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Specific Speed and Unit Conditions

The selection of Specific Speed and Unit Conditions for turbines with reference to Model Testing

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Contents

Introduction
Principles Of Similarity Applied To Turbines.
Specific Speed Of A Turbine
Specific Speeds For Differing Types Of Turbine
An Example Of The Use Of Specific Speed
Unit Conditions
The Performance Curves Of A Turbine
Characteristic Curves And Iso-efficiency Curves For A Turbine Under All Operating Conditions
An Example Of The Use Of Unit Conditions
Fundamental Similarity Conditions And Model Testing
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\tau$ ) = $I\omega\Omega$ (In the limit).

= Moment of Inertia.
$\omega$ = Angular velocity
$\Omega$ = Angular velocity of precession.

Introduction

In the selection and/or design of turbines for a particular application it is common to rely on model testing. The results are then scaled up using the following principles:

Principles Of Similarity Applied To Turbines.

A similar model means :

Geometrically similar - made from the same drawings but to a different scale.
Dynamically similar - Operating conditions and equal efficiencies.

Thus in comparing two similar turbines

All the linear dimensions will be in the same ratio.
All angles will be the same, the velocity triangles will be geometrically similar and all velocities will be in the same ratio.

Specific Speed Of A Turbine

The Specific Speed

of a turbine is the speed in rotations per minute (r.p.m.) at which a similar model of the turbine would run under a head of 1ft. when of such size as to develop 1 H.P.
(Note: The suffix "s" is used to denote the values associated with the Specific Turbine)

Each type of Turbine (Pelton Wheel, Francis etc.) has it's own characteristic limits of

Revolutions per minute (R.P.M.) is a measure of the frequency of a rotation. It annotates the number of full rotations completed in one minute around a fixed axis.

But $V = K\sqrt{2gH}$ . Therefore, $v\propto V_f\propto V\propto \sqrt{H}$

And $v = \displaystyle\frac{\pi D\,N}{60}$ . Therefore, $D\;\propto \displaystyle\frac{\sqrt{H}}{D}$

But,

= The Area of flow

The Velocity of flow

$Q = K\,\pi \;D\;b\;V_f$ and $b\;\propto D$ Or,

$Q\;\propto \;D^2\;\sqrt{H}$

But the weight of water per second is $W=w\times Q$ . Which is, $\propto \displaystyle\frac{H^\frac{3}{2}}{N^2}$
H.P.output of the Turbine $P=\displaystyle\frac{W\times H}{550}\times \eta$ . Which is $\propto \displaystyle\frac{H^\frac{5}{2}}{N^2}$

Note: The efficiencies are equal

$\therefore\;\;\;\;\;\frac{N\;\sqrt{P}}{H^{\frac{5}{4}}} = C = \frac{N_s\,\sqrt{P_s}}{H_s^{\frac{5}{4}}}$

(where

Constant)
But for the specific Turbine,

and

are 1.

$\therefore\;\;\;\;\;N_s = \frac{N\;\sqrt{P}}{H^{\frac{5}{4}}}$

23287/SpecSpeed-and-Unit-Conditions-0006.png

Notes On The Use Of The Specific Speed Of A Turbine

Horse power (H.P.) is the name of several units of measurement of power.
The mechanical horsepower (imperial horsepower), of exactly 550 foot-pounds per second is approximately equivalent to 745.7 watts.
Horse power was originally defined to compare the output of steam engines with the power of draft horses.

is based on the values of , and used at the design point. i.e. at maximum efficiency.
is NOT dimensionless and there are different values in each of the measurement systems.

Unless otherwise stated,

is in r.p.m. and

is in Brake Horse Power(b.h.p.) i.e. $\displaystyle \frac{ft.lb./sec}{550}$

The Dimensions of Specific Speed

The unit of

are

$\frac{1}{T}\left(\frac{LM}{T^2}\frac{L}{T} \right)}^{\frac{1}{2}}\div L^{\frac{5}{4}}=\frac{M^\frac{1}{2}}{T^\frac{5}{2}\;L^\frac{1}{4}}$

can be made dimensionless and still be a constant by dividing by $w^\frac{1}{2}\;g^\frac{3}{4}$ and this is called the The Speed Number.

