14.5 Fluid dynamics (Page 3/9)

University physics volume 1 Page 3 / 9

Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, so the equation must be applied with caution to gases if they are subjected to compression or expansion.

Calculating fluid speed through a nozzle

A nozzle with a diameter of 0.500 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle.

Strategy

We can use the relationship between flow rate and speed to find both speeds. We use the subscript 1 for the hose and 2 for the nozzle.

Solution

We solve the flow rate equation for speed and use $π r_{1}^{2}$ for the cross-sectional area of the hose, obtaining

$v = \frac{Q}{A} = \frac{Q}{π r_{1}^{2}} .$

Substituting values and using appropriate unit conversions yields

$v = \frac{(0.500 L/s) (10^{−3} m^{3} /L)}{3.14 {(9.00 \times 10^{−3} m)}^{2}} = 1.96 m/s .$
We could repeat this calculation to find the speed in the nozzle $v_{2}$ , but we use the equation of continuity to give a somewhat different insight. The equation states

$A_{1} v_{1} = A_{2} v_{2} .$

Solving for $v_{2}$ and substituting $π r^{2}$ for the cross-sectional area yields

$v_{2} = \frac{A_{1}}{A_{2}} v_{1} = \frac{π r_{1}^{2}}{π r_{2}^{2}} v_{1} = \frac{r_{1}^{2}}{r_{2}^{2}} v_{1} .$

Substituting known values,

$v_{2} = \frac{{(0.900 cm)}^{2}}{{(0.250 cm)}^{2}} 1.96 m/s = 25.5 m/s .$

Significance

A speed of 1.96 m/s is about right for water emerging from a hose with no nozzle. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.

The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.

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Mass conservation

The rate of flow of a fluid can also be described by the mass flow rate or mass rate of flow. This is the rate at which a mass of the fluid moves past a point. Refer once again to [link] , but this time consider the mass in the shaded volume. The mass can be determined from the density and the volume:

m = ρ V = ρ A x .

The mass flow rate is then

\frac{d m}{d t} = \frac{d}{d t} (ρ A x) = ρ A \frac{d x}{d t} = ρ A v,

where $ρ$ is the density, A is the cross-sectional area, and v is the magnitude of the velocity. The mass flow rate is an important quantity in fluid dynamics and can be used to solve many problems. Consider [link] . The pipe in the figure starts at the inlet with a cross sectional area of $A_{1}$ and constricts to an outlet with a smaller cross sectional area of $A_{2}$ . The mass of fluid entering the pipe has to be equal to the mass of fluid leaving the pipe. For this reason the velocity at the outlet $(v_{2})$ is greater than the velocity of the inlet $(v_{1})$ . Using the fact that the mass of fluid entering the pipe must be equal to the mass of fluid exiting the pipe, we can find a relationship between the velocity and the cross-sectional area by taking the rate of change of the mass in and the mass out:

\begin{matrix} {(\frac{d m}{d t})}_{1} & = & {(\frac{d m}{d t})}_{2} \\ ρ_{1} A_{1} v_{1} & = & ρ_{2} A_{2} v_{2} . \end{matrix}

[link] is also known as the continuity equation in general form. If the density of the fluid remains constant through the constriction—that is, the fluid is incompressible—then the density cancels from the continuity equation,

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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