Motor Pump-Horse Power Needed?

Question

If i needed to pump 900,000 litres of water per minute at 1,000 bars (14,500 psi), What horse power motor would I need.

Whats the calculation or equation that you use?

Answer 1

#Efficiency=shaft horse power/motor horse power

#Motor horse power
=shaft horse power/efficiency


# Shaft horsepower
=GPM * PSI / 1714
[900 000 liter = 237 754.85 gallon [US, liquid]]
=[237 754.85*14500]/1714
=2011344.99 hp
#The efficiency of most pressure washer pumps is about 85% or less,but we will consider 85% efficiency.
# Motor horse power
=2011344.99*100/85
=2366288.22 hp.
!!!!!!!!!!!!!!!!!!!
# It seems to be unrealistic.

# If you want to achieve this feet and make your system full-proof,then you have to go for some other arrangement.

# For flow you have to run the pumps in parallel[at least 3 pumps]
# For 14500 PSI PRESSURE you have to go for booster pumps or you have to run the pump in series.

# It means you have to run the pumps in parallel as well as each pumps have atleast 2 no. of booster pumps in series.

# If you are running this system for continuous plant , then you must have a stand-by system also and lot of inventories of pump spares.Also continuous power supply.

# There is no scope for positive displacement pump like reciprocating,gear,vane,piston ,screw pump or even regenerative turbine pump.These type of pumps are meant for high head-low flow applications.So, you have to go for multi-stages and series network of centrifugal pumps.

Answer 2

900,000 X .26417 = 237,753 gallons per minute
1000 X 33.4883 = 33,488 feet of head
Hp = (GPM X Ft. Head X Specific Gravity) / (3960 X Pump Efficiency)
For water, specific gravity = 1
Assume the pump is 100% efficient; efficiency = 1
237,753 X 33,488 / 3960 = 2 million horsepower

The largest pumps and motors in existence are probably the hydroelectric turbines and generators at the Bath County pumped storage power station in Virginia. When operating as a motor and pump, each unit uses 563,400 horsepower and pumps 1,833,333 gallons per minute. Those pumps raise the water only about 1260 feet. From those figures, it appears that the pumps are about 96% efficient.

The largest of the original pumps in the New Orleans pumping stations lifted about 450,000 gallons per minute about 10 feet. That required only about 1200 horsepower.

Pumping 237,753 gallons per minute at 14,500 psi is probably totally unrealistic.

Answer 3

That's about 237800 gal/min at a head of 33500 feet-H2O.

My reference book doesn't even go up that high. This is a *very* large pump, requiring a very large motor.

There are no straight formulas for figuring this out because different types of pumps work differently. There are displacement pumps, centrifugal pumps, and multi-stage pumps. Each has their own characteristics, each has their own inefficiencies, and the motors that would be attached have their own inefficiencies -- all of which cannot be captured in a simple formula. Besides all of that, the fluids' viscosity has to be taken into account.

Perhaps there are some heavy industries which could produce such a pump on a custom order.

Just to give you an idea of what you are dealing with. One of the pumps at the Grand Coulee Dam in Washington, USA, which pumps water from behind the dam up to a large irrigation canal 300 feet above, is rated for 2.72 million liters per minute (on order with your requirement), but only has a head pressure of 300 feet (your requirement is 100 times greater). That pump is rated for about 70,000 HP, so your pump will need to be rated for on the order of 2 to 5 megaHP.

.

Answer 4

For that much water per minute sounds like you need one of the pumps that was sold to the Army Corp of Engineers that were used to dewater New Orleans after Katrina hit a few years back. If I recall correctly it came from an oil field supply company in Oklahoma.

Answer 5

thr equation for horsepower is
hp = (quantity)*(unit weight of fluid)*(head or pressure)/550

The above units must all be english units (lbs-feet-seconds) or the 550 will change. you must also correct for efficiency of the pump. if you dont know this it is probably accurate enough to just use twice the horsepower that the equation gives you.