Question

(II)What is the volume rate of flow of water from a 1.85-cm-diameter faucet if the pressure head is 12.0 m?

Short answer

The volume rate of flow of water from a 1.85 cm diameter faucet is \(4.12 \times {10^{ - 3}}\;{{\rm{m}}^3}{\rm{/s}}\).

Step-by-step solution

Understanding volume flow rate

In this problem, the volume rate of flow of water from faucet will be calculated by using the product of area of the faucet and velocity of the faucet.

Given data

The diameter of the faucet is \(d = 1.{\rm{85}}\;{\rm{cm}}\).

The pressure head is \(h = 12.0\;{\rm{m}}\).

Calculation of the volume flow rate

The area of faucet is calculated below:

\(\begin{array}{c}A = \frac{\pi }{4}{d^2}\\ = \frac{\pi }{4}{\left( {1.85\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)^2}\\ = 2.69 \times {10^{ - 4}}\;{{\rm{m}}^2}\end{array}\)

The velocity at faucet is calculated below:

\(\begin{array}{c}v = \sqrt {2gh} \\ = \sqrt {2\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {12.0\;{\rm{m}}} \right)} \\ = 15.34\;{\rm{m/s}}\end{array}\)

Here, g is the gravitational acceleration.

The volume rate of flow water from faucet is calculated below:

\(\begin{array}{c}\dot V = Av\\ = \left( {2.69 \times {{10}^{ - 4}}\;{{\rm{m}}^2}} \right)\left( {15.34\;{\rm{m/s}}} \right)\\ = 4.12 \times {10^{ - 3}}\;{{\rm{m}}^3}{\rm{/s}}\end{array}\)

Hence, the volume rate of flow water from faucet is \(4.12 \times {10^{ - 3}}\;{{\rm{m}}^3}{\rm{/s}}\).