- In the hose.
- 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.
Example
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:
\[n_2 = \dfrac{(1)(\pi)(10\times 10^{-3}m)^2(0.27 m/s)}{(\pi)(4.0\times 10^{-6} m)(0.33\times 10^{-3} m/s)}\]
\[n_2 = 5.0\times 10^{9}\space capillaries\]