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Chapter 12 Fluid Dynamics and Its Biological and Medical Applications

85 12.i Menses Rate and Its Relation to Velocity

Summary

• Summate flow rate.
• Define units of volume.
• Describe incompressible fluids.
• Explain the consequences of the equation of continuity.

Flow rate [latex]\boldsymbol{Q}[/latex]is divers to be the volume of fluid passing by some location through an area during a period of fourth dimension, as seen in Effigy ane. In symbols, this tin be written as

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

where[latex]\boldsymbol{V}[/latex]is the volume and[latex]\boldsymbol{t}[/latex]is the elapsed time.

The SI unit for menses rate is[latex]\boldsymbol{\textbf{m}^3\textbf{/s}},[/latex]simply a number of other units for[latex]\boldsymbol{Q}[/latex]are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or yard cubic centimeters ([latex]\boldsymbol{10^{-3}\textbf{ m}^3}[/latex]or[latex]\boldsymbol{x^three\textbf{ cm}^iii}[/latex]). In this text we shall employ whatever metric units are most convenient for a given state of affairs.

Figure ane. Flow rate is the volume of fluid per unit time flowing by a signal through the area A . Hither the shaded cylinder of fluid flows past point P in a uniform pipage in fourth dimension t . The volume of the cylinder is Ad and the average velocity is v̄=d/t so that the period rate is Q=Ad/t=Av̄ .

Example one: Calculating Volume from Menses Rate: The Heart Pumps a Lot of Claret in a Lifetime

How many cubic meters of claret does the center pump in a 75-year lifetime, assuming the average flow rate is five.00 L/min?

Strategy

Time and period rate[latex]\boldsymbol{Q}[/latex]are given, and so the volume[latex]\boldsymbol{V}[/latex]tin can exist calculated from the definition of flow rate.

Solution

Solving[latex]\boldsymbol{Q=5/t}[/latex]for volume gives

[latex]\boldsymbol{V=Qt.}[/latex]

Substituting known values yields

[latex]\begin{array}{lcl} \boldsymbol{V} & \boldsymbol{=} & \boldsymbol{(\frac{5.00\textbf{ Fifty}}{1\textbf{ min}})(75\textbf{ y})(\frac{ane\textbf{ thousand}^iii}{10^3\textbf{Fifty}})(5.26\times10^five\frac{\textbf{min}}{\textbf{y}})} \\ {} & \boldsymbol{=} & \boldsymbol{two.0\times10^five\textbf{ m}^3.} \end{array}[/latex]

Discussion

This corporeality is about 200,000 tons of blood. For comparing, this value is equivalent to about 200 times the volume of water contained in a 6-lane l-chiliad lap pool.

Flow rate and velocity are related, but quite unlike, physical quantities. To brand the distinction clear, call back nigh the period rate of a river. The greater the velocity of the water, the greater the menstruum rate of the river. Just flow rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. The precise human relationship betwixt catamenia rate[latex]\boldsymbol{Q}[/latex]and velocity[latex]\boldsymbol{\bar{five}}[/latex]is

[latex]\boldsymbol{Q=A\bar{5}},[/latex]

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

[latex]\boldsymbol{5=Advertizement},[/latex]

which flows past the betoken[latex]\textbf{P}[/latex]in a time[latex]\boldsymbol{t}.[/latex]Dividing both sides of this relationship by[latex]\boldsymbol{t}[/latex]gives

[latex]\boldsymbol{\frac{V}{t}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{Ad}{t}}.[/latex]

Nosotros note that[latex]\boldsymbol{Q=5/t}[/latex]and the average speed is[latex]\boldsymbol{v\bar{v}=d/t}.[/latex]Thus the equation becomes[latex]\boldsymbol{Q=A\bar{v}}.[/latex]

Figure 2 shows an incompressible fluid flowing along a pipe of decreasing radius. Considering the fluid is incompressible, the same corporeality of fluid must flow past any bespeak in the tube in a given fourth dimension to ensure continuity of catamenia. In this case, because the cantankerous-exclusive area of the pipage decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must exist the same at all points along the pipe. In particular, for points 1 and two,

[latex]\begin{assortment}{c} \boldsymbol{Q_1=Q_2} \\ \boldsymbol{A_1\bar{five}_1=A_2\bar{five}_2.} \end{array}[/latex][latex]\rbrace[/latex]

This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity 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 terminate of a reservoir, the water slows considerably, perhaps picking upward speed again when it leaves the other end of the reservoir. In other words, speed increases when cantankerous-sectional surface area decreases, and speed decreases when cross-sectional expanse increases.

