Fluid in a Pipe the Pipe Is Narrow at Point a Widens at Point B and C and Narrows Again at D

Learning Objectives

By the end of this department, yous will be able to:

• Calculate flow rate.
• Define units of volume.
• Describe incompressible fluids.
• Explicate the consequences of the equation of continuity.

Flow rate Q is defined to exist the volume of fluid passing by some location through an area during a period of fourth dimension, equally seen in Figure 1. In symbols, this tin exist written as

[latex]Q=\frac{V}{t}\\[/latex],

where V is the book and t is the elapsed time. The SI unit of measurement for flow rate is m^three/southward, but a number of other units for Q are in common employ. For example, the middle of a resting adult pumps claret at a rate of v.00 liters per minute (L/min). Note that a liter (L) is ane/1000 of a cubic meter or 1000 cubic centimeters (10^-3 yard³ or 10³cm³). In this text we shall use whatever metric units are most convenient for a given situation.

Figure one. Flow rate is the volume of fluid per unit time flowing past a point through the area A. Here the shaded cylinder of fluid flows past bespeak P in a uniform pipage in time t. The volume of the cylinder is Ad and the average velocity is [latex]\overline{5}=d/t\\[/latex] and then that the flow rate is [latex]Q=\text{Ad}/t=A\overline{v}\\[/latex] .

Example ane. Calculating Book from Flow Rate: The Heart Pumps a Lot of Blood in a Lifetime

How many cubic meters of blood does the middle pump in a 75-year lifetime, assuming the average flow rate is 5.00 50/min?

Strategy

Time and catamenia rate Q are given, and then the volume V can be calculated from the definition of flow charge per unit.

Solution

Solving Q=V/t for volume gives

V = Qt.

Substituting known values yields

[latex]\begin{array}{lll}V& =& \left(\frac{5.00\text{ L}}{\text{1 min}}\right)\left(\text{75}\text{y}\right)\left(\frac{ane{\text{ m}}^{iii}}{{\text{ten}}^{3}\text{ L}}\right)\left(5.26\times {\text{10}}^{v}\frac{\text{min}}{\text{y}}\right)\\ \text{}& =& 2.0\times {\text{10}}^{5}{\text{m}}^{iii}\end{array}\\[/latex].

Give-and-take

This amount is virtually 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the book of water contained in a half dozen-lane l-g lap pool.

Menstruum charge per unit and velocity are related, just quite different, concrete quantities. To make the distinction clear, retrieve nearly the period charge per unit of a river. The greater the velocity of the h2o, the greater the flow rate of the river. But menstruum charge per unit likewise depends on the size of the river. A rapid mount stream carries far less water than the Amazon River in Brazil, for instance. The precise relationship between flow rate Q and velocity [latex]\bar{v}\\[/latex] is

[latex]Q=A\overline{v}\\[/latex],

where A is the cross-sectional expanse and [latex]\bar{five}\\[/latex] is the average velocity. This equation seems logical enough. The human relationship tells us that flow rate is directly proportional to both the magnitude of the boilerplate velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional expanse. Figure 1 illustrates how this relationship is obtained. The shaded cylinder has a volume

V = Ad,

which flows past the signal P in a fourth dimension t. Dividing both sides of this relationship by t gives

[latex]\frac{5}{t}=\frac{Advertizing}{t}\\[/latex].

We notation that Q=5/t and the average speed is [latex]\overline{v}=d/t\\[/latex]. Thus the equation becomes [latex]Q=A\overline{v}\\[/latex]. Figure ii shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the aforementioned amount of fluid must menses past any bespeak in the tube in a given fourth dimension to ensure continuity of flow. In this example, because the cantankerous-exclusive expanse of the pipe decreases, the velocity must necessarily increase. This logic can exist extended to say that the menstruum rate must exist the same at all points along the piping. In item, for points ane and two,

[latex]\begin{cases}Q_{1} &=& Q_{two}\\ A_{1}v_{1} &=&A_{two}v_{ii} \end{cases}\\[/latex]

This is chosen the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity tin can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the h2o slows considerably, perhaps picking up speed again when information technology leaves the other finish of the reservoir. In other words, speed increases when cross-exclusive area decreases, and speed decreases when cross-sectional area increases.

