• 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) v
2
1
2 g
 p

1
 
1
H pump
 v
2
2
2 g
 p

2 z
2
H
L
H
L
  h f
  h minor
  h f
 
K i v
2
2 g

• (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 discharge head
Total static head
Static suction head

• (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 discharge head
Total static head Static suction lift

Static discharge head Static discharge head
Total static head (both)
Static suction head
Static suction lift




Note: Suction and discharge head / lift measured from pump centerline

• (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
Total dynamic head
Dynamic discharge head
Dynamic suction lift

Example. Determine the static head, total dynamic head (TDH), and total headloss in the system shown below.
El = 730 ft
El = 640 ft p s
=

6 psig
El = 630 ft p d
=48 psig
Total static head
  
TDH


48
  psi
 

2.31 ft psi



124.7 ft
H

L
TDH Static head
    
24.7 ft

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 ft 3 /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 45 o bend in a 48-in pipe
0.30 for 90 o 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( v
2
2
 v
1
2 )/2 g
El = 6349 to
6357 ft
30
 to 48
 expansion
El = 6127 to
6132 ft
Short 30
 pipe w/30
 butterfly valve
4000
 of 48
 pipe w/two 45 o bends
8500
 of 36
 pipe w/one
90 o bend and eight butterfly valves

1.
Determine pipeline velocities from v = Q/A .
.
v
30
= 7.88 ft/s, v
36
= 5.47 ft/s, v
48
= 3.08 ft/s
2.
Minor losses, suction side:
Entrance: h
L

0.50
2 v
30
2 g

0.49 ft
Butterfly valve:
Expansion: o
Two 45 bends: h
L

0.16
2 v
30
2 g

0.16 ft h
L

0.25

 v
2
30
 v
2
48
2 g



0.21 ft h
L


2 * 0.18
2 v
48
2 g



0.05 ft
 h
L ,minor

0.91 ft

3.
Minor losses, discharge side:
8 Butterfly valves: o
One 90 bend: h
L


8* 0.35
2 v
36
2 g



1.30 ft h
L

0.30
2 v
36
2 g
 h
L ,minor

0.14 ft

1.90 ft

4.
Pipe friction losses:
S
 h
L f


Q
0.43
CD
2.63
 1.85
 h L f
Q
0.43
CD
2.63
 1.85
h h
 
4000




38.7
  
2.63


 1.85

2.76 ft
 
8500

 


38.7
  
2.63



 1.85

16.77 ft

5.
Loss of velocity head at exit:
Exit: h
L
 v
2
2
36 g

0.46 ft
6.
Total static head under worst-case scenario (lowest water level in aqueduct, highest in reservoir):
Static head
    
230 ft
7.
Total dynamic head required:
TDH

H static
  h
  h f
 h


230
       


252.8 ft

P

Q

TDH
 
E p
 
• P = Power supplied to the pump from the shaft; also called ‘brake power’ (kW or hp)
• Q = Flow (m 3 /s or ft 3 /s)
•
• TDH = Total dynamic head

= Specific wt. of fluid (9800 N/m 3 or 62.4 lb/ft 3 at 20 o C)
• CF = conversion factor: 1000 W/kW for SI, 550 (ft-lb/s)/hp for US
• E p
= pump efficiency, dimensionless; accounts only for pump, not the drive unit (electric motor)
Useful conversion: 0.746 kW/hp

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
= 0.003 ft, Efficiency P e
• Flow = 25 mgd = 38.68 ft 3 /s
• T = 60 o F
=80%
El = 230 ft 2
• Ignore minor losses
10 mi of 48
 pipe, e
=0.003 ft
El = 100 ft
1 v
1
2
2 g
 p
1

 
1
H pump
 v
2
2
2 g
 p
2

 
2
H
L
H pump

TDH z
2
 
1
H stat
H
L
 h f
TDH

H stat
 h f

El = 230 ft 2
10 mi of 48
 pipe, e
=0.0003 ft
El = 100 ft
1
TDH

H stat
 h f
H
 stat
 h f
 f
L

 v
2


D  2 g 
 
130 ft Find f from Moody diagram
Re
 vD



3.08 ft/s
 

5 2
1.22x10 ft /s

1.01x10
6 v

/

3.08 ft/s e
D

0.003 ft
4 ft

7.5x10

4

El = 100 ft
1
El = 230 ft 2
10 mi of 48
 pipe, e
=0.0003 ft
Re

1.01x10
6 e
D

7.5x10

4 f

0.0185

h f
 
0.0187
 10*5280 ft
4 ft


3.08 ft/s
2 32.2 ft/s
2

2


 
36.4 ft
TDH

H stat
 h f
   
166.4 ft
P

Q

TDH
 
E p
 


3
   
62.4 lb/ft
3



550 ft-lb/s hp



0.80


918 hp
  

Daily cost 918 hp 0.746
kW

$0.08
hp   kW-hr
24 hr d

$1315 / d

• System curve – indicates TDH required as a function of Q for the given system
– For a given static head, TDH depends only on H
L
, which is approximately proportional to v 2 /2 g
– Q is proportion to v , so H
L
Q 1.85
is approximately proportional to if H-W eqn is used to model h f
)
Q 2
– System curve is therefore approximately parabolic
(or

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 0.02. Coefficients for minor losses are shown below.
40 ft
6 ft
K values
Suction Discharge
1 @ 0.10
1 @ 0.12
1 @ 0.12
1 @ 0.20
1 @ 0.30
1 @ 0.60
2 @ 1.00
4 @ 1.00

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 v 2 /2 g .
An additional 1.0 v 2 /2 g of velocity head is lost when the water enters the receiving reservoir.
The frictional headloss is: h f
 f
L v
2
 
0.02


1000 ft

 v
1.33 ft

2
2 g v
2

15
2 g
Total headloss is therefore (8.04+1.0+15.0) v 2 /2 g = 24.04 v 2 /2 g . v can be written as Q / A , and A = p
D 2 / 4 = 1.40 ft 2 . The static head is 34 ft. So:
TDH
 v
2
H stat

H
L

34 ft

24.04
2 g

34 ft

24.04

Q / 1.40 ft
2
2


2


34 ft 0.19
s
2 ft
5


Q
2

180
160
140
120
100
80
60
40
20
0
0
TDH

34ft

0.19
s
2 ft 5


Q
2
5
Static head
10 15
Discharge, Q (ft
3
/s)
System curve
20 25

• 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)

Shutoff head
Rated hp
Rated capacity

• Pump curves depend on pump geometry (impeller D ) and speed

• 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

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

• 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

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?
120
110
100
90
80
70
60
50
40
30
20
10
0
0
200
70
60
50
40
400 600 800
Flow rate (gpm)
Pump B
Pump A
1000 1200

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.
120
110
100
90
80
70
60
50
40
30
20
10
0
0
200
70
60
50
40
400 600 800
Flow rate (gpm)
Pump B
Pump A
1000 1200

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.
120
110
100
90
80
70
60
50
40
30
20
10
0
0
200
70
60
50
40
400 600 800
Flow rate (gpm)
Pump B
Pump A
1000 1200

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.
120
110
100
90
80
70
60
50
40
30
20
10
0
0
200
70
60
50
40
400 600 800
Flow rate (gpm)
One A and one
B in parallel
Pump B
Pump A
1000 1200

At the interim conditions, a single type B pump would suffice.
A third type B pump would be required as backup.
120
110
100
90
80
70
60
50
40
30
20
10
0
0
One A and one
B in parallel
200
70
60
50
40
400 600 800
Flow rate (gpm)
Pump B
Pump A
1000 1200