Unsteady Tank Draining – Time to Lower Level from H1 to H2 A tank of uniform cross-sectional area A with a small orifice of area a at the bottom drains under gravity. The time required to drop the liquid level from H1 to H2 (including discharge coefficient Cd) is t = (2 * A / (Cd * a * sqrt(2 * g))) * (sqrt(H1) − sqrt(H2)).

Introduction:
Predicting drain-down time is a classic unsteady-flow problem. The solution uses Bernoulli’s principle with Torricelli’s efflux velocity and the continuity equation for a falling head in a reservoir-orifice configuration.

Given Data / Assumptions:

• Tank cross-section A; small orifice area a at bottom.
• Coefficient of discharge Cd accounts for contraction and velocity effects.
• Inviscid energy assumptions aside from Cd; negligible velocity of the tank free surface.

Concept / Approach:

Torricelli’s result gives exit speed V = Cd * sqrt(2 * g * H), so discharge Q = a * V = Cd * a * sqrt(2 * g * H). The falling head yields A * dH/dt = −Q. Separating variables produces an integral in 1/sqrt(H).

Step-by-Step Solution:

Verification / Alternative check:

Setting H2 = 0 gives complete emptying time, matching standard textbook expressions. Dimensional check confirms seconds for t.

Why Other Options Are Wrong:

Linear dependence on H (Option B, D) contradicts the square-root head relation. Logarithmic form (Option C) suits viscous laminar draining in tubes, not orifices. Option E has incorrect dependence on H and g.

Common Pitfalls:

Forgetting Cd; using A ≈ constant assumption where it is invalid (e.g., conical tanks); neglecting approach velocity corrections for large a/A.

Final Answer:

t = (2 * A / (Cd * a * sqrt(2 * g))) * (sqrt(H1) − sqrt(H2))