Example. For pump characterized on previous slide, is
cavitation a concern if a flow of 3500 gpm of 80oF water is
required, and the pump suction is 12 ft above the source
reservoir? Headloss in the inlet pipe is given by
hL≈6x10-7Q2, with hL in ft and Q in gpm.
zsl ,max 
patm ,abs

-  hL -
pvap

- NPSH R
lb  
in 2 

14.7 - 0.51 in 2  144 ft 2  
ft 
2


-7

6x10
3500
gpm
- 8.5ft



3
2 
62.4 lb/ft
gpm 

 16.9ft
zsl  zsl ,max ok!
Pump curve describes how much head the pump will
provide at a given Q. Now consider how much head is
needed to operate at that Q in a given system (hsys) .
If source and sink are at
different elevations,
head equal to Dz must be
provided, independent
of Q.
z2
hL,discharge
Dz
z1
K Qn
hsys = Dz + ShL ≈ Dz + Kh eq= Q
L,2
Frictional headlosses
also contribute. These
do depend on Q.
hL,1 = K1Qn
2
hL,valve
z ≡0
n
hsys = Dz + ShL ≈ Dz + KeqQn
hsys
ShL
Dz
z2
z1
= K Qn
L,2 =
hhL,2
2
hhL,1 == K Qn
L,1
1
z ≡0
2 2
2 2
22
11
V
pp1 1
VV
pp22
 z11 
  hpump
z2 z2  
hpump- -hLhL
2 g 
2 g2 g 
hpump  Dz   hL  hsys
(hpump is same as TDH; Munson is inconsistent, using ha
sometimes for hpump and sometimes for hsys)
For stable operation, the flow rate is such that the
pump supplies an amount of head equal to what the
system demands.
 Q, h  pump   Q, h sys
Pump curve (hpump)
Operating
point
System curve (hsys)
z2-z1=hmin
Pump curve
(TDH or hpump)
System curve
(hsys)
z2-z1=hmin
• hpump or TDH is the total head added by the pump, as expressed by the
pump curve (TDH vs. Q). This relationship characterizes the pump,
independent of the system in which it is installed.
• hsys is the head dissipated due to friction and changes in elevation when
fluid passes through the system at a flow rate Q. This relationship
characterizes the system, independent of how the flow is generated.
Example. A pump curve can be approximated by the
equation hpump= 180 - 6.1x10-4Q2, with hL in ft and Q in
gpm. If the pump is used to deliver water between two
reservoirs via a 4-in-diameter, 600-ft long pipe against a
static head of 50 ft, what flow rate will develop, if f=0.02?
Ignore minor headlosses.
VV1122 p1p1
VV2222 p22
 zz11 
  hpump
hLhL
z2z

hpump- 
-
2 
22gg  
22gg 
hpump  zsl   hL  hsys
hpump  zsl   hL  hsys
Pump curve
System geometry
D-W or H-W


Q


2



0.33
/
4


600 

-4
2
180 - 6.1x10 Q  50   0.02 
0.33
2  9.81
Q  365 gpm
2
Changes in system
capacity as pipes age
Dependence of system
behavior on impeller speed
Pump and system curves that don’t intersect
Composite pump curve for identical pumps in series
Two discharge pipes (and two system
configurations) supplied by a single pump
How do the discharge rate and head for the pump
relate to the flow rates and heads in the two pipes?
Composite system curve for two discharge
pipes from a single pump
Composite system curve for two discharge
pipes from a single pump
z2
Energy equation is sometimes
rearranged to equate a
‘modified’ pump curve with a
‘modified’ system curve
z1
hL,2 = K2Qn
hL,1 = K1Qn
z ≡0
Conventional: Head added by pump equals
head lost in rest of system
TDH  z2 - z1  K1Q n  K 2Q n   hL ,minor   hL ,minor  hsys
up stream
down stream
Modified: Head at some point in system is
same based on ‘path’ from either direction
TDH  z1 - K1Q n -  hL ,minor  z2  K 2Q n   hL ,minor
up stream
TDH mod  hsys ,mod
down stream
Similitude and Affinity Laws for Pump Performance
A
B
C
• Similitude is useful for
extrapolating experimental results
to uninvestigated systems
• Dimensional analysis allows us to
take maximum advantage of
similitude relationships by
identifying universally valuable
parameter combinations
• For membrane efficiency: rpart/rpore
2
4
Rˆi , pa  1 -  2 1 - i  - 1 - i  


