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""Model

team Turbines

HOW TO DESIGN AND BUILD THEM

A Practical Handbook for Model-Makers

BY

H. H. HARRISON

FULLY ILLUSTRATED

SPON & CHAMBERLAIN,

PUBLISHERS OF TECHNICAL BOOKS

23-125 LIBERTY STREET .... NEW YORK

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The "Model Engineer'' Series. No. 23

MODEL

STEAM TURBINES

HOW TO DESIGN AND BUILD THEM

BY

H. H. HARRISON

OF THE

UNIVERSTT

FULLY ILLUSTRATED

THIRD EDITION-REVISED

LONDON

PERCIVAL MARSHALL & CO.

26-29 POPPIN'S COURT, FLEET STREET, E.G.

&

PREFACE

THE steam turbine has rapidly come to the front

during the last few years on account of its many

advantages as compared with its older rival, the

reciprocating steam engine.

From a model point of view also, several advan-

tages may be claimed for it ; thus for model steamer

work its light weight for a given output, and the fact

that the centre of gravity is low, are two conditions

which make it specially applicable for high-speed craft.

In this small book it is proposed to lay down the

principles on which model steam turbines may be

designed ; and some examples, either built or pro-

jected, are given, being for the most part taken from

the columns of The Model Engineer.

H. H. HARRISON.

PREFACE TO THIRD EDITION

THE need for a third edition of this book having

arisen, advantage has been taken of the opportunity

to partially re-write it and to add further matter.

The reader is earnestly recommended to thoroughly

master Chapters I. and II., as he will then have a very

fair knowledge of the theory underlying the design

and construction of steam turbines in general.

H. H. HARRISON.

MODEL STEAM TURBINES

CHAPTEE I.

GENERAL CONSIDERATIONS.

STEAM turbines may be divided into two classes:

the impulse or action type, in which a jet of steam or

fluid at high velocity impinges on a number of

vanes or paddles fixed around the rim of a wheel

free to rotate, and the reaction type, more generally

known as Hero's Engine or Barker's mill.

Both these types are pretty familiar to us from

our childhood's days, in the windows of the toyshop,

and no doubt in many cases (the writer's also)

formed our first steam model. They are illustrated

in figs. 1 and 2, and frequently appear in patent

specifications, though, of course, with considerable

modification.

In order that the principles on which these

machines are designed and for which they depend

on their action may be thoroughly understood, the

following elementary mechanical considerations are

necessary.

A body has both mass and weight. The weight of

5

197984

MODEL STEAM TURBINES.

a body is the force of gravity acting on it ; whilst

mass is the quantity of matter which the body

contains.

Fig. 2. Hero's

Reaction Engine.

Fig. 1. Simple Impulse

Wheel.

Unit weight is the standard pound, and the unit

of mass is the quantity of matter in a body weighing

32-2 pounds.

Force, which is defined as so many pounds or tons

GENERAL CONSIDERATIONS. 7

(according to its magnitude), may be measured by

the velocity generated in one second on a body free

to move in any direction whatever. Unit force is

that force which, acting on a free body of one pound

weight, generates in it a velocity of one foot per

second. In our latitude, any body free to fall

acquires in doing so (neglecting the resistance of the

air) a velocity of 32-2 feet per second in each second

during the interval between its release and arrest by

the earth or any other body. This fact enables us

to compare the force acting on a moving body when

its weight in Ibs. and its initial and final velocities

in feet per second are known. An example will

serve to make this clear.

Assume that a body weighs say 8 Ibs. ; the pull

of the earth or the attraction of gravity would

generate a velocity, as we have seen, of 32*2 feet per

second after it is allowed to fall, and at the end of

the second time interval another 32*2 feet would be

added, and so on. The force acting in this case

is of course 8 Ibs. These successive increases in

velocity are termed the acceleration of the body, and

the figure 3 2' 2, the acceleration due to the force of

gravity, is symbolised by the letter g.

Now, suppose the body to be at rest but free to

move in any direction horizontally or vertically, and

that a force is suddenly applied in such a way that

the velocity of the body in its direction of motion is

found, after a time interval of one second, to be 40

feet per second. Since the body was at rest at the

beginning, and found to have the above velocity at

MODEL STEAM TURBINES.

the end of the second of time, the figure 40 represents

the acceleration or increase of velocity in the body.

It is evident that the unknown force, which we

will call F, is greater than 8 Ibs. (the force of

gravity) as the acceleration 40 exceeds 32'2, and we

can therefore find the value of F by rule of three

thus :

F : 8 : : 40 : 32'2 ;

and

T-, 8 X40 ! A 11 l

F = - = 10 Ibs. nearly.

oZ'Z

This may be stated in words as follows :

To find the force acting to generate motion in a

body, multiply the weight of the body (in pounds)

by the acceleration produced (feet per second), and

divide the product by the acceleration due to gravity

(32-2 =#). Calling the weight W, and the accelera-

tion attained a, the above can be represented sym-

bolically by

F = ^ ... (1)

g

By definition, unit mass is that quantity of matter

contained in a body weighing 32'2 Ibs. ; therefore

the mass of 1 Ib. is ' of 8 Ibs. -. and of

W pounds ; consequently in (1) represents

o 2i' 2i n

the mass of the body in question, and calling this

quantity M, (1) can be written

F = Ma ...... (2)

that is, Force = Mass x Acceleration ; pounds, feet

and seconds being understood.