The Specific Speed Of A Particular Form Of Turbine

For a particular type of Turbine

is constant.

$v = \displaystyle\frac{\pi \,D\,N}{60}$ and $v\propto \sqrt{H}$ . Therefore $N\propto \sqrt{H}$ . Or $\displaystyle\frac{N}{\sqrt{H}}$ (which is constant)

But $P = \displaystyle\frac{W\;H}{550}\times \;\eta$ $\propto W\;H$ $\propto \;w\;Q\;H$ $\propto D^2\;V_f\;H$ $\propto H^\frac{3}{2}$

Therefore, $\displaystyle\frac{P}{H^\frac{3}{2}}$ (which is constant)

Therefore, $\sqrt{\displaystyle\frac{P}{H^\frac{3}{2}}}\times \displaystyle\frac{N}{\sqrt{H}} = \displaystyle\frac{N\sqrt{P}}{H^\frac{5}{4}}= N_s$ (which is constant)

Specific Speeds For Differing Types Of Turbine

Specific speed

is a non-dimensional number used to classify pump impellers as to their type and proportions.

In Imperial units it is defined as the speed in revolutions per minute at which a geometrically similar impeller would operate if it were of such a size as to deliver one gallon per minute against one foot of hydraulic head.

In metric units flow may be in

and head in

, and care must be taken to state the units used.

For different types of Turbine

and a comparison of heads for a particular power and speed. The Turbine is required to develope100 b.h.p. at 1000 r.p.m.

An Example Of The Use Of Specific Speed

What type of turbine would be used if the supply head is of 10 cu.ft/sec with a head of 225 ft. ? Assume an efficiency of 80%.

Power Output = Water h.p.input

Efficiency

$= \displaystyle\frac{0.8\;w\;Q\,H}{550}= \displaystyle\frac{0.8\times 62.4\times 10\times 225}{550} = 204\;h.p.$

$N_s = \displaystyle\frac{N\sqrt{P}}{H^\frac{5}{4}} = \displaystyle\frac{600\sqrt{204}}{225^\frac{5}{4}} = 9.83$

It would therefore be necessary to use a Turgot Turbine. However it might be possible to use a Pelton Wheel with two jets.
Power per jet $= \displaystyle\frac{204}{2}h.p.$ . Therefore

per Jet $= \displaystyle\frac{9.83}{\sqrt{2}} = 6.95$

From the above table it can be seen that the value of

is too high. It is therefore worth considering a Pelton Wheel with four jets.
Now:

$N_s \text{ per Jet} = \frac{9.83}{\sqrt{4}} = 4.92$

This would be a practical proposition but would result in some loss of efficiency due to interference between the jets. Consequently, a better alternative would be to have two wheels on the same shaft with two jets per wheel.

Unit Conditions

The unit operating conditions for a turbine are those under which that particular turbine would run when working under a head of 1 ft. (or unit head in any other system) assuming there no change in efficiency.

This allows the performance of a given turbine to be compared when working under different heads and enables the characteristic curves to be drawn, showing the efficiency at all running conditions.

Unit Speed

is the Unit speed and

the speed under a head

$v = \displaystyle\frac{\pi DN}{60}$

And $v\propto V \propto H$ . Therefore $N \propto \sqrt{H}$ . Or $\displaystyle\frac{N}{\sqrt{H}}=C=\displaystyle\frac{N_u}{\sqrt{H_u}}$

(Where

represents unit conditions and

represents a Constant)
Unit speed:

$N_u=\frac{N}{\sqrt{H}}$

Unit Quantity

The Unit quantity of a Turbine is the flow through the turbine when operating under a head of 1 ft. assuming similar conditions.

Let,

be the flow under a head .
Therefore is the area of flow velocity.

