Question

A tube consisting of two vertical arms and a horizontal arm completely filled with water is held at rest. Inner diameter of the tube is much smaller than the length of each segment of the tube. The tube is made to move horizontally in its plane with a constant acceleration so that 1/16 of water flows out of the tube. Neglecting capillary action, deduce suitable expression for pressure at the closed end. Atmospheric pressure is $p_0$ and acceleration due to gravity is g.

Answer 1

$p_0 + \frac{29}{32}\rho g l - \frac{1}{2}\rho a l$

Answer 2

Let the acceleration be $\vec{a}$ in the x-direction and gravity $\vec{g}$ in the -y direction. The effective gravitational acceleration in the frame of the tube is $\vec{g}_{eff} = \vec{g} - \vec{a}_{apparent}$. If the tube accelerates with $\vec{a}$, the apparent acceleration is $-\vec{a}$. Thus, $\vec{g}_{eff} = -a\hat{i} - g\hat{j}$.

The pressure gradient is $\nabla P = \rho \vec{g}_{eff}$, so $\frac{\partial P}{\partial x} = -\rho a$ and $\frac{\partial P}{\partial y} = -\rho g$. Integrating, we get $P(x,y) = -\rho a x - \rho g y + C$.

Let the horizontal arm have length $l$ (from $x=0$ to $x=l$) and the vertical arms have length $l$ (from $y=0$ to $y=l$). The water surfaces in the vertical arms are at atmospheric pressure $p_0$. Let the heights of water in the left and right vertical arms be $h_L$ and $h_R$ respectively. $P(0, h_L) = p_0$ and $P(l, h_R) = p_0$. Using the pressure relation between $(0, h_L)$ and $(l, h_R)$: $P(0, h_L) = P(l, h_R) - \rho a (0-l) - \rho g (h_L - h_R)$ $p_0 = p_0 + \rho a l - \rho g (h_L - h_R) \implies h_L - h_R = \frac{a}{g}l$.

Initially, the volume is $V_{init} = A(l+l+l) = 3Al$. After $\frac{1}{16}$ flows out, $V_{final} = \frac{15}{16}V_{init} = \frac{45}{16}Al$. The final volume is also $V_{final} = A h_L + A l + A h_R$. $h_L + h_R = (\frac{45}{16} - 1)l = \frac{29}{16}l$.

Solving the system: $h_L - h_R = \frac{a}{g}l$ $h_L + h_R = \frac{29}{16}l$ Adding them: $2h_L = (\frac{29}{16} + \frac{a}{g})l \implies h_L = \frac{l}{2}(\frac{29}{16} + \frac{a}{g})$. Subtracting them: $2h_R = (\frac{29}{16} - \frac{a}{g})l \implies h_R = \frac{l}{2}(\frac{29}{16} - \frac{a}{g})$.

The pressure at the closed end (bottom of the horizontal arm, at $(l,0)$) is $P(l,0)$. $P(l,0) = P(l, h_R) - \rho g (0 - h_R) = p_0 + \rho g h_R$. Substituting $h_R$: $P(l,0) = p_0 + \rho g \left( \frac{l}{2}(\frac{29}{16} - \frac{a}{g}) \right) = p_0 + \frac{29}{32}\rho g l - \frac{1}{2}\rho a l$.