GENERAL CONSIDERATIONS. 9

It is necessary here to distinguish carefully between

mere velocity or speed, and acceleration or increase

of velocity.

Supposing a body travels 100 feet in five seconds,

its average velocity is evidently l-g- = 20 feet per

second, or

v -f < 3 >

where V is the average velocity in feet per second,

s the distance covered, and t the time taken in

traversing s.

If the body started from rest under the action of

a constant force of such magnitude that its velocity

increased at the rate of 20 feet per second in each

second, its successive velocities at the end of the

1st, 2nd, 3rd, 4th, and 5th seconds would be 20, 40,

60, 80, and 100 feet per second. Since its initial

velocity was zero, and the velocity at the end of five

seconds 100, the average velocity is given by

+M = 50 feet.

2

From (3) we see that the distance travelled is

equal to velocity x time, and as V is obtained by

dividing the final velocity by 2, the distance

traversed by a body not moving at a constant

velocity as in (3), but at a uniformly increasing one,

is equal to

Final velocity at any instant x time

~2~

or s=^vt .... (4)

from which 2s = vt . . . . . (5)

10 MODEL STEAM TURBINES.

We have now to obtain a value for the accelera-

tion. This is, by definition, the increase of velocity

per second ; and, as the body has no initial velocity

and the velocity it attains in five seconds is 100

feet, the acceleration is obviously = 20 ; or

o

a =- t <>

By combining (5) and (6) in the manner shown

below,

we get finally

v z = 2as (7)

If the body is moving under the action of gravity

instead of any other force, h is usually substituted

for s and g for a, and we get the well-known expres-

sion

v* = 2gh__. . (8)

from which v = JZyh,

or v =8-027/1,

giving us the relation between the velocity attained

by a body under a constant force, gravitational or

otherwise, when travelling over a certain space s or

falling through a height h.

We can use the result in (8) to find the energy in

a moving mass, W Ibs. in weight and having a

velocity of v feet per second. If this weight is

GENERAL CONSIDERATIONS. 11

raised to a height li feet, the work done is evidently

wh foot pounds, and this amount is called the

potential or stored energy in the body, since it is

evident that if the weight be let fall through the

same height it will give out Wh foot pounds in doing

so.

From equation (8) the height through which a

body has fallen is given in terms of its velocity by

7 ,2

*' : '.':.' (9)

and since the energy given out in falling is Wh foot

pounds,

(10)

The right-hand expression is known as the kinetic

energy of the body or energy of motion, and will be

found useful in calculating the pressure set up by

a jet when impinging on the wheel vanes of a

turbine.

Momentum, or quantity of motion, is given by

"VV x v

the expression Mass x Velocity = - .

/

If a body travel along a plane surface without

friciicn at a velocity v, and is suddenly struck so

that its velocity is increased to v z> the momentum

is increased.

The initial moment um = - * ,

final

12 MODEL STEAM TURBINES.

and the increase or change of momentum

The change of momentum in (11) as the result

of the blow is termed an impulse, and the action of

a constant force may be regarded as an indefinitely

large number of impulses gradually changing the

momentum of the -body.

If the force which we will call F acts for t

seconds, then

9

-p _ whole change of momentum ,., 9 x

tinie~

Where the initial velocity is zero, the whole

change of velocity is V feet per second, and (12) =

W

F = V, and if t, the time of application of the

t7

force, is unity,

*-* <>

OF TH E

UNIVERSITY

OP

CHAPTER

PRESSURE DEVELOPED ON SURFACES BY AN IMPINGING

JET. VELOCITY AND FLOW OF STEAM THLOUGH

ORIFICES.

IT has been seen from Chapter I. that if a body of

W

mass travelling at velocity of v feet per second,

9

receives a push or impulse P during time t seconds,

the velocity, and consequently the energy, of the

body is increased.

In fig. 3 a piston is shown mounted in a cylinder,

the back end of which terminates in a nozzle. The

cylinder is filled with water which, the piston being

moved to the right, leaves the nozzle with a certain

velocity V, and impinges on the flat stationary sur-

face. On striking the plate its velocity in the

direction of flow is nil, and the whole change of

velocity is V. Hence

W

P.|V ...... (1)

where P is pressure in Ibs. per square inch, W weight

of fluid or gas discharged per second, and V the

.original velocity of the jet.

13

14

MODEL STEAM TUKBINES.

If the flat surface in fig. 3 forms one of a series

of vanes round the rim of a wheel, it will move away

from the jet with a certain velocity which is some

fraction of the velocity of the jet. Calling this fraction

Fig. 3.

-' the vane velocity is the jet velocity divided by x ;

X

thus

V (the jet velocity) x - = (the vane velocity).

Fig. 4.

It is evident th^t in this case the velocity of the

jet will not be completely destroyed, as it is still able

to follow the vane up (see fig. 4). The change of

-r~r

velocity of the jet is therefore only V and the.

PRESSURE ON SURFACES BY AN IMPINGING JET. 15

pressure P becomes

' ..... (2)

g

The work done per second is, obviously, the pressure

in Ibs. on the vane multiplied by the distance in feet

through which the pressure is exerted ; and the

distance in feet traversed by the vanes in a second

V

is the velocity of these in feet per second, or -.

x

Therefore the work is equal to

P x - foot pounds.

x

inserting the value of P from (2);

W / V\ V

the work in foot pounds = x ( V - ) x . . . (3)

g \ xJ x

The efficiency is given by the ratio of the work

done on the vanes to the energy possessed by the jet.

