1. Pumps and Pumping Stations
Pumps add energy to fluids and therefore are accounted for in the energy equation
Energy required by the pump depends on:
Discharge rate
Resistance to flow (head that the pump must overcome)
Pump efficiency (ratio of power entering fluid to power supplied to the pump)
Efficiency of the drive (usually an electric motor)

2. Pump Jargon
(Total) Static head – difference in head between suction and discharge sides of pump in the absence of flow; equals difference in elevation of free surfaces of the fluid source and destination
Static suction head – head on suction side of pump in absence of flow, if pressure at that point is >0
Static discharge head – head on discharge side of pump in absence of flow
Static suction head
Static discharge head
Total static head

3. Pump Jargon
(Total) Static head – difference in head between suction and discharge sides of pump in the absence of flow; equals difference in elevation of free surfaces of the fluid source and destination
Static suction lift – negative head on suction side of pump in absence of flow, if pressure at that point is <0
Static discharge head – head on discharge side of pump in absence of flow
Static suction lift
Static discharge head
Total static head

4. Pump Jargon
Static suction head
Static discharge head
Static suction lift
Static discharge head
Total static head (both)
Note: Suction and discharge head / lift measured from pump centerline

5. Pump Jargon
(Total) Dynamic head, dynamic suction head or lift, and dynamic discharge head – same as corresponding static heads, but for a given pumping scenario; includes frictional and minor headlosses
Energy Line
Dynamic discharge head
Total dynamic head
Dynamic suction lift

6. Example. Determine the static head, total dynamic head (TDH), and total headloss in the system shown below.
El = 730 ft
ps =6 psig
El = 640 ft
pd =48 psig
El = 630 ft

7. H-W ‘C’ values are 120 on suction side and 145 on discharge side
Example. A booster pumping station is being designed to transport water from an aqueduct to a water supply reservoir, as shown below. The maximum design flow is 25 mgd (38.68 ft3/s). Determine the required TDH, given the following:
H-W ‘C’ values are 120 on suction side and 145 on discharge side
Minor loss coefficients are
0.50 for pipe entrance
0.18 for 45o bend in a 48-in pipe
0.30 for 90o bend in a 36-in pipe
0.16 and 0.35 for 30-in and 36-in butterfly valves, respectively
Minor loss for an expansion is 0.25(v22  v12)/2g
Short 30 pipe w/30 butterfly valve
El = 6349 to 6357 ft
El = 6127 to 6132 ft
30 to 48 expansion
4000of 48 pipe w/two 45o bends
8500of 36 pipe w/one 90o bend and eight butterfly valves

8. Determine pipeline velocities from v = Q/A..
v30= 7.88 ft/s, v36= 5.47 ft/s, v48= 3.08 ft/s
Minor losses, suction side:

9. Minor losses, discharge side:

10. Pipe friction losses:

11. Loss of velocity head at exit:
Total static head under worst-case scenario (lowest water level in aqueduct, highest in reservoir):
Total dynamic head required:

12. Pump Power
P = Power supplied to the pump from the shaft; also called ‘brake power’ (kW or hp)
Q = Flow (m3/s or ft3/s)
TDH = Total dynamic head
 = Specific wt. of fluid (9800 N/m3 or 62.4 lb/ft3 at 20oC)
CF = conversion factor: 1000 W/kW for SI, 550 (ft-lb/s)/hp for US
Ep = pump efficiency, dimensionless; accounts only for pump, not the drive unit (electric motor)
Useful conversion: kW/hp

13. Diameter D = 48 in; Roughness e = 0.003 ft, Efficiency Pe =80%
Example. Water is pumped 10 miles from a lake at elevation 100 ft to a reservoir at 230 ft. What is the monthly power cost at $0.08/kW-hr, assuming continuous pumping and given the following info:
Diameter D = 48 in; Roughness e = ft, Efficiency Pe =80%
Flow = 25 mgd = ft3/s
T = 60o F
Ignore minor losses
El = 100 ft
10 mi of 48 pipe, e =0.003 ft
El = 230 ft 2 1

14. Find f from Moody diagram
El = 100 ft
10 mi of 48 pipe, e = ft
El = 230 ft 2 1
Find f from Moody diagram

15. El = 100 ft
10 mi of 48 pipe, e = ft
El = 230 ft 2 1

17. Pump Selection
System curve – indicates TDH required as a function of Q for the given system
For a given static head, TDH depends only on HL, which is approximately proportional to v2/2g
Q is proportion to v, so HL is approximately proportional to Q2 (or Q1.85 if H-W eqn is used to model hf)
System curve is therefore approximately parabolic

