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Question 12.40: A single cylinder horizontal engine runs at 120 rpm with a s...

A single cylinder horizontal engine runs at 120 rpm with a stroke of 400 mm. The mass of the revolving parts assumed concentrated at the crankpin is 100kg and mass of the reciprocating parts is 150 kg. Determine the magnitude of the balancing mass to be placed opposite to the crank at a radius of 150 mm which is equivalent to all the revolving and 2/3 rd of the reciprocating parts. If the crank turns 30° from the inner dead centre, find the magnitude of the unbalanced force due to the balancing mass.

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N=120 rpm , L=400 mm , M=100 kg , R=150 kg , b=150 mm , c=2 / 3, \theta=30^{\circ} , Bb = (M = c R) r .

0.15 B=(100+2 \times 150 / 3) \times 0.2 .

B=266.67 kg .

\omega=2 \times 120 / 60=12.57 rad / s .

\text { Residual unbalanced force }=R \omega^{2} r\left[(1-c)^{2} \cos ^{2} \theta+c^{2} \sin ^{2} \theta\right]^{0.5} .

=150 \times(12.57)^{2} \times 0.2\left[(1-2 / 3)^{2} \times \cos ^{2} 30^{\circ}+4 \times \sin ^{2} 30^{\circ} / 9\right]^{0.5} .

= 2090.2 N.

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