Solution 16: Rotating Pipe in Lake
Given: Pipe length $l = 80 \text{ m}$, Lake depth $h = 100 \text{ m}$, Initial air column $x_0 = 9 \text{ m}$.
State 1: Horizontal Pipe
The pipe lies at the bottom (depth $h$). The pressure of the trapped air $P_1$ balances the water pressure. $$P_1 = \rho g h$$ Volume of air $V_1 = A x_0$.
The pipe lies at the bottom (depth $h$). The pressure of the trapped air $P_1$ balances the water pressure. $$P_1 = \rho g h$$ Volume of air $V_1 = A x_0$.
State 2: Vertical Pipe
The pipe is rotated about the closed end. The air is now trapped at the bottom. Let the new length of the air column be $x$. The piston is at height $x$ from the bottom, so its depth from the surface is $(h – x)$. For equilibrium at the piston: $$P_{inside} = P_{outside} \implies P_2 = \rho g (h – x)$$ Volume of air $V_2 = A x$.
The pipe is rotated about the closed end. The air is now trapped at the bottom. Let the new length of the air column be $x$. The piston is at height $x$ from the bottom, so its depth from the surface is $(h – x)$. For equilibrium at the piston: $$P_{inside} = P_{outside} \implies P_2 = \rho g (h – x)$$ Volume of air $V_2 = A x$.
Calculation:
Assume the process is isothermal ($P_1 V_1 = P_2 V_2$). $$(\rho g h)(A x_0) = [\rho g (h – x)](A x)$$ Canceling $\rho g A$: $$h x_0 = (h – x)x \implies x^2 – hx + h x_0 = 0$$ Solving for $x$: $$x = \frac{h \pm \sqrt{h^2 – 4 h x_0}}{2}$$ Substituting $h = 100$ and $x_0 = 9$: $$x = \frac{100 \pm \sqrt{10000 – 3600}}{2} = \frac{100 \pm \sqrt{6400}}{2}$$ $$x = \frac{100 \pm 80}{2}$$ Two solutions: 1. $x = 90 \text{ m}$ (Rejected as $x > l=80 \text{ m}$) 2. $x = 10 \text{ m}$
Assume the process is isothermal ($P_1 V_1 = P_2 V_2$). $$(\rho g h)(A x_0) = [\rho g (h – x)](A x)$$ Canceling $\rho g A$: $$h x_0 = (h – x)x \implies x^2 – hx + h x_0 = 0$$ Solving for $x$: $$x = \frac{h \pm \sqrt{h^2 – 4 h x_0}}{2}$$ Substituting $h = 100$ and $x_0 = 9$: $$x = \frac{100 \pm \sqrt{10000 – 3600}}{2} = \frac{100 \pm \sqrt{6400}}{2}$$ $$x = \frac{100 \pm 80}{2}$$ Two solutions: 1. $x = 90 \text{ m}$ (Rejected as $x > l=80 \text{ m}$) 2. $x = 10 \text{ m}$
Answer: The piston will be at a height of 10 m above the closed end.