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 byhL≈6x10-7Q2, with hL in ft and Q in gpm.

,, RNPSHvapatm abs

sl max L

ppz h

- - -

2

2 227

3 2

lb in14.7 0.51 144in ft ft6x10 3500 gpm 8.5ft

62.4 lb/ft gpm-

- - -

16.9ft , ok!sl sl maxz z

hL,2 = K2Qn

Dz

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.

Frictional headlosses also contribute. These do depend on Q.

hL,1 = K1Qn

z2

z1

z ≡ 0

hL,discharge

hL,valve

hsys = Dz + ShL ≈ Dz + KeqQn

hsys = Dz + ShL ≈ Dz + KeqQn

hsys

Dz

ShL

2 22 2 1 1

2 12 2 pump LV p V pz z h hg g

- 2

22 2

Vzg

2p

2

11 2Vzg

1p

pump Lh h -

pump Lh z hD

hL,2 =hL,2 = K2Qn

hL,1 =hL,1 = K1Qn

z2

z1

z ≡ 0

pump L sysh z h hD (hpump is same as TDH; Munson is inconsistent, using ha

sometimes for hpump and sometimes for hsys)

, ,pump sys

Q h Q h

Pump curve (hpump)

System curve (hsys)

z2-z1=hmin

Operating point

For stable operation, the flow rate is such that the pump supplies an amount of head equal to what the

system demands.

• 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.

Pump curve (TDH or hpump) System curve

(hsys)

z2-z1=hmin

2 22 2 1 1

2 12 2 pump LV p V pz z h hg g

- 2

22 2

Vzg

2p

2

11 2Vzg

1p

pump Lh h -

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.

pump sl L sysh z h h

pump sl L sysh z h h

2

2

4 2 0.33 / 4600180 6.1x10 50 0.020.33 2 9.81

Q

Q-

-

Pump curve System geometry D-W or H-W

365 gpmQ

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

z1

z ≡ 0

1 1 , 2 2 ,

,

TDH

TDH

n nL minor L minor

up downstream stream

mod sys mod

z K Q h z K Q h

h

- -

- -

2 1 1 2 , ,TDH n nL minor L minor sys

up downstream stream

z z K Q K Q h h h- -

-

Conventional: Head added by pump equals head lost in rest of system

Modified: Head at some point in system is same based on ‘path’ from either direction

Energy equation is sometimes rearranged to equate a

‘modified’ pump curve with a ‘modified’ system curve

hL,2 = K2Qn

hL,1 = K1Qn

Similitude and Affinity Laws for Pump Performance• 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

• For partially full pipe flow: hliq/D

A

B

C 

 

2 4,

ˆ 1 2 1 1i pa i iR - - - -

• Procedure: Identify important dependent parameters and make assumptions about independent parameters that control them, i.e.,

1 1 1 2

2 2 1 2

, ,...

, ,... , etc.n

n

Dep fcn In In In

Dep fcn In In In

1 1 , , , , ,LDep h fcn l D V

*1 1 Re, ,Lh lDep fcn

l D D

• 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.:

• 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.:

• One set of dimensionless groups is:

TDH , , , , , ,ifcn D l Q w

2*12 2 3

2*23 5 3

2*3 3

TDH, , ,

, , ,

, , ,

i

shaft i

i

g l Q DfcnD D D D

P l Q DfcnD D D D

l Q DfcnD D D

ww w

ww w

whw

2*12 2 3

2*23 5 3

2*3 3

TDH, , ,

, , ,

, , ,

iH

shaft iP

i

g l Q DC fcnD D D D

P l Q DC fcnD D D D

l Q DfcnD D D

ww w

ww w

whw

Head rise coefficient

Power coefficient

Efficiency

• 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:

2*12 2 3

2*23 5 3

2*3 3

TDH, , ,

, , ,

, , ,

iH

shaft iP

i

g l Q DC fcnD D D D

P l Q DC fcnD D D D

l Q DfcnD D D

ww w

ww w

whw

* * *1 2 3H Q P Q QC fcn C C fcn C fcn Ch

• 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.

• Tentatively select a pump• Obtain CH, CP, and h as fcn of CQ

for a geometrically similar pump from manufacturer

• Determine CQ for the proposed Q, w, and D in your system

• Determine CH and CP for your system

• Assess whether anticipated performance is satisfactory; if not, modify w or D to improve performance at given Q

CPCP

CH

CH

CQ

h

h

CPCP

CH

CH

CQ

h

h

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?

CPCP

CH

CH

CQ

h

h

Under initial conditions (1), h =hmax, so CQ≈ 0.065 and CH≈ 0.19.

2 2

TDHH

gC

Dw

2 2

2 2

2

TDH

0.19 25/s 0.75m9.81 m/s

6.8 m

HC Dgw

CPCP

CH

CH

CQ

h

h

After the increase in impeller rotation speed (2), Q and D remain the same, and w increases by 33%, so:

1,2 ,1

2

1 0.065 0.0491.33

Q QC Cww

At this value of CQ, h ≈ 62% and CH ≈ 0.21.

32 2 2,2 1

3,1 21 1 1

Q

Q

Q DCC Q D

w www

2 2

2 21

22

TDH

0.21 2000 min 0.75 m13.4 m

9.81 m/s 60 s min

HC Dgw

-

Increasing w by 33% almost doubles TDH.