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For a steam engine, the following data is given: Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m calculate inertia force at θ=30 degrees from IDC.

For a steam engine, the following data is given: Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m calculate inertia force at θ=30 degrees from IDC. Correct Answer 19064 N

l1 = l – G.C = 1.5 – 0.5 = 1m Fi = (Mr + G.C,Mc/l).ω2.r(cosθ + cos2θ/n) Mr=300 Kg Mc=250 kg ω=13.1 rad/s r=0.3m substituting these values will give Fi = 19064 N.

Related Questions

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find angular acceleration of connecting rod in rad/s2.
Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find the equivalent length L of a simple pendulum swung about an axis.
From the data given: Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find the correcting couple in N-m?
The primary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by (where m = Mass of reciprocating parts, $$\omega $$ = Angular speed of crank, r = Radius of crank, $$\theta $$ = Angle of inclination of crank with the line of stroke and n = Ratio of the length of connecting rod to radius of crank)
The secondary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by (where m = Mass of reciprocating parts, $$\omega $$ = Angular speed of crank, r = Radius of crank, $$\theta $$ = Angle of inclination of crank with the line of stroke and n = Ratio of the length of connecting rod to radius of crank)
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An Internal combustion engine has a connecting rod of mass 2 kg and the distance between the centre of crank and centre of gudgeon pin is 25 cm. The Center of Gravity lies at a point 10 cm from the gudgeon pin along the line of centres. The radius of gyration of this system about an axis through the Center of Gravity perpendicular to the plane of rotation is 11 cm. If two masses are used instead of the connecting rod, one at the gudgeon pin and other at the crank pin. What will be the new radius of gyration?
In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine linear velocity of the midpoint of the connecting rod in m/s. Crank angle = 45° from IDC.
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