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Cyclopedia of engineering : a general reference work on steam boilers, pumps, engines, and turbines, gas and oil engines, automobiles, marine and locomotive work, heating and ventilating, compressed air, refrigeration, dynamos motors, electric wiring, electric lighting, elevators, etc. (Volume 2) online

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Online LibraryAmerican Technical SocietyCyclopedia of engineering : a general reference work on steam boilers, pumps, engines, and turbines, gas and oil engines, automobiles, marine and locomotive work, heating and ventilating, compressed air, refrigeration, dynamos motors, electric wiring, electric lighting, elevators, etc. (Volume 2) → online text (page 18 of 30)
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OF the maximum valve displacement is again reached and the
valve moves in the opposite direction until at O'D its displacement



from mid-position, is again equal to OQ = OP = the inside lap,
and compression takes place. At Oil' the valve is again in mid-
position. At OX the displacement of the valve is OI, but since
the valve has to move the distance OJ before the port begius to
open IJ must represent the port opening when the crank is on
dead center and by definition we know T that lead is the amount of
port opening at this position. Therefore IJ represents the lead.

At the position K, the port is open an amount equal to TO,
at E the opening is a maximum equal to EX. At C the opening
is nothing. If LAV represents the total width of the steam port,
the exhaust port will be open wide when the displacement of the
valve is equal to OWand it will remain wide open while the crank
swings from OW to OK.

Fig. 23.

If the width of steam port in addition to the outside lap were
laid oft' on the other valve circle it would fall at E'. For the ad-
mission port to bo wide open, the displacement of the valve would
have to be equal to OE', which is more than the maximum dis-
placement. This shows that in this case the steam port is never
fully open and that the left-hand edge of the valve overlaps the
right-Land edge of the port by an amount equal to EE' when the
valve has reached its maximum displacement.



Fig. 22, with its two valve circled, shows the diagram for the
head end of the cylinder only. The crank-end diagram would be
similar except that the laps might not be equal to those of the
head end.

We are now in a position to consider more in detail the effect
of changing in any way either the valve or the setting. Let us con-
sider Fig. 23, which is in every way like Fig. 22 except that all
unnecessary letters and lines are omitted to avoid confusion. If
the outside hip is increased an amount equal to KM, the ad-
mission will take place later, at crank position OA'; the lead will

Fig. 24.

be reduced to 1(1 and cut-off will fake place earlier at OC'. If
the outside lap is 'reduced a like amount the contrary effects will
be observed. Tf the inside lap is increased an amount equal to LS,
the release will take place later at the crank position OB' and com-
pression will take place earlier at OD'. The contrary effect will
be observed by decreasing the inside lap.

If the angular advance is increased (see Fig. 24) all the events
will occur earlier. This is evident from the fimire: the crank


revolves in the direction indicated by the arrow and O A' (new posi-
tion of admission) is ahead of OA, the old position.




It' the eccentricity is increased, Fig. 25, the valve travel will
increase and admission will take place earlier at OA'; the lead
will he increased an amount equal to II', and cut-off will take
place later at OC'. Eelease will he earlier at ' OI3' and com-

Fig. 2S.

pression will be later at OD'. The upper valve circle will no^
cut the arc drawn from O as a center, with a radius equal to the
outside lap plus the width of steam port, in the points W and II'.
and the admission port will be open wide while the crank is mov-
ing from OW to Oil'. Similarly, the lower valve circle cuts the
arc drawn from () as a center, with a radius equal to the inside
lap plus the width of steam port, in the points "NV and II. The
steam port is then wide open to exhaust while the crank is moving
from W to II. From the above it will he seen that the periods
are all changed by changing the travel; thus, admission and ex-
haust begin sooner and last longer, while expansion and com-
pression begin later and cease sooner. With change in the angular
advance, however (see Fig. 24), the periods are neither increased
nor decreased.



For convenience, these results are collected in the following
table which shows the effect of changing the laps, travel, and
angular advance:

Outside Lap.

Inside Lap.


Angular Ad van c?o.


Is later.
Ceases sooner.

Not changed.

Begins earlier.

Begins earlier
Same period.


Is earlier.


Begins later.
Ceases sooner.

Begins earlier.
Same period.



Occurs later.
Ceases sooner.

Begins earlier.'
Ceases later.

Begins earlier.
Same period.


Begins at same

Begins sooner.

Begins later.
Ceases sooner.

Begins earlier.
Same period.


All the problems on valve gears involve the relations between
certain variables which are :

The valve travel.
Angle of lead.
Outside lap.
Inside lap.

