FLUIDS CYU 16

Problem 16: Time for a Floating Tank to Sink

Analysis:
Let the mass of the tank be $M$ and the area of the base be $A = ab$. Initially, the tank floats with height $h$ remaining out of the water. By Archimedes’ principle, the weight of the tank is balanced by the buoyant force. Let $d$ be the initial depth of immersion. $$Mg = \rho A d g$$ Since the total height is $H$ and the part outside is $h$, the submerged depth is $d = H – h$. Therefore, $M = \rho A (H – h)$.

As water enters the hole, the tank becomes heavier and sinks further. Let $y$ be the depth of the tank bottom below the outside water surface, and $y’$ be the depth of water inside the tank. The condition for floating equilibrium at any instant is: $$Mg + (\text{Weight of water inside}) = \text{Buoyant Force}$$ $$Mg + \rho A y’ g = \rho A y g$$ Dividing by $g$: $$M + \rho A y’ = \rho A y \implies \rho A (y – y’) = M$$ Substituting $M = \rho A (H – h)$: $$\rho A (y – y’) = \rho A (H – h) \implies y – y’ = H – h$$ This result is crucial. It tells us that the difference between the outside water level and the inside water level (the “head” driving the flow) remains constant throughout the process. The driving head is $\Delta h_{eff} = H – h$.

Calculations:
According to Torricelli’s law, the velocity of influx $v$ depends on the pressure head difference: $$v = \sqrt{2g(\Delta h_{eff})} = \sqrt{2g(H – h)}$$ Since the head difference is constant, the velocity of influx is constant.

The tank begins to sink completely when the top rim of the tank touches the water surface. At this point, the tank is submerged to depth $y = H$. Using our relation $y – y’ = H – h$, when $y = H$, the water level inside the tank is: $$H – y’ = H – h \implies y’ = h$$ So, the tank sinks when the volume of water inside reaches $V_{in} = A \cdot h$.

The time taken $t$ is simply the total volume required divided by the rate of flow (Volume flux $Q = S v$): $$t = \frac{V_{in}}{S \cdot v} = \frac{abh}{S\sqrt{2g(H – h)}}$$