.729) = 0.271 i.e 0.271 x100 = 27.1%

DHAYANANDHAN.S

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#6

Re: Pump VFD

08/03/2015 2:55 PM

You need to look at your process requirements, first. A throttling valve will be cheaper than a VFD.

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#7
In reply to #6

Re: Pump VFD

08/03/2015 3:19 PM

Agreed.

A quick look at the pump curve should give a ball park indication of power savings.

Unless he's pumping mud, 10% throttling won't over-heat anything either.

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#8
In reply to #6

Re: Pump VFD

08/03/2015 6:19 PM

Assuming of course that it is a centrifugal pump.

Not a good idea to throttle the output of a PD pump, a recirculating valve would reduce the output but would do nothing for power consumption.

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#9

Re: Pump VFD

08/03/2015 8:49 PM

It depends on the design and type of pump as well as the process requirements.

Measure the pump flow and electrical current on the VFD input side and on the VFD output at 100% speed.

Decrease the speed by 10% take flow and current readings at same locations then compare the differences to identify any efficiency gains realized.

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