• For partially full pipe flow: hliq/D
• Procedure: Identify important
dependent parameters and make
assumptions about independent
parameters that control them, i.e.,
Dep1  fcn1  In1 , In2 ,...Inn 
Dep2  fcn2  In1 , In2 ,...Inn  , etc.
• Use dimensional analysis to convert these parameters into
dimensionless terms, Depi* and Inj*, such that each Depi* term
contains only one Depi and one or more Inj. Elimination of
dimensions causes the number of Inj* terms to be less than the
number of Inj terms; e.g.:
Dep1  hL  fcn1  l , D, V ,  ,  ,  
hL
l  

*
Dep1 
 fcn1  Re, , 
l
D D

• In hydraulics, similitude typically
requires geometric similarity
• For pumps, TDH, Pshaft, and h are
logical Depi’s; D, li, , Q, w, , and
 are logical Ij’s, e.g.:
TDH  fcn  D, li ,  , Q, w ,  ,  
• One set of dimensionless groups is:
Head rise coefficient
22
ggTDH
TDH



l
l


Q
Q
w
w
D
D
** i i
CH  22 22  fcn
fcn11 , , , , 33, ,

ww DD
D
D
D
D
w
w
D
D




Power coefficient
22



l
l


Q
Q
w
w
D
D
**
ii
CP  33 55  fcn
fcn22 ,, ,, 33 ,,

w
w D
D
D
D
D
D
w
w
D
D




Efficiency
PPshaft
shaft
2
2




l
l

Q

Q
w
D
w
D
i *
h  fcn
h 3* fcn
,3  i, , 3, , 3 , 

D
D
D
w
D
D
w
D






g  TDH 
 Q w D 2 
*  li
CH 
 fcn1  , ,
,

2 2
3
w D
D
D
w
D



 li  Q w D 2 
CP 
 fcn  , ,
,

3 5
3
w D
D
D
w
D



Pshaft
*
2
 li  Q w D 2 
h  fcn  , ,
,

3
D
D
w
D



*
3
• First and second independent parameters are constant for
geometrically similar pumps, and fourth is unimportant for
high-Re flow (which is typical in pumps). Q/(wD3) is called the
flow coefficient, CQ. So:
CH  fcn1*  CQ 
CP  fcn2*  CQ 
h  fcn3*  CQ 
• CH, CP, and h can be determined for a pump as a
function of CQ. Then these relationships will apply to
all geometrically similar pumps.
h
CH
CH
CP
CQ
• Tentatively select a pump
• Obtain CH, CP, and h as fcn of
CQ for a geometrically similar
h
pump from manufacturer
• Determine CQ for the proposed
Q, w, and D in your system
• Determine CH and CP for your
system
CP
• Assess whether anticipated
performance is satisfactory; if
not, modify w or D to improve
performance at given Q
A series of homologous pumps is characterized by the following curves. One
such pump has an impeller diameter of 0.75 m and operates with maximum
efficiency at 1500 rpm (= 25 rps). What is TDH?
If the impeller speed is adjusted to 2000 rpm, and a valve is adjusted to
maintain the original discharge rate, what will the new efficiency and TDH be?
h
CH
h
CH
CP
CQ
CP
Under initial conditions (1), h =hmax, so CQ≈ 0.065 and
CH≈ 0.19.
g  TDH 
CH 
w 2 D2
h
CH
CH w 2 D 2
TDH 
g
h
CH
0.19  25/s   0.75m 


2
9.81 m/s 2
CP
CQ
CP
 6.8 m
2
After the increase in impeller rotation speed (2), Q and D
remain the same, and w increases by 33%, so:
CQ ,2
h
CH
CQ ,1
h
CQ ,2
CH
CP

Q2
3
D
 2 w2 
Q1  D13w1 
w1

w2
w1

CQ ,1
w2
1

 0.065   0.049
1.33
CP
At this value of CQ,
h ≈ 62% and CH ≈ 0.21.
CQ
CH w 2 D 2
TDH 
g
 0.21  2000 min   0.75 m 

2
2
 9.81 m/s   60 s min 
-1 2
2
 13.4 m
Increasing w by 33% almost doubles TDH.