Effigy 2. When a tube narrows, the same volume occupies a greater length. For the aforementioned volume to laissez passer points 1 and 2 in a given time, the speed must be greater at point 2. The process is exactly reversible. If the fluid flows in the opposite direction, its speed will subtract when the tube widens. (Notation that the relative volumes of the 2 cylinders and the corresponding velocity vector arrows are not drawn to calibration.)

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

Instance 2. Calculating Fluid Speed: Speed Increases When a Tube Narrows

A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The period rate through hose and nozzle is 0.500 L/south. Summate the speed of the water (a) in the hose and (b) in the nozzle.

Strategy

Nosotros can apply the human relationship between flow charge per unit and speed to find both velocities. Nosotros will use the subscript 1 for the hose and 2 for the nozzle.

Solution for (a)

Showtime, nosotros solve [latex]Q=A\overline{five}\\[/latex] for v ₁ and note that the cross-sectional area is A=πr ^two, yielding

[latex]{\overline{v}}_{1}=\frac{Q}{{A}_{one}}=\frac{Q}{{{{\pi r}}_{one}}^{2}}\\[/latex].

Substituting known values and making appropriate unit conversions yields

[latex]\bar{v}_{1}=\frac{\left(0.500\text{ 50/southward}\right)\left(10^{-3}\text{ m}^{iii}\text{L}\right)}{\pi \left(9.00\times 10^{-3}\text{ grand}\right)^{two}}=1.96\text{ one thousand/due south}\\[/latex].

Solution for (b)

We could echo this adding to find the speed in the nozzle [latex]\bar{v}_{two}\\[/latex], but we will use the equation of continuity to requite a somewhat dissimilar insight. Using the equation which states

[latex]{A}_{ane}{\overline{five}}_{1}={A}_{two}{\overline{5}}_{2}\\[/latex],

solving for [latex]{\overline{v}}_{2}\\[/latex] and substituting πr ² for the cross-exclusive surface area yields

[latex]\overline{five}_{2}=\frac{{A}_{1}}{{A}_{2}}\bar{5}_{1}=\frac{{\pi r_{one}}^{ii}}{{\pi r_{2}}^{2}}\bar{v}_{1}=\frac{{r_{one}}^{2}}{{r_{two}}^{ii}}\bar{v}_{i}\\[/latex].

Substituting known values,

[latex]\overline{v}_{2}=\frac{\left(0.900\text{ cm}\right)^{2}}{\left(0.250\text{ cm}\correct)^{ii}}1.96\text{ m/s}=25.5 \text{ g/s}\\[/latex].

Discussion

A speed of 1.96 thousand/s is near correct for h2o emerging from a nozzleless hose. 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 tin blow out a candle at quite a distance, for instance, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective. In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the eye into arteries that subdivide into smaller arteries (arterioles) which co-operative into very fine vessels called capillaries. In this situation, continuity of catamenia is maintained but it is the sum of the period rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general class becomes

[latex]{n}_{one}{A}_{one}{\overline{v}}_{ane}={n}_{2}{A}_{2}{\overline{five}}_{2}\\[/latex],

where n _ane and due north ₂ are the number of branches in each of the sections forth the tube.

Instance 3. Calculating Flow Speed and Vessel Diameter: Branching in the Cardiovascular Arrangement

The aorta is the main blood vessel through which blood leaves the heart in order to circulate around the torso. (a) Calculate the average speed of the claret in the aorta if the menses rate is 5.0 L/min. The aorta has a radius of x mm. (b) Blood besides flows through smaller blood vessels known as capillaries. When the rate of blood period in the aorta is v.0 L/min, the speed of blood in the capillaries is almost 0.33 mm/due south. Given that the average diameter of a capillary is 8.0 μ grand , calculate the number of capillaries in the blood circulatory organization.

Strategy

Nosotros can use [latex]Q=A\overline{v}\\[/latex] to calculate the speed of flow in the aorta and so use the full general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.

