SAFETY-VALVES.
BY
RICHARD H. BUEL, 0. E.
(REPRINTED FROM THE "RAILROAD GAZETTE.")
NEW  YORK:
D. VAN NOSTRAND, PUBLISHER,
23 MURRAY AND 27 WARREN STREET.
1 875.








INTRODUCTION.
The writer, in presenting these remarks to
engineers, does not pretend to offer much that
is original, but has aimed to gather what is
valuable from the great mass of material to be
found in scientific periodicals and in publications that are not generally accessible. An
endeavor has been made to systematize the
treatment of the subject, and to give such
varied solutions of the problems that arise in
proportioning the parts of safety-valves as to
render them plain to those who have only an
elementary education.  The importance of
having the general principles of safety-valves
understood by those who are charged with the
care of steam machinery cannot well be overestimated. With a safety-valve that is in reality all which its name implies, a large proportion of the risks incident to the use of boilers
will be avoided; while on the other hand, a
safety-valve that is only such in name is one of
the readiest assistants to a disastrous boiler
explosion.
NEW YORK, August, 1875.








SAFETY-VALVES.
I. THE REQUISITE QUALIFICATIONS OF
A SAFETY-VALVE.
As a safety-valve is designed to prevent the accumulation of pressure in a
steam boiler beyond a certain point, it
is necessary that the parts should be so
proportioned that the valve will rise
when the given pressure is attained.
Until the valve rises it is subjected to
the pressure of the steam at rest, so that
this part of the subject involves the
statical condition of a safety-valve. As
soon as the pressure of the steam in a
boiler lifts the valve, new conditions are
introduced, because the steam is in motion, escaping through the orifice between the valve and the seat. It will




6
thus be evident that from the time the
valve is raised until it is again seated by
the reduction of the steam pressure the
dynamical conditions are to be regarded.
A good safety-valve should be so constructed that not only will it lift when
the required pressure is attained, but so
that it will also prevent the further increase of pressure, and will close promptly as soon as that pressure is reduced.
II. PROPORTIONING THE PARTS OF SAFETY - VALVES, IN ORDER THAT THEY
MAY RISE WITH GIVEN PRESSURES.
This part of the subject, as already
remarked, deals with the statical condition of safety-valves. In other words,
it is a question of the equilibrium of two
forces acting in contrary directions-one,
a weight or the tension of a spring,
tending to hold the valve down; and
the other, the pressure of the steam,
tending to raise it. When these opposing forces balance each other the valve
is ready to lift, and any slight increment




7
of pressure will raise it. It is, then,
the conditions of the balancing or equilibrium of the steam pressure and the
spring or weight that are to be considered. In some forms of safety-valves, a
spring or weight is placed directly above
the valve, and resists the upward pressure of the steam; or a weight is suspended directly under the valve, passing
into the boiler.  In other forms, the
spring or weight is attached to a lever,
to which the valve is also connected, the
weight or spring being ordinarily at a
greater distance from the fulcrum than
the valve is. In the first form of construction, in which the steam pressure is
opposed directly by the force of a weight
or spring, without the intervention of a
lever, these two forces will evidently
balance when they are equal to each
other. It is only necessary, therefore, to
multiply the pressure of the steam in
pounds per square inch by the area of
the valve in square inches to find what
weight must be attached, or what tension put upon the spring, to prevent the




8
valve from rising before this pressure is
reached. For example, suppose that a
valve has a diameter of 4 inches, and is
required to rise when the steam pressure
is  100  pounds per square inch, what
weight must be attached to it?
The area of a valve having a diameter
of four inches is about..12 9-16 square inches.
Multiply by............  100
Weight to be attached to
valve.............. 1,2561 pounds.
In the second case, where a lever is
employed, it is evident that the forces
will not balance if they are equal, since
they act at different points of the lever.
Fig. 1 is a sketch of an ordinary lever......l..  __
y       Fig 1        w
safety-valve, and the forces acting in a




9
case of this kind are represented graphically by arrows, the directions in which
they point corresponding to the directions in which the forces act. It will be
seen that there are four forces:
1. The weight, represented by the arrow W.
2. The weight of the lever, represented
by the arrow L.
3. The weight of the valve and stem,
represented by the arrow V.
4. The pressure of the steam, represented by the arrow P.
The latter force, the pressure of the
steam, is the pressure per square inch
mtltiplied by the area of the valve in
square inches. The lever is arranged so
that it can turn about the point A as a
fulcrum.  The first three forces tend to
keep the valve down, the fourth force
tends to raise it, and all the forces act
vertically. The lever arm of a force is
the distance from  the force to the fulcrum, measured on a line that is drawn
perpendicular to the direction  of the
force.  As these forces act vertically,




10
the lever arms must be measured on horizontal lines-and it will be seen that
there are three lever arms to be considered:
1. The lever arm of the weight, represented by w.
2. The lever arm of the lever, represented by 1.
3. The lever arm of the valve, represented by p.
The lever arms of these forces are the
horizontal distances from the centers of
gravity of the weight, lever, and valve,
respectively, to the fulcrum. The centers of gravity of the weight and valve
are ordinarily in vertical lines passing
through their centers, and the center of
gravity of the lever is most readily determined by balancing it upon a knife
edge, the center of gravity being on a
vertical line passing through the point
on which it balances.
To recapitulate, there are eight quantities that may be varied, in proportioning a safety-valve, viz.:
1. The weight.




11
2. The weight of the lever.
3. The weight of the valve.
4. The diameter of the valve.
5. The lever arm of the weight, or its
position on the lever.
6. The lever arm of the lever, depending upon its form.
7. The lever arm of the valve.
8. The pressure of the steam, in
pounds per square inch.
Any seven of these parts may be proportioned at pleasure, and the remaining
one can be determined, the sole condition being that the valve shall rise as
soon as a given pressure is exceeded.
Ordinarily, however, there are only two
cases that require solution:
1st. With what steampressure will a given
valve rise?  2d. Where must the weight
be placed on the lever of a given safetyvalve in order that the valve may rise
when a certain steam pressure is reached?
These problems can be solved either by
experiment or calculation. Both methods will be explained.




12
A. —Experimental Method.
Ascertain the weights of the ball, lever
and valve, the diameter of the valve, the
center of gravity of the lever, and the lever arms of the weight, lever and valve.
The lever arms of the weight and valve
can be measured directly, and the lever
arm of the lever, or the horizontal distance
from its center of gravity to the fulcrum, can be determined by balancing it
upon a knife-edge, as already explained.
In measuring the valve, if the seat is
beveled and the valve fits well, the
smallest diameter is to be taken. Then
provide a bar, A B, Fig. 2, of uniform
section, at least twice as long as the distance from the weight to the fulcrum of
the given valve. Mark the center-point,
C, of the lever, and points E, F, G, to
the right of the center, such that the
distance C Eis the same as the horizontal distance of the center of the valve
from the fulcrum, C F is the horizontal
distance from the center of gravity of
the lever to the fulcrum, and C G the
horizontal distance from the center of




13
6^
to t
the weight to the fulcrum, if this is
known. Also lay off to the left of the
center a distance, C H, equal to CE.
Balance the bar upon a knife-edge, D,
and the apparatus will be ready for
making the required determinations.




14
1st. To find with what steam pressure
the valve will rise. —Hang on the ball at
the point G, the lever at F, and the
valve at E, or attach equivalent weights,
as may be most convenient. Then hang
on weights at the point H, until they
just balance the weights on the' other
side of the center, and bring the bar to
a horizontal position.  Ascertain the
area of the valve in square inches, and
divide the weight at H  by this area.
The result will be the required pressure
of the steam, in pounds per square inch.
2nd. To find where to place the weight
on the lever of a given safety-valve, in
order that the valve may rise with a given
steam pressure.-Multiply the area of the
valve in square inches by the pressure of
the steam in pounds per square inch, and
attach a weight at H equal to this product. Also hang on the valve and lever,
or equivalent weights, at E and F, as
before, and move the ball along the bar
until it balances the other weights. The
distance from C, of the point at which it
balances will be the horizontal distance




15
from the fulcrum to the weight, on the
lever of the given safety-valve.
To facilitate the only calculation required in this experimental method, the
areas of valves are given below, for the
majority of cases that occur in practice:
Table of Areas of Valves of Diferent Diametres
Diameter of Valve in Area of Valve in square
inches.               inches.
i or 0.5            13-64 or 0.19635
i or 0.625           5-16 or 0.30680
or 0.75             7-16 or 0.44179
or 0.875           19-32 or 0.60132
1                    25-32 or 0.7854
1i or 1.25          1 15-64 or 1.2272
1 or 1.5           1 49-64 or 1.7671
1 or 1.75         2 13-32 or 2.4053
2                   3 9-64 or 3.1416
2J or 2.5          4 29-32 or 4.9087
3                  7 1-16 or 7.0686
3 or 3.5           9 5-8  or9.6211
4                 12 9-16 or 12.5664
4j or 4.5         15 29-32 or 15.9043
5                 19 41-64 or 19.635
5~ or 5.5         23 49-64 or 23.7583
6                 28 9-32 or 28.2744




