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.

A Nozzle With A Radius

  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:

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

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

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