FLUIDS CYU 20

Problem 20: Bubble Shift in Accelerated Tube

Concept: Effective Mass and Inertia
An air bubble in a liquid behaves as if it has an “effective mass” related to the mass of the liquid it displaces. Because the bubble is nearly massless compared to the liquid, when the tube is given a sudden velocity $u$, the bubble tends to lag behind due to the inertia of the surrounding fluid that must flow around it to allow the motion.

Step 1: Determine the Drag Coefficient ($k$)
In the vertical case, the bubble rises with terminal velocity $v_0$. The buoyant force balances the viscous drag (assuming bubble mass $\approx 0$). $$F_B = k v_0$$ The buoyant force is equal to the weight of the displaced liquid, $m_w g$. $$m_w g = k v_0 \implies k = \frac{m_w g}{v_0}$$

Step 2: Horizontal Motion
When the tube is given a velocity $u$, the liquid moves with the tube. In the frame of the tube, the liquid is stationary, but the bubble has an initial relative velocity $-u$ (it wants to stay behind). The effective inertia opposing the motion of the bubble through the fluid is approximated by the mass of the displaced fluid, $m_w$. (This is a standard result in fluid dynamics for a massless void accelerating through fluid). The equation of motion for the bubble (relative to the tube) is: $$m_{eff} \frac{dv}{dt} = -F_{drag} = -kv$$ Using $m_{eff} \approx m_w$: $$m_w \frac{dv}{dt} = -k v$$ We want to find the total shift $S$. We can write $v = \frac{dx}{dt}$. $$m_w \frac{dv}{dt} = -k \frac{dx}{dt}$$ Integrating from initial velocity $u$ to final relative velocity $0$: $$\int_{u}^{0} m_w \, dv = \int_{0}^{S} -k \, dx$$ $$m_w [0 – u] = -k [S – 0]$$ $$-m_w u = -k S \implies S = \frac{m_w u}{k}$$