Pumps and Lift Pumps and Lift StationsStations

BackgroundBackgroundFluid Moving Equipment

Fluids are moved through flow systems using pumps, fans, blowers, and compressors. Such devices increase the mechanical energy of the fluid. The additional energy can be used to increase

• Velocity (flow rate)• Pressure• Elevation

Why are most water towers 130 feet high?

Because a column of water

2.31 ft high exerts a

pressure of 1 psi, and most

city water systems

operate at 50 to 60 psi

Static Head Difference in height

between source and destination

Independent of flow

destination

source

Statichead

Statichead

Flow

Friction Head Resistance to flow in pipe and fittings

Depends on size, pipes, pipe fittings, flow rate, nature of

liquid

Proportional to square of flow rate

Frictionhead

Flow

The allowable limit to the suction head on a

pump

33 - friction losses through the pipe, elbows, foot valves and other

fittings on the suction side of the pump

Net Positive Suction Net Positive Suction HeadHead

Determine the power required to pump 2000

gpmagainst a head of 14 ft, if the efficiency of the pump

is75% or 50%.

Power Requirement Power Requirement Example Example

Power Power Requirement Requirement

Example Example

Lower efficiency means bigger motor and electrical controls, higher operating costs

DC = 0.375” per dayArea drained = 28 acresLift (from tile outlet in sump todischarge pipe outlet) = 6 ftLength of discharge pipe = 25 ftPump cycles per hour = 5 and 20.

Determine suitable sump size

Sump Sizing Sump Sizing Example Example

Storage Volume (ft3) = 2 x Q (gpm)

n (cycles/hr)

Sump Sizing Sump Sizing Example Example

Area of a Circle = 3.14 x (Diameter)2

4 .

Design Flowrate

Q = 18.9 x DC x A

Q = 18.9 x 0.375 x 28 = 198.5 gpm

Say use Q = 200 gpm

Sump Volume Calculation (5 cycles/hr)

Vol = 2 x Q / n

Vol = 2 x 200 / 5 = 80 ft3

Min Required Volume = 80 ft3

-Try 4 ft diameter well

Area = 3.14 x 42 / 4 = 12.6 ft2

Storage Depth = 80 /12.6 = 6.35 ft

Try 6 ft diameter well

Area = p x 62 / 4 = 28.3 ft2

Storage Depth = 80 /28.3 = 2.83 ft

Flowrate varies with rotational speed:

Q1/Q2 = N1/N2

Head varies with rotational speed squared:

H1/H2 = (N1/N2)2

Power varies with rotational speed cubed:

P1/P2 = (N1/N2)3

AFFINITY LAWSAFFINITY LAWS

A pump with an efficiency of 80%, connected to a diesel engine, pumps 200 gpm against a head of 12 ft. What is the power output of the engine? What will be the flow rate, head and power output if the motor speed is increased from 500 rpm to 600 rpm?

Affinity Laws Affinity Laws Example Example

P =(200 x 12)/(3960 x 0.8) = 0.76 hp

Q2 = 200 x (600/500) = 240 gpm

H2 = 12 x (600/500)2 = 17.3 ft

P2 = 0.76 x (600/500)3 = 1.3 hp

Affinity Laws Affinity Laws Example Example