motor size calculation from pump required power 1

[gilles56]

Hi, I am comfused with the power consumed by the pump and the power deliver to it.
Could you tell how to get the right motor size (kW) knowing pump total Head.

Ta,.

 

[electricpete]

Here's an example calculation for converting 1600gpm and 465 psid into horsepower (unit conversion). I did this myself and I'm not 100% sure it's correct:
[1600 gal / min] * [465 lbf / in^2] * [144in^2/ft^2] * [1FT^3 / (7.48 GAL)] *(1hp /[33,000 ft-lbf/min]) ~ 434 HP.

The above amount must also be increased by (1/pump efficiency). There may be other factors in pump selection.

(As a side note, motor people use the quantity 5250 ft-lbf/min per horsepower when converting between Power (hp), speed (rpm) and Torque (ft-lbf). This conversion differs from the 33,000 ft-lbf/min per horsepower by a factor of 2*pi).

Now I'll cut and paste my response to your other question (motor bulleting board) where you asked how to determine motor size if you already know required pump horsepower.

From a sizing standpoint, you don't need to take the motor efficiency into account. The rated horsepower of a motor is expressed in terms of motor output power delivered to the shaft.

If motor will operate continuously at full load, I believe it is common practice to select motor approx 133% larger than required. This allows some margin for errors in the calculation or unexpected factors which may increase motor load. (better to get one a little to big than a little too small and have to run it overloaded). Also this means that if calculated load is correct, motor will be operating approx 75% - which is often the most efficient point (from motor standpoint) within the motor range.

 

[TBP]

Hi Gillies56 - You don't say what the liquid is that you're pumping, or the flow.

The formula I use is:

Bhp = QHSg/3960E

Bhp = brake horsepower
Q = flow of fluid in GPM
H = total head in feet
Sg = specific gravity of the fluid at the operating temperature
E = pump efficiency at the selected operating conditions

For example, if you have to move 60 GPM of brine (spec grav of 1.2), at a total head of 180 feet, using a pump with 60% eff:

60 x 180 x 1.2/3960 x 0.6 = 5.45 horsepower

It never hurts to talk to the pump manufacturer. They supply drives with their pumps all the time, and can give you info like efficiency, and the RPM a given pump needs to do the job.

 

[electricpete]

I tried using TBP's formula on my own example and got the same result as using my calc (approx 433HP).

One interesting aspect is that using my own formula, I did not need to know the fluid density (only the differential pressure). This works well for my particular situation since I'm dealing with an already-installed system and trying to evaluate power consumed based on measured flow rate (gpm) and measure differential pressure (psid). None of my meters give me head in feet or any indication of density.

At first it sounded like a paradox that one method requires me to know the density for calculation and the other does not. I realized the answer lies in the fact that in TBP's formula H depends on density (H = dp/rho) and Sg depends on density (dp = rho/rho_standard), but the product
(H*Sg) = dp/rho * rho/rho_standard = dp/rho_standard does not depend on density.

(I "depends on" is a relative term - determined by what info you start with..... in design situations density and head are more directly known and differential pressure is calculated from these).

If anyone's interested, here's how I reworked my example using tbp's formula given psid and gpm but no density:

H(ft-lbf/lbm) = (465 lbf/in^2) / (rho lbm/ft^3)*(144in^2/ft^2) = 66960/rho (ft-lbf/lbm)
[where rho is numerical value of density in lbm/ft^3]

H*Sg = H * (rho/rho_standard)
= 66960/rho (ft-lbf/lbm)* rho/62.4
= 1073 (ft-lbf/lbm)

SHP=
Q(HSg)/3960E = 1600*(1073)/3960 / E = 434/E (hp)

 

[TBP]

That's interesting, electricpete. There's almost always more than one way to skin a cat.

The formula (and example) I used is from an out-of-print book I have:

"Handbook of Data Sheets For Solution Of Mechanical Systems Problems" edited by Roose. It's great for all kinds of things - a couple of pages on about 150 different kinds of problems, including weird ones like "water flow in an above ground pipeline to prevent freezing". A page of write-up including a formula and examply, and a page that's a nomograph, showing the solution to the same example.

I find it really handy for problems I don't normally deal with. You don't need an in depth knowledge of a particular subject to get a quick answer. If nothing else, you at least get a feel for what's worth investigating further, and what is not.

 

[GreyGoose]

a few things to keep in mind:

Once you arrive at the size required from calculations, select a standard motor size (i.e. if your calculations result in a 433 hp pump, specify a 500 hp motor for your application). Refer to NEMA MG 1 or National Electrical Code (NFPA 70), Article 430.

Design for worse-case operating temperature (i.e. coldest expected) for the fluid; it may make a difference in total head and therefore change HP requirements.

The conversion from hp to kW is as follows:
kW = (746 * motor hp)/(motor efficiency)
Note, the motor efficiency used above is different form the pump efficiency in TBP's post.

the ratio of brake hp to motor rated hp is the load factor for that motor:
load factor = (Bhp)/(hp)