Points of stroke at which admission cut-oft", release and compression
take place.

In designing a Slide Valve, a few of these variables depend
upon the conditions under which the engine is to run. For instance,
the valve travel is limited, cut-off must be at a certain point and
the engine must have a certain lead. Then, with the aid of a Zenn-
IM-'S diagram, the remaining proportions of the valve may be deter-

Let us consider a few examples:

Given the valve travel = ;*. inchea.
Inside hip = % inch.

A 1 1 u 1 a r ad van ce = :!o

Angle at out-oil = 115



To determine the laps, the lead and the crank angles at
admission, compression and release.

In Fig. 2G, let XY represent the valve travel = 3 inches.
Draw OM perpendicular to XY, and on XY as a diameter draw
the circle XMYF representing the path of the center of the eccen-
tric as it revolves about the shaft. Lay off the angle MOE = the
angular advance = 35 so that the angle XOE is e^ual to DIP

Fig. 26.

minus the angular advanceo Produce EO to F. Then on OE
and OF as diameters draw the valve circles. The eccentricity
OE or OF, if no rocker is used, will be half the valve travel. Lay
off the crank angle XOC = angle of crank at cut-off = 115, and
OK will then represent the distance of the valve from mid-posi-
tion when cut-off takes place. This distance we know is the out-
side lap. Draw the arc KI, known as the lap circle, and it will
cut the valve circle again at Y. When the valve is again the dis-
tance OV the outside lap from mid-position, admission will take
place. Draw the line OYA and this will represent the position
of the crank at admission.



When the crank is at OX, the valve displacement is equal to
OJ. This is at dead center and the valve is open the amount IJ,
for it has moved this distance more than the outside lap. There-
fore IJ is the lead for this end.

Xow on the other valve circle, draw the arc PQ with the
inside lap (^ inch) as a radius. It will cut the valve circle at
P and Q. AVhen the valve displacement is equal to OQ, the
exhaust port has just closed and the engine is at compression. In
the same way OP is the valve displacement at release when the
port begins to open. OQD represents the crank position at com-
pression and OPI3 the crank position at release.

The results then are as follows :

Given :

Valve travel = XY = 3 inches.

Angular advance = angle MOE = 35.

Inside lap = OP = % inch.

Crank angle at cut-off = angle XOC 115
Found :

Outside lap = OK = % inch.

Angle of lead = XOA = 5.

Linear lead = IJ = 1( \ inch.

Max. port opening for admission = HE = % inch.
Crank angle at compression = XOD = 185
Crank angle at release = XOB = 65

Max. port opening for exhaust = FN = % inch.

Fig. 26 is drawn full size, and all of these measurements may
readily be verified. This figure is drawn for the head end only,
if the crank angle at cut-off is the same on both ends, the Zeuner's
diagram for the crank end will be exactly like Fig. 26.

Given :

The valve travel = 3 inches.

The lead angle = 6".

Crank angle at cut-off = 70.

Crank angle at compression = 75.
To Find :

Angular advance.


Linear lead.

Crank angle at release.

As before, let XY represent the valve travel = 3 inches and
draw OM and the circle XMYF. See Fig. 27, Lay off the lead



angle XOA = J . Then ()A represents the crank position at
admission. Next lay off the crank angle XOC, the angle at cut-
off 70. Bisect the angle COA by the line OE and on OE draw
the valve circle. Angle MOE = the angular advance. The valve
circle will cut the crank lines OC and OA at Kand Y respectively.
If the work has been carefully done, OK will be exactly equal to
OV and will represent the outside lap. The lead is IJ as before.
Draw OD at position of the crank at compression so that angle
XOD = 75. Continue OE to cut the eccentric circle at F. On

Fig. 27.

OF draw the second valve circle. It will cut OD at Q, and OQ
will represent the inside lap. Draw the lap circle OP, and the
crank position OPB. This will be the crank position at release.

The angular advance in this problem is large and all the
events of the stroke are early. Compression and release are excess-
ively early and the outside lap is unusually large. In the previ-
ous problem, with cut-off at about two-thirds stroke, the results
were nearly normal. Cut-off with the plain slide valve, earlier
than half stroke cannot be had without sacrificing the steam dis-
tribution on the other events.



To sum up we have
liven :

Valve travel = XY =.3 inches.

Lead angle = XOA = 6.

Crank angle at cut-off = XOO = 70.

Crank angle at compression = XOD = 75.

To find :

Angular advance MOE = 58.

Outside lap = OV = 1 & inches.

Lead = IJ & inch.

Inside lap = OQ = j., inch.

Crank angle at release = XOB = 130.

