A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm² cross

Q: Wait, why is the area $A(h)$ not just $\pi R^2$? Why are we using $r$?

That is a fantastic question! Remember, the tank is a hemisphere, not a cylinder. As the water level drops, the surface of the water is a circle, but its radius $r$ shrinks as we move down from the top to the bottom. The cross-sectional area of the water surface is $\pi r^2$, and since $r$ depends on $h$, the area is a function of $h$. We must account for this changing geometry!

Q: Why is there a negative sign in the continuity equation $\frac{dV}{dt} = -a \cdot v(t)$?

Think about what is happening to the water. The volume $V$ is decreasing as water flows out. The rate of change $\frac{dV}{dt}$ must be negative because the final volume is less than the initial volume. The outflow rate $a \cdot v(t)$ is a positive quantity (area times speed). To make the equation balance, we must include that negative sign to indicate the loss of volume.

Q: Can we just use Torricelli's law directly? Why is there a $0.6$ in the velocity formula?

Torricelli's law, $v = \sqrt{2gh}$, assumes an ideal fluid with no energy loss. In real-world scenarios, or specific problems like this, the flow is often restricted by the shape of the outlet or fluid friction, represented by a discharge coefficient. The $0.6$ is that coefficient, accounting for the actual velocity being less than the theoretical ideal. Always use the formula provided in the problem statement!

Question

A hemispherical tank of radius 2 metres is initially full of water and has an outlet of $12 	ext{ cm}^2$ cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $v(t) = 0.6 \sqrt{2gh(t)}$, where $v(t)$ and $h(t)$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t, and g is the acceleration due to gravity. Find the time it takes to empty the tank.

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Surface

Our first challenge is to understand the shape of the water surface. Because the tank is a hemisphere, the surface is a circle, but its radius $r$ is not constant.

If you look at a vertical cross-section of the tank, you can form a right-angled triangle. The hypotenuse is the tank's radius $R=2$, one leg is the vertical distance from the center of the sphere to the water surface, which is $R-h$, and the other leg is the surface radius $r$.

By the Pythagorean theorem, we have $r^2 + (R-h)^2 = R^2$. Substituting $R=2$, this simplifies to:

$$r^2 = 4h - h^2$$

The area of the water surface at any height $h$ is given by:

$$A(h) = \pi r^2 = \pi (4h - h^2)$$

### The Continuity Equation

The principle of continuity states that the rate at which the volume of water in the tank changes must equal the rate at which it leaves through the outlet. Mathematically, this is expressed as:

$$\frac{dV}{dt} = -a \cdot v(t)$$

We know the outlet area $a = 12 	imes 10^{-4} 	ext{ m}^2$ and the velocity law $v(t) = 0.6 \sqrt{2gh}$. Combining these, we get:

$$\frac{dV}{dt} = -a \cdot 0.6 \sqrt{2gh}$$

Since the change in volume for an infinitesimal drop in height $dh$ is $dV = A(h) dh$, we can write:

$$\frac{dV}{dt} = \pi (4h - h^2) \frac{dh}{dt}$$

### The Calculus of Emptying

Equating the two expressions for $\frac{dV}{dt}$ gives us our differential equation:

$$\pi (4h - h^2) \frac{dh}{dt} = -7.2 	imes 10^{-4} \sqrt{2gh}$$

By separating the variables, we divide both sides by $\sqrt{h}$ and move $dt$ to the right:

$$\pi (4h^{1/2} - h^{3/2}) dh = -7.2 	imes 10^{-4} \sqrt{2g} dt$$

Now, we integrate the left side from the initial height $h=2$ to the final height $h=0$, and the right side from time $t=0$ to $t=T$:

$$\pi \int_{2}^{0} (4h^{1/2} - h^{3/2}) dh = \int_{0}^{T} -7.2 	imes 10^{-4} \sqrt{2g} dt$$

### Final Calculation

The integration yields the following expression:

$$\pi \left[ \frac{8}{3} h^{3/2} - \frac{2}{5} h^{5/2} ight]_2^0 = -7.2 	imes 10^{-4} \sqrt{2g} T$$

Evaluating the limits, the upper limit at $h=0$ vanishes. The lower limit at $h=2$ provides a negative value that cancels the negative sign on the right side:

$$\pi \sqrt{2} \left( \frac{16}{3} - \frac{8}{5} ight) = 7.2 	imes 10^{-4} \sqrt{2} \sqrt{g} T$$

The $\sqrt{2}$ terms cancel out, leading to the final result for the time $T$:

**$$T = \frac{14\pi}{27\sqrt{g}} 	imes 10^4 	ext{ seconds}$$**

Visualized Solution

Visualizing the System

The Principle of Continuity

Geometry of the Water Surface

Relating $d V$ and $d h$

Equating the Rates

Separating the Variables

Setting Up the Integral

Executing the Integration

Evaluating the Limits

Solving for Time $T$

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Geometry of the Surface

The Continuity Equation

The Calculus of Emptying

Final Calculation

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