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)

A

$${\text{m}}{\omega ^2}{\text{r}}\sin \theta $$

B

$${\text{m}}{\omega ^2}{\text{r}}\cos \theta $$

C

$${\text{m}}{\omega ^2}{\text{r}}\frac{{\sin 2\theta }}{{\text{n}}}$$

D

$${\text{m}}{\omega ^2}{\text{r}}\frac{{\cos2\theta }}{{\text{n}}}$$

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.

A

16.782

B

17.824

C

15.142

D

17.161

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.

A

19000 N

B

19064 N

C

19032 N

D

20064 N

Evaluate the following:
$$\frac{{\cos 2\theta \cdot \cos 3\theta - \cos 2\theta \cdot \cos 7\theta + \cos \theta \cdot \cos 10\theta }}{{\sin 4\theta \cdot \sin 3\theta - \sin 2\theta \cdot \sin 5\theta + \sin 4\theta \cdot \sin 7\theta }}$$

A

cot6θ.cot5θ

B

cos6θ.cos5θ

C

cos6θ.cot5θ

D

-cot6θ.cot5θ

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.

A

1.35 m

B

1.42 m

C

1.48 m

D

1.50 m

The tractive force in a locomotive with two cylinders is given by (where c = Fraction of reciprocating parts per cylinder, m = Mass of reciprocating parts, $$\omega $$ = Angular speed of crank, r = Radius of crank and $$\theta $$ = Angle of inclination of crank to the line of stroke)

A

$${\text{m}}{\omega ^2}{\text{r}}\cos \theta $$

B

$${\text{cm}}{\omega ^2}{\text{r}}\sin \theta $$

C

$$\left( {1 - {\text{c}}} \right){\text{m}}{\omega ^2}{\text{r}}\left( {\cos \theta - \sin \theta } \right)$$

D

$${\text{m}}{\omega ^2}{\text{r}}\left( {\cos \theta - \sin \theta } \right)$$

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?

A

52.7

B

49.5

C

59.5

D

56.5

The velocity of piston in a reciprocating steam engine is given by (where $$\omega $$ = Angular velocity of crank, r = Radius of crank pin circle, $$\theta $$ = Angle turned by crank from inner dead center and n = Ratio of length of connecting rod to the radius of crank)

A

$$\omega {\text{r}}\left( {\sin \theta + \frac{{\sin 2\theta }}{{\text{n}}}} \right)$$

B

$$\omega {\text{r}}\left( {\cos\theta + \frac{{\cos2\theta }}{{\text{n}}}} \right)$$

C

$${\omega ^2}{\text{r}}\left( {\sin \theta + \frac{{\sin 2\theta }}{{\text{n}}}} \right)$$

D

$${\omega ^2}{\text{r}}\left( {\cos\theta + \frac{{\cos2\theta }}{{\text{n}}}} \right)$$

The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$ is equal to:

A

2cosθ

B

cosecθsecθ

C

2sinθ

D

sinθcosθ

In the below figure, PC is the connecting rod and OC is the crank making an angle $$\theta $$ with the line of stroke PO and rotates with uniform angular velocity at $$\omega $$ rad/s. The Klien's acceleration diagram for determining the acceleration of the piston P is shown by quadrilateral CQNO, the acceleration of the piston is _________ when the crank OC and connecting rod PC are at right angles to each other.

A

Infinity

B

Zero

C

Any +ve value

D

Any -ve value

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