18. Example. Generate the system curve for the pumping scenario shown below. The pump is close enough to the source reservoir that suction pipe friction can be ignored, but valves, fittings, and other sources of minor losses should be considered. On the discharge side, the 1000 ft of 16-in pipe connects the pump to the receiving reservoir. The flow is fully turbulent with D-W friction factor of Coefficients for minor losses are shown below.
K values
Suction
Discharge
0.10
0.12
0.20
0.30
0.60
1.00
1.00
40 ft
6 ft

19. The sum of the K values for minor losses is 2
The sum of the K values for minor losses is 2.52 on the suction side and 5.52 on the discharge side. The total of minor headlosses is therefore 8.04 v2/2g.
An additional 1.0 v2/2g of velocity head is lost when the water enters the receiving reservoir.
The frictional headloss is:
Total headloss is therefore ( )v2/2g = v2/2g. v can be written as Q/A, and A = pD2/ 4 = 1.40 ft2. The static head is 34 ft. So:

20. System curve
Static head

21. Pump Selection
Pump curve – indicates TDH provided by the pump as a function of Q;
Depends on particular pump; info usually provided by manufacturer
TDH at zero flow is called the ‘shutoff head’
Pump efficiency
Can be plotted as fcn(Q), along with pump curve, on a single graph
Typically drops fairly rapidly on either side of an optimum; flow at optimum efficiency known as “normal” or “rated” capacity
Ideally, pump should be chosen so that operating point corresponds to nearly peak pump efficiency (‘BEP’, best efficiency point)

22. Pump Performance and Efficiency Curves
Shutoff head
Rated capacity
Rated hp

23. Pump Selection

24. Pump Efficiency
Pump curves depend on pump geometry (impeller D) and speed

25. Pump Selection
At any instant, a system has a single Q and a single TDH, so both curves must pass through that point;  operating point is intersection of system and pump curves

26. Pump System Curve
System curve may change over time, due to fluctuating reservoir levels, gradual changes in friction coefficients, or changed valve settings.

27. Pump Selection: Multiple Pumps
Pumps often used in series or parallel to achieve desired pumping scenario
In most cases, a backup pump must be provided to meet maximum flow conditions if one of the operating (‘duty’) pumps is out of service.
Pumps in series have the same Q, so if they are identical, they each impart the same TDH, and the total TDH is additive
Pumps in parallel must operate against the same TDH, so if they are identical, they contribute equal Q, and the total Q is additive
Adding a second pump moves the operating point “up” the system curve, but in different ways for series and parallel operation

28. Example. A pump station is to be designed for an ultimate Q of 1200 gpm at a TDH of 80 ft. At present, it must deliver 750 gpm at 60 ft. Two types of pump are available, with pump curves as shown. Select appropriate pumps and describe the operating strategy. How will the system operate under an interim condition when the requirement is for 600 gpm and 80-ft TDH?
TDH (ft)
Flow rate (gpm)
Pump A only
Pump B only
System curve
Efficiency, %
Pump B
Pump A 10
100 40 50 30 20 90 80 70 60
120
110
200
400
1200
1000
800
600

29. Either type of pump can meet current needs (750 gpm at 60 ft); pump B will supply slightly more flow and head than needed, so a valve could be partially closed. Pump B has higher efficiency under these conditions, and so would be preferred.
TDH (ft)
Flow rate (gpm)
Pump A only
Pump B only
System curve
Efficiency, %
Pump B
Pump A 10
100 40 50 30 20 90 80 70 60
120
110
200
400
1200
1000
800
600

30. The pump characteristic curve for two type-B pumps in parallel can be drawn by taking the curve for one type-B pump, and doubling Q at each value of TDH. Such a scenario would meet the ultimate need (1200 gpm at 80 ft), as shown below.
TDH (ft)
Flow rate (gpm)
A only
B only
Two B’s
System curve
Efficiency, %
Pump B
Pump A 10
100 40 50 30 20 90 80 70 60
120
110
200
400
1200
1000
800
600

31. A pump characteristic curve for one type-A and one type-B pump in parallel can be drawn in the same way. This arrangement would also meet the ultimate demand. Note that the type-B pump provides no flow at TDH>113 ft, so at higher TDH, the composite curve is identical to that for just one type-A pump. (A check valve would prevent reverse flow through pump B.) Again, since type B is more efficient, two type-B pumps would be preferred over one type-A and one type-B.
TDH (ft)
Flow rate (gpm)
A only
B only
Two B’s
System curve
Efficiency, %
Pump B
Pump A 10
100 40 50 30 20 90 80 70 60
120
110
200
400
1200
1000
800
600
One A and one B in parallel

32. At the interim conditions, a single type B pump would suffice.
A third type B pump would be required as backup.
TDH (ft)
Flow rate (gpm)
A only
B only
Two B’s
System curve
Efficiency, %
Pump B
Pump A 10
100 40 50 30 20 90 80 70 60
120
110
200
400
1200
1000
800
600
One A and one B in parallel