And since the area is constant, and velocity is $\propto \;\sqrt{H}$ $Q\propto \sqrt{H}$ . Or $\displaystyle\frac{Q}{\sqrt{H}} = C$
(where

is a Constant)

$\therefore\;\;\;\;\;\frac{Q}{\sqrt{H}} = \frac{Q_u}{\sqrt{H_u}}$

Unit Power

The Unit Power of a given turbine is the power output of the turbine when operating under a head of 1 ft. assuming no change in efficiency .

is the output under a head

Then:

$P = \frac{W\;H}{550}\times \eta$

If $\eta$ is unchanged

$\therefore\;\;\;\;\;W\propto \sqrt{H}$

And:

$P\propto \sqrt{H}\times H\;\;\;\;\propto H^{\frac{3}{2}}\;\;\;\;\;\;\;\therefore\;\;\;\frac{P}{H^{\frac{3}{2}}} = C = \frac{P_u}{H_u^{\frac{3}{2}}}$

(where

is a Constant)
But

Unit Power

$P_u = \frac{P}{H^{\frac{3}{2}}}$

Note:

$N_u\sqrt{P_U} = \frac{N}{\sqrt{H}}\times \frac{\sqrt{P}}{H^{\frac{3}{2}}} = \frac{N\sqrt{P}}{H^{\frac{5}{4}}} = N_s$

The Performance Curves Of A Turbine

Performance Curves are plotted for a constant head and a constant Gate opening (or needle valve setting) and are on the basis of speed in r.p.m.

Francis Turbine

23287/SpecSpeed-and-Unit-Conditions-0016.png

23287/SpecSpeed-and-Unit-Conditions-0012-1.png

Pelton Wheel

Characteristic Curves And Iso-efficiency Curves For A Turbine Under All Operating Conditions

A turbine is a rotary engine that extracts energy from a fluid flow and converts it into useful work.

The Turbine is tested under a constant head

for each of several gate openings, and the values of Power output

and speed

are reduced to unit conditions ( Equations (125) (132)). Suitable values for efficiency are then marked on the curve for the differing Gate openings and lines of iso-efficiency are drawn.

These graphs enable the best running speed or the best gate opening to be chosen,as well as the corresponding power output for that speed and gate setting to be found for any particular head.

An Example Of The Use Of Unit Conditions

A Francis Turbine develops 3240 h.p. at 120 rpm when under a head of 36 ft. What would be the speed and output under a head of 25 ft. assuming no loss in efficiency.

$N_u = \frac{N}{\sqrt{H}} = \frac{N_1}{\sqrt{H_1}}\;\;\;\;\;\;\;\therefore\;\;\frac{120}{\sqrt{36}} = \frac{N_1}{\sqrt{25}}\;\;\;\;\;\;\;\;\therefore\;\;N_1 = 100\,r.p.m.$

But since

, $Q_u = \displaystyle\frac{Q}{\sqrt{H}}$

Unit Power, $P_u = \displaystyle\frac{P}{H^\frac{3}{2}} = \displaystyle\frac{P_1}{H_1^\frac{3}{2}} = \displaystyle\frac{3240}{36^\frac{3}{2}} = \displaystyle\frac{P_1}{25^\frac{3}{2}}$ . Therefore $P_1 = 1875\; h.p.$

Fundamental Similarity Conditions And Model Testing

The following theory assumes equal efficiencies for both the Model and the Prototype. For geometrically similar turbines operating under dynamically similar conditions, the velocity triangles will be similar and:

$v\propto V_f\propto V\propto \sqrt{H}$

But, $v = \displaystyle\frac{\pi DN}{60}$ . Therefore, $\sqrt{H} = DN$ . Or $\displaystyle\frac{\sqrt{H}}{DN} = C}$

And, $Q = K\pi DbV_f$ , $b\propto D$ . Therefore, $Q\propto D^2\sqrt{H}$ . Or $\displaystyle\frac{Q}{D^2\sqrt{H}} = C$ .

