Pumps and Lift Stations

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# Pumps and Lift Stations - PowerPoint PPT Presentation

Pumps and Lift Stations. Background. Fluid 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.

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Presentation Transcript

Background

Fluid 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

destination

source

Static

Difference in height between source and destination

Independent of flow

Static

Flow

Resistance to flow in pipe and fittings

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

Proportional to square of flow rate

Friction

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

Power Requirement Example

Determine the power required to pump 2000 gpm

against a head of 14 ft, if the efficiency of the pump is

75% or 50%.

Power Requirement Example

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

Sump Sizing Example

DC = 0.375” per day

Area drained = 28 acres

Lift (from tile outlet in sump to

discharge pipe outlet) = 6 ft

Length of discharge pipe = 25 ft

Pump cycles per hour = 5 and 20.

Determine suitable sump size

Sump Sizing Example

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

n (cycles/hr)

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

AFFINITY LAWS

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 Laws Example

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 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