Steam Engineering; A Text Book

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1913 edition. Excerpt: ... of 14 X 0.05 =0.7 inch to the stroke. Neglecting compression and cylinder condensation, we shall have, since the cut-off is at 0.5 stroke, 63.62 X 7.7 _ 28, H CuDic feet 1728 00 of steam supplied to the H.P. cylinder per stroke, which is the initial volume vi. The L.P. clearance of 3 per cent is equivalent to the addition of 14 X 0.03 = 0.42 inch to the stroke. The initial volume vi must eventually occupy the L.P. cylinder and its clearance one end, so that we shall have for the final volume 226.98 X 14.42 o u. c v2 = 2 = 1.894 cubic feet. 1728 Therefore r = =-4 = 6 681 Dl O.2835 To apply the formula r =-we have, by denoting the L.P. clearance by k, _ Volume L.P. cylinder + its clearance volume Volume H.P. cylinder + its clearance volume _ Area L.P. cylinder (1 + k) _ (17)2 X 1.03 _ Area H.P. cylinder (1 + c) fo)2 X 1.05 3'5' Then r = 3-5(i+-5)=0.681, 0.5 + 0-5 The specific volume of steam at 150 pounds pressure is 3.012 cubic feet, so that in the above example the weight of steam used per stroke was-2 = 0.0945 pound. At the commence3.012 ment of the return stroke of the H.P. piston the delivery of this steam into the receiver will begin and all of it will have been delivered when the return stroke is completed. Its withdrawal from the receiver will begin with the commencement of the stroke of the L.P. piston and must all be drawn out by the time the L.P. cut-off takes place. If it were not all drawn out the pressure in the receiver would increase as the engine ran until no exhaust from the H.P. cylinder could take place; if more were drawn out a point would soon be reached at which a vacuum would exist in the receiver. Both these suppositions being impossible, it follows that all the steam admitted to the H.P. cylinder eventually finds its way...