Solution for (a)

The flow rate is given past [latex]Q=A\overline{v}\\[/latex] or [latex]\overline{5}=\frac{Q}{{\pi r}^{2}}\\[/latex] for a cylindrical vessel. Substituting the known values (converted to units of meters and seconds) gives

[latex]\overline{v}=\frac{\left(5.0\text{ Fifty/min}\correct)\left(10^{-three}{\text{ g}}^{three}\text{/L}\right)\left(1\text{ min/}60\text{s}\right)}{\pi {\left(0.010\text{ k}\right)}^{two}}=0.27\text{ thou/south}\\[/latex].

Solution for (b)

Using [latex]{n}_{1}{A}_{1}{\overline{v}}_{i}={north}_{two}{A}_{2}{\overline{v}}_{1}\\[/latex], assigning the subscript one to the aorta and 2 to the capillaries, and solving for northward _ii (the number of capillaries) gives [latex]{n}_{ii}=\frac{{n}_{1}{A}_{1}{\overline{v}}_{1}}{{A}_{2}{\overline{v}}_{2}}\\[/latex]. Converting all quantities to units of meters and seconds and substituting into the equation above gives

[latex]{n}_{2}=\frac{\left(1\correct)\left(\pi \correct){\left(\text{10}\times {\text{10}}^{-iii}\text{1000}\right)}^{2}\left(0.27 \text{ k/s}\correct)}{\left(pi \right){\left(four.0\times {\text{10}}^{-half-dozen}\text{thou}\right)}^{two}\left(0.33\times {\text{10}}^{-3}\text{m/southward}\correct)}=5.0\times {\text{10}}^{9}\text{capillaries}\\[/latex].

Discussion

Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the full cantankerous-sectional area at the capillaries. This low speed is to permit sufficient fourth dimension for constructive substitution to occur although it is as of import for the catamenia not to become stationary in order to avert the possibility of clotting. Does this large number of capillaries in the torso seem reasonable? In agile musculus, i finds about 200 capillaries per mm³, or nigh 200 × 10⁶ per one kg of muscle. For 20 kg of muscle, this amounts to about 4 × 10⁹ capillaries.

Section Summary

• Menstruation charge per unitQ is defined to exist the volume Vflowing past a point in time t, or [latex]Q=\frac{V}{t}\\[/latex] where V is volume and t is time.
• The SI unit of volume is yard³.
• Another mutual unit is the liter (L), which is 10^-3 one thousand³.
• Menses charge per unit and velocity are related by [latex]Q=A\overline{v}\\[/latex] where A is the cross-sectional area of the flow and[latex]\overline{v}\\[/latex] is its average velocity.
• For incompressible fluids, menstruum charge per unit at diverse points is constant. That is,

[latex]\begin{cases}Q_{one} &=& Q_{2}\\ A_{1}v_{1} &=&A_{two}v_{ii}\\ n_{1}A_{1}\bar{five}_{ane} &=& n_{ii}A_{2}\bar{v}_{two}\end{cases}\\[/latex].

Conceptual Questions

1. What is the deviation between menses rate and fluid velocity? How are they related?

two. Many figures in the text testify streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the human relationship between fluid velocity and the cross-exclusive area through which it flows.)

3. Place some substances that are incompressible and some that are not.

Bug & Exercises

1. What is the average period rate in cm³/s of gasoline to the engine of a auto traveling at 100 km/h if it averages ten.0 km/L?

two. The heart of a resting adult pumps claret at a charge per unit of 5.00 L/min. (a) Convert this to cm³/s . (b) What is this charge per unit in k³/s ?

3. Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of claret through the aorta.

4. Blood is flowing through an artery of radius ii mm at a rate of 40 cm/due south. Decide the menses charge per unit and the volume that passes through the avenue in a period of 30 s.

five. The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions (see Effigy 3). On boilerplate the river has a flow rate of well-nigh 300,000 Fifty/s. At the gorge, the river narrows to xx m wide and averages xx thousand deep. (a) What is the average speed of the river in the gorge? (b) What is the average speed of the water in the river downstream of the falls when it widens to 60 1000 and its depth increases to an average of twoscore 1000?