Suppose in this last problem the cut-off had been given at
half stroke instead of having the crank angle given, and that the
compression had been given in the same way. We should, of
course, need to know the ratio of length of connecting rod to
crank. Let this be given as 4, that is, the connecting rod is four
times the length of the crank.

In Fig. 28 let XY represent the valve travel. Extend XY
to the left to the point Z, and make OZ equal to four times OX.
With Z as a center and OZ as a radius, strike an arc OC that will
cut the eccentric circle at C; then draw OC, which will represent
the crank when the piston is at half-stroke, which is assumed to
be the point of cut-off.

To find the crank angle at compression, lay off YH equal
to .8 of the distance YX. From II lay off II W = OZ = four times
OX. From W r as a center with a radius WII, draw an arc cut-
ting the eccentric circle at D. Draw OI), which will represent
the position of the crank at compression.

The student is advised to read over agaiu pages 13 to 14 if this expla-
nation of finding the crank angle does not seem perfectly clear.

Given :

Cut off at .6 stroke.

Lead = ^ inch

Maximum port opening = % inch.

Ratio of crank to connecting rod = 4.

To find :

The eccentricity.
Lead angle.
Angular advance.




In Fig. 29 assume an eccentricity that will if possible be a
little too large. Let us take for trial 2 inches and draw XY
equal to twice the assumed eccentricity equal to 4^ inches. Lay
off XC' equal to .6 of XY,
and with a radius equal to
four times OX draw the arc
C'C as already explained.

Then draw OC, which
will represent the position of
the crank at cut-off, and
XOCwill be the crank-angle
at cut-off. Assume a lead
angle of about 7 and draw
OA, which, if this assump-
tion be true, will represent
the crank-angleatadmission.
Bisect the angle CO A by tho
line OE, and on OE draw
the valve circle. Draw the
lap circle VNK. With this
assumed eccentricity we find
a maximum port opening of
NE = .75 inch, which is
larger than the conditions of
the problem demand. We
may then form a proportion,

The actual port opening de-
sired : the port opening with the
assumed eccentricity : : probable
eccentricity: assumed eccentric-

Substituting the figures
we have .5 : .75 : : a? : 2- .-.
re = the probable eccentric-
ity; equals 1.42 inches.

Now draw on OE, a new
valve circle (dotted) with a diameter equal to the required eccen-
tricity of 1.42 inches. See Fig. 290. It will cut the crank line



()0 at K', and OK' will be the new outside lap and I'J' will be
the new lead (assuming the lead angle to be 7). This lead I'J' is
i inch, while the required lead is only -^ inch. Now decrease the
angular advance enough to correct one half of this difference, by
drawing a new lap circle J"K" of -^ inch greater radius. This
will make the valve circle cut OO at K", so that OK" will now be


the final lap, and I"J" the final lead, which is equal to the required
/,. inch. The lead angle is now XO A' instead of XOA. The port
opening at Nil' is 4 inch (nearly) as required, but the change in
angular advance necessitates an increase of lap if cut-off is to
remain the same. This reduces the port opening by the amount
Hir, so that the maximum opening is only .46 inch. JBy increas-
ing the eccentricity this port opening mav be increased.


X <

.46 : .50 : : 1.36 : x.

x = 1.48, the true eccentricity.

Now draw the valve circle on OE' with a diameter of 1.48
inches. It will cut OC in K'" and OX in J'". The lap will be
OK'" = .97 inch, the lead will be I'" J'" = T \, inch, the angular
advance will be MOE* and the eccentricity ON'.

To sum up we have

Given :

The cut-off = .6 stroke.

The lead = -jV i ucn

Max. port opening = Yi, iuch.



Obtained all of the above conditions together with:

Lap = .07 inch.

Lead nielli =XOA'

Angular advance MOIS'

Compression, and inside laps aro found as in the pre-
vious problems.

There are of course all sorts of combinations that would make
up different problems, but they can all be solved in the same gen-
eral way, as they are modifications of the above.


In designing a slide valve some of the variables are assumed
and the others are found by means of diagrams as we have already
seen. These diagrams show only the dimensions of the inside and
outside laps and travel of valve; the other dimensions of the valve
and seat must be calculated.

Area of Steam Pipe. Pipes that supply the steam chest
should be large enough to prevent an excessive loss of pressure due
to friction. Jf the pipes are long they should be of such size that
the mean velocity of steam in them does not exceed 100 feet per
second or 0,000 feet per minute. For this calculation it is usual
to assume steam admitted to the cylinder during the whole stroke.