"\Y-y2

From Chapter I. the jet energy is ' and the

40

efficiency is therefore

W<?? 2

Equation (3) + ~ . . (4)

<J

For example :

Supposing the velocity of the jet is 300 feet

per second, and the size of the nozzle is such that

2 Ibs. of water or gas are discharged in one second,

the energy of the jet = ^ = 2 J12|| = 2795

foot pounds, and assuming that the vane velocity is

16

MODEL STEAM TURBINES.

one-third that of the jet, or - =A, - =^ U - =

x x 3

per second.

Work done on the vanes is from (3)

2

feet

32-2

- x (300 - 100) X 100 = 1240 foot pounds,

/ 2 3 4- 3-

Values ofjc

Fig. 5. Variations of Efficiency with Different Speeds.

and the efficiency which is equal to the ratio

Work got out

g

the ed - ca i culations

work put in

1240 . K

___=44'5 percent.

Taking the same figures as to velocity of jet and

weight of fluid discharges, but varying the vane

velocities, the curve, fig. 5, is obtained, showing

PRESSURE ON SURFACES BY AN IMPINGING JET. 17

that the efficiency is a maximum when the vane

speed is half that of the jet. It will also be seen

that the maximum efficiency is only 50 per cent.,

which is all that can be attained with a flat plate.

This loss of efficiency is due to shock, and to the

fact that the jet moves on with the wheel, and the

energy which it still contains is not utilised. The

B

Fig. 6. Action of Curved Blades.

first loss can be avoided by curving the vane as

shown at A, fig. 6, which gradually changes the direc-

tion of flow of the jet. If matters are so arranged

that the direction of flow, of the jet is completely

reversed as at B, fig. 6, it is clear that the energy

abstracted will be the maximum possible ; and if

the vane speed is, as before, half that of the jet,

the efficiency becomes 100 per cent. In practice,

however, owing to the necessity of arranging for

232

18

MODEL STEAM TURBINES.

the entrance and exit of the jet, the angle between

the initial and final directions is somewhat less than

180. If a Pelton type of wheel is employed, as in

the Bateau single wheel turbine and that of Messrs

Eiedler & Stumpf, this angle becomes 180, but

the steam nozzle has to be inclined, so that the

5 4- S

of JC

Fig. 7. Variations of Pressure on Blades with Different Speeds.

effective velocity of the jet is less than its absolute

Velocity, i.e. the velocity with which it leaves the

nozzle. Fig. 7 shows the increase of pressure on the

vanes as the peripheral speed falls below half the jet

velocity.

From the foregoing it is clear that to design a

turbine to do a certain amount of work all that is

required to be known is the weight discharged in

PRESSURE ON SURFACES BY AN IMPINGING JET. 19

unit time and the velocity attained by the working

agent, whether it be a gas, as steam or compressed

air, or a fluid such as water.

The extended development of the turbine in recent

years has directed the attention of experimenters to

Fig. 8.

the at one time obscure subject of the velocity and

outflow of steam through nozzles, and much valuable

information is now available in the various technical

journals. The following approximate method of treat-

ment will, however, be sufficient for the purpose of

designing model steam turbines.

Let T (fig. 8). be a tank in which the water is kept

20

MODEL STEAM TURBINES.

at a constant level, and let it be provided with a

discharge pipe terminating in a nozzle as shown.

Suppose the relation between the height H and

the area of the nozzle is such that one pound of

water per second is discharged; each pound of

water in falling from the tank to the nozzle does

1 X II foot pounds of work, this work is expended in

giving velocity to the jet as it leaves the nozzle.

Now the kinetic energy of the jet is, as has been

Fig. 9.

proved, foot pounds, or since W = 1, - is equal

47 *9

to the 1 x H foot pounds of work done in falling

from the tank. Assuming no losses by friction in

piping or nozzle, then

v 2

1 x H or H (foot pounds) = (foot pounds), and there-

ty

fore v 2 = 2g (H foot pounds), or the velocity acquired

v= v / 1g x the work done in falling through H feet (5)

In exactly the same way we may calculate the

velocity of steam in falling from an initial pressure

p to final pressure p 2 . The steam does work on

PEESSUKE ON SURFACES BY AN IMPINGING JET. 21

itself, as each layer may be considered as acting as

a perfectly free piston to the steam behind it, as in

fig. 9. Similarly a projectile in a gun acquires velocity

by the expansion of the gases generated by explosion.

If, therefore, the work done on itself by steam in fall-

ing through a certain difference of pressure (the ' H '

of the hydraulic analogy) is calculated, the velocity

can be obtained as indicated by equation (5).

T

Fig. 10.

For example :

The work done on itself by steam expanding from

50 Ibs. absolute (gauge pressure 35 Ibs.) to J Ib.

absolute (a vacuum of 29J" mercury) is 276,300 foot

pounds approximately, and (5) becomes

v=

32-2x276,300;

or since the square root of 2 x 32'2 is 8 nearly,

v = 8^27(3,300 = 4200 feet per second.

There are two methods of evaluating the work done

by steam in expanding from one pressure to another.

One is to construct an ' indicator diagram ' for one

Ib. of steam, as in fig. 10; or, alternatively, to

22

MODEL STEAM TURBINES.

calculate it. Both are tedious, and the latter is both

tedious and difficult to those not accustomed to such

calculations.

