Pump HP Vs Energy Balance
Pump HP Vs Energy Balance
(OP)
Good evening all,
I have a Goulds 6x8-15/11.5 731+ pump running at 1750 rpm. Looking at the curve the pump is rated for 2000 gpm at 100 ft head. The curve shows our 75 hp motor should be more than sufficient.
We are planning on switching a condenser over to the cooling tower loop this pump supplies. The condenser only requires 250 gpm, and is approximately 50 ft in elevation.
With out doing extensive friction loss calculations on the run of 4 & 8 inch pipe between the pump and the condenser, I wanted to do just a simple double check to see how much power was required to move a quantity of 250 gal of water, 50 ft in the air every minute. What I came up with doesn't make sense, what did I do wrong?
250 gal = 2090 lbm (pound mass)
Height = 50 ft
Weight x gravity x height = change in potential energy
(2900 lbm) x (32 ft/s2) x (50 ft) = 3,344,000 lbm*ft2/s2
1 lbs*ft/s2 = 1 lbf (pound force)
3,344,000 lbf*ft, this is actually per minute, so divide by 60.
55,700 lbf*ft/s
And 550 lbf*ft/s = 1 hp
So 100 hp is required to move 250 gallons up 50 ft. This can't be the case as 75 hp is capable of doing nearly x20 as much!
What did I miss? Was my assumption that the increase in potential energy would be equal to the work done wrong?
I have a Goulds 6x8-15/11.5 731+ pump running at 1750 rpm. Looking at the curve the pump is rated for 2000 gpm at 100 ft head. The curve shows our 75 hp motor should be more than sufficient.
We are planning on switching a condenser over to the cooling tower loop this pump supplies. The condenser only requires 250 gpm, and is approximately 50 ft in elevation.
With out doing extensive friction loss calculations on the run of 4 & 8 inch pipe between the pump and the condenser, I wanted to do just a simple double check to see how much power was required to move a quantity of 250 gal of water, 50 ft in the air every minute. What I came up with doesn't make sense, what did I do wrong?
250 gal = 2090 lbm (pound mass)
Height = 50 ft
Weight x gravity x height = change in potential energy
(2900 lbm) x (32 ft/s2) x (50 ft) = 3,344,000 lbm*ft2/s2
1 lbs*ft/s2 = 1 lbf (pound force)
3,344,000 lbf*ft, this is actually per minute, so divide by 60.
55,700 lbf*ft/s
And 550 lbf*ft/s = 1 hp
So 100 hp is required to move 250 gallons up 50 ft. This can't be the case as 75 hp is capable of doing nearly x20 as much!
What did I miss? Was my assumption that the increase in potential energy would be equal to the work done wrong?





RE: Pump HP Vs Energy Balance
I don't like working in english units in instances like this, so it's easier for me to convert to metric, do the calculation, and then convert back to english terms.
RE: Pump HP Vs Energy Balance
250 gpm at 50 ft?
250 x 50 / 3960 = 3.1 hp if 100 % efficient, and SG = 1.0
RE: Pump HP Vs Energy Balance
Sorry for bugging yall over something so trivial, but I did learn something out of it.
Thanks again.
RE: Pump HP Vs Energy Balance
The full equation TenPenny references is a good one to keep for your files...
(FLOW,gpm) X (HEAD,ft.) X (Specific gravity)
HP = ----------------------------------------------
3960 X Pump Efficiency,%
RE: Pump HP Vs Energy Balance
For interest, what are you / were you going to do with the surplus flow and head - throttle it with a discharge valve?
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.)
RE: Pump HP Vs Energy Balance
YOU WILL NEED A LITTLE ENERGY.
BUT YOUR PUMP WILL BE VERY FAR FROM ORIGONINAL TO YOUR DUTY POINT, SO YOU WILL HAVE ANOTHER HYDRAULIC AND MECHANICAL PROBLEMS.
YOU WOULD LIKE TO HANDLE 250 GPM WITH 2000 GPM PUMP.
IT SOUNDS THAT YOU HAVE TO CHANGE THE PUMP, FOT LITTLE ONE.
RE: Pump HP Vs Energy Balance
These pumps are used to supply cooling to a pair of 8000 gallon reactors and to overhead condensers. As time progresses, we are seeing higher run time for these applications.
I do not throttle the discharge of the pump. Rather, I have a bypass loop at the end of the line to provide an outlet for the excess flow from the pumps.
DubMac, I had seen that equation before and that is something I had been looking for. Thankyou!!