For example. Suppose an engine is 10" X 18", and makes
180 revolutions per minute. What is the diameter of the steam
pipe 2

The piston displacement or volume of the cylinder is :
- 3.141(3 X 10 2

4 4


-~ - X 18 = 1413.72 cubic inches.

a,-* = .818 cubic feet.

If the engine makes 180 revolutions it would use 2 X 180 X
.818 = 294.48 cubic feet per minute.

The area would be = (> ' = .04908 sq. ft. = 7.0675 sq. in.

The diameter corresponding to 7.0075 square inches is 3

A three-inch pipe would be large enough, especially if the
engine cut off early in the stroke.



For a very large engine cutting off early, the allowable veloc-
ity may be taken as 8,000 feet per minute instead of 6,000 feet.

Width of Steam Port. The port opening at admission should
give nearly as great an area as the steam pipe in order to prevent
loss of pressure due to wire-drawing, but the actual width of the
port should be great enough for the free exhaust of steam. It
is well to have the steam port a little larger than the area of the
steam pipe, then with a port opening of .0 to .9 of the port area
for admission and full port opening at exhaust, satisfactory condi-
tions will result.

The length of the ports is usually made about .8 the diameter ol
the cylinder. Then in the 10" X 18" engine the steam ports would
be 8 inches long. If the area for admitting steam is 8.0675 sqnan
inches and the length of port is 8 inches, the width will be

= -^- = .8834 inch, or about ^ inch.

The width of port opening would be about .9 X .8834 = .79500
inch or about -i-g- inch.

Width of Exhaust Port. When the slide valve is at its
maximum displacement, the valve overlapping the exhaust port
as shown in Fig. 7 reduces the area more or less. In designing
the valve, the exhaust port should be of such a width that the
maximum displacement of the valve does not reduce the area of
the exhaust port to less than the area of the steam port. It is not
advisable to make the exhaust port too large for this increases
the size of the valve and thus causes excessive friction.

The height of the exhaust cavity should never be less than the
width of the steam port, and may be made much higher to advantage.

Width of Bridge. The bridge must be of sufficient width so
that outside edges of the valve cannot uncover the exhaust port.
The width of the steam port plus the width of the outside lap plus
the width of the bridge must bo greater than the maximum dis-

The width of the bridges should be not less than the thickness
of the cylinder wall in order to make a good casting.

The Point of Cut-off. In the study of Indicators, it was shown
that if the point of cut-off is early, the oilier events are not good.
If a plain slide valve is used with an automatic cut-oft, the cut-oil'



is hastened either by clianging the eccentricity or by changing
the angular advance. Either of these methods will accomplish the
result at the expense of the compression which consequently will
be earlier and excessive. Except for locomotives and high-speed
engines, where compression is an advantage, the plain slide valve
is not arranged to cut-off earlier than -| or stroke. If an earlier
cut-off is desired, large outside laps are necessary. The cut-offs
may be equalized by giving the head end a greater lap than the
crank end, but this will cause an inequality of lead.

Lead. The lead of stationary engines varies from zero to 2
inch according to the style of engine. An engine having high
compression that compresses the steam nearly to boiler pressure.
will give good results with little or no lead. If the ports are small,
and the clearance large, there should be considerable lead in order
to insure full initial pressure on the piston at the beginning of the
stroke. Valves that open slowly require more lead than quick-
act in ff valves.


Let us design and lay out the valve and valve seat for the fol-
lowing engine:

Diameter of cylinder = 10 inches.

Stroke = 18 inches.

Revolutions = 180 per minute.

Lead angle = 3.

Cut-off to be equal at both ends and to take place at .75 stroke.

Max. port opening = .9 area of steam pipe.

Compression to bo .85 of the stroke at both ends.

Length of connecting rod = 3 feet.

The piston displacement, or cylinder volume, will be
_ X 18 = 1413.7 cubic inches or .818 cubic feet. If the

engine makes ISO revolutions, it will use 2 X 180 X .818 = 294.48

cubic feet of steam per minute. Steam pipe area = = .0491

square feet = 7.07 square inches.

This 7.07 square inches would also be the least possible area
of the steam ports. If the length of port is made .8 the diam-
eter of cylinder, the width will be - - .88 inches or about

inch. The width of maximum port opening will be .9 X .88 = .792
or naarly -{ inch.



It will be necessary to draw a separate valve circle for each
end of the cylinder. First consider the head end.

