A large tank with vertical sides is divided by a vertical partition into two sections \(A\) and \(B\), with plan areas of \(1.5 \mathrm{~m}^{2}\) and \(7.5 \mathrm{~m}^{2}\) respectively. The partition contains a \(25 \mathrm{~mm}\) diameter orifice \(\left(C_{d}=0.6\right)\) at a height of \(300 \mathrm{~mm}\) above the base. Initially section \(A\) contains water to a depth of \(2.15 \mathrm{~m}\) and section \(B\) contains water to a depth of \(950 \mathrm{~mm}\). Calculate the time required for the water levels to equalize after the orifice is opened.

In fluid mechanics, understanding orifice flow is crucial for solving problems like the given tank partition exercise. An orifice is a small hole or opening, typically in a vertical partition, that allows fluid to pass through. The flow rate through an orifice can be found using the formula:
\[ Q = C_d \times A \times \text{√}(2gh) \]
where \(Q\) is the discharge rate, \(C_d\) is the discharge coefficient, \(A\) is the area of the orifice, \(g\) is the acceleration due to gravity, and \(h\) is the head difference (height difference) of the fluid levels. Understanding this flow helps us calculate how quickly water moves from one tank to another, ultimately leading to equal water levels.

The continuity equation in fluid dynamics states that the mass flow rate must remain constant from one cross-section of a stream to another. In simpler terms, what flows into a section must flow out. Mathematically, it's expressed as:
\[ A_1 \times v_1 = A_2 \times v_2 \]
where \(A\) represents the cross-sectional area, and \(v\) represents the flow velocity. For the given tank problem, this means the rate of water flowing out from one section has to be equivalent to the rate at which it fills another. This principle ensures we balance the volumes and calculate the changing heights in both sections over time.

Differential equations are essential in modeling the changes in fluid levels over time. For the tank problem, we use differential equations to represent how the heights of water, \(h_A\) and \(h_B\), change. The basic form used is:
\[ \frac{dh_A}{dt} = \frac{-Cd \times A \times \text{√}(2g(h_A - h_B))}{A_A} \]
and
\[ \frac{dh_B}{dt} = \frac{Cd \times A \times \text{√}(2g(h_A - h_B))}{A_B} \]
These equations depict the rate of change of the water levels in sections \(A\) and \(B\). Solving these differential equations helps us understand how quickly the water levels equalize.

Integration comes into play when solving the differential equations for fluid dynamics. To find the total time required for the water levels to equalize, we integrate the rate equations. The goal is to solve:
\[ \frac{d}{dt}(h_A - h_B) = k \times \text{√}(h_A - h_B) \]
We integrate over time to determine how long it takes for the head difference, \((h_A - h_B)\), to become zero. Specifically, we evaluate:
\[ \int_{0}^{t} dt = \int_{h_{A}(0) - h_{B}(0)}^{0} \frac{1}{k \times \text{√}(h_A - h_B)} \times d(h_A - h_B) \]
This integration gives us the time t in seconds or minutes, required for the levels to equalize.