A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.750 cm. The flow rate through hose and nozzle is 0.0009. Calculate the speed of the water.

  1. In the hose.
  2. In the nozzle.

This problem aims to familiarize us with the relationship between flow rate and speed of a liquid from a specific cross-sectional area. The concept required to solve this problem is as mentioned, but it would be a plus if you are familiar with Bernoulli’s principle.

Now the flow rate $Q$ is described to be the volume $V$ of liquid passing through a cross-sectional area during a given specific time $t$, its equation is given by:

\[ Q = \dfrac{V}{t} \]

If the liquid is passing through a cylindrical shape, then we can represent $V$ as the product of area and unit distance i.e. $Ad$, $= \dfrac{Ad}{t}$. Where,

$\vec{v} = \dfrac{d}{t}$, so the flow rate becomes $Q = \dfrac{Ad}{t} = A \vec{v}$.

Expert Answer

Part a:

For better understanding, we are going to use subscript $1$ for the hose and $2$ for the nozzle when using the relationship between flow rate and velocity.

First, we will solve for $v_1$, and keeping in view that the cross-sectional area of a cylinder is $A = \pi r^2$, gives us:

\[ \vec{v_1} = \dfrac{Q}{A_1} \]

Substiuting $A = \pi r^2$:

\[ \vec{v_1} = \dfrac{Q}{\pi r_1^2} \]

Given the following information:

The flow rate $Q = 0.500 L/s$ and,

The radius of the hose $r_1 = 0.750 cm$.

Plugging in the values after making the appropriate unit conversions gives us:

\[\vec{v_1} = \dfrac{(0.500 L/s)(10^{-3} m^3/L)}{\pi (7.50\times 10^{-3} m)^2} \]

\[\vec{v_1} = 8.96 m/s\]

Thus, the speed of water through the hose is $8.96 m/s$.

Part b:

The radius of the nozzle $r_2 = 0.250 cm$.

For this part, we are going to use the equation of continuity to calculate $v_2$. We could have used the same approach, but this will give you a different insight. Using the equation:

\[A_1\vec{v_1} = A_2\vec{v_2}\]

Solving for $v_2$ and substituting $A = \pi r^2$ for the cross-sectional area gives us:

\[\vec{v_2} =\dfrac{A_1}{A_2}\vec{v_1}\]

\[\vec{v_2} =\dfrac{ \pi r_1^2}{ \pi r_2^2}\vec{v_1}\]

\[\vec{v_2} =\dfrac{r_1^2}{r_2^2}\vec{v_1}\]

Plugging in the given values in the above equation:

\[\vec{v_2} =\dfrac{(0.750 cm)^2}{(0.250 cm)^2} 8.96 m/s\]

\[\vec{v_2} =80.64 m/s\]

Numerical Result

A speed of about $8.96 m/s$ is required for the water to emerge from the nozzle-less hose. When the nozzle is attached, it offers a much faster stream of water by tightening the flow to a narrow tube.


The flow rate of blood is $5.0 L/min$. Calculate the average speed of the blood in aorta when it has a radius of $10 mm$. The speed of blood is about $0.33 mm/s$. The average diameter of a capillary is $8.0 \mu m$, find the number of capillaries in the circulatory system.

Part a:

The flow rate is given as $Q = A\vec{v}$, rearranging the expression for $\vec{v}$:

\[\vec{v} =\dfrac{Q}{\pi r^2}\]

Substituting the values yields:

\[\vec{v} =\dfrac{5.0\times 10^{-3} m^3/s }{\pi (0.010 m)^2}\]

\[\vec{v} =0.27 m/s\]

Part b:

Using the equation:

\[n_1A_1 \vec{v_1} = n_2A_2 \vec{v_2}\]

Solving for $n_2$ gives us:

\[n_2 = \dfrac{(1)(\pi)(10\times 10^{-3}m)^2(0.27 m/s)}{(\pi)(4.0\times 10^{-6} m)(0.33\times 10^{-3} m/s)}\]

\[n_2 = 5.0\times 10^{9}\space capillaries\]

5/5 - (15 votes)