The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 30° from T.D.C., the difference between the driving and the back pressures is 0.45 N/mm^2. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate the net load on piston.

Question

The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 30° from T.D.C., the difference between the driving and the back pressures is 0.45 N/mm^2. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate the net load on piston.

asked Sep 22, 2022 in Machine Dynamics by Grayson (34.3k points)

The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 30° from T.D.C., the difference between the driving and the back pressures is 0.45 N/mm^2. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate the net load on piston.

(a) 88000 N

(b) 90560 N

(c) 78036 N

(d) 88357 N

This question was posed to me in a job interview.

This key question is from Forces on the Reciprocating Parts of an Engine Neglecting Weight of the Connecting Rod topic in portion Inertia Forces in Reciprocating Parts of Machine Dynamics

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Fred · Answer 1 · 2022-10-13T21:40:35+0000

The correct choice is (d) 88357 N

The best explanation: Net piston load is given by F(l) = (p1-p2)πD^2÷4

p1-p2 = 0.45 N/mm^2

D = 500 mm

Therefore F(l) = 88357 N.

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