1. Pumps and Lift Stations

2. Background Fluid Moving Equipment Velocity (flow rate) Pressure
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

3. 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

4. .

7. Difference in height between source and destination
Static Head
destination
source
Static
head
Difference in height between source and destination
Independent of flow
Static
head
Flow

8. 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
Friction
head
Flow

10. Net Positive Suction Head
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

11. 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%.

12. Power Requirement Example
Lower efficiency means bigger motor and electrical controls, higher operating costs

16. Sump Sizing Example Determine suitable sump size 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

17. Sump Sizing Example 4 . Storage Volume (ft3) = 2 x Q (gpm)
n (cycles/hr)
Area of a Circle = 3.14 x (Diameter)2

18. Design Flowrate
Q = 18.9 x DC x A
Q = 18.9 x x 28 = gpm
Say use Q = 200 gpm
Sump Volume Calculation
(5 cycles/hr)
Vol = 2 x Q / n
Vol = 2 x 200 / 5 = 80 ft3

19. 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

21. 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

22. 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?

23. 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