Figure 3. The Huka Falls in Taupo, New Zealand, demonstrate menses rate. (credit: RaviGogna, Flickr)

vi. A major artery with a cross-sectional area of 1.00 cm^two branches into 18 smaller arteries, each with an boilerplate cross-exclusive area of 0.400 cm². By what factor is the average velocity of the blood reduced when it passes into these branches?

7. (a) As blood passes through the capillary bed in an organ, the capillaries join to grade venules (small veins). If the blood speed increases past a gene of four.00 and the total cross-sectional surface area of the venules is 10.0 cm^two, what is the total cantankerous-exclusive expanse of the capillaries feeding these venules? (b) How many capillaries are involved if their average diameter is x.0 μ m?

eight. The human circulation organisation has approximately 1 × x^nine capillary vessels. Each vessel has a diameter of about 8 μ m. Bold cardiac output is 5 Fifty/min, determine the average velocity of claret menstruation through each capillary vessel.

9. (a) Estimate the time it would take to fill a private pond pool with a capacity of 80,000 L using a garden hose delivering lx 50/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at 5000 m³/south, into information technology?

10. The flow rate of blood through a 2.00 × 10^-6-radius capillary is 3.80 × 10⁹. (a) What is the speed of the blood flow? (This small speed allows time for improvidence of materials to and from the blood.) (b) Bold all the claret in the body passes through capillaries, how many of them must there exist to carry a total flow of ninety.0 cm^three/s? (The large number obtained is an overestimate, but it is notwithstanding reasonable.)

11. (a) What is the fluid speed in a burn hose with a ix.00-cm diameter carrying 80.0 50 of water per second? (b) What is the menstruum rate in cubic meters per second? (c) Would your answers be different if salt h2o replaced the fresh water in the fire hose?

12. The main uptake air duct of a forced air gas heater is 0.300 m in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the house's interior every 15 min? The inside volume of the house is equivalent to a rectangular solid 13.0 m wide by 20.0 m long past 2.75 m loftier.

13. H2o is moving at a velocity of 2.00 m/s through a hose with an internal diameter of 1.sixty cm. (a) What is the flow rate in liters per second? (b) The fluid velocity in this hose's nozzle is 15.0 m/southward. What is the nozzle's inside diameter?

fourteen. Bear witness that the speed of an incompressible fluid through a constriction, such as in a Venturi tube, increases by a factor equal to the foursquare of the factor by which the bore decreases. (The converse applies for menstruation out of a constriction into a larger-diameter region.)

fifteen. H2o emerges straight down from a faucet with a one.80-cm diameter at a speed of 0.500 g/s. (Because of the construction of the faucet, there is no variation in speed across the stream.) (a) What is the flow rate in cm³/s? (b) What is the diameter of the stream 0.200 yard below the faucet? Neglect whatever effects due to surface tension.

16. Unreasonable ResultsA mountain stream is 10.0 k wide and averages 2.00 m in depth. During the leap runoff, the flow in the stream reaches 100,000 grand³/s. (a) What is the boilerplate velocity of the stream under these conditions? (b) What is unreasonable about this velocity? (c) What is unreasonable or inconsistent about the premises?

Glossary

flow rate:

abbreviated Q, it is the volume Five that flows past a particular indicate during a time t, or Q = V/t

liter:

a unit of volume, equal to 10⁻³ m³

Selected Solutions to Bug & Exercises

1. two.78 cm^three/s

3. 27 cm/s

5. (a) 0.75 m/due south (b) 0.13 m/s

vii. (a) 40.0 cm² (b) 5 . 09 × 10⁷

ix. (a) 22 h (b) 0.016 s

11. (a) 12.6 m/south (b) 0.0800 m³/s (c) No, independent of density.

xiii. (a) 0.402 L/s (b) 0.584 cm

15. (a) 128 cm³/s (b) 0.890 cm

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Source: https://courses.lumenlearning.com/physics/chapter/12-1-flow-rate-and-its-relation-to-velocity/

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