Question

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 8.05 mm into 4 pipes, Each with a radius of 5 mm. If water flows through the single pipe at 1.95 m/s, calculate the speed in meters per second of the water in the narrower pipes.

Answer 1

We are asked to determine the velocity of flow in a pipe. To do that we must determine first the volumetric rate of fluid in the larger pipe. We use the following equation:

$Q=Av$

Where:

$\begin{gathered} Q=\text{ flow rate} \\ A=\text{ cross-sectional area} \\ v=\text{ velocity} \end{gathered}$

Now, we plug in the values for the larger pipe. We use the area of a circle:

$Q=(\pi r_1^2)v$

Now, we substitute the values:

$Q=\pi(0.00805m)^2(1.95\frac{m}{s})$

Solving the operations:

$Q=0.000396\frac{m^3}{s}$

Since there are 4 pipes with the same radius this means that the flow in a single pipe is 1/4 of the flow of the larger pipe:

$Q_0=\frac{0.000396\frac{m^3}{s}}{4}=0.0000992\frac{m^3}{s}$

Now, to determine the velocity we use the same equation:

$Q_0=A_0v_0$

Now, we divide both sides by the area:

$\frac{Q_0}{A_0}=v_0$

Now, we plug in the values:

$\frac{0.0000992\frac{m^3}{s}}{\pi(0.005m){}^2}=v_0$

Solving the operations:

$1.26\frac{m}{s}=v_0$

Therefore, the velocity in each individual pipe is 1.26 m/s.