Given,
stroke of the steam engine r = 0.6m
length of the connecting rod l = 1.5m
speed of the crank N = 180 rpm
n = l/r = 1.5/0.6 = 2.5
a) velocity of the piston at angle = 400 is
v = l (1/square root 1-(sin thita/n)2 * (r/l)2 2sin thita cos thita *w +rsin thita *w
v = 1.5 (1/square root 1-(sin 400/2.5)2 *(0.6/1.5)2 2sin400 cos400*18.84 +0.6 sin 400 *18.84
v = 6.217m/s
acceleration of the piston is given by
a = w2r (cos thita + cos 2 thita/n)
a = 18.842 * 0.6 ( cos 400 + cos 2 (400)/2.5 )
a = 9.44 m2/sec
b) position of the crank when the acceleration is zero
a = w2r (cos thita + cos 2 thita/n)
thita = 18.842 *0.6 ( cos thita + cos 2 thita /2.5)
thita = 71.400
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