Substitute for

in the above equation: $Q\propto ND^3$ . Or $\displaystyle\frac{Q}{DN^3} =C$

Or substitute for

: $Q\propto \displaystyle\frac{H}{N^2}\sqrt{H}$ Or $\displaystyle\frac{QN^2}{H^\frac{3}{2}} = C$

Power, $P = \displaystyle\frac{WH}{550}\times \eta$ if $\eta$ is unchanged $P\propto QH$

From Equation (146), $P\propto D^2H^\frac{3}{2}$ . Or $\displaystyle\frac{P}{D^2H^\frac{3}{2}} = C$

From Equation (148), $P\propto \displaystyle\frac{H^\frac{3}{2}}{N^2}\times H$ . Or $\displaystyle\frac{PN^2}{H^\frac{5}{2}} = C$

i.e. $\displaystyle\frac{N\sqrt{P}}{H^\frac{5}{4}} = N_s = C$ (where

is a Constant)

From equations (144) and (150) $P\propto D^2\propto D^3N^3$ . Or $\displaystyle\frac{P}{N^3D^5} = C$

These seven expressions allow the performance of the prototype turbine to be estimated from tests on the model. Note that there are, in fact, only three independent equations.

The efficiency predicted for a large Turbine from test carried out on a model are usual lower than that obtained from the actual prototype. This is because of the relatively greater frictional losses in the smaller passages of the model.

Strictly speaking, the surface finish of the model should be geometrically similar to that of the prototype. The reduction in efficiency is said to be due to scale effects and is correcter for in practice by the use of empirical equations such as:

The Moody equation

$\frac{1 - \eta _p}{1 - \eta _m} = \left(\frac{D_m}{D_p} \right)^\frac{1}{4}\left(\frac{H_m}{H_p} \right)^\frac{1}{10}$

Example:

[imperial]

Example - Example 1

Problem

A quarter scale Turbine is tested under a head of 36 ft.The full scale Turbine is required to work under a head of 100 ft.and to run at 428 r.p.m.

At what speed must the model be run and if it develops 135 h.p. and uses 38cu.ft.of water per second at this speed, what power will be obtained from the full scale Turbine, assuming that it's efficiency is 3% better than that of the model?

Workings

$\left(\frac{\sqrt{H}}{ND} \right)_m = \left(\frac{\sqrt{H}}{ND} \right)_p$

$\therefore\;\;\;\;\;\frac{N_m}{N_p} = \sqrt{\frac{H_m}{H_p}}\times \frac{D_p}{D_m} = \sqrt{\frac{36}{100}}\times 4 = \frac{24}{10}$

$\therefore\;\;\;\;\;N_m = 2.4\times 428 = 1027\;r.p.m.$

W.H.P. of the model $= \displaystyle\frac{62.4\times 38\times 36}{550} = 155\;h.p.$

$\therefore\;\;\;\;\;\eta _m = \left(\frac{P_m}{W.H.P.} \right) = \frac{135}{155} = 87.1\%$

$\therefore\;\;\;\;\;\eta _p = 90.1\%$

But

$\left(\frac{Q}{ND^3} \right)_m = \left(\frac{Q}{ND^3} \right)_p$

$\frac{(WHP)_p}{(WHP)_m} = \frac{Q_p\,H_p}{Q_m\,H_m} = \frac{N_pD_p^3}{N_mD_m^3}\times \frac{H_p}{H_m} = \frac{10}{24}\times 4^3\times \frac{100}{36}$

$= \frac{2000}{27}$

Therefore, WHP $= \displaystyle\frac{2000}{27}\times 155 = 11.500\;h.p.$

$P= 1150\times 0.901 = 10.360\;h.p.$

$N_s = \frac{N\sqrt{P}}{H^\frac{5}{4}} = \frac{428\sqrt{10.360}}{100^\frac{5}{4}} = 138$

With a value for $N_s$ of 138 the Turbine must be a Propeller Turbine.

Solution

The specific speed is $N_s=138$
The power is $10.360\;h.p.$