A tank \(1.5 \mathrm{~m}\) high is \(1.2 \mathrm{~m}\) in diameter at its upper end and tapers uniformly to \(900 \mathrm{~mm}\) diameter at its base (its axis being vertical). It is to be emptied through a pipe \(36 \mathrm{~m}\) long connected to its base and the outlet of the pipe is to be \(1.5 \mathrm{~m}\) below the bottom of the tank. Determine a suitable diameter for the pipe if the depth of water in the tank is to be reduced from \(1.3 \mathrm{~m}\) to \(200 \mathrm{~mm}\) in not more than 10 minutes. Losses at entry and exit may be neglected and \(f\) assumed constant at \(0.008\).

In this problem, we are dealing with a tank that tapers, forming a frustum. A frustum is essentially a truncated cone. To calculate the volume of liquid flowing out, we need to determine the volume of this frustum. The formula to find the volume of a frustum is:
\[ V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2) \] where:

• \( h \) is the height of the frustum (the difference in height before and after emptying)
• \( r_1 \) is the radius at the upper end
• \( r_2 \) is the radius at the base

For our problem:
\( h = 1.3 \text{ m} - 0.2 \text{ m} = 1.1 \text{ m} \)
\( r_1 = \frac{1.2}{2} = 0.6 \text{ m} \)
\( r_2 = \frac{0.9}{2} = 0.45 \text{ m} \)
Plugging these values in: \[ V = \frac{1}{3} \pi \times 1.1 \times (0.6^2 + 0.45^2 + 0.6 \times 0.45) \]
Simplify to find the exact volume of the water to be emptied. This volume is critical for the next calculations.

The continuity equation in fluid mechanics ensures that the mass flow rate remains constant across sections of a streamline. We use it to connect the frustum's volume to the flow rate through the pipe. The formula is:
\[ Q = \frac{V}{t} \] where:

• \( Q \) is the flow rate (volume per time)
• \( V \) is the volume (from the frustum calculation)
• \( t \) is the time duration (in seconds)

We need to empty a volume \( V \) in 10 minutes (converted to seconds, 600 seconds). So, we divide the frustum volume by this time span to find the required flow rate. Let's assume, using our prior volume calculation result, that \( V = 0.858 \) cubic meters (m³). Then,
\[ Q = \frac{0.858 \, \text{m}^3}{600 \, \text{s}} = 0.00143 \, \text{m}^3/\text{s} \] This flow rate will be used to determine other factors like velocity and diameter of the pipe.

Bernoulli's equation relates the pressure, velocity, and height along a streamline, enabling us to calculate the velocity of the flow. The simplified form of Bernoulli's equation, assuming negligible losses, is:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \] Given that we are calculating for the tank (\( P_1 \)) vs. the exit (\( P_2 \)), assuming atmospheric pressure at both ends simplifies to:
\[ \frac{1}{2} \rho v_1^2 + \rho g h_1 = \frac{1}{2} \rho v_2^2 + \rho g h_2 \] If we set the velocity at the top (\( v_1 \)) as nearly zero due to the large area, we solve for the velocity at the pipe exit \( v_2 \). Given the height drop (\( h = 1.5 \) m),
\[ \rho g \Delta h = \frac{1}{2} \rho v_2^2 \]
\[ v_2 = \sqrt{2 g h} = \sqrt{2 \times 9.81 \times 1.5} \]
Solving this gives us the exit velocity, crucial for the next steps.

The Darcy-Weisbach equation helps us calculate head losses due to friction in a pipe, allowing us to determine the required pipe diameter. The equation is:
\[ h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \] where:

• \( h_f \) is the head loss
• \( f \) is the friction factor (given as 0.008)
• \( L \) is the length of the pipe (36 m)
• \( D \) is the diameter of the pipe
• \( v \) is the velocity from Bernoulli's equation
• \( g \) is the gravitational constant (9.81 m/s²)

We rearrange to solve for \( D \):
\[ D = \sqrt{ \frac{f L v^2}{2 g h_f} } \] Plug in the known values for \( f \), \( L \), velocity, and \( h_f \), and find the diameter \( D \). This diameter is necessary to ensure the water empties within the specified time.