Unsteady outflow – Time to empty a hemispherical tank A hemispherical tank of radius R has a small orifice of cross-sectional area a at its bottom and is initially full of liquid. Assuming a constant coefficient of discharge Cd and neglecting viscous head losses other than at the orifice, what is the time required to empty the tank completely?

Introduction:
Unsteady tank-draining problems combine hydrostatics (head–depth relation) with continuity through an orifice governed by Torricelli’s law corrected by the discharge coefficient Cd. For curved tanks like hemispheres, the surface area and cross-section change with depth, leading to integrals that must be evaluated to obtain the emptying time.

Given Data / Assumptions:

• Hemispherical tank, radius R, orifice area a at the bottom centre.
• Coefficient of discharge Cd is constant over the draining process.
• Outflow obeys Q = Cd * a * sqrt(2 * g * h), with h measured from the orifice.
• Neglect additional head losses and velocity of approach in the tank (large plan area relative to orifice).

Concept / Approach:

Let h be instantaneous liquid height from the orifice. For a hemisphere, the cross-sectional area at height h is A(h) = pi * (2 * R * h − h^2). The unsteady mass balance gives A(h) * dh/dt = − Q = − Cd * a * sqrt(2 * g * h). Rearranging: dt = − A(h) / (Cd * a * sqrt(2 * g * h)) * dh. Integrate h from R down to 0 (or 0 up to R with a sign change).

Step-by-Step Solution:

Verification / Alternative check:

Dimensional analysis: numerator has length^(5/2); denominator contributes length^(3/2)/time, yielding time. The result reduces to known flat-tank results when the geometry is changed appropriately.

Why Other Options Are Wrong:

Scaled variants with 8/15 or altered powers of R do not match the integral of A(h) / sqrt(h).Forms mixing R inside the square root in the denominator incorrectly combine variables.Expressions without R^(5/2) cannot capture the geometry of the hemisphere.

Common Pitfalls:

Using A = constant (as if the tank were prismatic) or forgetting the square root on 2 * g * h in Torricelli’s outflow law, both of which lead to incorrect time estimates.

Final Answer:

t = (14 * pi * R^(5/2)) / (15 * Cd * a * sqrt(2 * g))