Question

The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in length. The crank rotates at 180 rpm. Clock wise. Determine the velocity and acceleration of the piston when the crank is at 40 degrees from the inner dead centre position. Also determine the position of the crank for zero acceleration of the piston.

Answer 1

Question #81296

The crank and connecting rod of a steam engine are $0.3\mathrm{m}$ and $1.5\mathrm{m}$ in length. The crank rotates at 180 rpm. Clock wise. Determine the velocity and acceleration of the piston when the crank is at 40 degrees from the inner dead centre position. Also determine the position of the crank for zero acceleration of the piston.

Answer:

Figure 1 shows the position of the mechanism at the given moment.

FIGURE 1. Crank-piston mechanism

The rotational velocity of the crank is

$$\omega_ {1} = \frac {1 8 0 \pi}{3 0} = 1 8. 8 5 \mathrm {r a d / s}.$$

The velocity $\nu_{A}$ of the point $A$ is directed perpendicularly to the crank and has magnitude of

$$v _ {A} = O A \cdot \omega_ {1} = 0. 3 \cdot 1 8. 5 = 5. 6 5 \mathrm {m / s}.$$

The velocity $\nu_{B}$ of the piston $B$ equals to the horizontal component of $\nu_{A}$

$$v _ {B} = v _ {A} \cdot \sin 4 0 ^ {\circ} = 5. 6 5 \cdot 0. 6 4 2 8 = 3. 6 3 \mathrm {m / s}.$$

The acceleration $a_1$ of the point $A$ is directed along the crank to $O$ and has magnitude of

$$a _ {1} = O A \cdot \omega_ {1} ^ {2} = 0. 3 \cdot 1 8. 5 ^ {2} = 1 0 6. 6 \mathrm {m / s ^ {2}}.$$

In triangle ADC we have

$$A D = O A \cdot \sin 4 0 ^ {\circ} = 0. 3 \cdot 0. 6 4 2 8 = 0. 1 9 3 \mathrm {m},$$ $$A C = O A \cdot \tan 4 0 ^ {\circ} = 0. 3 \cdot 0. 8 3 9 = 0. 2 5 2 \mathrm {m},$$ $$C D = A C \cdot \sin 4 0 ^ {\circ} = 0. 2 5 2 \cdot 0. 6 4 2 8 = 0. 1 6 2 \mathrm {m}.$$

The distances $BD$ and $BC$ equal to

$$BD = \sqrt{AB^2 - AD^2} = \sqrt{1.5^2 - 0.193^2} = 1.49 \text{ m},$$ $$BC = BD + CD = 1.49 + 0.162 = 1.65 \text{ m}.$$

Now, we can find the distances $AP$ and $BP$ to the current centre $P$ of the rotation of the conrod $AB$ as follow

$$\frac{AC}{CP} = \frac{CD}{BC},$$ $$CP = AC \cdot \frac{BC}{CD} = 0.252 \cdot \frac{1.65}{0.162} = 2.57 \text{ m},$$ $$AP = CP - AC = 2.57 - 0.252 = 2.31 \text{ m},$$ $$\frac{AD}{BP} = \frac{CD}{BC},$$ $$BP = AD \cdot \frac{BC}{CD} = 0.193 \cdot \frac{1.65}{0.162} = 1.97 \text{ m}.$$

Thus, the current rotational acceleration $\varepsilon_{2}$ of the conrod is

$$\varepsilon_{2} = \frac{a_{1}}{AP} = \frac{106.6}{2.31} = 46.1 \text{ rad/s}^2.$$

Finally, the acceleration of the piston is

$$a_{2} = BP \cdot \varepsilon_{2} = 1.97 \cdot 46.1 = 90.5 \text{ m/s}^2.$$

The piston has zero acceleration at the dead points. The positions of the crank at the piston dead points are 0 and 180 degrees.