For model turbines it will not be possible, as a

rule, to apply a condenser, and the expansion will

not therefore be anything like so high as in the fore-

going example. The following table, calculated on

the assumption of atmospheric exhaust, gives the

velocities attained by steam jets under most condi-

tions as to initial or boiler pressures likely to be

met with.

TABLE I.

Gauge pressure, Ibs. per sq. i n. ,

15

20

40

50 j 60

80

Velocity of steam at throat

of nozzle,

1430

1450

1460

1470 1475

1480

Terminal velocity,

1600

2050

2256

2368 2512

2720

In fig. 11 a steam nozzle of the De Laval type is

shown, and by reference to this it will be seen

that the nozzle consists of two parts, the inlet part

to the left of the figure being convergent up to

the minimum area or * throat ' A m , after which the

nozzle diverges.

On reference to Table I. it will be noticed that the

velocity of the stearn at the throat of the nozzle is

almost constant, whilst the terminal velocities in-

crease as the initial pressure is raised. At 15 Ibs.

pressure (30 Ibs. absolute) the ratio of expansion is

30

TV = 2, and the final velocity when leaving the

J.O

PRESSURE ON SURFACES BY AN IMPINGING JET. 23

nozzle 1600 feet per second ; whilst at 80 Ibs. with

a ratio of expansion of - ^ = 6, the velocity is

15

2720, or 1*7 times as great. From the very close

agreement of the throat velocities we should expect

that up to the throat the ratio of expansion is

the same whatever the initial pressure may be ; and

this is so, it being a well-known fact that steam

discharging through a converging nozzle expands to

58 per cent, of the initial pressure. This peculiarity

Fig. 11. De Laval Nozzle.

(which will be useful for calculating the size of the

orifice for passing a given weight of steam) may be

explained shortly as follows :

In expanding to any extent, the volume of steam

increases directly as the ratio of expansion. Thus at

20 Ibs. absolute the volume of a pound of steam is

19 '73 cubic feet, and at 10 Ibs. absolute (ratio of

expansion 2) 37'83 cubic feet. With a fixed dis-

charge orifice, if the volume of steam is doubled, say,

by increasing the ratio of expansion, the velocity

required to discharge it in the same time must be

24 MODEL STEAM TURBINES.

doubled also. Assume the velocity is 1000 feet per

second, and that one Ib. of steam passes per second.

The energy in foot pounds

Wv* 1x1000x1000

~-^~~ 2x32-2 =15,500 foot pounds.

If the ratio of expansion is twice that in the first

instance, the energy developed will be proportion-

ately increased, but the energy required to double

the velocity is equal to

1x2000x2000

- ;- =62,000 foot pounds,

A X oZi'A

or more than twice times the amount actually avail-

able, consequently the steam cannot get away, and

accumulates until its pressure (and consequently its

volume) is such that the available energy suffices to

discharge the steam from the nozzle, and this point

is reached, as before stated, when the throat pressure

is 58 per cent, of the initial. This explains why a

boiler does not instantaneously discharge the whole

of the steam when the safety-valve lifts. We are now

in a position to understand the function of the diverg-

ing part of the nozzle. By progressive enlargement

of the nozzle area as the expansion of the steam is

extended, the jet is able to expand laterally as well

as in the direction of flow. The increase of area of

the jet along the nozzle as the volume of steam grows,

makes the velocity required to discharge it at any

given point the same as the velocity actually pro-

duced by the degree of expansion at that point. If

the nozzle were convergent, only, the steam would

PRESSURE ON SURFACES BY AN IMPINGING JET. 25

spread laterally as at A, fig. 12, and the velocity

would be about 1400 feet only, and would not be

increased even if the nozzle were discharging into a

perfect vacuum ; but by providing a taper extension

piece, the discharge is as sketch B, and the energy

developed is much greater. The following table

A

it

B

Fig. 12. Difference between Simple Orifice and Taper Nozzle.

gives the weights of steam passed through a De Laval

nozzle having a throat area of 1 square inch, and also

shows the increased area required at the discharge

end, atmospheric pressure in the turbine casing being

assumed. The taper of the flaring end should be

1 in 12, which, with the throat and discharge dimen-

sions, fixes the length of the nozzle.

2G

MODEL STEAM TURBINES.

TABLE II.

737

H3

F.C BERKELEY LIBRARY

. Price, 25 cts.

""Model

team Turbines

HOW TO DESIGN AND BUILD THEM

A Practical Handbook for Model-Makers

BY

H. H. HARRISON

FULLY ILLUSTRATED

SPON & CHAMBERLAIN,

PUBLISHERS OF TECHNICAL BOOKS

23-125 LIBERTY STREET .... NEW YORK

50 Cent Books.

PRACTICAL DYNAMO AND MOTOR CONSTRUCTION. A

handbook of Constructive Details and Workshop Methods used in

Building Small Machines. By ALFRED W. MARSHALL. Contents

of Chapters: 1. Field Magnets. 2. Winding Field Magnets. 3.

Drum Armature Building. 4. Ring Armature Building. 5. How

to Wind Armatures. General Notes. Siemens or H Armatures.

Polar Armatures. 6. How to Wind Armatures (continued). Drum

and Ring Armatures. Binding Wires and Repairs. 7. Commutator

Making. 8. Brush Gears. 9. Mechanical Details of Dynamos and

Motors. 10. Terminals and Connections. 131 pages, 133 illustra-

tions, 12mo., boards. 50c.*

SMALL ACCUMULATORS. How made and used. An elementary

handbook for the use cf amateurs and students. By PERCIVAL

MARSHALL, A.I.M.E. Contents of Chapters: 1. The Theory of the

Accumulator. 2. How to Make a 4-Volt Pocket Accumulator.

