To explain this, suppose ABCDEF to be the curve given by the indicator (which, it may be remarked, is described continuously in the direction ABCDEFA), AG the atmospheric line; draw PNQ any double ordinate, then PN represents the excess of the steam pressure above that of the atmosphere when the ascending piston is at a certain point, and NQ represents the defect of the vacuum pressure below that of the atmosphere when the descending piston is at the same point. Now the effective pressure of the steam is the excess of the steam pressure above the vaccum pressure; but PN=steam pressure -atmospheric pressure, NQ=atmospheric pressure-vacuum pressure, ...PN+NQ=steam pressure -vacuum pressure, therefore PQ represents the effective pressure of the steam when the ascending piston is at the point corresponding to N, i. e. assuming the vacuum pressure at any point of one stroke to be the same at the same point of the next stroke. If, then, for the sake of distinctness,* we suppose each inch of the ordinate to denote a pressure of 1 lb. and each inch of the abscissa (i. e. of the atmospheric line) to denote a foot of the stroke, the area of the curve will give the number of units of work done during a single stroke by the steam on an area equal to that of the plunger, and if the area of the piston of the steam engine be n times that of the plunger, the work done by the steam during a single stroke will be n times that given by the curve. The area of the curve may be found by Simpson's rule, viz. Divide AG into any even number of equal parts, and draw the corresponding ordinates; take the sum of the extreme ordinates, four times the sum of the even ordinates, and twice the sum of the odd ordinates (i. e. ex * In practice the scale would be considerably less than this. cepting the first and last), add them together, and multiply the sum by one of the parts of the abscissa; the product will be three times the area of the curve.* Ex. 147. Let the curve shown in the figure be that given by a stroke of 5 ft.; let an be divided into 10 equal parts, and let the ordinates 1, 2, 3, 4, FIG. 10. be drawn; suppose them to represent respectively 19, 22, 22, 17-5, 13, 11, 9, 7.5, 6, 5.5, 4 lbs. pressures per square inch. The radius of the piston being 20 in., determine the units of work done per stroke, and the mean effective pressure per square inch on the piston-i. e. the constant pressure that would do the same work. 1 2 3 4 5 6 7 8 9 10 11 Ans. (1) 79,000 units. (2)12.6 lbs. Ex. 148. Determine the number of units of work and the mean pressure per square inch on a piston 3 feet in diameter having a stroke of 5 feet, if the ordinates measured at intervals corresponding to three inches of the stroke give the following pressures 5.03, 12:57, 18:04, 2073, 21:03, 21-11, 21-25, 20-72, 2014, 18.63, 15.45, 13-24, 10.83, 8.53, 6-49, 4.87, 3.99, 3-74, 3-52, 3-25, 2.75. Ans. (1) 87,600 units. (2) 12.65 lbs. per sq. in. 21. Work expended on the Elongation of Bars. It is plain that if a rod be lengthened by a gradually increasing strain, the strain at any degree of elongation will be proportional to that elongation; so that if the abscissæ represent the degree of elongation, and the ordinates the strain, the area which gives the units of work will be a triangle. Hence: Ex. 149. There is a bar the length of which is L and section k; it is gradually elongated by a length 1; if its modulus of elasticity be E, show that the work expended on its elongation will be given by the formula 12 Ex. 150. The pumping apparatus of a mine is connected with the engine by means of a series of wrought-iron rods 200 ft. long; the section of each rod is of a square inch; the strain is estimated at 6 tons; how many units of work are expended at every stroke upon the elongation of the bars? Ans, 830 units. * The curve given by the indicator is useful in other ways beside that mentioned in the text. - Bourne on Steam Engine, p. 246. Ex. 151. A bar of wrought iron 100 ft. long with a section of 2 square inches has its temperature raised from 32° F. to 212° F.; how many units of work has the heat done? Ans. 3875 units. 22. The Work expended in raising Weights through various Heights. The questions arising out of this important part of the present subject are solved by means of the following proposition. Proposition 2. When any weights are raised through different heights, the aggregate of the work expended is equal to the work that would be expended in lifting a weight equal to the sum of the weights through the same height as that through which the centre of gravity of the weights has been raised. 