SIMILARITY RULES(unit 4)

Sec. 12.3

UNIT 6

Performance Prediction• So far we have used vector diagram to predict the

performance of fan/pump/turbine• A simpler but practical alternative is to use dimensional

analysis. It helps predict the performance of geometrically similar machines.

• Using dimensional analysis we can resolve this relationship into one amongst several dimensionless groups.

• For pump head developed is important. So we write• gH = φ1(Q,D,ω,ρ,µ, e);

• For turbine power produced is important So we write• P = φ2(Q,D, w, ρ,µ, e)

where w is revolution in radian per seconds, e is friction coefficient.

Dimensionless equation• Dimensional analysis gives from First relation

• Ornote: u = ωD; Re = uDρ/µ

• Second relation gives

=

∆DewD

wDQ

DwHg ,,

2

3122 µρφ

=

DeCC QH Re,,1φ

=

=

DeCC

DewD

wDQ

DwP

Qp Re,,

,,

2

2

3253

φ

µρφ

ρ

Coefficients• Since friction losses is small compared to other losses like

eddy losses the Reynolds number (Re) and the roughness factor (e/D) are not considered for similitude studies.

• So, two machines will have similar performance if the following dimensionless numbers are equal

• Power coefficient (turbine), Cp =

• Head coefficient (Pump) CH =

• Capacity coefficient, CQ=

53DwP

ρ

22DwHg∆

3wDQ

Similarity rules• For two pumps or fans to have similar performance

these numbers must be equal• From CQ1= CQ2 we get

• From CH1= CH2 we get

• From CP1= CP2 we get

322

2311

1

DwQ

DwQ

=

222

22

211

21

DwH

DwH

=

522

32

511

31

DwP

DwP

=

Fan Laws• Effect of change in speed (N)

1. Q ~ N; H ~ N2; P~ N3

• Effect of blade dia (D) change at Constant tip speed, U

1. Q~D2; 2. H=k; 3. N~ D-1; 4. P~ D2

• Effect of blade Dia (D) change at constant N1. Q~D3; 2. H~D2; 3. U ~D; 4. P~D5

• Effect of gas density (ρ) change at Const Q1. P ~ ρ; 2. Η ~ ρ

• Effect of gas density ρ change . Const. P1. Q~ ρ−0.5; 2. Ν∼ ρ−0.5; 3. P∼ ρ−0.5

Specific Speed• To help selection of a fan/pump we need a parameter

which will include all items except for the size. Specific speed is such a parameter.

• Specific speed is always related to the best efficiency efficiency pointpoint. Graphs of best efficiency and specific speed are provided.

• Specific speed allows one to represent the whole family of machines by a single plot.

• Specific speed ΩP for pump

• Specific speed ΩT for turbine

75.075.0

5.0

)(gHQw

CC

H

QP ==Ω

25.125.1

5.0

)(gHPw

CC

P

HT ρ

==Ω

Range of dimensionless Specific speeds for pump & turbine

0 – 1.0Impulse

1.0 – 3.50 – 1.0Radial flow

3.5 – 7.01.0 – 4.0Mixed flow

7.0 – 14.0> 4.0 Axial flow

Turbine (ΩT)Pump (ΩP)Types

Here w is in rad/s, Q, H, P, ρ are in m3/s, m, watt and kg/m3. It gives a dimensionless specific speed Ω. You may find dimensional specific speed Nsp elsewhere where w is in rpm, P in kW etc

Efficiency

HQ

PT CC

CQgHP

==ρ

η• Turbine efficiency =•

• Pump efficiency = • P

HQp C

CCP

QgH==

ρη

Specific speed (Nsp) for pumps

Values are tentative

Practical specific speed with dimensions

• Specific speed Nsp= is dimensionless provided n is in revolution/s or radian/s, Q in m3/s, g in m2/s and Hin m. Sometime g or ρ is dropped making Sp. speed dimensional N’sp

• For pumps N’sp = where N - rpm, H- m, Q –m3/s

• For turbines N’sp = where N - rpm, P- kW, H –m

QN

25.1

PN

75.0)(H

H

Example Problem

• Find the total head delivered by a prototype pump with an impeller size of 1.2 m operating at a speed of 1750 rpm and delivering 1.3 m3/s flow?

• To help build this large pump a small model of the pump has been tested in the laboratory to give performance curves as shown. Consider that dynamic similarity can be achieved between model flow and prototype flow

1. What is its mechanical efficiency?2. What is the dimensionless specific speed?

[Shames, p-772]

Head coefficient vs Flow coefficient

Home work (Douglas-p795)

• A centrifugal pump will operate at 300 rpm delivering 6 m3/s against 100 m head.

• Laboratory facilities for a model are: maximum flow 0.28 m3/s and maximum power available 225 kW. Using water and assuming that the efficiencies of model and prototype are the same, find the speed of the model and the scale ratio. Also calculate the specific speed.

• [1196 rpm, 4.4, 0.439 (rad)]

Problems solutionsUsing Cq1=Cq2