Problem for Practice - The connecting rod of an internal combustion engine is \( 225 \mathrm{~mm} \) long and has a mass \( 1.6 \mathrm{~kg} \). The mass of the piston and gudgeon pin is \( 2.4 \mathrm{~kg} \) and the stroke is \( 150 \mathrm{~mm} \). The cylinder bore is \( 112.5 \mathrm{~mm} \). The center of gravity of the connecting rod is \( 150 \mathrm{~mm} \) from the small end. Its radius of gyration about the centre of gravity for oscillations in the plane of swing of the connecting rod is \( 87.5 \mathrm{~mm} \). Determine the torque on acankshaft to overcome the inertia of moving parts, by ANALYTICAL method, when the crank is at \( 40^{\circ} \) and the piston is moving away from inner dead center under an effective gas pressure of \( 1.8 \mathrm{MN} / \mathrm{m}^{2} \). The engine speed is 1200 rpm. - [Check your answer: 927.15 N.m]
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First, we need to find the angular velocity (ω) of the crankshaft. We are given the engine speed as 1200 rpm. To convert it to radians per second, we can use the following formula: ω = (2 * π * n) / 60 where n is the engine speed in rpm. ω = (2 * π * Show more…
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The connecting rod $A B$ of a certain internalcombustion engine weighs 1.2 lb with mass center at $G$ and has a radius of gyration about $G$ of 1.12 in. The piston and piston pin $A$ together weigh $1.80 \mathrm{lb}$ The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is $3000(2 \pi) / 60=100 \pi$ rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin $A$ for the crank angle $\theta=90^{\circ} .$ (Suggestion: Use the alternative moment relation, Eq. $6 / 3,$ with $B$ as the moment center.
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