16
The method here described is exceedingly simple and accurate, and will, it is
hoped, find favor with those who have
difficulty in using the ordinary modes of
calculation. It is not uncommon for the
graduations on many safety-valves to be
wrongly marked, and it is important for
steam users to know certainly whether
or not they are correct.
A simple experimental test of the accuracy of the adjustment of a safetyvalve may be made without removing it
from its position on the boiler. The
method is described below, the account
being taken from  an article by the
writer published in the Scientific American for Oct. 31, 1874.
"Secure the valve stem of the safetyvalve to the lever with wire or string,
and attach a loop to the lever, into
which pass the hook of an accurate
spring balance, arranging the loop so
that it is directly over the center of the
valve stem. Then take hold of the upper part of the spring balance and lift
the valve slightly, noting the reading of




1T
the balance. Measure the lower diameter of the safety-valve, and find its
area; divide the reading of the spring
balance by the area of the valve, and
the result will be the pressure, in pounds
per square inch, at which the steam will
raise the safety-valve. Suppose, for instance, that the diameter of the safetyvalve is 1 inch; its area will be about
RM4- of an inch. Now, if the tension of
the spring balance in raising the valve is
120 pounds, the pressure at which the
valve will rise is the quotient arising
from dividing 120 by %ioM, or 153 pounds
per square inch."
B.-Graphical Method.
The statical problems in connection
with safety-valves can also be solved
graphically or by construction. A force
can be represented in intensity and direction by a straight line, by assuming
some unit of length for the unit of intensity. For instance, if the force is 5
pounds, it may be represented by a line
i of an inch in length. To simplify the




18
graphical solution, the several constructions upon which it depends will be presented separately.
1. Two or more parallel forces, acting
in the same direction on a lever, being
given, to find the direction, intensity, and
point of application of a single force
which can replace them andproduce the
same effect.
In Fig. 3, the lever, A B, is acted
upon by two parallel forces, C D, and
E F. Lay off horizontally to the right
of E, and the left of C, any two distances,
Ea, Cb, equal to each other. From a
draw a c parallel to EF, and from F
draw Fc parallel to Ea, forming a rectangle, Ea F c, on the two adjacent
sides, Ea, E F.  Also complete the
rectangle, C b d D, of which C b, C D,
are the adjacent sides.  Through the
points C and E draw diagonals, d C, cE,
to the two rectangles, and produce them
until they meet in the point, e. Through
e draw a line, e H, parallel to CD and
EF, and the point G, in which it cuts
the lower line of the lever, will be the




19
A,,      \            8
L-  j      0',  \: ~.
F ~a8
point of application of a force which will
singly produce the same effect as the
two forces, CD, EF., Make G H equal
to the sum of CD, and E F, and it will
represent the intensity of the required
force, which is called the resultant of
the other two forces.
If there are three forces to be replaced by a single one, after finding the
force that is equivalent to two of them,
consider this and the third force as two
single forces, and proceed as before. In
this way any number of parallel forces
acting in the same direction can be replaced by their resultant, which is a
single force, equivalent in effect to all
the others.




20
1Ai         (/   <\D-<tBN
d i'             Fig 4
2. The intensity and point of application of a vertical force acting on a lever
being given, also the position of the fulcrum, to find the intensity of a parallel
force, acting at a given point of the lever,
which will balance the first force.
D E, Fig. 4, is the given force, C the
fulcrum, and F  the point at which the
required force must act and balance D E.
Draw  a vertical line, C e, from  the fulcrum. Lay off, to the right of D and
the left of  _\ any two distances, D a,
_Fb, equal to each other. Complete the
rectangle, D a c E, and draw a vertical
line, b d, from the point b. Through D,
draw a diagonal to the rectangle D a cE,
and produce it until it meets the vertical




21
line through the fulcrum in the point e.
From  e, draw  a line through F, the
point of application of the required
force, continuing it until it meets the
vertical line drawn from b in the point d.
Complete the rectangle of which b F,
b d, are the adjacent sides, and F G will
be the force required to balance D E.
I       e        I'dB
o A^ 
/                      E
3. The intensity and point of application of one vertical force, the intensity of
a second parallelforce, and the position
of the fulcrum, being given, to find the
point of application of the second force,
in order that it may balance the first one.
F G, Fig. 5, is the force whose intensity and point of application are given.
C is the fulcrum, and a b is the intensity




22
of the second force. Draw a vertical
line, Ce, from  the fulcrum.  Lay off
from F, towards the left, any distance,
Fc, and complete the rectangle Fc d G,
Through F draw a diagonal to the rectangle, continuing it to e, where it meets
the vertical line drawn from the fulcrum. In any convenient place, lay off
a horizontal line, fg, equal to F c, and
from the point g draw the perpendicular
g h, equal to a b. Through the pointsf
and h, draw the line fh, and from e draw
a line, eD, parallel to fh.  The point
D, in which the line last drawn meets
the lower side of the lever, is the point
of application of the second force; and
D E, equal to a b, represents that force
applied at such a point as to balance
F G.
By the application of the graphical
methods just explained, it will be easy to
solve the problems previously determined
by experiment.
1st. Tofind with what steam pressure
a safety valve will rise.-On the lever A
B, Fig. 6, mark the fulcrum C. To the




23
/
/1'' i /gF,  6
~I,'
right of   lay off the distance C D, C G)
C I, equal to the lever arms of the valve,
the lever and the weight respectively;
and to the left of C lay off the distance
C L, equal to C D. Also draw vertical lines, D F, G H, I K, to represent
the weights of the valve, lever and ball.




24
Find by construction a single force, P
R, which can replace these three forces
and produce the same effect.  This is
done by replacing G If  and I     by a
single force, N 0, and then finding a
force, P R, which is equivalent to N O
and 1D F. Then construct the force L
M, which acts at the point L  and
balances P R, or the three forces, D F,
G H, 1 K.  L M  represents the total
pressure on the valve in pounds, and this,
being divided by the area of the valve in
square inches, will give the pressure of
the steam, in pounds per square inch, at
which the valve will rise.
2nd. To find where to place the weight
on the lever of a given safety-valve, in order that the valve may rise with a given
steam pressure.-In Fig. 7, L M represents the required pressure in pounds per
square inch multiplied by the area of the
valve in square inches, D F is the weight
of the valve, G Tf is the weight of the
lever, and a b is the weight of the ball,
the points of application of all the
weights, except that of the ball, being




25
/  I.... 
I________ j    DI  p\. G',  I
gle force, PzA which will produce the
at L, and balancing P. Subtracting
L V   from the whole force l, there
z ~ ^           l~q7'0
given, Replace. ons and G pr by a single force, P R, which will produce the
same effect, and find L 2V, a force acting
at I, and balancing PR_. Subtracting
L N from the whole force L My there
remains L O, which must be balanced
by the weight of the ball a B. Then, by
the third construction, previously given,
find the point I at which the ball must
be placed in order to balance L 0.




26
In the use of the graphical method
for testing the accuracy for proportioning the parts of safety-valves, the scale
on which the construction is made should
be as large as possible, in order that apparently trivial errors on a drawing may
not be magnified greatly in the practical
applications.
C.- Analytical Method.
The mode of calculating  the parts
of safety - valves from given or observed  data will now  be presented.
And first it may be well to say something in regard to the means of ascertaining the weights of the various
parts, and the position of the center of
gravity of the lever. The most obvious
method of determining the weight of any
thing is to weigh it, and the center of
gravity of a lever is most readily found
by balancing a lever upon a knife-edge.
It is not always convenient, however, to
make these experiments. Sometimes no
facilities for weighing the parts are at
hand, and in certain cases it may be nee



27
essary to make the calculations from such
data as can be obtained from a drawing.
When the parts can be removed, their
weights  can be determined approximately by experiment.  Immerse each
part in water, using a shallow vessel, and
mark how much the water rises, which
will give, by a simple calculation, the
volume of the  part in cubic inches.
Having ascertained the number of cubic
inches in a part, the weight can be obtained by multiplying this number
By 0.2778, if the part is of wrought iron.
By 0.2604, "   "   " cast iron.
By 0.2917, "   "   " brass.
The weights so found will generally be
only approximate, but will be tolerably
accurate, if the experiment is carefully
performed.
When only the dimensions of the parts
are given, their volumes must be determined by calculation.  The valve and
weight will ordinarily present no difficulties, as they have forms for which
rules are given in most works on mensuration, and their centers of gravity usu



