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)) ].
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At = (pi / (Cd * a * sqrt(2 * g))) * [ (4 * R / 3) * (H1^(3/2) − H2^(3/2)) − (2 / 5) * (H1^(5/2) − H2^(5/2)) ]
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Bt = (2 * A / (Cd * a * sqrt(2 * g))) * (sqrt(H1) − sqrt(H2))
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Ct = (A / (Cd * a)) * (H1 − H2)
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Dnone of the above
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Et = (pi R^2 / (Cd * a * sqrt(2 * g))) * ln(H1 / H2)
Answer
Correct 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)) ]
Explanation
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:
1) Write A(H) = pi (2 R H − H^2).2) Use A(H) dH/dt = −Cd a sqrt(2 g H).3) Rearrange and integrate from H1 to H2: t = (pi / (Cd a sqrt(2 g))) ∫_{H2}^{H1} (2 R H − H^2) / sqrt(H) dH.4) Evaluate: ∫(2 R H^{1/2} − H^{3/2}) dH = (4 R / 3) H^{3/2} − (2 / 5) H^{5/2}.5) Apply limits to obtain the final expression shown in Option A.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)) ]