Question

Water is flowing through a tube of radius r at constant velocity with power P. What happens to power when radius of the tube doubled?

Hint:

When dealing with fluid flow power, always check which parameter is constant. If velocity $v$ is constant, $P \propto r^2$. If mass flow rate $\frac{dm}{dt}$ is constant, $v$ changes, leading to a very different relationship.

Answer 1

Step 1: Understanding the Concept:
Power is the rate at which work is done or energy is transferred. In the context of fluid flowing through a tube, if a pump is maintaining the flow, the power delivered is equal to the rate at which kinetic energy is imparted to the fluid.
Step 2: Key Formula or Approach:
The kinetic energy of mass \(m\) moving at velocity \(v\) is \(\frac{1}{2}mv^2\).
Power (\(P\)) is the rate of change of kinetic energy: \(P = \frac{d(K.E.)}{dt} = \frac{1}{2} \frac{dm}{dt} v^2\).
The mass flow rate (\(\frac{dm}{dt}\)) is density (\(\rho\)) times volume flow rate (\(Q = \text{Area} \times \text{velocity} = A \cdot v\)).
So, \(\frac{dm}{dt} = \rho \cdot A \cdot v = \rho \cdot (\pi r^2) \cdot v\).
Step 3: Detailed Explanation:
Substitute the mass flow rate into the power equation:
\[ P = \frac{1}{2} (\rho \pi r^2 v) v^2 \]
\[ P = \frac{1}{2} \rho \pi r^2 v^3 \]
The problem states the fluid is flowing at a constant velocity \(v\). The fluid density \(\rho\) and \(\pi\) are constants.
Therefore, we can see the relationship between Power and radius:
\[ P \propto r^2 \]
Let the initial power be \(P_1 = P\) when radius is \(r_1 = r\).
If the radius is doubled, the new radius is \(r_2 = 2r\).
The new power \(P_2\) will be proportional to \((r_2)^2\):
\[ \frac{P_2}{P_1} = \frac{(r_2)^2}{(r_1)^2} = \frac{(2r)^2}{r^2} \]
\[ \frac{P_2}{P} = \frac{4r^2}{r^2} = 4 \]
\[ P_2 = 4P \]
The power required increases by a factor of 4.
Step 4: Final Answer:
The power becomes 4 times its original value (\(4P\)).