=

or if

by HP piston and drives the MP piston, but as the HP moves much faster than the MP the pressure is increased until the volume begins to decrease. This it will do when the vel. of MP piston _HP piston area vel. of HP piston MP piston area' a = angle that the MP crank makes with the line of centres; when X, xv, sin av, cos a, or when cot a =λ1. The volume now occupied by the steam is v, vers a + v, vers (90 − a) +w2

=

2

λη,

and its pressure is given by Pav, vers a + v, vers (90 − a) + w ̧

After this the steam expands until its volume is w1 +

V2

2

At this point the HP cylinder opens to exhaust, and we have an HP cylinder full of steam at pressure Pc admitted to the reservoir, and the final

pressure po is given by

1

P12 λιν,

2

:{

w1 + v1 (1 − m) }

1-m}

+1

λ

2

Thus all points for the first reservoir are determined.

The calculations for the other diagrams are similar to the above, remembering that now there are two

2

steam lines of the two LP engines and MP A one back-pressure line of MP diagram

to be considered in conjunction.

We now want an expression for

ND CK

when L,P is at point of

cut-off (Fig. 403), and this is evidently the same expression as found before for the MP cut-off

=

C

N

K

D

F/L2

I say.

273

And when L,P is just opening to steam the volume of steam is

The volume 23 is now cut off from the reservoir in the cylinder L,P, and

At cut-off in LP we have this same steam in a volume

We can now obtain the cut-off pressures in L,P and LP in terms of the one unknown quantity, PDI

The weight of steam entering and leaving the engine is the same in one revolution. And this weight of steam is proportional to pv.

Let Kpv the weight per revolution,

=

:.K (P+P) = weight passed to condenser,

and K P, 21 = weight entering engine... P + P

2

=

which gives an equation for P + P2, and knowing P and P, in terms of PD we can find PD, and thence P, P, and P,. We have thus all the principal pressures. Thus PD, being known, the pressure falls to E, cut-off pressure of L,P, and then still falls to K, where pressure initial pressure of L1 and then falls still quicker to N, when L,P cuts off. As the MP piston now moves faster than the L,P piston, we get a rise of pressure to Q, which is obtained as before by equating the rate of diminution of volume due to the MP piston to the rate of increase of volume due to the LP piston. The pressure then falls to L,, when the MP again exhausts and the pressure at which point is equal to P,.

It is easy then to fill in the steam curves for the two LP diagrams Thus: 3, 4, 5, 1, 2, is the steam-line for the L,P cylinder, and 13, 12, 11, 9, 10, is the steam line for the LP cylinder, found in exactly the same way as described for the steam line of the MP diagram. The mean LP diagram is the mean of these two, viz. A11, Z1, H1, B11, C11. The back-pressure line is 7.8.

Four cylinder triple-expansion engine with two LP cranks opposite one another. The next case to be considered is that of the same engine as in

the previous case, with the cranks differently arranged, that is, having the intermediate crank opposite the high pressure and the two low pressures opposite each other, and at right angles to the other cranks, which is the usual arrangement. The two low-pressure cylinders therefore take steam and exhaust together. The diagram lines between the high pressure and the intermediate are evidently the same as the second case considered in this chapter, viz. a compound engine with cranks at 180° and with an intermediate receiver, and will be as shown at FEDC and G H K in Fig. 387. The diagram for the exhaust line of intermediate pressure and steam line of low pressure is the same as another of our previously investigated cases, viz. that of a compound engine with cranks at right angles, and will therefore be similar to G FEDC and H KLW M of Fig. 389, if, as is usually the case, the cut-off in the low pressure is after half-stroke. The diagram for this case can therefore be easily drawn from previous investigations.

In all these investigations the effect of clearance, always considerable, has been neglected for shortness and simplicity, but it is easily taken account of when the principles worked out above are understood, and it will be a useful exercise for the student to draw the last two diagrams, making the 4C necessary allowances for cylinder clearance and compression.

41

APPENDIX

(A.) APPLICATION OF THE INDICATOR DIAGRAM TO DETERMINE THE STRESSES ON CRANK-SHAFTS. CURVES OF TWISTING MOMENTS.

FOR the sake of simplicity, suppose the obliquity of the connecting-rod and the weights of the reciprocating parts to be neglected.

Let Fig. 404 represent the crank-circle, O being the centre of the shaft, and suppose the forward pressure to be constant throughout the stroke, and equal to P. Then, it is clear that when the crank is in any position, O C, making an angle with the line of dead points, the twisting moment exerted will be = P x L sin ; where L represents the length of the crank. If, therefore, the crank-circle be divided into any number of equal parts, each subtending an angle a, the successive twisting moments will be, P L sin a, PL sin 2a, PL sin 3a, &c.

The base line of the diagram of twisting moments is taken to represent the

circumference of the crank-circle, and is divided into a number of equal parts representing equal angles of the crank with the line of dead points.

Ordinates are set up at these divisions equal to the twisting moment (P x L sin ), for the corresponding angle, and a fair curve is drawn through the ends of the ordinates thus obtained.

In the case under consideration, the curve will be symmetrical, the obliquity of the connecting-rod having been neglected, so that the piston is supposed to have exact harmonic motion. This curve of sines is shown by the full lines in Fig. 405.

When the obliquity of the connecting-rod is taken into account the difference of speed of piston at the opposite ends of the stroke will be found to destroy the symmetry of the curve of twisting moments. By reference to Fig. 406, it will be seen that, if o be the angle of the connecting-rod when the crank makes an angle with the line of dead points, the twisting moment, instead of being = P x L sin 0 simply, is

=Q× O C sin OCD=Q × L sin (8 + $)
P

where Q thrust on the connecting-rod- cos o

.. the twisting moment is P x L

sin (8+) - P.L (sin + cos 0 tan p)

cos