28
ally coincide with their geometrical centers. Having found their volumes in
cubic inches, their weights can be obtained by multiplying these volumes by
the weight of a cubic inch of the material as given above. The lever, not unfrequently, is quite irregular in form, and
the center of gravity is scarcely ever in
the middle of the bar. Whatever its
shape, however, its volume can be found
with great accuracy by the method to be
explained. Make a drawing of the lever
in plan and elevation, as shown in Fig.
8, C D being the plan, and E F the elevation, the form of the lever being
much more irregular than is usual in
practice. To determine the volume of
this lever in cubic inches, it will be necessary to find the area of one face, as,
for instance, the elevation and the mean
width of the other. In finding the mean
width, the area of the face represented
on the plan will first be calculated, and
then divided by the length. To calculate the area of either of these faces, divide it into any even number of parts,




29
A f!. 2.        i'  6i.f3   4i 7 j 8.' z.I. 
ar to any line, assumed as may be most
by equi-distant lines drawn perpendicular to any line, assumed as may be most
convenient. Thus, in Fig. 8, the elevation is so irregular that a straight line
would not lie wholly within it, and A B
is drawn without, and the elevation is
divided into 20 parts, by equi-distant
lines perpendicular to A B. In the plan,
the line C D is within the figure, and
the same number of equi-distant lines
are drawn, perpendicular to C D. These
perpendicular lines are called ordinates,




30
and it will be observed that as the number of divisions of a face is even, the
number of ordinates is uneven. Number the ordinates from first to last, 1, 2,
3, 4, etc., and measure their lengths,
either between the boundaries of the
faces or between the boundary on one
side, and the straight line, which is called
the base. Then the area of a face, or
the area between the base and the boundary on one side (according to the manner in which the ordinates are measured),
can be found by the following rule:
Add together the first and last ordinates, four times the sum of the even
numbered intermediate ordinates, twice
the sum of the odd numbered ordinates,
and multiply the sum so obtained by onethird of the distance between any two
consecutive ordinates.
In Fig. 8, the lever is supposed to be
20 inches long, and the distance between
ordinates is one inch, so that there are
20 divisions, and 21 ordinates. This example is merely for the sake of illustration. Where great accuracy is required,




31
the  ordinates  should  be  taken  much
closer together. The lengths of the ordinates are shown on the drawing, the
dimensions being in inches. The diameter of the hole in the lever is 9-10 of an
inch.  To find the area of the elevation,
the area included between the base A  B
and the boundary of the figure nearest
to it will be calculated and subtracted
from the whole area between the base
and the outside boundary.  The following are the calculations:
AREA BETWEEN BASE AND NEAREST BOUNDARY.
Extreme      Even-numbered    Odd-numbered
ordinates.      ordinates.       ordinates.
1.....3.45   2........1.74   3.... 1.41
21..    5.40   4......1.68   5........2.04
6....... 2.25   7........2.52
8.85   8.......2.90   9......3.45
10........4.10  11........4.84
12........5.56  13.......6.84
14........7.04  15.......7.67
16......8.12  17.......8.40
18.....8.48  19.......8.32
20........7.72
44.99
49.59




32
Sum  of even-numbered ordinates...... 49.59
Multiply by.......................                  4
198.36
Sum  of odd-numbered ordinates........ 44.99
Multiply by....................                  2
89.98
Add................            198.36.
i 198.36
Add(   8.85
297.19
Multiply by....................
Area between base and nearest boundary
(square inches).................... 99.06
AREA BETWEEN BASE AND OUTSIDE BOUNDARY.
Extreme          Even-numbered         Odd-numbered
ordinates.          ordinates.            ordinates.
1......3.45    2........5.13    3.......5.44
21....7.23    4........5.12    5.......4.75
6.......4.76    7........4.91
10.68    8......5.15    9........5.50
10....5.91   11.......6.41
12........6.92   13.......7.48
14.......8.01'5....8.44
16........8.89   17......9.05
18........9.14   19.......8.94
20........8.77
---   -60.92
67.80




33
67.80               60.92
4                   2
271.20              121.84
271.20
10.68
403.72
_
Area between base and outer boundary
(square inches)..............134.57
Subtract............................ 99.06
Area of elevation of lever, square inches 35.51
AREA OF PLAN OF LEVER........2.00    2......2.00    3........2.00
21.......1.00   4........2.00    5........1.70
6.......1.82    7.......1.97
3.00    8........2.11    9.......2.23
10....... 2.30  11.......2.40
12.......2.43  13....2.45
14......2.52  15.......2.50
16.......2.45  17........2.32
18......2.15  19.......1.83
20........1.44
-.... 19.40
21.22                   2
4
-~~-         38.80
84.88              84.88




34
3.00
126.68
Area of plan of lever, in square inches.. 42.23
Divide by length of lever............. 20)42.23
Average width of plan..............   2.11
Multiply by area of elevation........  35.51
Volume of lever incubic inches...... 74.9261
Deduct volume of hole..............  1.2723
73.6538
Multiply by weight of a cubic inch of
iron............................    0.2778
Weight of lever in pounds...........  20.47
This method of finding the area of an
irregular figure is very useful in many
cases that arise in practice. It will be
a good plan for the reader to calculate
the weight of a lever in the manner here
explained, and  compare  it with  the
weight as ascertained by trial.
The position of the center of gravity
of any lever can also be found by calculation, and the manner of doing it will
now  be explained.  It may be well to




35
give, in the first place, a few definitions
and illustrations.
The resultant of two or more forces is
a single force, whose action is equal to
the combined effect of the several forces
which are called components.
The moment of a force with reference
to any point is the product arising from
multiplying the intensity of the force by
the distance of the point from the force
measured in a straight line drawn from
the point perpendicular to the direction
of the force. This distance is called the
arm of the force.
When any number of forces act in the
same plane, the moment of the resultant
is equal to the sum of the moments of
those forces which tend to turn a body in
the same direction that the resultant
does, diminished by the sum of the moments of those forces that tend to turn a
body in the opposite direction. This is
generally more concisely expressed by
saying that the moment of the resultant is
equal to the algebraic sum of the moments
of the components. In Fig. 9 an illustra



36
Ag 9
F;g 9
tion of this principle is given.   Four
forces in the same plane, represented by
the darts a, b, c, d, are acting at the
points B, C, D, E of a body which is secured so that it can only turn around
the point A, and r is a single force which
will produce the same effect as these
four forces, or it is the resultant of a, b,
c, d.  The perpendicular lines drawn
from the point A to the directions of the
several forces represent the arms of these
forces. An inspection of the figure will
show that the forces a, b, tend to turn
the body in the same direction that the
resultant does, while the forces c, d, tend
to turn it in the opposite direction. The
analytical statement of the  principle
given above will, then, be:




37
rx AL=a A Gx bX AK
-(c X A  + d X AI)
From this equation it appears that if
all but one of the parts in any example
of this kind are known, the other can
readily be ascertained. Suppose, for instance, that the intensities of all the
forces are known, and all the arms, except that of the resultant, as follows:
a = 108 pounds A G =  75 inches.
b=114    "   AK= 170 
c =  72   "   AH=  65  "
d=  81   "   AI=  75  "
r = 150   "
From these values the following equation can be formed:
150 x AL= 8,100+19,380-(4,680+6075)
=27,480-10,755
=16,725
and AL=16,725 - 150
=-111 inches.
The lever of a safety-valve is acted
upon at every point by a force which is
the weight of the particle at that point,
and the resultant of all these forces is
their sum, or the weight of the whole




38
lever, and the point at which it acts is
the center of gravity of the lever. This
point, then, can be found with sufficient
accuracy, by supposing that the lever is
divided up into such small parts that
the center of gravity of each can be estimated at its geometrical center, without serious error, and its weight calculated as if its faces were trapezoids-so
that an equation can be formed with the
movements of these forces.  In Fig. 8
the lever is divided into 20 parts, and the
weight of each is calculated, being shown
by the darts on the elevation.
The moments of the force are referred
to the point E, and the calculation by
which the center of gravity is found, is
given below:
MOMENT OF WEIGHT OF PART.
- 2, —0.98 x  0.5 =  0.49
2- 3,-1.95 x  1.5 =  2.93
3- 4,-2.09 x  2.5 =  5.22
4- 5,-1.63 x  3.5 =  5.71
5- 6,-1.33 x  4.5 =  5.98
6- 7,-1.34 x  5.5 =  7.37
7- 8,-1.36 x  6.5 =  8.84
8- 9,-1.35 x  7.5 = 10.13




