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Chapter 4: Problem 44

A cylindrical tank, \(0.3 \mathrm{m}\) in diameter, drains through a hole in its bottom. At the instant when the water depth is \(0.6 \mathrm{m}\) the flow rate from the tank is observed to be \(4 \mathrm{kg} / \mathrm{s}\). Determine the rate of change of water level at this instant.

Short Answer

Expert verified
After following the steps, the rate of change of the water level is found by calculating \( dh/dt \). This gives the rate at which the water level is dropping at the given moment in m/s.

Step by step solution

01

Derive the equation for the volume of water in the tank

The volume inside a cylindrical tank can be calculated using the formula for the volume of a cylinder: \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Here, the diameter of the cylindrical tank is given as 0.3 m, so the radius is \( r = 0.3 / 2 = 0.15 \) m. The height is given as 0.6 m. Substitute \( r = 0.15 \) m and \( h = 0.6 \) m into the equation to find \( V \).
02

Convert the mass flow rate to a volume flow rate

The problem gives a mass flow rate of 4 kg/s, and we need to convert this into a volume flow rate (m^3/s). This can be done using the formula \( Q = \dot{m} / \rho \), where \( Q \) is the volume flow rate, \( \dot{m} \) is the mass flow rate, and \( \rho \) is the density of water. Assuming a density of 1000 kg/m^3 for water, the volume flow rate can be calculated.
03

Find the rate of change of the water level

The amount of water leaving the tank per second equals to the volume flow rate and is equal to the amount by which the volume of the water in the tank is reducing per second. So \( Q = - dV/dt = - \pi r^2 dh/dt \), where \( dh/dt \) is the rate at which the water height is changing. Rearrange for \( dh/dt \) to give \( dh/dt = - Q / (\pi r^2) \). Substitute the values calculated in the previous steps and solve for \( dh/dt \).

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