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

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 Static head Static head 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 Friction head 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 Head

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

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 (ft 3 ) = 2 x Q (gpm) n (cycles/hr) Sump Sizing Example Area of a Circle = 3.14 x (Diameter) 2 4.

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 ft 3

Min Required Volume = 80 ft 3 -Try 4 ft diameter well Area = 3.14 x 4 2 / 4 = 12.6 ft 2 Storage Depth = 80 /12.6 = 6.35 ft Try 6 ft diameter well Area = p x 6 2 / 4 = 28.3 ft 2 Storage Depth = 80 /28.3 = 2.83 ft

Flowrate varies with rotational speed: Q 1 /Q 2 = N 1 /N 2 Head varies with rotational speed squared: H 1 /H 2 = (N 1 /N 2 ) 2 Power varies with rotational speed cubed: P 1 /P 2 = (N 1 /N 2 ) 3 AFFINITY 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 Example

P =(200 x 12)/(3960 x 0.8) = 0.76 hp Q 2 = 200 x (600/500) = 240 gpm H 2 = 12 x (600/500) 2 = 17.3 ft P 2 = 0.76 x (600/500) 3 = 1.3 hp Affinity Laws Example