3. How to Make a 32-Ampere Hour Accumulator. 4. Types of

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6. Applications of Small Accumulators, Electrical Novelties, etc.

Glossary of Technical Terms. 80 pages. 40 illustrations. 12mo.,

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THE MAGNETO-TELEPHONE. Its construction, fitting up and

adaptability to everyday use. By NORMAN HUGHES. Contents of

Chapters': Some Electrical Considerations: 1. Introductory. 2.

Construction. 3. Lines, Indoor Lines. 4. Signaling Apparatus.

5. Batteries. Open Circuit Batteries. CLsed Circuit Batteries.

G. Practical Operations. Circuit with Magneto Bells and Lightning

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ELECTRIC GAS LIGHTING. How to install electric gas igniting

apparatus, including the jump spark and multiple system for use in

houses, churches, theatres, halls, schools, stores or any large building.

Also the care and selection of suitable Batteries, Wiring and Re -

pairs. By H. S. NORRIE (author of Induction Coils and Coil Mak-

ing). Contents of Chapters: 1. Introduction. Means of Producing

Sparks, Induction, Induction Coils. 2. Application of Induction

Coils to Gas Lighting. Forms of Burners used, Pendant, Rachet,

Stem, Welsbach, Automatic Burners for Gasolene and Acetylene.

3. How to Connect up Apparatus. Wiring a House. Locating

Breaks or Short Circuits. Wiring in finished Houses. General

Remarks. 4. Primary Coils and Safe Devices. 5. How to Wire

ar.i Fit up Different Systems for Lightning Large Buildings. 6. The

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The "Model Engineer'' Series. No. 23

MODEL

STEAM TURBINES

HOW TO DESIGN AND BUILD THEM

BY

H. H. HARRISON

OF THE

UNIVERSTT

FULLY ILLUSTRATED

THIRD EDITION-REVISED

LONDON

PERCIVAL MARSHALL & CO.

26-29 POPPIN'S COURT, FLEET STREET, E.G.

&

PREFACE

THE steam turbine has rapidly come to the front

during the last few years on account of its many

advantages as compared with its older rival, the

reciprocating steam engine.

From a model point of view also, several advan-

tages may be claimed for it ; thus for model steamer

work its light weight for a given output, and the fact

that the centre of gravity is low, are two conditions

which make it specially applicable for high-speed craft.

In this small book it is proposed to lay down the

principles on which model steam turbines may be

designed ; and some examples, either built or pro-

jected, are given, being for the most part taken from

the columns of The Model Engineer.

H. H. HARRISON.

PREFACE TO THIRD EDITION

THE need for a third edition of this book having

arisen, advantage has been taken of the opportunity

to partially re-write it and to add further matter.

The reader is earnestly recommended to thoroughly

master Chapters I. and II., as he will then have a very

fair knowledge of the theory underlying the design

and construction of steam turbines in general.

H. H. HARRISON.

MODEL STEAM TURBINES

CHAPTEE I.

GENERAL CONSIDERATIONS.

STEAM turbines may be divided into two classes:

the impulse or action type, in which a jet of steam or

fluid at high velocity impinges on a number of

vanes or paddles fixed around the rim of a wheel

free to rotate, and the reaction type, more generally

known as Hero's Engine or Barker's mill.

Both these types are pretty familiar to us from

our childhood's days, in the windows of the toyshop,

and no doubt in many cases (the writer's also)

formed our first steam model. They are illustrated

in figs. 1 and 2, and frequently appear in patent

specifications, though, of course, with considerable

modification.

In order that the principles on which these

machines are designed and for which they depend

on their action may be thoroughly understood, the

following elementary mechanical considerations are

necessary.

A body has both mass and weight. The weight of

5

197984

MODEL STEAM TURBINES.

a body is the force of gravity acting on it ; whilst

mass is the quantity of matter which the body

contains.

Fig. 2. Hero's

Reaction Engine.

Fig. 1. Simple Impulse

Wheel.

Unit weight is the standard pound, and the unit

of mass is the quantity of matter in a body weighing

32-2 pounds.

Force, which is defined as so many pounds or tons

GENERAL CONSIDERATIONS. 7

(according to its magnitude), may be measured by

the velocity generated in one second on a body free

to move in any direction whatever. Unit force is

that force which, acting on a free body of one pound

weight, generates in it a velocity of one foot per

second. In our latitude, any body free to fall

acquires in doing so (neglecting the resistance of the

air) a velocity of 32-2 feet per second in each second

during the interval between its release and arrest by

the earth or any other body. This fact enables us

to compare the force acting on a moving body when

its weight in Ibs. and its initial and final velocities

in feet per second are known. An example will

serve to make this clear.

Assume that a body weighs say 8 Ibs. ; the pull

of the earth or the attraction of gravity would

generate a velocity, as we have seen, of 32*2 feet per

second after it is allowed to fall, and at the end of

the second time interval another 32*2 feet would be

added, and so on. The force acting in this case

is of course 8 Ibs. These successive increases in

velocity are termed the acceleration of the body, and

the figure 3 2' 2, the acceleration due to the force of

gravity, is symbolised by the letter g.

Now, suppose the body to be at rest but free to

move in any direction horizontally or vertically, and

that a force is suddenly applied in such a way that

the velocity of the body in its direction of motion is

found, after a time interval of one second, to be 40

feet per second. Since the body was at rest at the

beginning, and found to have the above velocity at

MODEL STEAM TURBINES.

the end of the second of time, the figure 40 represents

the acceleration or increase of velocity in the body.

