Question 10.45: A vertical engine running at 1200 rpm with a stroke of 120 m...

Angular speed        \omega=\frac{2 \pi \times 1200}{60}=126.66 rad / s .

L=120 mm \text { or } r =\frac{L}{2}=60 mn .

l=300 mm , m=1.5 kg , \theta=35^{\circ}, n=\frac{l}{r}=\frac{300}{600}=5 .

(a) Radius of gyration of the rod
Distance of the centre of gravity of rod from the point of suspension,

l_{1}=300-100=200 mm .

Frequency of oscillation of a compound pendulum,

f_{n}=\left(\frac{1}{2 \pi}\right)\left[\frac{g l_{1}}{K^{2}+l_{1}^{2}}\right]^{0.5} .

\frac{20}{20}=\left(\frac{1}{2 \pi}\right)\left[\frac{9.81 \times 200 \times 10^{3}}{K^{2}+200^{2}}\right]^{0.5} .

4 \pi^{2}=\frac{9.81 \times 200 \times 10^{3}}{K^{2}+200^{2}} .

or  K^{2}+40000=49698 .

or  K^{2}=9698 .

or  K=98.47 mm.

(b) Acceleration of the connecting rod:

Angular acceleration of the rod,    f_{r}=-\omega^{2} \frac{\sin \theta}{n} .

=-(126.66)^{2} \frac{\sin 35^{\circ}}{5} .

=-1840.35 rad / s ^{2} .

Acceleration of the piston,      f_{p}=\omega^{2} r\left[\cos \theta+\frac{\cos 2 \theta}{n}\right] .

=(126.66)^{2} \times 0.06\left[\cos 35^{\circ}+\frac{\cos 70^{\circ}}{5}\right] .

=854.33 m / s ^{2} .

Inertia torque exerted on the crankshaft

\text { Mass of the rod at the gudgeon pin, } m_{g}=\frac{m\left(l-l_{1}\right)}{l}=\frac{1.5(300-200)}{300}=0.5 kg .

\text { Vertical inertia force due to } m_{g}, \quad F_{i}=m_{g} f_{p}=0.5 \times 854.33=427.16 N.

Now          \sin \phi=\frac{\sin \theta}{n}=\frac{\sin 35^{\circ}}{5}=0.1147 .

\phi=6.587^{\circ} .

\frac{O M}{\sin (\theta+\phi)}=\frac{r}{\cos \phi} .

O M=\frac{60 \sin 41.587^{\circ}}{\cos 6.587^{\circ}}=40.1 mm .

\text { Torque due to } F_{i} ,          T=-F_{i} \cdot O M=-427.16 \times 0.0401 .

=-17.125 Nm \text { or } 17.125 Nm \text { (counter-clockwise) } .

Equivalent length of a pendulum, l_{e}=\frac{K^{2}+l_{1}^{2}}{l_{1}}=\frac{9698+200^{2}}{200}=248.49 mm .

Correction couple,    T_{o}=-m l_{1}\left(l-l_{e}\right) \alpha_{r} .

=-1.5 \times 200(300-248.49) 10^{-6} \times 1840.35 .

= –28.44 N m.

\text { Corresponding torque on the crankshaft, } T_{c s}=\frac{T_{o} \cos \theta}{n}=\frac{-28.44 \cos 35^{\circ}}{5}=-4.66 Nm .

=4.66 N m (counter-anticlockwise).

Torque due to the mass at the gudgeon pin,

T_{g}=m_{g} g \cdot O M=0.5 \times 9.81 \times 0.0401 .

=0.1967 N m (clockwise).

Equivalent mass of the rod at the crank pin,  m_{c}=\frac{1.5 \times 200}{300}=1 kg .

Torque due to this mass,    T_{c}=m_{c} g \cdot l \sin \phi=1 \times 9.81 \times 0.3 \times \sin 6.587^{c} .

= 0.3376 N m (clockwise).

\text { Inertia torque exerted on the crankshaft }=T+T_{c s}-T_{g}-T_{c} .

=17.125+4.66-0.1967-0.3376 .

= 21.25 N m (counter-clockwise).