Equalising levels between two identical tanks via a small orifice Two identical cylindrical vessels (cross-sectional area A each) are interconnected by an orifice of area a and coefficient of discharge C_d. If the initial difference in liquid levels is H, what is the time T for the levels to equalise (neglecting losses other than the orifice)?

Introduction / Context:
Interconnected-tank transients are common in laboratory hydraulics and process engineering. When two identical tanks are connected by a small orifice, the level difference decays in time, and the equalisation time can be derived from unsteady continuity combined with the orifice equation.

Given Data / Assumptions:

• Two tanks, each of area A, connected by an orifice of area a with discharge coefficient C_d.
• Head loss confined to the orifice; inviscid elsewhere; free discharge between tanks.
• Initial level difference Δh(0) = H; final equalised difference Δh(T) = 0.

Concept / Approach:

Flow through the orifice is Q = C_d a √(2 g Δh). Because one tank empties while the other fills, the rate of change of difference is doubled: d(Δh)/dt = −(2Q)/A. This yields a first-order, separable differential equation in √Δh, leading to a closed-form time to reach Δh = 0 from H.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional analysis: [T] = [A]/([a] √(g) √[H]) → seconds, consistent.

Why Other Options Are Wrong:

(b) and (c) have incorrect factors of 2; (d) is dimensionally inconsistent; (e) omits √2 in the orifice velocity.

Common Pitfalls:

Forgetting the factor of 2 because both tanks change level; misapplying Torricelli’s law without the discharge coefficient.

Final Answer:

T = (A / (C_d a √(2 g))) * √H