The variation on torque against crank angle for one cycle of...

The variation on torque against crank angle for one cycle of a double-acting engine rotating with a mean speed of 270 rev/min is as follows: 0°-30° Zero torque 30°-45° Torque increases uniformly to 1800 Nm 45°-90° Torque constant at 1800 Nm 90°-180° Torque reduces uniformly to zero 180°-210° Zero torque 210°-240° Torque increases uniformly to 1600 Nm 240°-285° Torque constant at 1600 Nm 285°-360° Torque reduces uniformly to zero Sketch one cycle of the turning-moment diagram and calculate; (a) The mean torque (b) The power developed (c) The mass of the flywheel of radius of gyration 580 mm to keep the fluctuation of speed within ± 0.5% of the mean speed.

Transcript

00:01 Here the curve delivered by the push to pin is represented by that is a t is equal to 1200 plus 1400 sin theta plus 210 sin 2 theta plus 21 sin 3 theta and here that is the engine speed that is n is equal to 10 rotation per minute and here the radius of direction here that is a that is equal to 800 mm and here the fluctuation and here that is k is equal to 0 .015 plus 0 .015 and here that is equal to 0 .0203 and here now the p mean value here the p mean that is equal to 1 and here integration of 0 to 2 pi and here p d theta now for the value here that is 1 divided by 2 pi and here integration of 0 to 2 pi and here that is 1200 plus 1400 sin theta 1400 sin theta plus 210 sin 2 theta plus 21 sin 3 theta d theta so here calculated we get this 1 by 2 pi and here that is 1200 theta plus 1400 cos theta plus here 210 cos 2 theta divided by 2 plus 21 cos 3 theta divided by 3 and here it is 0 to 2 pi now simply we get the value that is equal to 1200 newton meter now here the power of engine here the power of engine here that is equal to t into w that is equal to 1200 into that is 2 pi into 210 divided by 16 and here calculate it we get the value that is 26 .39 lower now the minimum mass here the minimum mass the minimum mass of the flywheel here the minimum mass of flywheel flywheel here that is equal to del a p max and here that is equal to i square c s now here the kinetic energy maximum when the del p max is maximum so here for del t max here that is maximum it's p minus t mu so here that is equal to 1200 plus 1400 sin theta plus 210 sin 2 theta plus 21 sin 3 theta 21 sin 3 theta and here minus the t mu that is 1200 so here that is equal to 1400 sin theta plus 210 sin 2 theta plus 21 sin 3 theta now here you have to do the derivative of the with respect to theta and here the limit and here the limit theta tends to 0 so here that is equal to 1400 sin 0 that is 1 plus 210 sin 0 is here sin the left one plus here 21 into 3 because sin 3 theta so here that is equal to 1400 plus 210 into 2 plus 21 into 3 and that is equal to 1400 plus 420 plus 16 and that is equal to here 2823 newton meter now as we said that is del a p max here that is equal to i square c s and here we know that i is equal to m k square so here we can write here that is here the value of m into 0 .8 whole square into 2 pi into 210 divided by 60 whole square into c s that is 0 .03 and here that is equal to here we can write that and that imply the value of k max that is 2823 and here that is equal to m into 0 .8 whole square plus 2 pi into 210 divided by 60 whole square into 0 .03 and here that imply the value of m we get from the equation which is a 2823 divided by 0 .8 whole square into 2 pi into 210 divided by 60 whole square into 0 .03 and calculated we get the value that is 303 .9 kg...

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