It is evident that the unknown force, which we

will call F, is greater than 8 Ibs. (the force of

gravity) as the acceleration 40 exceeds 32'2, and we

can therefore find the value of F by rule of three

thus :

F : 8 : : 40 : 32'2 ;

and

T-, 8 X40 ! A 11 l

F = - = 10 Ibs. nearly.

oZ'Z

This may be stated in words as follows :

To find the force acting to generate motion in a

body, multiply the weight of the body (in pounds)

by the acceleration produced (feet per second), and

divide the product by the acceleration due to gravity

(32-2 =#). Calling the weight W, and the accelera-

tion attained a, the above can be represented sym-

bolically by

F = ^ ... (1)

g

By definition, unit mass is that quantity of matter

contained in a body weighing 32'2 Ibs. ; therefore

the mass of 1 Ib. is ' of 8 Ibs. -. and of

W pounds ; consequently in (1) represents

o 2i' 2i n

the mass of the body in question, and calling this

quantity M, (1) can be written

F = Ma ...... (2)

that is, Force = Mass x Acceleration ; pounds, feet

and seconds being understood.

GENERAL CONSIDERATIONS. 9

It is necessary here to distinguish carefully between

mere velocity or speed, and acceleration or increase

of velocity.

Supposing a body travels 100 feet in five seconds,

its average velocity is evidently l-g- = 20 feet per

second, or

v -f < 3 >

where V is the average velocity in feet per second,

s the distance covered, and t the time taken in

traversing s.

If the body started from rest under the action of

a constant force of such magnitude that its velocity

increased at the rate of 20 feet per second in each

second, its successive velocities at the end of the

1st, 2nd, 3rd, 4th, and 5th seconds would be 20, 40,

60, 80, and 100 feet per second. Since its initial

velocity was zero, and the velocity at the end of five

seconds 100, the average velocity is given by

+M = 50 feet.

2

From (3) we see that the distance travelled is

equal to velocity x time, and as V is obtained by

dividing the final velocity by 2, the distance

traversed by a body not moving at a constant

velocity as in (3), but at a uniformly increasing one,

is equal to

Final velocity at any instant x time

~2~

or s=^vt .... (4)

from which 2s = vt . . . . . (5)

10 MODEL STEAM TURBINES.

We have now to obtain a value for the accelera-

tion. This is, by definition, the increase of velocity

per second ; and, as the body has no initial velocity

and the velocity it attains in five seconds is 100

feet, the acceleration is obviously = 20 ; or

o

a =- t <>

By combining (5) and (6) in the manner shown

below,

we get finally

v z = 2as (7)

If the body is moving under the action of gravity

instead of any other force, h is usually substituted

for s and g for a, and we get the well-known expres-

sion

v* = 2gh__. . (8)

from which v = JZyh,

or v =8-027/1,

giving us the relation between the velocity attained

by a body under a constant force, gravitational or

otherwise, when travelling over a certain space s or

falling through a height h.

We can use the result in (8) to find the energy in

a moving mass, W Ibs. in weight and having a

velocity of v feet per second. If this weight is

GENERAL CONSIDERATIONS. 11

raised to a height li feet, the work done is evidently

wh foot pounds, and this amount is called the

potential or stored energy in the body, since it is

evident that if the weight be let fall through the

same height it will give out Wh foot pounds in doing

so.

From equation (8) the height through which a

body has fallen is given in terms of its velocity by

7 ,2

*' : '.':.' (9)

and since the energy given out in falling is Wh foot

pounds,

(10)

The right-hand expression is known as the kinetic

energy of the body or energy of motion, and will be

found useful in calculating the pressure set up by

a jet when impinging on the wheel vanes of a

turbine.

Momentum, or quantity of motion, is given by

"VV x v

the expression Mass x Velocity = - .

/

If a body travel along a plane surface without

friciicn at a velocity v, and is suddenly struck so

that its velocity is increased to v z> the momentum

is increased.

The initial moment um = - * ,

final

12 MODEL STEAM TURBINES.

and the increase or change of momentum

The change of momentum in (11) as the result

of the blow is termed an impulse, and the action of

a constant force may be regarded as an indefinitely

large number of impulses gradually changing the

momentum of the -body.

If the force which we will call F acts for t

seconds, then

9

-p _ whole change of momentum ,., 9 x

tinie~

Where the initial velocity is zero, the whole

change of velocity is V feet per second, and (12) =

W

F = V, and if t, the time of application of the

t7

force, is unity,

*-* <>

OF TH E

UNIVERSITY

OP

CHAPTER

PRESSURE DEVELOPED ON SURFACES BY AN IMPINGING

JET. VELOCITY AND FLOW OF STEAM THLOUGH

ORIFICES.

IT has been seen from Chapter I. that if a body of

W

mass travelling at velocity of v feet per second,

9

receives a push or impulse P during time t seconds,

the velocity, and consequently the energy, of the

body is increased.

In fig. 3 a piston is shown mounted in a cylinder,

the back end of which terminates in a nozzle. The

cylinder is filled with water which, the piston being

moved to the right, leaves the nozzle with a certain

velocity V, and impinges on the flat stationary sur-

face. On striking the plate its velocity in the

direction of flow is nil, and the whole change of

velocity is V. Hence

W

P.|V ...... (1)

where P is pressure in Ibs. per square inch, W weight

of fluid or gas discharged per second, and V the

.original velocity of the jet.