The valve travel not being known, we shall lay off XY on an
assumption of (> inches travel and draw the eccentric circle as
shown in Fig. 30. Lay off the lead angle XOA 3\ Lay off
XC' .75 of the assumed valve travel = 4-i inches. Draw the
arc CC' as previously explained and draw OC which will be the
crank angle at cut-off. The radius of the arc C'C will be equal to


Fig. 30.

i times the radius of the eccentric circle, or 12 inches, because
the connecting rod is 4: times the length of the crank. Bisect
the angle AOC by the line OK, and on OE draw the valve circle.
OV = OK is then the outside lap, with these assumed 1 condi-




tions. Draw the lap circle; then EN will be the maximum port
opening. EN 1 ^ inches, while -i-jj- inch is all that is necessary.
The assumed eccentricity is 3 inches, therefore the probable eccen-
tricity = x : 3 ::][: 1^. x = !]- inches.

Now draw a new eccentric circle with a radius of 1|J inches
and a new valve circle with OE' = lj-J inches as a diameter. OK'
is now the outside lap and the maximum port opening is equal to
E'N', which from actual measurement is found to be -{ ^ inch. The
outside lap = OK' = OV = J-J- inch and the lead is IJ = -^ inch.

Produce EO to F and draw another valve circle. We shall
use this valve circle to determine the outside laps and lead for the
crank end of the cylinder. Since the cut-off is to be .75 of the
stroke, we may lay off OH' OC', and with a radius of 12 inches

Fig. 31.

draw the arc UK'. Then, as already explained, Oil will be the
crank angle at cut-off on the return stroke. OB will be the out-
side lap J!! inch. Draw the lap circle intersecting the valve
circle at D. Then ODA' is the crank angle at admission on the
return stroke and LM = | inch is the lead on the crank end of the
cylinder. The maximum port opening will always be greater at the
crank end than at the head end because the crank end lap is les.s
in order to get the equal cut-off. If the laps were equal, of course
the port openings would be equal.

Now lay off YG' = .85 of XY and find the crank position
O(i. This is the compression on the brad end of the cylinder and
gives an inside lap on this end of 3 7 2 - inch, which is equal to Ol\



Draw the lap circle PQ, which allows us to draw through Q the
crank line OK, which is the release on the forward stroke.

Lay off XS' == YG' .85 of XY, and construct the crank
line OS, which is the crank position at the crank end compression.
OS intersects the valve circle at T, which gives OT = ^\. inch =
inside lap on the crank end. Draw this lap circle, which will
intersect the valve circle at U. This enables us to draw OUAV, the
crank angle at release, on the return stroke.

From the data determined by means of these diagrams the
valve may now be laid out. For convenience let us tabulate the
results obtained as follows:

Data. Head End. Crank End.

Cut-off, per cent of Stroke 75 75

Outside Lap J-J" -| f "

Inside Lap ^-" J^' 1

Lead ^ I"

Port Opening A-" l T y
Width of Port .?"

8 8

Fig. 81 shows tliis valve in .section. Let us begin at the end
having the largest inside lap, or in this case at the crank end.
Lay out the steam port | inch wide, and the crank-end outside la])
= -i-|- inch. The bridge will be, say, j| inch wide. From the
inner edge of the steam port, lay off the crank-end inside lap =
, 7 ( . inch. When the valve moves to the left, the point E' will
travel ![/. inches, a distance equal to the eccentricity, and in this
position of extreme displacement the exhaust port EF must be open
an amount at least equal to the steam port, | inch. Therefore we
lay off EF equal to 1 [ J" + ^" 2^,.". The inside lap overlaps the
bridge nearly inch, so that we shall have to make the exhaust
port opening equal to 2g- inches. Lay off | inch again for the
bridge and measure back -^V inch, equal to the head-end inside
lap. The port is | inch wide, and the head-end inside lap of ||-
inch completes the outline of the valve seat.


The principles of valve diagrams are useful in setting valves
as well as in designing them. The valve is usually set as accu-
rately as possible, and then, after indicator cards have been taken,




the final adjustment can be made to correct slight irregularities.
The slide valve is so designed that the laps cannot be altered
without considerable labor, and the radius of the eccentric, which
determines the travel of the valve, is usually fixed. The adjust-
able parts are commonly the length of the valve spindle and the
anmilar advance of the eccentric.


]]y lengthening or shortening the valve spindle, the valve is
made to travel an equal distance each side of the mid-position.

Online LibraryAmerican Technical SocietyCyclopedia of engineering : a general reference work on steam boilers, pumps, engines, and turbines, gas and oil engines, automobiles, marine and locomotive work, heating and ventilating, compressed air, refrigeration, dynamos motors, electric wiring, electric lighting, elevators, etc. (Volume 2) → online text (page 18 of 30)