39
9-10,-1.27 x  8.5 = 10.79
10 —, —1.14 x  9.5 = 10.83
11-12,-1.03 x 10.5    10.82
12-13,-0.87 x 11.5 = 10.00
13-14,-0.76 x 12.5     9.50
14-15,-0.63 x 13.5=  8.51
15-16,-0.55 x 14.5=  7.97
16-17,-0.49 x 15.5=  7.60
17-18,-0.43 x 16.5 =  7.09
18-19,-0.36 x 17.5 =  6.30
19-20,-0.40 x 18.5=  7.40
20-21,-0.51 x 19.5 =  9.95
153.43
Calling Effthe distance from the end
of the lever to the center of gravity, and
recollecting that the resultant is equal
to the weight of the lever,
20.47 x E H= 153.43:
Whence, E H = 7~ inches.
This mode of determining the center
of gravity of a lever is quite tedious,
if  the  parts  are taken  small enough
to secure great accuracy, but it is useful in case the plan by trial is impossible.
The condition that a safety-valve shall




40
rise at a given pressure of steam requires,
as has already been stated, that the
forces tending to hold the valve down
shall be just balanced by the tendency of
the opposing forces to raise it. In Fig.
1 there are three forces holding the valve
down, the weight of the ball, W; the
weight of the lever, L, and the weight
of the valve  V and these must be balanced by the pressure of the steam, P.
When any number of forces balance
each other, the sum of the moments of
the forces which act in one direction is
equal to the sum of the moments of the
forces which act in the other direction.
Hence the following equation will represent the action of the forces when the
valve is just balanced:
Pxp= WxwL + L x I + VXp.
If any seven of the quantities in this
equation are known, the other can be determined. The equation can, however,
be changed a little, so as to be more generally useful:
Let S= pressure of steam in pounds per




41
square inch, D = diameter of valve in
inches; then,
P = 0.7854 X D2 X S
and the equation becomes:
0.7854 X D2X Sxp
=         X   +L X l+ VTXp.
For the ordinary cases that arise in practice the equation will take the following
forms;
1st.  To find the pressure per square
inch at which a given valve will open:
FWXw+Lx C + Vxp
S=
0.7854 X D' Xp
2d. To find where to place the weight on
a given safety valve, so that it shall open
at a given pressure per square inch:
0.7854 X D  X SXp --    X l —Vxp
3d.  To find what diameter a safety
valve must have, all the other parts being
known, to open at a given steam pressure 




42
D IV(  w+X LXl+ VXpv
D-V        0.7854X SXp 
As many who are familiar with arithmetic, but do not understand algebraic
expressions, may like to use these rules,
they are translated below:
1st.  To find the pressure per square
inch at which a given valve will open.
Measure the following distances horizontally from the fulcrum to:
1. The center of the valve stem.
2. The center of the weight.
3. The center of gravity of the lever,
or the point on which it will balance if
placed upon a knife-edge.
Measure the diameter of the valve,
and determine its area, either by a table
or by multiplying the square of the di.
ameter by 0.7854.
Find the weight of:
1. The valve.
2. The lever.
3. The ball.
Multiply(1). The weight of the ball by its
horizontal distance from the fulcrum.




43
(2). The weight of the lever by its
horizontal distance from the fulcrum.
(3). The weight of the valve by its
horizontal distance from the fulcrum.
(4). The area of the valve by its horizontal distance from the fulcrum.
Add together the first three products
and divide the sum by the fourth product.
Example.-A  given safety-valve has
a weight of 50 pounds 24 inches from
the fulcrum, the lever weighs 6 pounds,
and its center of gravity is 15 inches
from  the fulcrum; the weight of the
valve is 2 pounds, and its center is 4
inches from the fulcrum.  The diameter
of the valve is 2 inches. At'what pressure will the valve begin to rise?
Square of diameter...............  4
Multiply by....................0.7854
Area of valve in square inches...3.1416
(1). 50  times 24 is.....  1,200
(2). 6    "   1" 5.....    90
(3). 2    "    4".....         8
(4.) 3.1416"    4"...... 12.5664
The sum  of (1), (2) and (3)is 1,298,




44
which divided by 12.5664 (the fourth
product) is 103.03, the  pressure  in
pounds per square inch at which the
valve will open.
2. To find where to place the weight
on a safety-valve, so that it shall open at
a given pressure of steam:
Multiply(1). The weight of the lever by the
horizontal distance of its center of gravity from the fulcrum.
(2). The weight of the valve by its
horizontal distance from the fulcrum.
(3). The area of the valve by the pressure of steam in pounds per square inch,
and by the horizontal distance of the
valve from the fulcrum.
Add together the first two products,
subtract their sum from  the third product, and divide the difference by the
weight of the ball.
Example. —The ball of a safety-valve
weighs 100 pounds, the lever weighs 10
pounds, the valve weighs 2 pounds and
has a diameter of 3 inches. The distance of the center of gravity of the lever




45
from the fulcrum is 25 inches, and the
distance of the center of the valve from
the fulcrum is 5' inches. How far from
the fulcrum must the valve be placed, in
order that the lever may open at a pressure of 100 pounds?
Area of valve........ 7.07 square inches.
(1). o0 times 25 is........ 250
(2).   2"      5"......  10
(3). 7.07  "   100"....... 707
707 "      5 "....3,535
Adding together products (1) and (2)
we have as their sum  260; subtracting
this from 3,535, the third product, we
have 3,275. Dividing this difference by
100, the weight of the ball, we have
32.75, or 321 inches as the distance from
fulcrum to ball.
3d.  To find what diameter a safetyvalve must have, the other parts being
known, to open at a given steam pressure:
Multiply,
(1). The weight of the ball by its horizontal distance from the fulcrum.
(2). The weight of the lever by the




46
horizontal distance of its center of gravity from the fulcrum.
(3). The weight of the valve by the
horizontal distance of its center from the
fulcrum.
(4). The pressure of steam in pounds
per square inch by the horizontal distance of the valve from the fulcrum, and
by the number 0.7854.
Add together the first three products,
divide their sum by the fourth product,
and take the square root of the quotient.
Example.-Weight of ball, 60 pounds;
lever, 7 pounds; valve, 3 pounds. Distances from fulcrum: ball, 30 inches;
center of gravity of lever, 16 inches;
center of valve, 3 inches. Pressure of
steam, 70 pounds per square inch. What
should be the diameter of the valve?
(1). 60 times  80 is..... 1,800
(2).  7 "     16 "        112
(3).  3 "      3 ".....     9
(4). 70 "      3.....  210
210 " 0.7854 "....164.934
The sum  of the first three products
(1,800, 112 and 9) is 1,921.  Dividing




47
this sum by 164.934 (the fourth product),
we have 11.647 inches.  The  square
root of this number is 3.41 +inches, which
by the rule is the required diameter of
the valve.
Instead of weights, springs are sometimes employed to hold down safetyvalves. The calculations involving merely the condition that the valve shall
open at a given pressure, are the same
as those  previously employed, except
that the tension of the spring is to be
substituted for the weight of the ball,
this tension being first determined by
experiment.  Having ascertained the dimensions of a spring most suitable for a
given case, it is easy to calculate the
dimensions of a spring for any other case.
Knowing, for instance, the cross-section
of a spring adapted to one pressure on
the valve, make a proportion, thus:
Cross    Cross sec-  Pressure  Pressure
section of. tion of     on. on
given   ~ required     first    second
spring.    spring.    valve.    valve.
The pitch of the second spring, or the




48
distance from the center of one coil to
the center of the next, is found by the
following proportion:
Side of a     Side of a
Pitch   Pitch      square         square
of.   of    jequal             equal
given- required:   in area to j   in area to
spring  spring.      section of  section of
Lfirst spring   L2d spring
Exrample.-It is a common practice in
proportioning the parts of direct-loaded
spring safety-valves to use a spring of
the following dimensions: for a valve 3
inches in diameter and 100 pounds steam
pressure, a spring made of square steel,
having a cross section of j of a square
inch, and a pitch of one inch, the side of
a square equal in area to the cross section being, of course, i inch. What
should be the proportions of a similar
spring for a valve 5 inches in diameter,
and a steam pressure of 50 pounds per
square inch?
Area of first valve.............. 7.07 in.
Multiply by steam pressure...... 100
Total pressure on first valve... 707




49
Area of second valve............19.64 in,
Multiply by steam pressure.....  50
Total pressure on second valve. 982
To  find  cross  section  of required
spring:
0.Q25.Cross section.
5 of second spring' 707  982
and by the rule for proportion, the product of the two extremes (245.5), divided
by the third term (707), will give the
second term.  Now 245.5 divided by 707
is 0.34724 in., equals cross section of second spring. The square root of 0.32724
is 0.589 +, and this is the side of a square
equal in area to cross section of second
spring.
To find pitch of second spring:
ePitc  of. s   0.   50.589.
second spring'
Here we solve the problem  by multiplying 1 by 0.589 and dividing the product
by 0.5, which gives 1.178 inches for the
pitch of second spring.