13

14

MODEL STEAM TUKBINES.

If the flat surface in fig. 3 forms one of a series

of vanes round the rim of a wheel, it will move away

from the jet with a certain velocity which is some

fraction of the velocity of the jet. Calling this fraction

Fig. 3.

-' the vane velocity is the jet velocity divided by x ;

X

thus

V (the jet velocity) x - = (the vane velocity).

Fig. 4.

It is evident th^t in this case the velocity of the

jet will not be completely destroyed, as it is still able

to follow the vane up (see fig. 4). The change of

-r~r

velocity of the jet is therefore only V and the.

PRESSURE ON SURFACES BY AN IMPINGING JET. 15

pressure P becomes

' ..... (2)

g

The work done per second is, obviously, the pressure

in Ibs. on the vane multiplied by the distance in feet

through which the pressure is exerted ; and the

distance in feet traversed by the vanes in a second

V

is the velocity of these in feet per second, or -.

x

Therefore the work is equal to

P x - foot pounds.

x

inserting the value of P from (2);

W / V\ V

the work in foot pounds = x ( V - ) x . . . (3)

g \ xJ x

The efficiency is given by the ratio of the work

done on the vanes to the energy possessed by the jet.

"\Y-y2

From Chapter I. the jet energy is ' and the

40

efficiency is therefore

W<?? 2

Equation (3) + ~ . . (4)

<J

For example :

Supposing the velocity of the jet is 300 feet

per second, and the size of the nozzle is such that

2 Ibs. of water or gas are discharged in one second,

the energy of the jet = ^ = 2 J12|| = 2795

foot pounds, and assuming that the vane velocity is

16

MODEL STEAM TURBINES.

one-third that of the jet, or - =A, - =^ U - =

x x 3

per second.

Work done on the vanes is from (3)

2

feet

32-2

- x (300 - 100) X 100 = 1240 foot pounds,

/ 2 3 4- 3-

Values ofjc

Fig. 5. Variations of Efficiency with Different Speeds.

and the efficiency which is equal to the ratio

Work got out

g

the ed - ca i culations

work put in

1240 . K

___=44'5 percent.

Taking the same figures as to velocity of jet and

weight of fluid discharges, but varying the vane

velocities, the curve, fig. 5, is obtained, showing

PRESSURE ON SURFACES BY AN IMPINGING JET. 17

that the efficiency is a maximum when the vane

speed is half that of the jet. It will also be seen

that the maximum efficiency is only 50 per cent.,

which is all that can be attained with a flat plate.

This loss of efficiency is due to shock, and to the

fact that the jet moves on with the wheel, and the

energy which it still contains is not utilised. The

B

Fig. 6. Action of Curved Blades.

first loss can be avoided by curving the vane as

shown at A, fig. 6, which gradually changes the direc-

tion of flow of the jet. If matters are so arranged

that the direction of flow, of the jet is completely

reversed as at B, fig. 6, it is clear that the energy

abstracted will be the maximum possible ; and if

the vane speed is, as before, half that of the jet,

the efficiency becomes 100 per cent. In practice,

however, owing to the necessity of arranging for

232

18

MODEL STEAM TURBINES.

the entrance and exit of the jet, the angle between

the initial and final directions is somewhat less than

180. If a Pelton type of wheel is employed, as in

the Bateau single wheel turbine and that of Messrs

Eiedler & Stumpf, this angle becomes 180, but

the steam nozzle has to be inclined, so that the

5 4- S

of JC

Fig. 7. Variations of Pressure on Blades with Different Speeds.

effective velocity of the jet is less than its absolute

Velocity, i.e. the velocity with which it leaves the

nozzle. Fig. 7 shows the increase of pressure on the

vanes as the peripheral speed falls below half the jet

velocity.

From the foregoing it is clear that to design a

turbine to do a certain amount of work all that is

required to be known is the weight discharged in

PRESSURE ON SURFACES BY AN IMPINGING JET. 19

unit time and the velocity attained by the working

agent, whether it be a gas, as steam or compressed

air, or a fluid such as water.

The extended development of the turbine in recent

years has directed the attention of experimenters to

Fig. 8.

the at one time obscure subject of the velocity and

outflow of steam through nozzles, and much valuable

information is now available in the various technical

journals. The following approximate method of treat-

ment will, however, be sufficient for the purpose of

designing model steam turbines.

Let T (fig. 8). be a tank in which the water is kept

20

MODEL STEAM TURBINES.

at a constant level, and let it be provided with a

discharge pipe terminating in a nozzle as shown.

Suppose the relation between the height H and

the area of the nozzle is such that one pound of

water per second is discharged; each pound of

water in falling from the tank to the nozzle does

1 X II foot pounds of work, this work is expended in

giving velocity to the jet as it leaves the nozzle.

Now the kinetic energy of the jet is, as has been

Fig. 9.

proved, foot pounds, or since W = 1, - is equal

47 *9

to the 1 x H foot pounds of work done in falling

from the tank. Assuming no losses by friction in

piping or nozzle, then

v 2

1 x H or H (foot pounds) = (foot pounds), and there-

ty

fore v 2 = 2g (H foot pounds), or the velocity acquired

v= v / 1g x the work done in falling through H feet (5)

In exactly the same way we may calculate the

velocity of steam in falling from an initial pressure

p to final pressure p 2 . The steam does work on

PEESSUKE ON SURFACES BY AN IMPINGING JET. 21

itself, as each layer may be considered as acting as

a perfectly free piston to the steam behind it, as in

fig. 9. Similarly a projectile in a gun acquires velocity

by the expansion of the gases generated by explosion.

