Pump Performance

To determine pump performance charts are prepared to show the results of pump

testing. For centrifugal pumps and compressors performance charts are plotted for constant

shaft rotation with the speed expressed in rpm. The basic independent variable considered is

the discharge Q, with the head H, power P and efficiency taken as the dependent variables.

Typical performance curves for centrifugal pumps show that the head is approximately

constant at low discharge and then drops to zero at Qmax. This indicates that at the given speed

and impeller size, the pump has maximized its delivery rate to Qmax when H=0.

Test results show that efficiency is always zero at no flow and at Qmax, and it reaches a

maximum, approximately 85%, at about 0.6Qmax. This is the design flow rate Q or best

efficiency point (BEP), where = max, and with the head and brake power expressed as H and

P, respectively. Note that is not independent of H and P but rather is calculated from the

relation = QH/P.

Affinity Laws

Affinity or Similarity Laws predict the performance of a given machine under different conditions of operation from those under which it has been tested. Note that there are two sets of affinity laws,

a) Affinity laws for a specific centrifugal pump - to approximate head, capacity and power curves for different motor speeds and /or different impeller diameters.

b) Affinity laws for a family of geometrically similar or homologous (means same in structure, i.e. therefore, having the same velocity diagrams at entrance and exit) centrifugal pumps - to approximate head, capacity and power curves for different motor speeds and /or different diameter of impellers

Affinity Laws for a Specific Centrifugal Pump:

Where:

22 H Head developed

D Impeller diameter

32 Q Capacity

33 T Torque developed

NOTE: The efficiency remains constant for any change in speed or small changes in diameter

Affinity Laws for a Family Geometrically Similar Pumps:

Where:

22 H Head developed

3 D Impeller diameter

52 Q Capacity

53 T Torque developed

P Power, = = 2

Problem No.1

Tests by P Brosnan Pump Co. for a 37-cm diameter centrifugal pump conveying water at

2150 rpm shows the following results:

Q, L/s 0 50 100 150 200 250 300

H, m 100 100 97 90 81 65 40

P, kW 90 95 120 160 185 200 210

Determine the BEP and estimate the maximum discharge possible.

Solution:

Convert L/s to m3

s

Use = QH/P

BEP, 0 51.63% 79.3% 82.2% 85.9% 74.7% 56.06%

Use highest value for BEP,

BEP, is at Q=200 L/s

H = 81m

P = 185 kw

H=85.7%

Qmax =Q

0.6=

200/

0.6=333.333L/s

Problem No.2

A pump from the same family as problem 1 is built to deliver gasoline at 20OC with a 50-

cm impeller and a BEP power of 190 kW. Using the affinity laws determine the resulting speed,

flow rate and head at BEP.

Problem No.3

An axial flow pump delivers 300 L/s at a head of 6.0 m when rotating at 2000 rpm. If its

efficiency is 80%, how many kilowatts of power must the shaft deliver to the pump? If this same

pump were operated at 2400 rpm, what would be the H, Q, and the power delivered by the

shaft for homologous conditions?

Given: Q1 = 300 L/s

Hp1 = 6.0 m

n1 = 2000 rpm

p = 80%

n2 = 2400 rpm

Reqd: a) Pp1

b) Hp2, Q2, and Pp2

Solution:

) 1 =

1 = (9.807

3) (0.3

3

) (6 ) [

] [

]

1 = 17.65

1=

1

=17.65

0.8

= .

) 22 1 = 1

21 2

22 2 = 2 22

2

21

= (21

)2

(21

)2

; 1 = 2

2 = (6 ) (2400

2000 )

2

= .

21

= (21

) (21

) ; 1 = 2

2 = (300

) (

2400

2000 )

=

2 = 22 = (1000

3) (9.807

3

) (8.64 ) (0.36

3

) (

1

1000)

2 = 30.5

2=

2

=30.5

0.8

= .

Check:

33

21

= (21

) (21

)3

(21

)3

; 1 = 2 ; 1 = 2

2 = 1 (21

)3

= (22.06 ) (2400

2000 )

3

= .

Problem No.4

The performance of a particular pump running with various impeller diameters indicates that while running at 1650 rpm the capacity is 280 gpm, the head is 180 ft, and the Bhp is approx. 25 hp. This performance data has been determined by actual tests by the manufacturer using a 12" diameter impeller. Determine the results of the performance data if the impeller diameter were changed to 10 and the pump were driven at 2000 rpm.