50
III. PROPER DIAMETER FOR A SAFETYVALVE.
There are seven  rules, commonly
quoted by different authorities, for determining the area of a safety-valve.
They are given below, with an example
to illustrate their use:
Let
G=square feet of grate surface in
boiler.
H= square feet of heating surface in
boiler.
C=pounds of coal burned per hour.
W=  "   " water evaporated per
hour.
P=pressure of steam, as shown by
gauge.
A =area of safety - valve in square
inches.
When the area is known, the diameter
of the valve can be found by dividing
the area by the number 0.7854, and extracting the square root of the quotient.
1. United States lRule.-Allow  one




51
square inch of area in the valve for 25
square feet of heating surface in the
boiler, or
H
A=25
2. English  Rule.-For boilers with
natural draft, allow half a square inch
of area in the valve for each square foot
of grate surface, or
G
2
3..French Rule. —1. Multiply the grate
surface by the number 22.5.
2. Add the number 8.62 to the steam
pressure.
3. Divide the first quantity by the second. The quotient will be the area of
valve, or
GX22.5
A=P+8.62
4. Molesworth's -Rule.-Allow a valve
area of eight-tenths of an inch for each
square foot of grate surface, or




52
A= GX0.8.
5. Professor Thurston's First Rule.1. Multiply the pounds of coal burned
per hour by the number 4.
2. Add the number 10 to the steam
pressure.
Divide the first quantity by the second, or
4 C
A=P+1o
6. Professor Thurston's Second Rule.
-1. Multiply the heating surface by the
number 5.
2. Add the number 10 to the steam
pressure, and multiply the sum by the
number 2.
Divide the first quantity by the second, or,
5lH
A=
2(P+ 20)
7. Professor Rankine's Rule. —Allow
a valve area of six-thousandths of an
inch for each pound of water evaporated
per hour, or,
A=o.oo6 IW




53
In many cases these rules give widely
different results, as will be illustrated by
their application to an example.
It is required to find the area of a
safety-valve for a boiler having 15 square
feet of grate surface, 472.5 square feet
of heating surface, carrying 40 pounds
of steam, burning 210 pounds of coal,
and evaporating 1,470 pounds of water
per hour..
1. United States Rule:
472.5+25=18.9 square inches=area required.
2. English Rule:
15 —2=7.5 square inches=area required.
3. French Rule:
40+8.62=48.62.
15 x 22.5=337.5.
837.5+48.62=6.94 square inches-area required
4. Molesworth's Rule:
15 x 0.8 —12 square inches=area required.
5. Professor Thurston's First Rule:
40+10=50.
210 x 4=840.
840+50=16.8 square inches=area required.
6. Professor Thurston's Second Rule:
40+10 —50.




50 x 2-100.
472.5 x 5=2,362.5.
2,362.54-100=23.63 sq. inches-area required.
7.T1Professor Rankine's Rule:
1,470 x 0.006 in.=8.82 sq. inches=area required
It is not remarkable that these rules
should give such varying results, when
it is remembered that the performance
of different boilers of precisely the same
dimensions varies greatly.  A  safetyvalve should have such proportions that
it can permit all the steam to escape
that a boiler can generate when forced
to the utmost extent.  Hence its area
depends upon:
1. The amount of steam  to be discharged in a given time.
2. The lift of the valve.
3. The velocity with which the steam
escapes.
There  is such  a difference in the
amount of coal burned per square foot
of grate per hour in different boilers,
and the amount of water evaporated per
pound of coal, that it would seem impossible to use the same constants in a




55
formula for all cases. It is believed,
however, that the following estimates
give a most liberal allowance on the side
of safety:
Pounds of coal Pounds of
burned per    water
square foot evaporated
of grate per  per pound
hour.    of coal.
Stationary and marine boil-   15         9
ers with natural draft....
Stationary and marine boil-   30         7
ers with forced draft.... 
Locomotive boilers.......... 100         6
On these assumptions, the pounds of
water evaporated, and consequently the
weight of steam that the safety-valve
would have to release per hour for each
square foot of grate surface would be:
For stationary and marine boilers with
natural draft..........................  135
For stationary and marine boilers with
forced draft.......................... 210
For locomotive boilers................. 600
These figures, as has been stated, represent much better than  average performance, and can readily be modified to
suit any particular case. Of course, in




56
constructing a formula for general use,
in the absence of precise data, and in
view  of the varying results obtained
from boilers of the same kind, the estimate should be taken high enough to include all cases.
One element of the proposed formula
for the diameter of a safety-valve for
any given boiler, then, will be the number of square feet of grate surface in the
boiler, multiplied by a constant.
The amount of opening afforded by a
safety-valve for the escape of steam depends not only on its diameter, but also
on the distance that the valve lifts. In
order that the area for escape shall be
equal to the area of the valve, it is necessary that the lift should be one-quarter
of the diameter of the valve.
A valve without bevel, Fig. 10, eviFig io




dently gives an opening, when raised,
equal to the surface of a cylinder having
the same diameter, a b, as that of the
valve, and a height equal to the lift, c b.
Hence the rule for finding the opening of
aflat valve, due to a giver lift, will be as
follows: multiply the diameter of the
valve by the lift, and by the number
3.1416.
Examnple.-A  flat valve 3 inches in
diameter lifts 1-16 of an inch.  What is
the area of the opening?
Lift.....................     0.0625
Multiply  by....................... 3.1416
0.196+
Multiply  by......................  3
Area of opening in square inches.... 0.589+
If the valve has a bevel, as shown in
Fig. 11, until it lifts clear of the seat the
opening will be equal to the surface of
the frustum  of a cone, of which the
diameter, a b, of the upper base is the
same as that of the valve, the slant
height, b e, is the perpendicular distance
between the lower edge of the valve and




58
0           4
Fig 1
the seat, and the diameter of the lower
base, f e, is the diameter of the seat,
measured at the intersection of perpendiculars b e, af, from opposite points of
the lower edge of the valve to the seat.
The bevel or inclination of the valve is
the angle of inclination to a vertical line,
fca, or e d b.
To find the amount of opening afforded
by a valve with beveled seat for any lift
less than the depth of the seat.
(1). Multiply the diameter of the valve
by the lift, by the sine of the angle of
inclination, and by the number 3.1416.
(2). Multiply the square of the lift by
the square of the sine of the angle of in



59
clination, by the cosine of this angle, and
by the number 3.1416.
Add these two products.
Example.-The diameter of a safetyvalve is 21 inches, the seat is I of an
inch deep, and has a bevel of 25 degrees.
What is the area of opening, for a lift of
i of an inch?
Sine of 25~........................ 0.423
Multiply by lift.................. 0.25
0.106Multiply by.......................... 3.1416
0.333
Multiply by diameter of valve........ 2.5
1st Product.................... 0. 83+
Square of sine of 25~............... 0.179
Multiply by cosine of 25 ~............ 0.9060.162+
Multiply by square of lift...........0.0625
0.0101+
Multiply by................. 3.1416
2d product...................... 0.03+
Add 1st product............... 0.83+
Area of opening in square inches.... 0.86+




60
It often happens that the depth of the
bevel is very slight, so that the valve
lifts clear of the seat. In such a case
the  opening  must be  computed for a
beveled valve with  a lift equal to the
depth of the seat, and for a flat valve
with the remainder of the lift.
Example.-The diameter of a valve is
4 inches, the bevel is 35~, and the depth
of seat i of an inch.  What is the area
of opening for a lift of I of an inch?
First calculate the the area of opening
for a beveled valve, with a lift of 4 inch,
the depth of the seat.
Sine of 35......................... 0.5745
Multiply by...................... 3.1416
1.8
Multiply by lift.................... 0.25
0.45
Multiply by diameter of valve........   4
First product............ 1.80
Square of sine of 35~................ 0.329
Multiply by cosine of 35..... 0.819
0.269
Multiply by........................ 3.1416
0.85




61
Multiply by square of lift............ 0.0625
Second product................ 0.05
Add first product.................... 1.80
Area of opening, in square inches,
in lifting clear of seat........... 1.85
Next calculate the amount of opening
due to the lift above the seat, which in
this case is * of an inch:
Lift...............................  0.125
Multiply by........................ 3.1416
0.3927
Multiply by diameter of valve........   4
Area of opening due to lift above
seat, in square inches.......... 1.57
Add opening due to lifting clear of seat 1.85
Total area of opening, in square
inches......................... 3.42
The most usual bevel for valves is an
angle of 45~, and 300 is also frequently
adopted.  As a matter of convenience,
the rules have been somewhat simplified
for these cases.
Tofind area of opening for a lift less
than depth of seat, when the bevel of the
valve is 45 degrees.