If, therefore, the work done on itself by steam in fall-

ing through a certain difference of pressure (the ' H '

of the hydraulic analogy) is calculated, the velocity

can be obtained as indicated by equation (5).

T

Fig. 10.

For example :

The work done on itself by steam expanding from

50 Ibs. absolute (gauge pressure 35 Ibs.) to J Ib.

absolute (a vacuum of 29J" mercury) is 276,300 foot

pounds approximately, and (5) becomes

v=

32-2x276,300;

or since the square root of 2 x 32'2 is 8 nearly,

v = 8^27(3,300 = 4200 feet per second.

There are two methods of evaluating the work done

by steam in expanding from one pressure to another.

One is to construct an ' indicator diagram ' for one

Ib. of steam, as in fig. 10; or, alternatively, to

22

MODEL STEAM TURBINES.

calculate it. Both are tedious, and the latter is both

tedious and difficult to those not accustomed to such

calculations.

For model turbines it will not be possible, as a

rule, to apply a condenser, and the expansion will

not therefore be anything like so high as in the fore-

going example. The following table, calculated on

the assumption of atmospheric exhaust, gives the

velocities attained by steam jets under most condi-

tions as to initial or boiler pressures likely to be

met with.

TABLE I.

Gauge pressure, Ibs. per sq. i n. ,

15

20

40

50 j 60

80

Velocity of steam at throat

of nozzle,

1430

1450

1460

1470 1475

1480

Terminal velocity,

1600

2050

2256

2368 2512

2720

In fig. 11 a steam nozzle of the De Laval type is

shown, and by reference to this it will be seen

that the nozzle consists of two parts, the inlet part

to the left of the figure being convergent up to

the minimum area or * throat ' A m , after which the

nozzle diverges.

On reference to Table I. it will be noticed that the

velocity of the stearn at the throat of the nozzle is

almost constant, whilst the terminal velocities in-

crease as the initial pressure is raised. At 15 Ibs.

pressure (30 Ibs. absolute) the ratio of expansion is

30

TV = 2, and the final velocity when leaving the

J.O

PRESSURE ON SURFACES BY AN IMPINGING JET. 23

nozzle 1600 feet per second ; whilst at 80 Ibs. with

a ratio of expansion of - ^ = 6, the velocity is

15

2720, or 1*7 times as great. From the very close

agreement of the throat velocities we should expect

that up to the throat the ratio of expansion is

the same whatever the initial pressure may be ; and

this is so, it being a well-known fact that steam

discharging through a converging nozzle expands to

58 per cent, of the initial pressure. This peculiarity

Fig. 11. De Laval Nozzle.

(which will be useful for calculating the size of the

orifice for passing a given weight of steam) may be

explained shortly as follows :

In expanding to any extent, the volume of steam

increases directly as the ratio of expansion. Thus at

20 Ibs. absolute the volume of a pound of steam is

19 '73 cubic feet, and at 10 Ibs. absolute (ratio of

expansion 2) 37'83 cubic feet. With a fixed dis-

charge orifice, if the volume of steam is doubled, say,

by increasing the ratio of expansion, the velocity

required to discharge it in the same time must be

24 MODEL STEAM TURBINES.

doubled also. Assume the velocity is 1000 feet per

second, and that one Ib. of steam passes per second.

The energy in foot pounds

Wv* 1x1000x1000

~-^~~ 2x32-2 =15,500 foot pounds.

If the ratio of expansion is twice that in the first

instance, the energy developed will be proportion-

ately increased, but the energy required to double

the velocity is equal to

1x2000x2000

- ;- =62,000 foot pounds,

A X oZi'A

or more than twice times the amount actually avail-

able, consequently the steam cannot get away, and

accumulates until its pressure (and consequently its

volume) is such that the available energy suffices to

discharge the steam from the nozzle, and this point

is reached, as before stated, when the throat pressure

is 58 per cent, of the initial. This explains why a

boiler does not instantaneously discharge the whole

of the steam when the safety-valve lifts. We are now

in a position to understand the function of the diverg-

ing part of the nozzle. By progressive enlargement

of the nozzle area as the expansion of the steam is

extended, the jet is able to expand laterally as well

as in the direction of flow. The increase of area of

the jet along the nozzle as the volume of steam grows,

makes the velocity required to discharge it at any

given point the same as the velocity actually pro-

duced by the degree of expansion at that point. If

the nozzle were convergent, only, the steam would

PRESSURE ON SURFACES BY AN IMPINGING JET. 25

spread laterally as at A, fig. 12, and the velocity

would be about 1400 feet only, and would not be

increased even if the nozzle were discharging into a

perfect vacuum ; but by providing a taper extension

piece, the discharge is as sketch B, and the energy

developed is much greater. The following table

A

it

B

Fig. 12. Difference between Simple Orifice and Taper Nozzle.

gives the weights of steam passed through a De Laval

nozzle having a throat area of 1 square inch, and also

shows the increased area required at the discharge

end, atmospheric pressure in the turbine casing being

assumed. The taper of the flaring end should be

1 in 12, which, with the throat and discharge dimen-

sions, fixes the length of the nozzle.

2G

MODEL STEAM TURBINES.

TABLE II.

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