American Technical Society. # 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 Library → American Technical Society → 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 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

258

VALVE GEARS 19

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.

259

20 VALVE GEARS

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

t>

revolves in the direction indicated by the arrow and O A' (new posi-

tion of admission) is ahead of OA, the old position.

260

VALVE GEARS

21

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.

261

VALVE GEARS

For convenience, these results are collected in the following

table which shows the effect of changing the laps, travel, and

angular advance:

Increasing

Outside Lap.

Increasing

Inside Lap.

Increasing

Travel.

Increasing

Angular Ad van c?o.

Admission.

Is later.

Ceases sooner.

Not changed.

Begins earlier.

Continues

longer.

Begins earlier

Same period.

Expansion.

Is earlier.

Continues

longer.

Beginning

unchanged.

Continues

longer.

Begins later.

Ceases sooner.

Begins earlier.

Same period.

Exhaust.

Unchanged.

Occurs later.

Ceases sooner.

Begins earlier.'

Ceases later.

Begins earlier.

Same period.

Compression

Begins at same

point.

Begins sooner.

Continues

longer.

Begins later.

Ceases sooner.

Begins earlier.

Same period.

PROBLEMS.

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-

mined.

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

262

VALVE GEARS

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.

263

24 VALVE GEARS

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.

ANOTHER PROBLEM.

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.

Laps.

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

VALVE GEAKS

25

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.

265

26 VALVE GEARS

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.

ANOTHER PROBLEM.

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.

Laps.

266

VALVE GEARS

27

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,

namely:

The actual port opening de-

sired : the port opening with the

assumed eccentricity : : probable

eccentricity: assumed eccentric-

ity.

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

267

VALVE GEARS

()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.

VALVE GEARS

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.

269

30 VALVE GEARS

Obtained all of the above conditions together with:

Lap = .07 inch.

Lead nielli =XOA'

Angular advance MOIS'

Compression, ivlea.sc 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.

DESIGN OF THE SLIDE VALVE.

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

1413.72

-~ - 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.

294.48

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

The diameter corresponding to 7.0075 square inches is 3

inches.

A three-inch pipe would be large enough, especially if the

engine cut off early in the stroke.

270

VALVE GEARS 31

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-

placement.

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'

271

VALVE GEARS

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.

r}

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

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.

272

VALVE GEARS

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

H

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-

273

34

VALVE GEARS

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\

274

VALVE GEARS 35

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.

VALVE SETTING.

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,

275

36

VALVE GEARS

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.

ri

]]y lengthening or shortening the valve spindle, the valve is

made to travel an equal distance each side of the mid-position.

valve moves in the opposite direction until at O'D its displacement

258

VALVE GEARS 19

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.

259

20 VALVE GEARS

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

t>

revolves in the direction indicated by the arrow and O A' (new posi-

tion of admission) is ahead of OA, the old position.

260

VALVE GEARS

21

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.

261

VALVE GEARS

For convenience, these results are collected in the following

table which shows the effect of changing the laps, travel, and

angular advance:

Increasing

Outside Lap.

Increasing

Inside Lap.

Increasing

Travel.

Increasing

Angular Ad van c?o.

Admission.

Is later.

Ceases sooner.

Not changed.

Begins earlier.

Continues

longer.

Begins earlier

Same period.

Expansion.

Is earlier.

Continues

longer.

Beginning

unchanged.

Continues

longer.

Begins later.

Ceases sooner.

Begins earlier.

Same period.

Exhaust.

Unchanged.

Occurs later.

Ceases sooner.

Begins earlier.'

Ceases later.

Begins earlier.

Same period.

Compression

Begins at same

point.

Begins sooner.

Continues

longer.

Begins later.

Ceases sooner.

Begins earlier.

Same period.

PROBLEMS.

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-

mined.

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

262

VALVE GEARS

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.

263

24 VALVE GEARS

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.

ANOTHER PROBLEM.

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.

Laps.

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

VALVE GEAKS

25

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.

265

26 VALVE GEARS

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.

ANOTHER PROBLEM.

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.

Laps.

266

VALVE GEARS

27

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,

namely:

The actual port opening de-

sired : the port opening with the

assumed eccentricity : : probable

eccentricity: assumed eccentric-

ity.

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

267

VALVE GEARS

()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.

VALVE GEARS

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.

269

30 VALVE GEARS

Obtained all of the above conditions together with:

Lap = .07 inch.

Lead nielli =XOA'

Angular advance MOIS'

Compression, ivlea.sc 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.

DESIGN OF THE SLIDE VALVE.

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

1413.72

-~ - 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.

294.48

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

The diameter corresponding to 7.0075 square inches is 3

inches.

A three-inch pipe would be large enough, especially if the

engine cut off early in the stroke.

270

VALVE GEARS 31

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-

placement.

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'

271

VALVE GEARS

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.

r}

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

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.

272

VALVE GEARS

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

H

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-

273

34

VALVE GEARS

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\

274

VALVE GEARS 35

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.

VALVE SETTING.

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,

275

36

VALVE GEARS

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.

ri

]]y lengthening or shortening the valve spindle, the valve is

made to travel an equal distance each side of the mid-position.

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