3: Let W1, W2, W3 be the weights of each separate body; conceive a horizontal plane to pass below them all; let h1, h2, h3 be the heights of these bodies above the plane before they are lifted, and let u be the height of their common centre of gravity; then (Prop. 16) .... H (W1+2+W3....)=W1h1+W2h2 + W3h3+... Also, let k1, k2, k3 .... (1) be the heights of the weights respectively, after they have been lifted, and K the height of their common centre of gravity; then K (W1+W2+W3... :)=W1k1+W2k2+W3k3+... hence, subtracting (1) from (2), we obtain (2) (K-H) (W+W2 + W3...)=W1 (k1-h1)+W2 (k2-h2) + W3 (k3-h3)... (3) Now, W1, W2, W3 • are severally raised through the heights k1-h1, k2-h2, kh3....; therefore the right hand side of equation (3) gives the aggregate work expended in lifting them; hence that work is equal to (к-н) (W1+2+W3... .), i. e. to the work that must be expended in lifting a weight W1+2+3+ through a height к-н. ... (Q. E. D.) Cor. In the case of the transport of bodies along any parallel lines, the principle enunciated in the theorem will hold good, since the resistances are in a constant ratio to the weights. Ex. 152. How many units of work must be expended in raising the materials for building a column of brickwork 100 ft. high and 14ft. square; and in how many hours will an engine of 2 horse-power raise them? Ans. (1) 109,760,000 units. (2) 27.71 hours. [Since the material has to be raised from the ground, the common centre of gravity will have to be raised from the ground to the centre of gravity of the column, i. e. to its middle point 50 ft. above the ground.] Ex. 153. A shaft has to be sunk to the depth of 130 fathoms through chalk; the diameter of the shaft is 10 ft.; how many units of work must be expended in raising the materials? In how long a time could this be done by a horse walking in a whim gin? How many men working in a capstan would do it in the same time? Determine the expense of the work supposing the horse to cost 3s. 6d. a day, and the wages of a labourer to be 2s. 6d. a day. Ans. (1) 3,457 million units. (2) 409.6 days. (3) 5.62 min. (4) Cost of horse 711. 14s. Cost of men 2887. Ex. 154. If the work in the last example is to be done in 24 weeks by a steam engine working 8 hours a day, 6 days a week, what must be the horse-power of the engine? Ans. 1.521 H.-P. Ex. 155. In Ex. 153 suppose the box in which the material is raised to weigh cwt., the rope to be 3 in. in diameter, and each load to be 4 cwt. of chalk, also suppose the box to take as long in ascending as in descending and that of a minute is lost in unhooking and hooking at the bottom of the shaft and the same at the top; when the shaft is 100 ft. deep determine the time that elapses between the starting of one load and the starting of the next; the engine working at 11⁄2 horse-power. Ans. 2.62 min. Ex. 156. Determine the same as in the last example when the shaft is z ft. deep. + 0.5 min. Ans. 112x+0.045x2 Ex. 157. Determine the whole time of raising the materials of the shaft in Ex. 153 under the conditions of Ex. 155. Ans. 3331 hours. Ex. 158. Referring to Ex. 153, 155, suppose the drum of the winding machine to have two ropes wound round it in contrary directions, so that it unwinds one rope while winding up the other, and that consequently an empty box descends while a full one is being raised (as in Ex. 141); determine the time that must elapse between two consecutive lifts of 4 cwt. when the shaft is 100 ft. deep. Ans. 1·155 min. Ex. 159. Obtain a determination similar to that in the last example, when the shaft is æ ft. deep. Ans. 448x + 0.25 min. 49,500 Ex. 160. Obtain the whole time of lifting the materials from the shaft under the circumstances of Ex. 158. Ans. 1246 hours. Ex. 161. In how long a time would a 15 horse-power engine empty a shaft full of water, the diameter of the shaft being 8 ft. and the depth 200 fathoms? If the engine has a duty of 30 millions determine the amount of coal consumed in emptying the shaft. Ans. (1) 76 hours. (2) 75.4 bushels. Ex. 162. There is a certain railway 200 miles long; it may be assumed that in the course of 10 years there will be 50,000 tons of iron railing laid down, and that it will be equally distributed along the line. How many units of work must be expended in conveying the rails (neglecting the weight of the trucks), if the depot is at one end of the line? And how many if the depot is in the middle of the line? The resistances being reckoned at 8 lbs. per ton. (1) 211,200 million units. (2) 105,600 million units. Ex. 163. How many journeys of 200 miles performed by a train weighing 50 tons does the difference of the results in the last example represent? Resistances 8 lbs. per ton. Ans. 250 journeys. : |