A To find where the Velocity of the Piston is greatest. Rule 4.-Divide the length of the stroke by the distance. travelled by the piston before the steam is cut off; take the hyperbolic logarithm of the quotient, add unity to this logarithm for a divisor, and for a dividend take the length of the stroke. The quotient will be the distance moved over by the piston when the velocity is greatest. Example 1.-The pressure of steam upon the piston is 60 lbs. per square inch, the length of the stroke is 10 feet, the steam is cut off at of the stroke: find the number of units of work done upon each square inch of the piston. 60 × 2 120 units of work done before expansion begins. Hyp. log. of 10 = hyp. log. of 5 = 1·6094379. 2 1-6094379 × 120 193·132548, units of work done after expansion begins. 120.00 units of work before expansion. 193.13 units of work after expansion. 313.13 whole work done on one square inch in one stroke. In the above example, to find the load. 313.13 10 piston. = 31.313, load in lbs. upon one square inch of the If the cylinder be 40 inches in diameter, then the area will be 40 × 40 × 7854 = 1256·64 square inches. The whole work done on one inch in one stroke with and without expansion being 313.13 lbs., the whole work on the piston of the above cylinder will be 1256 64 x 313·13 = 393491.68 lbs. The load upon each square inch of the piston being 31.313 lbs., the whole load upon the whole piston will therefore be 1256 64 × 31.313 = 39349.17 lbs. If the engine makes 20 strokes per minute, to find the horse power. 1256.64 × 313.13 393491.68 lbs. 393491.68 × 20 33000 =238-48 horse power. Example 2.-At what pressure per square inch must the steam be admitted when the load is 22 lbs. per square inch, the length of the stroke 5 feet, and the steam is cut off at 2 feet? 5 feet, length of the stroke. 2, number of feet described by the piston before the steam is cut off. 110 3.8325814 29 lbs. per square inch pressure nearly. Example 3.-The pressure of steam upon the piston is 40 lbs. per square inch; the resistance arising from imperfect condensation, 3 lbs. per square inch; the length of the stroke, 12 feet; and the steam is cut off at of the stroke: find the number of units of work done upon each square inch of the piston, and the number of units of work gained by working expansively. Also find the load per square inch, and the position of the piston when the velocity is greatest. 40 × 280= work done before expansion begins. 12 Hyp. log. of hyp. log. of 61-7917594, 1.7917594 × 80 expansion begins. 143·340752, units of work done after 80.00 units of work before expansion. 143.34 units of work after expansion. 223.34 whole work done on one square inch in one stroke. But as the resistance from uncondensed vapour is 3 lbs. per square inch, for the whole length of the stroke, it will be 12 times 3; because 3 lbs. resisting through 12 feet gives 36 lbs. for the whole. We must subtract this from the whole work done to obtain the effective work. 223.34 whole work done on one square inch. resistance of the uncondensed vapour on one square inch. Diff. 187.34 = effective work on one square inch. To find the advantage gained by working expansively; when the engine works without expansion, we have 12 x 40 480 units of work done upon each square inch. But, as in this case the steam is not cut off till the end But working expansively, we have obtained 223.34 units of work; hence 223.3480 143.34 lbs. gained in this 80143.34 example. Now, 223.34 2.8 times as much work done by the same quantity of steam when worked expansively. To find the load. By rule 2. 223.34 12 186 lbs., the load per square inch. To find where the velocity of the piston is greatest. By rule 4. The hyp. log. of 12 hyp. log. of 6 = 1·7917594 Add 1 Hence the divisor mentioned in the rule. 12 2.7917 piston before acquiring its greatest volocity. 2.7917594 4·3 the number of feet travelled by the When it is stated that the steam is cut off at 1, 3, 1 1 &c., of the stroke, there is no necessity for dividing the distance moved by the piston before expansion by the length of the stroke, whatever may be the fractional part; we need only multiply the hyperbolic logarithm of the denominator by the work done before expansion begins, and the product is the work done by expansion: for example, suppose the length of the stroke to be 12 feet, and the steam is cut off at of the stroke, the pressure upon the piston being 60 lbs. per square inch. Here the denominator is 3; its hyperbolic logarithm is 1.0986123. The steam being cut off at 4 feet, the work done before expansion is 4 x 60 240; and 1.0986123 × 240 = 263-6669520 = the units of work done by expansion, and 263.6669520 240 503.6669520 the whole work done upon each square inch in one stroke. The following rule is a good approximation. Rule. Divide that part of the stroke through which the expansion takes place, into any even number of equal parts, and calculate the pressure per square inch upon the piston at each division of the stroke; take the sum of the extreme pressure in pounds per square inch, four times the sum of the even pressures, and twice the sum of the odd pressures; multiply the sum of all these by one-third of the common distance between the positions of the piston, and the result will be the work done upon each square inch of the piston after expansion begins. The work done before the expansion begins is evidently equal to the pressure per square inch multiplied by the number of feet described before expansion. The whole work done during a single stroke is equal to the sum of the works done before and after expansion. Example. The pressure of steam upon the piston is 40 lbs. per square inch; the resistance arising from imperfect condensation, 3 lbs. per square inch; the length of the stroke, 12 feet; and the steam is cut off at of the stroke: find the number of units of work done upon each square inch of the piston. The steam being cut off at 2 feet, divide the remaining part of the stroke, viz., 10 feet, into 10 equal parts. 40+ 6.666 46·666 sum of extreme pressures. 26.666 16.000 11.428 8.888 7.273 70.255 sum of even pressures. 4 281.020 four times the sum of even pressures. |