Cost of energy for pressure drop across filters 2

[mussayab]

Hi Everyone,

I want to estimate the cost of pressure loss across filters in a pumping system. These filters are located ahead of the heat exchangers in an open loop. Right now we are use 25 micron filters and I am suggesting that we do not need that much filtration for the cooling tower and that we can change the filters to 100 microns at least which will reduce the pressure drop.
Is there a specific formula with which I can calculate the the cost by reducing from 25 psi to lets say 5 psi.

Thanks

 

[1gibson]

Convert the psi to ft of head (psi*.434/sg = ft) run the pump curve after subtracting the difference, see what the difference in HP is, get a new amperage draw of the motor, calculate electricity cost.

You'll need to fill in some blanks, I assume the pump is existing and you won't trim the impeller, so that may limit the savings. How you control the flow of the pump will be a major factor, it might not help at all with saving money, but you would get some extra flow through the system.If you gave a flowrate, the difference in "water horsepower" could be calculated and ignoring changes in pump/motor efficiency, someone could come up with a rough guess.

If you don't save money due to lack of ability to control the pump, and you don't benefit from the extra flow, then just leave it alone. I guess a 100 micron filter will be cheaper than a 25 micron filter so if there are no negatives, then go for it next time you buy filters.

 

[mussayab]

Thanks 1gibson for you reply.

I already tried to use that by converting the psi to ft. of head and calculate the horsepower for the reduction, but these are huge pumps I am talking about 200 hp each (3 pumps), 2500 gpm per pump. With the reduced head it does not fall in the pump operating region and there is no way for me to estimate the horse power then. Yes, we are putting VFD's on all three pumps to control the flow.

 

[1gibson]

Ok, you can calculate it based on the concept of water horsepower. Here is a basic calculator:

http://easycalculation.com/physics/fluid-mechanics/water-horsepower.php

You think you can reduce the system resistance by 20 psi (20 / .434 * 1.0 = 46 ft of head.)
(Note my error in previous post, to get head from psi you divide by .434, not multiply.)

46 ft at 2500 gpm = 29.04 water horsepower that you are saving. So in the real world, because you don't have to deal with the (let's assume 75%) pump + motor combined efficiency needed to create the 29.04 water hp, you save 29.04/0.75 = 38.72 "real world" hp.

With the reduced head requirement, at a reduced speed, it will absolutely fall into the pump operating region. Head drops off to the square of the RPM difference, and flow is a 1:1 ratio with RPM. Google "Affinity laws" for details.

With the flexibility of the VFD, you will be able to capture a large portion of those energy savings, probably at least 30hp. It looks viable to me, although your pressure drop estimate does seem high, so don't spend that money yet.

 

[MortenA]

Two things:

1) If its an open loop system VFD could be an inefficient way of controlling your flow.
2) If you should control your flow with a valve it wont matter what the dP is across the filer (wrt to energy) - it will just be compensated by the FCV.

Best regards,Morten

 

[LittleInch]

If you've got filters working at 25 psi differential you've either got blocked filters or they are too small. Normally filter DP is allowed to go form 0 to 10 psi max before they need cleaning.

Hence filter pressure drop is normally neglected in any power saving or such calculations. Optimising your pump to match your system curve is a good idea, but there are many ways of doing that.

Filtration is a protection device for the d/s equipment. Before you go around changing it you need to understand why 25 micron was chosen in place of a higher mesh size. What is D/stream that requires clean fluid? how often do you change the filters? what is the cleanliness of the fluid, what is the potential impact of 100micron particles?

Only you know so we can't make a judgement on that.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Artisi]

Think the simple approach would be to obtain the pressure drop across the filter at the flow rate under consideration(filter manufacturer for detail),then simply apply the formula for calculating power, l/s x m X SG. / 102 x E = kW or USgpm x ft x SG / 3960 x E = HP. This should give you a good estimate to calculate the power costs.
You will need to balance this against any increased flow you might achieve but bear in mind that any increased flow will increase the head thru the overall system and possibly lower power input - so who knows what the real answer will be.

Ideally just do what 1gibson has advised, buy a set of 100# filters and try it out.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[bimr]

Use the online calculator:

http://www.engineeringtoolbox.com/water-pumping-costs-d_1527.html

The costs of pumping water can be calculated as


C = 0.746 Q h c / (3960 μp μm) (1)

where

C = cost per hour

Q = volume flow (gpm)

h = head (ft) (Note: 20 psi equals 46.2 FT)

c = cost rate per kWh

μp = pump efficiency

μm= motor efficiency

Note that when the filter has lower head loss, the pump curve will move to the right causing more flow to be pumped (with higher pumping cost).

I agree with you that 25 microns is too much filtration and 100 microns filter will suffice.

 

[btrueblood]

I agree with LittleInch, the cost in power savings must be balanced against the recurring maintenance/replacement costs of wear/clogging of the downstream equipment due to larger particles.

 

[Compositepro]

What kind of filters are being used? In a cooling tower loop it is most common to use basket strainers to remove the large trash and debris before the pump, and use a back-washable sand filter. A sand filter is a depth filter which will catch both large and very small particles. The fine particles may pass through the filter a number of times before finally being caught. There really is no reason to use a troublesome absolute filter that will not let any particles through that are above its rating, but will quickly plug and require manual change-out.

 

[mussayab]

Thanks bimr. This is the formula I was looking for. Increase in flow with the reduction in the pressure loss is not going to be a problem as we are putting VFDs on the motors.

Thank you guys for your responses. Really appreciate it. Keep up the good work!