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HOW TO DESIGN AND BUILD THEM

A Practical Handbook for Model-Makers



BY
H. H. HARRISON

FULLY ILLUSTRATED




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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.