62
1. Multiply the diameter of the valve
by the lift, and by the number 2.22.
2. Multiply the square of the lift by
the number 1.11.
Add these two products.
Example.-What is the area of opening of a 2-inch valve, for i-inch  lift;
depth of seat, j inch; bevel of valve 45
degrees?
1. Opening in lifting clear of seat, i
inch=0.25 inch:
(1). 2 x 2.22 x =1.11.
(2). 0.25x0.25 x.11-0.07.
Adding the products (1) and (2)
1. 11+0.07=-1.18 square inches in lifting
clear of seat.
2. Opening for lifting above seat, ~
inch:
3.1416 x 4 x 2=1.57 square inches, in lifting above seat.
1.57 + 1.18 = 2.75 square inches-total
area of opening.
To find area of opening for a lift less
than depth of seat, when the bevel of the
valve is 30~.




63
1. Multiply the diameter of the valve
by the lift, and by the number 1.57.
2. Multiply the square of the lift by
the number 0.68.
3. Add the two products.
Example.-A 4-inch valve has a bevel
of 30~, and the depth of seat is j inch.
What is the area of opening for a lift of
3-16=0.1875 of an inch?
(1). 4xo.18'75X1.57= 1.18
(2.) 0.1875 X 0.1875 X 0.68 = 0.035 X
0.68=0.02
(3). 1.18+0.02 = 1.2 square inches=
area of opening.
The analytical expressions for the preceding rules are given below, together
with the values of the sines and cosines
of angles from 20" to 50~.
Let
D=diameter of valve in inches.
-=lift       "       "
a=angle of bevel.
d=depth of seat, in inches
A=area of opening, in square inches,
1. VWhen there is no bevel:




64
A=3.1416 x xD
2. Beveled valve, lift less than depth of
seat.
A =3.1416X[DXlX sin a+ l'X(sin a)'
Xcos a]
3. Beveled valve, lift greater  than
depth of seat.
A = 3.1416X [DXd X sin a + d X (sin a)2
Xcos a+DX( -d)]
4. Bevel of 45~, lift less than depth of
seat:
A=2.22 X-DX/+ 1.11 X2
5. Bevel of 450, lift greater than depth
of seat.
A=2.22XDx d+ 1.11 x d+ 3.1416XD
x ( —d)
6. Bevel of 30~, lift less than depth of
seat:
A  1.57X.DXl+ 0.68X 1
7. Bevel of 30~, lift greater than depth
of seat:
A =1.5 7XDx d+ 0.68X d + 3.141.6 XD
x ( —d)




65
Angle. Sine. Cosine. Angle. Sine. Cosine.
200.342.940      36~.588.809
210.358.934      37~.602.799
220.375.927      38~.616.788
230.391.921      39~.629.777
240.407.914      40~.643.766
250.423.906      41~.656.755
26~.438.899      420.669.743
27~.454.891      43~.682.731
280.469.883      44~.695.719
29~.485.875      45~.707.707
30~.500.866      46~.719.695
31~.515.857      47~.731.682
320.530.848      48~.743.669
33~.545.839      49~.755.656
34~.559.829      50~.766.643
35~.574.819
It will now  be necessary to find the
velocity  with  which   steam   escapes
through  the  opening  afforded  by  a
valve.   Professor Rankine  determined
that the number of pounds of steam
that will flow through an orifice having
an area of one square inch, into the atmosphere, in a second, is one-seventieth
of the absolute pressure of the steam;
and this result has been confirmed by
the  experiments  of Mr. Napier.   It
should be remarked that this rule is only




66
approximate, as the  constants  vary
slightly for different pressures of steam,
but it is sufficiently accurate for practical purposes. The absolute pressure of
steam is the pressure as shown by a steam
gauge, inreased by the pressure of the
atmosphere, which is, on an average,
14.7 pounds per square inch. from these
principles, a rule can be constructed for
the area ofopening required in any given
case.
Multiply the absolute pressure of steam
in the boiler by the number 51.43, and
divide the number of pounds of water
evaporated per hour by this product.
Example.-Suppose that a boiler evaporates 2,000 pounds of water per hour,what
should be the area of opening afforded
by the safety-valve, so that the pressure
shall not exceed 100 pounds per square
inch?
Pressure by gauge............... 100.0 lbs.
Add........................ 14.7
Absolute pressure............ 114.7 lbs.
Multiply  by.....................   51.43
5,899+




67
2,000 -5,899=0.339+ =area of opening
in square inches.  Knowing the area of
opening required, if the lift is fixed, the
diameter of the valve that will furnish
the requisite opening with the given lift
can be calculated. There are three general cases for which rules will be given:
1st. When the valve has no bevel —
Multiply the lift of the valve by the
number 3.1416, and divide the area of
opening by this product.
Example-A  safety valve is required
to have one inch area of opening, with a
lift of i inch. What should be its diameter?
Lift........................    0.25  in.
Multiply  by.......................3.1416
0.7854
0.7854) 1.0000
Diameter of valve.......... 1.27+inches.
2d. When the valve is beveled, and the
lift is less than the depth of seat:
1. Multiply the square of the lift by
the square of the sine of the angle of inclination, by the cosine of this angle, and
by the number 3.1416.




68
2. Multiply the lift by the sine of the
angle of inclination, ahd by the number
3.1416.
3. Subtract the first product from the
area of opening, and divide the difference by the second product.
Example-A safety valve has a bevel
of 32~, and the depth  of seat is ~  inch.
It is  to  lift  - of  an  inch, and  give  an
opening  of  1i  square   inches.    What
should be its diameter!
Square of lift.......................  0.141
Square of sine of 32.............   0.281
Their product........................   0.04
Cosine of 32........................  0.848
Product of last two............  0.034
Multiply by........................ 3.1416
And we have........................  0.107
Area of opening.....................  1.500
Subtract............................  0.1.07
1.393
Lift.................................  0.375
Sine of 32..........................   0.53
Their product......................  0.199




69
Multiply by...................... 3.1416
And we have........................ 0.625
Dividing 1.393 by 0.625 we have
Diameter of valve==2.23+inches.
3d. When the valve is beveled, and the
lift is greater than the depth of seat:
1. Multiply the square of the depth of
seat by the square of the sine of the
angle of inclination, by the cosine of the
angle, and by the number 3.1416.
2. Multiply the depth of seat by the
sine of the angle of inclination, and by
the number 3.1416.
3. Multiply the difference between the
lift and depth of seat by the number
3.1416.
4. Subtract the first product from the
area of opening, and divide the difference by the sum of the second and third
products.
Examlpe —A safety valve has a bevel
of 33~, a depth of seat of i inch, and is
required to give an area of opening of 2
inches,with a lift of j inch. What should
be its diameter?




70
Square of depth  of seat.............. 0.0625
Square of sine of 33~................ 0.297
Product.......................  0.019
Cosine of 33~....................... 0.839
Product........................ 0.016
Multiply by........................ 3.1416
Product.....................                       0.05
Area of opening....................  2.00 in
Subtract........................ 0.05
Difference is................... 1.95
Depth  of seat........................ 0.25
Sine of 33..........................   0.545
Their product.................  0.136
Multiply by.........................   3.1416
And we have.................... 0.427
Lift................................. 0.50
Subtract depth of seat................  0.25
The difference  is............... 0.25
Multiply by......................   3.1416
And the product is............. 0.785
Add................................. 0.427
1.212
Now 1.95 divided by 1.212 gives the
diameter of valve=1.61 inches, nearly.




71
It may be useful to give a summary of
the foregoing rules.
lb'find the proper diameterfor a safetyvalve:
1. The number of pounds of water
evaporated per hour is equal to the area
of the grate in square feet multiplied by,
135.  For stationary and marine
boilers with natural draft;
210.  For stationary and marine
boilers with forced draft;
600. For locomotive boilers.
2. Find the area of opening required
to discharge this weight of steam per
hour, the pressure of steam being given.
3. Find the diameter of a valve that
will afford the required opening, with a
given lift.
Example.-A  stationary boiler with
natural draft has 36 square feet of grate
surface, and the pressure of steam is 90
pounds. What should be the diameter
of the safety-valve, the valve to have a
bevel of 45~. a depth of seat of I of an
inch, and to lift i inch?




