Hemispherical Tank Draining – Time to Lower Level from H1 to H2 A hemispherical tank of radius R drains through a small orifice of area a at its bottom. Including coefficient of discharge Cd, the time to reduce the liquid head from H1 to H2 (measured from the orifice) is t = (pi / (Cd * a * sqrt(2 * g))) * [ (4 * R / 3) * (H1^(3/2) − H2^(3/2)) − (2 / 5) * (H1^(5/2) − H2^(5/2)) ].

Introduction:
Unsteady draining of curved tanks requires accounting for how the cross-sectional area varies with head. For a hemisphere, the area changes quadratically with head, leading to an integral that produces powers 3/2 and 5/2 in the time expression.

Given Data / Assumptions:

• Hemispherical tank of radius R, orifice of area a at the bottom center.
• Head H measured vertically above the orifice (0 ≤ H ≤ R).
• Torricelli efflux with coefficient Cd; inviscid elsewhere; free surface velocity negligible.

Concept / Approach:

The free-surface area at head H is A(H) = pi * (2 * R * H − H^2). Continuity gives A(H) dH/dt = −Cd * a * sqrt(2 * g * H). Separating variables yields an integral of the form ∫ (2 R H − H^2) / sqrt(H) dH, producing the stated closed form for time.

Step-by-Step Solution:

Verification / Alternative check:

Setting H2 = 0 yields the emptying time, matching known textbook forms for hemispherical tanks. Units reduce to seconds.

Why Other Options Are Wrong:

Option B is for a prismatic tank of constant area A. Option C assumes linear head dependence. Option E is for laminar tube draining or viscous head-loss models. Hence Option A is the correct hemispherical result; Option D is false.

Common Pitfalls:

Using the constant-area formula; misdefining H from the free surface; neglecting Cd which shortens predicted times if omitted.

Final Answer:

t = (pi / (Cd * a * sqrt(2 * g))) * [ (4 * R / 3) * (H1^(3/2) − H2^(3/2)) − (2 / 5) * (H1^(5/2) − H2^(5/2)) ]