Question

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each with a radius of 5.00 mm. If water flows through the single pipe at 1.45 m/s, calculate the speed (in m/s) of the water in the narrower pipes.

Answer 1

1.24 m/s

Step-by-step explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

`\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s`

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

`\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74*10^(-5) m^3/s`

So the flow speed of each of the narrower pipe is:

`v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}`

`v_n = (9.74*10^(-5))/(\pi 0.005^2) = 1.24 m/s`