72
Grate surface..................   3.6
Multiply by........................ 135
Pounds of water evaporated per hour.4,860
Steam  pressure, per gauge..........   90
Add................................   14.7
Total pressure..............  104.7
Multiply by.......................   51.43
5,384.7
4,860+5,384.7
Gives area of opening in square inches_0.903
Square of lift.....................  0.0625
Square of sine of 45~...............  0.5
Their product.......................  0.031
Multiply by cosine of 45~...........    0.707
And we have......................  0.022
Which, multiplied by................  3.1416
Gives..............................   0.069
Area of opening...................  0.903
Subtract...................  0.069
The difference is..............  0.83
Lift..............................    0.25
Multiply by sine of 45~...........    0.707
Product....................                           0.177
Multiply by.......................  3.1416
Product...........................   0.556




73
0.834-0.556 = 1.5 inches =diameter of
valve.
The algebraic expressions for the preceding rules are as follows:
Let
)D=diameter of valve, in inches.
I=lift      "        "
a=angle of bevel.
d=depth of seat, in inches.
A=area of opening in square inches.
W=pounds of steam to be discharged
through the safety-valve per hour.
P=pressure of steam, in pounds per
square inch, as shown by gauge.
p=absolute pressure of steam, in
pounds per square inch.
p=P+14.7
W
A= 
51.43 Xp
1. When the valve has no bevel:
A
D —=
3.1416Xl
2. When the valve is beveled, and the
lift is less than the depth of seat:




74
A- 3.1416 X   X (sin a)'X cos a
3.1416 X IX sin a
3. When the valve is beveled, and the
lift is greater than the depth of seat:
A —3.1416 X d" X (sin a)'X cos a
D=
3.1416 X (dX sin a + —d)
4. When the bevel of the valve is 45~,
and the lift is less than the depth of seat:
A-i. llxl
J=~
2.22 X 
5. When the bevel of the' valve is 45~,
and the lift is greater than the depth of
seat:
A-1.1 x xd
DD=
2.22 X d+3.1416 X ( —d)
6. When the bevel of the valve is 30~,
and the lift is less than the depth of seat:
A —0.68XP
D= —
1.57X1
7. When the bevel of the valve is 300,




75
and the lift is greater than the depth of
seat:
A-0.68 X d'
1.57xd+3.1416 X (l-d)
It would be a good plan for those who
have charge of steam boilers to experiment with the safety - valves, and see
whether they conform to these rules. It
will be easy to measure the lift of a
valve, when it opens under the highest
pressure of steam that is to be carried
on the boiler to which it is attached,
and the other data needed can be readily
ascertained. To determine the bevel of
a valve, measure its greatest and least
diameters, together with the depth of
seat. Then divide half the difference
between the diameters by the depth of
seat, and the quotient is the tangent of
the angle of inclination. The algebraic
expression is as follows:
Let
T    greatest diameter of valve, in inches.
t =least     "      "       "
d - depth    "      "   "




76
a - angle of inclination.
T-t
tan a = -
2 d
The tangents of angles from 20' to
50~ are given in the accompanying table,
and it will be sufficiently accurate in any
example to take the angle corresponding
to the number in the table that is nearest to the calculated quotient.
Example. - The  greatest and  least
diameter of a valve are 4 6-10 and 4
inches, respectively, and the depth is S
inch. What is the bevel?
Greatest diameter...................   4.6
Least      "....................   4.
2)0.6
0.5)0.3
Tangent of angle of inclination......   0.6
From  the table it appears that the
angle corresponding to this is nearly 31~.




Table of Tangents from 20 to 50 Degrees.
Angle.   Tangent.   Angle.   Tangent.
20~.364        360.727
21~.384       37~.754
22~.404        38~.781
230       424        39~.810
240.445        400.839
25~.466        41~.869
26~.488        42~.900
27~.510        43~.933
28~.532        44~.966
29~.554       45~       1.000
30~.577        46~       1.036
31~.601        47~       1.072
32~.625        48~       1.111
33~.649        49~       1.150
34".675        50~       1.192
350.700
IV. PROPER FORM FOR A SAFETY-VALVE
IN ORDER THAT IT SHALL HAVE  A
GIVEN LIFT.
When a safety-valve is raised by the
pressure of the steam in a boiler it usually exposes a greater area to the pressure.  Hence, if the valve were loaded
with a weight it might be expected that
it would open wide at once from the unbalanced pressure due to the action of




78
the steam on the increased area. It is
found  in practice, however, that the
ordinary safety-valve only rises very
slightly when the pressure for which it
is set is reached, and that it is necessary
for this pressure to be increased considerably to raise the valve much higher. In
a previous part of this paper it was stated
that as soon as a safety-valve was opened
a new set of conditions was to be considered. With the opening of the valve,
steam begins to escape from the boiler
with a high velocity, and in its escape it
has to overcome the resistance afforded
by the opening. In this way its pressure
is reduced, near the orifice, so that,
although there is a greater area of valve
to be acted on by the steam, the pressure
is reduced. How much the pressure will
be reduced in any given case can only be
determined by experiment, since the data
for a theoretical calculation cannot be
given. Some experiments made in England showed that in the case of a threeinch valve, relieving a boiler at a pressure
of 100 pounds per square inch, the




79
pressure under the valve was 25 pounds
per square inch. It is probable that the
pressure under a valve, when opened,
will vary with the form of valve and the
pressure in the boiler, as a change in the
valve may vary the resistance opposed to
the escape of steam, and a change of
steam pressure effects a corresponding
change in the velocity with which the
steam issues from the orifice.
In the case of a safety-valve kept
down by a spring, it must be evident
that unless arrangements are made for
increasing the upward pressure, when
the valve is opened it will lift but
slightly, since every increase of lift requires additional force for the extension
or compression of the spring. But in
whatever manner the valve is loaded, it
would seem that the first step in proportioning it would be to find what
pressure is acting when the valve is
opened. It will then be possible, knowing the resistance that must be overcome
to give the desired lift, to afford such an
increase of area in the open valve that




80
the steam pressure will balance the resistance. It must be remembered in this
connection, that it is desirable to have
the valve close promptly as soon as the
pressure is reduced, and unless some
special provision is made for this, the
valve, though it may open wide from the
pressure due to the increased area, will
be prevented, from the same cause, from
closing until the pressure is greatly reduced. As before remarked, experiment
is necessary in order to give data by
which the desired results can be effected,
and experiments in this direction have
not been sufficiently extended to enable
general rules to be deduced. It may be
interesting and instructive, however, to
call attention to examples of valves that
give good results in practice, and explain, as far as possible, the principles
on which they act. It is not considered
necessary to illustrate all the safetyvalves in the market, but only such
prominent ones as embody distinct principles. And first it may be well to give
the results of some experiments made by
a commission appointed by Congress to




81
examine life-saving inventions, in 1868.
This  commission  received  14  safetyvalves for examination, and tested them
by attaching them to a steam boiler and
observing their action. It is reasonable
to suppose that the valves offered for
competition represented the most prominent and successful ones in the market;
but it will be seen that many of them
were far from fulfilling all the conditions
required of a good safety-valve. The
accompanying table gives the essential
features of the test, the different valves
being represented by numbers.
Locked' 0
(W Blew off
Valve.    or                  C Bw of   losed at
Open.          at
No. 1.  Open.  107    115         106
Locked. 107    117       114
No. 2.  Open.   53      59         58
Locked. 53     60         57
No. 3.  Open.   40      58         52
Open.   45     49      25 or 24,
but continued blowing,
No. 4.                        until shut
off byhand.
Open.   45     46         35
Locked. 45     45         35




82
Locked ^ o  Tlpw nff
Valve.       d       Blew off  Closed at
Open. j ~
Open.   42      50         50
No. 5.  Open.   82       91         89
Locked. 82      91         89
Open.   35  45, began
to blow a
little, but
was  shut
~To. 6.              off at 54,
~No. 6.        not having
blown
freely, even at that
point.
Open.   90      93     74, and did
not cease
blowing.
No. 7.  Locked. 90       90         85
Open.   50      50         45
Open.   50      50         46
Locked. 50      50         46
No. 8.  Open.   50       60         50
Locked. 42        45         38
No. 9.   Open.  103    106          92
No. 9 Locked. 103    106            97
No. 10I Open.   75       80         74
Locked. 75        80         73
Open.  100      99      Failed to
close, shut
No. 11                         off at 85.
Locked. 100     98     95, leaking
badly.




83
Locked,   Blew off
Valve.   or                   Closed at
Open. H       a
r Open.  130    133      Shut off at
I~~~ ~~~114.
No. 12  Locked. 130    136     Shut off at
115.
Locked. 135    137      121
No. 13  Open.  136    137         138
Locked. 136    136       137
No. 14  Open.  30       31         30.   Locked. 30      31         30*
In  some experiments made by the
Academy of Sciences, in France, and the
Franklin Institute in this country, a valve
of the form  shown in Fig. 12 was employed. It will be apparent that as soon
as the valve is raised a little, the weights
will move on the lever, causing the valve
to open wide at once. A valve of this
kind, however, is not suitable for general
practice, since a sudden opening to the
full capacity is apt to empty the boiler
of both steam and water, and, moreover,
there is no provision made for the automatic closing of the valve by a slight reduction of the pressure.