Given: n1 = 1650 rpm Reqd:

Q1 = 280 gpm Q2, Hp2, Bhp2

Hp1 = 180 ft

Bhp1 = 25 hp

D1 = 12 ; D2 = 15

n1 = 2000 rpm

Problem No.5

A model centrifugal pump is made with a model ratio of 1:10. The model was tested at 3600 rpm and delivered 0.085 m3/s at a head of 38 m with an efficiency of 88%. Assuming a prototype having an efficiency of 91% develops the same head, what will be its speed, capacity, and the required shaft power? Liquid pumped is water.

Given:

=

1

10 Reqd: np, Qp, Pp

nm = 3600 rpm

Q = 0.085 m3/s

Hpm = 38 m

m = 88%

p = 91%

Hpp = 38 m

Solution:

22

= (

)

2

(

)

2

= (

) = (3600 ) (1

10)

= 360

3

= (

)

3

(

)

= (

)3

(

) = (0.085 3

) (

10

1)

3

(360

3600 )

= .

=

=(9.807

3

) (8.5 3

)(38 ) [

] [

]

0.91

= .

Problem No.6

Determine the scale ratio for a water pump if tests were made on a model using oil; the water pump is to be driven by a 95 kW motor running at 1450 rpm while a 150 kW motor is used to drive the oil pump at 1800 rpm. Use SGoil = 0.85. Given: Pw = 95 kW Reqd: Dw/Do

nw = 1450 rpm

Po = 150 kW

no = 1800 rpm

SGoil = 0.85

Solution:

53

= (

) (

)5

(

)3

= [(

) (

) (

)3

]

15

= [(95

150 ) (

0.85

1) (

1800

1450 )

3

]

15

= .

Problem No. 7

At its optimum point of operation, a given centrifugal pump with an impeller diameter of 50 cm delivers 3200 L/s of water against a head of 25 m when rotating at 1450 rpm. (a). If its efficiency is 82%, what is the brake power of the driving shaft? (b). If a homologous pump with an impeller diameter of 80 cm is rotating at 1200 rpm, what would be the discharge, head, and shaft power? Assume both pump operate at the same efficiency. (c). Compute the specific speed of both pumps.

Given: D1 = 50 cm (a). 1 = 82% (b). D2 = 80 cm

Q1 = 3200 L/s (b). n2 = 1200 rpm

Hp 1 = 25 m

n1 = 1450 rpm

Reqd: (a). P = ? (b). Q2 = ? (c). Ns 1 = ?

(a). Hp 2 = ? (c). Ns 2 = ?

(a). P2 = ?

Solution:

(a) = = (9.807

3) (3.2

3

) (25 ) = 784.56

=

=

=

784.56

0.82

= .

(b). 3

21

= (21

)3

(21

)

2 = 1 (21

)3

(21

) = (3200

) (

80

50 )

3

(1200

1450 )

= .

22

21

= (21

)2

(21

)2

2 = 1 (21

)2

(21

)2

= (25 ) (80

50 )

2

(1200

1450 )

2

= .

2 =222

=

(9.807 3

) (10.8473 3

)(43.8335 )

0.82

P2 = .

(c). 1 =11

12

1

34

=(1450 )(3.2 3/)

12

(25 )34

=

2 =22

12

2

34

=(1200 )(10.8473 3/)

12

(43.8335 )34

=

Problem No. 8

A model centrifugal pump has a scale ratio of 1:15. The model was tested at 3200 rpm and delivered 100 L/s of water at a head of 40 m with an efficiency of 85%. Assuming a prototype has an efficiency of 88%, what will be its speed, capacity and power equivalent at a head of 50 m?

Given: Dm/Dp = 1/15 Reqd: np = ?

nm = 3200 rpm Qp = ?

Qm = 100 L/s Pp = ?

Hpm = 40 m

m = 85%

Hpp = 50 m

p = 88%

Solution:

22

= (

)

2

(

)

2

40

50 = (

1

15)

2

(3200

)

2

= 238.5139

3

= (

)

3

(

)

100 /

= (

1

15)

3

(3200

238.5139 )

= .

=

=

=

(9.807 3

) (25.1558 3

)(50 )

0.88

= .

References:

Douglas, J. F./ Gasiorek, J. M./ Swaffield, J. A. Fluid Mechanics, 4th Edition, Prentice-Hall Streeter, V. L./ Wylie, E. B./ Bedford, K. W. Fluid Mechanics, 9th Edition, McGraw-Hill Publishing

Co. White, F. M. Fluid Mechanics, 3rd Edition, McGraw-Hill Publishing Co. http://udel.edu/~inamdar/EGTE215/Laminar_turbulent. http://pessoal.utfpr.edu.br/mannich/arquivos/whi38447_appA http://www.engineeringtoolbox.com/affinity-laws-d_408.html