Problem 30: Terminal Speeds of Pistons
1. Equation of Continuity:
Let $v_1$ and $v_2$ be the speeds of the pistons in sections with radii $r_1$ and $r_2$. Since the fluid is incompressible: $$ A_1 v_1 = A_2 v_2 $$ $$ \pi r_1^2 v_1 = \pi r_2^2 v_2 \implies v_1 = \frac{r_2^2}{r_1^2} v_2 $$
Let $v_1$ and $v_2$ be the speeds of the pistons in sections with radii $r_1$ and $r_2$. Since the fluid is incompressible: $$ A_1 v_1 = A_2 v_2 $$ $$ \pi r_1^2 v_1 = \pi r_2^2 v_2 \implies v_1 = \frac{r_2^2}{r_1^2} v_2 $$
2. Work-Energy Principle (Bernoulli’s Equation):
Since the pistons move at constant speed, the work done by the external pressure difference is equal to the change in kinetic energy of the fluid. $$ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 $$ $$ P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2) $$
Since the pistons move at constant speed, the work done by the external pressure difference is equal to the change in kinetic energy of the fluid. $$ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 $$ $$ P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2) $$
3. Solve for $v_2$:
Substitute $v_1 = (\frac{r_2^2}{r_1^2}) v_2$: $$ P_1 – P_2 = \frac{1}{2}\rho \left[ v_2^2 – \left(\frac{r_2^2}{r_1^2} v_2\right)^2 \right] $$ $$ P_1 – P_2 = \frac{1}{2}\rho v_2^2 \left( 1 – \frac{r_2^4}{r_1^4} \right) $$ $$ P_1 – P_2 = \frac{1}{2}\rho v_2^2 \left( \frac{r_1^4 – r_2^4}{r_1^4} \right) $$ $$ v_2^2 = \frac{2(P_1 – P_2) r_1^4}{\rho (r_1^4 – r_2^4)} $$ $$ v_2 = \sqrt{\frac{2r_1^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} $$
Substitute $v_1 = (\frac{r_2^2}{r_1^2}) v_2$: $$ P_1 – P_2 = \frac{1}{2}\rho \left[ v_2^2 – \left(\frac{r_2^2}{r_1^2} v_2\right)^2 \right] $$ $$ P_1 – P_2 = \frac{1}{2}\rho v_2^2 \left( 1 – \frac{r_2^4}{r_1^4} \right) $$ $$ P_1 – P_2 = \frac{1}{2}\rho v_2^2 \left( \frac{r_1^4 – r_2^4}{r_1^4} \right) $$ $$ v_2^2 = \frac{2(P_1 – P_2) r_1^4}{\rho (r_1^4 – r_2^4)} $$ $$ v_2 = \sqrt{\frac{2r_1^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} $$
4. Solve for $v_1$:
Using $v_1 = \frac{r_2^2}{r_1^2} v_2$: $$ v_1 = \frac{r_2^2}{r_1^2} \sqrt{\frac{2r_1^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} = r_2^2 \sqrt{\frac{2(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} = \sqrt{\frac{2r_2^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} $$
Using $v_1 = \frac{r_2^2}{r_1^2} v_2$: $$ v_1 = \frac{r_2^2}{r_1^2} \sqrt{\frac{2r_1^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} = r_2^2 \sqrt{\frac{2(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} = \sqrt{\frac{2r_2^4(P_1 – P_2)}{\rho (r_1^4 – r_2^4)}} $$
Answer:
$$ v_1 = \sqrt{\frac{2r_2^4 (p_1 – p_2)}{\rho (r_1^4 – r_2^4)}} \quad \text{and} \quad v_2 = \sqrt{\frac{2r_1^4 (p_1 – p_2)}{\rho (r_1^4 – r_2^4)}} $$
$$ v_1 = \sqrt{\frac{2r_2^4 (p_1 – p_2)}{\rho (r_1^4 – r_2^4)}} \quad \text{and} \quad v_2 = \sqrt{\frac{2r_1^4 (p_1 – p_2)}{\rho (r_1^4 – r_2^4)}} $$