84
L..
An ordinary valve, if not kept constantly in use frequently sticks fast to
the seat, so that the pressure must be
raised beyond the point for which it is
set before it will open. The friction of
the joints on the lever also frequently
increases the resistance to opening very




85
considerably. A valve called by the inventor the "positive  safety-valve" is
sketched in Fig. 13. The seat is spherical, and the weight is within the boiler,
Fi 13 
a tube on the end of the valve-stem extending down into the water. There are
no joints to cause friction in the valve,
and the motion of the water produces a
movement in the valve, so that it is continually changing its position on the
seat, and is not liable to stick. By a
proper proportion of seat it is possible to
expose enbugh additional area of valve,
when opened, to give any desired lift.




86
Instead of pin connections for the
levers of safety-valves, it is common to
make the levers turn on knife edges,
thus reducing the friction. In the case
of valves loaded by springs, it has been
mentioned that every increase of lift requires an increase of force to overcome
the resistance of the spring. In Naylor's
" compensating safety-valve," Fig. 14,
the spring is attached to the point A of
the bent lever, so that as the lift of the
valve increases, the lever arm of the resistance diminishes, which has the same
effect as increasing the pressure on the
valve.
Fig 14
An arrangement for giving a large
opening with a comparatively small lift,
by having a double-seated valve, has
been proposed. A sketch of the plan is
shown in Fig. 15, A  and B being the




87
Fig 15
two seats. It is found difficult, however, to keep such a valve tight.
Mr. Thomas Adams, of Scotland, constructs a valve with a projecting lip, A,
Fig. 16. By applying steam gauges at
the points shown on the sketch, he observed that they gave the indicated
Fig t6




88
readings at the time the valve opened,
and that then the pressures began to
lower, around the valve first, and lastly
in the boiler, when the valve suddenly
closed. He proportions a 3-inch valve
for 100 pounds pressure to have a lift of
five-sixteenths of an inch. Finding by
experiment that the spring used requires
a load of 138 pounds to compress it this
amount, he makes the area of the lip
sufficient to produce this load when
acted upon by steam having a pressure
of 25 pounds per square inch. All good
safety-valves must be proportioned somewhat after this manner. It is better to
make the direct experiment with steam
gauges for any particular case, as Mr.
Adams has done; but if this is not convenient, different forms of valves must
be tried, until the right proportions of
the parts are determined.
There are several valves made on the
same general principle as that just described. The Richardson valve, Fig. 17,
is very well known, and largely used in
this country. The valve has a curved




89
lip, A, and there is a recess, B, surrounding the seat. It is claimed by the inventor that the recoil of the escaping steam
produced by this arrangement gives a
higher lift to the valve. Some experiments were made with this valve by orFi.1
der of the Navy Department, in 1873.
The boiler to which it was attached was
first fitted with a common valve 4 inches
in diameter, which let the steam pressure
increase in 4 minutes 5 pounds beyond
the point for which it was set, when the
furnace doors were closed, and a 3-inch
common valve allowed the pressure to
increase 9 pounds in 6 minutes, under
the same circumstances. A Richardson
valve, 3 inches in diameter, being at



90
tached, the pressure could not be increased beyond the point for which the
valve was set, and it closed promptly on
a slight reduction of the steam pressure.
The common safety-valve being set to
open at different pressures gave the same
amount of lift in each case, as far as
could be observed, while the Richardson
valve gave an increased lift for each increase of pressure at which it was set.
A competitive trial of safety-valves
was recently conducted in England, for
a prize of $500 offered by the Editor of
the Nautical Mfagazine.  A  sketch of
the Rochford valve, to which the prize
was awarded, is shown in Fig. 18. It
will be seen that it has a projecting lip,
A, and the spring is enclosed in a case,
B, which can be filled with oil if desired,
for the preservation of the spring. In
the trial the valve was attached to.a
Lancashire boiler, 33 feet long, 7 feet 2
inches in diameter, flues 2 feet 8 inches
in diameter, grate surface 30 square feet;
being set to blow off at, 60 pounds, the
pressure could not be increased beyond




91
pounds by the most severe firing,
633  pounds by the most severe firing,
and when it fell to 60 pounds, the valve
closed promptly.
Instead of having a projecting lip on
the valve, a piston is sometimes attached
to it, which is exposed to the action of
steam at rest. The Rochow valve, Fig.
19, furnishes a good illustration of this
principle. A piston, A, is attached by a
stem to the valve, and the tube in which
it works is extended into the boiler by a
pipe, C, so as to be acted upon by quiescent steam. A ring, B, nearly closes the
orifice by which steam is admitted to the




cm
A- g  ~      1 I
19    "                                     ^ 




93
valve proper, when it is seated; and as
the valve rises, the ring opens the orifice
more and more. From this description
it will be evident that when the valve
rises a little, the pressure above the piston will be reduced, and consequently
there will be an unbalanced pressure on
the under side of the piston to raise the
valve higher, and by giving a proper
proportion to this piston any required
lift can be obtained. In the form represented in the sketch, the valve is arranged to be loaded with a weight, and
there is a stop at D to check the valve
when it has lifted high enough.  The
piston and ring do not work steam tight,
but are fitted loosely, so that they can
move without friction.
In these illustrations, while only a few
of the most prominent valves have been
noticed, it is believed that no important
principle of construction has been overlooked. It is a matter of regret that
more information cannot be given on this
part of the subject, but, as before remarked, the record of experiments from




94
which general rules can be deduced is
very slight. Enough has been said, how:
ever, to show that there are good safety
valves in the market.
There is one experiment that everyone
who uses a safety valve can make, to see
whether he has a valve that is proportioned to give the proper lift. Let him
secure a cord to the lever or stem of the
valve, so that it can be opened by hand
if necessary. (Indeed, it may be said, in
passing, that some convenient arrangement should alwaysbe fitted for opening
the valve by hand, and it should be used
at least once a day, to keep the valve in
working order.) Then, by shutting off
steam from the engine or wherever else
it is used, and making up a good fire in
the boiler, he can determine in a very
short time whether or not he has a safety
valve; and if it will not relieve the
boiler automatically it will be easy to
give a larger opening by hand, so that
the experiment will not be attended with
danger. This simple experiment is earnestly recommended to every steam user,




95
for with a good safety-valve in working
order the chances of a disastrous boiler
explosion are greatly diminished.
V. THE RELATIVE MERITS OF VALVES
LOADED WITH WEIGHTS AND SPRINGS.
The best manner of loading a safetyvalve has been the subject of animated
discussion among engineers. The opinion of the majority can be summed up as
follows:
Safety-valves for the boilers of locomotives and steamers, and in all cases in
which they will be subjected to oscillations and jars, should be loaded with
springs. For stationary boilers, either
weights or springs can be used at pleasure. In employing a spring it is generally considered best to arrange it so that
it shall be compressed, rather than extended, when the valve is raised.








APPENDIX.
The demonstrations of the rules given
in a former part of this book, for finding
the opening due to a given lift of a valve,
when it does not rise clear of the seat,
and for determining the area of opening
necessary to discharge a given weight of
steam of known pressure, are appended.
a.-The area of opening of a valve with
a beveled seat, for a lift A E=l, Fig. 20,
is the area of the frustum of a cone whose
upper base A B = -, has the same diameter as the valve,'and whose lower
base, C7 D, measured from points, C, D,
where perpendiculars from A and B intersect the seat, is equal to
D + CG+  )-D=D+ 2 CG
the slant height being A C or B D. The




98
tt.
area of the surface is equal to the half
sum of the circumferences of the upper
and lower bases multiplied by the slant
height, or the area of opening due to the
3.1416 (2 D + C G)
lift I =X A C.
2
The angle of bevel, A BE   = a, and




99
the angle A C G is also equal a. From
trigonometry, in the triangle A E C,
AC  AC
Sine a =        = 
A E   I
A C = I X sine a.
In the triangle A C G,
CG  CG
Cosine a      =
A' I X sine a.
C G =I X sine a X cosine a
Substituting these values in the previous equation:
Area of opening3.1416 (2 D x 2  x sine a x cosine a)
____=       ~~__ — X Ix sine a
2
= 3.1416 [D x Ix sine a+l2 x (sine a)2 x cosine a]
If a = 45~;
Sine a = 0.707, cosine a = 0.707
(Sine a)O = 0.5
(Sine a)' X cos a =  0.3535
Whence, area of opening= 3.1416 (D X l X 0.707 + 1' X 0.3535)
= 2.22 X D X I +  1.11 1




100
If a = 30~,
Sine a = 0.5, cosine a = 0.866
(Sine a)' = 0.25
(Sine a)' X cos a = 0.2165.
Area of opening3.1416 (D X I X 0.5 + 1  X 0.2165
-1.57 X D X + + 0.68 X 12
b.-Let
A  - area of opening required to discharge Wpounds of steam at the pressure
P per hour. The quantity discharged
by this opening per second will be, approximately,
PXA
70
Hence the quantity discharged per hour,
P X  A X 3600
W=70
= 51.43 P X A, nearly,
AndW
A =
51.43 X P




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8