Water flows into a cylindrical vessel of large cross - sectional area at a rate of 10 – 4 m3/s. It flows out from a hole of area 10 – 4 m2, which has been punched through the base. How high (in m) does the water rise in the vessel?

Engineering

Physics

BernoulIis Equation and Equation of Continuity

Question

Water flows into a cylindrical vessel of large cross - sectional area at a rate of 10^{– 4} m³/s. It flows out from a hole of area 10^{– 4} m², which has been punched through the base. How high (in m) does the water rise in the vessel?

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This involves Torricelli's theorem and continuity equation. The water rises until inflow equals outflow. Outflow velocity from hole is $v = \sqrt{2 gh}$ .

Flow rate out is $Q_{out} = A \sqrt{2 gh}$ .

At equilibrium, $Q_{in} = Q_{out}$ .

Given $Q_{in} = 10^{- 4} \frac{m^{3}}{s}$ and $A = 10^{- 4} m^{2}$ .

So, $10^{- 4} = 10^{- 4} \sqrt{2 \times 9.8 \times h}$ .

Solving:

$1 = \sqrt{19.6 h} \to 1 = 19.6 h$

$\to h = \frac{1}{19.6} \approx 0.051 m$ .

Final answer: The water rises to a height of approximately 0.051 meters.