Figure 2. When a tube narrows, the same book occupies a greater length. For the same volume to pass points one and 2 in a given fourth dimension, the speed must be greater at point ii. The process is exactly reversible. If the fluid flows in the opposite management, its speed volition decrease when the tube widens. (Note that the relative volumes of the two cylinders and the respective velocity vector arrows are not fatigued to scale.)

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.

Example ii: Calculating Fluid Speed: Speed Increases When a Tube Narrows

A nozzle with a radius of 0.250 cm is fastened to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 Fifty/southward. Calculate the speed of the h2o (a) in the hose and (b) in the nozzle.

Strategy

We tin can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and two for the nozzle.

Solution for (a)

First, we solve[latex]\boldsymbol{Q=A\bar{v}}[/latex]for[latex]\boldsymbol{\bar{5}_1}[/latex]and annotation that the cantankerous-sectional expanse is[latex]\boldsymbol{A=\pi{r}^2},[/latex]yielding

[latex]\boldsymbol{\bar{v}_1\:=}[/latex][latex]\boldsymbol{\frac{Q}{A_1}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{Q}{\pi{r}_1^2}}.[/latex]

Substituting known values and making appropriate unit of measurement conversions yields

[latex]\boldsymbol{\bar{5}_1\:=}[/latex][latex]\boldsymbol{\frac{(0.500\textbf{ L/s})(10^{-3}\textbf{ thou}^3\textbf{/Fifty})}{\pi(9.00\times10^{-3}\textbf{m})^2}}[/latex][latex]\boldsymbol{=\:1.96\textbf{ m/s}}.[/latex]

Solution for (b)

We could repeat this adding to find the speed in the nozzle[latex]\boldsymbol{\bar{v}_2},[/latex]but nosotros volition employ the equation of continuity to give a somewhat different insight. Using the equation which states

[latex]\boldsymbol{A_1\bar{5}_1=A_2\bar{v}_2},[/latex]

solving for[latex]\boldsymbol{\bar{five}_2}[/latex]and substituting[latex]\boldsymbol{\pi{r}^two}[/latex]for the cross-sectional area yields

[latex]\boldsymbol{\bar{v}_2\:=}[/latex][latex]\boldsymbol{\frac{A_1}{A_2}}[/latex][latex]\boldsymbol{\bar{5}_1\:=}[/latex][latex]\boldsymbol{\frac{\pi{r}_1^2}{\pi{r}_2^2}}[/latex][latex]\boldsymbol{\bar{5}_1\:=}[/latex][latex]\boldsymbol{\frac{r_1^2}{r_2^2}}[/latex][latex]\boldsymbol{\bar{v}_1}.[/latex]

Substituting known values,

[latex]\boldsymbol{\bar{v}_2\:=}[/latex][latex]\boldsymbol{\frac{(0.900\textbf{ cm})^2}{(0.250\textbf{ cm})^two}}[/latex][latex]\boldsymbol{1.96\textbf{ m/south}=25.v\textbf{ m/south}}.[/latex]

Word

A speed of i.96 thou/s is about correct for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream simply by constricting the period to a narrower tube.

The solution to the last function of the example shows that speed is inversely proportional to the foursquare of the radius of the tube, making for big effects when radius varies. Nosotros 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.

In many situations, including in the cardiovascular system, branching of the catamenia occurs. The claret is pumped from the centre into arteries that subdivide into smaller arteries (arterioles) which co-operative into very fine vessels called capillaries. In this situation, continuity of flow is maintained simply it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes

[latex]\boldsymbol{n_1A_1\bar{5}_1=n_2A_2\bar{v}_2},[/latex]

where[latex]\boldsymbol{n_1}[/latex]and[latex]\boldsymbol{n_2}[/latex]are the number of branches in each of the sections forth the tube.

Case 3: Computing Menstruum Speed and Vessel Diameter: Branching in the Cardiovascular System

The aorta is the principal blood vessel through which blood leaves the heart in order to broadcast around the body. (a) Calculate the average speed of the blood in the aorta if the menses rate is 5.0 Fifty/min. The aorta has a radius of x mm. (b) Blood likewise flows through smaller blood vessels known as capillaries. When the rate of blood menses in the aorta is v.0 Fifty/min, the speed of claret in the capillaries is about 0.33 mm/south. Given that the boilerplate diameter of a capillary is[latex]\boldsymbol{viii.0\:\mu},[/latex]calculate the number of capillaries in the claret circulatory system.

Strategy

We tin can utilise[latex]\boldsymbol{Q=A\bar{v}}[/latex]to calculate the speed of period in the aorta and so apply the 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 by[latex]\boldsymbol{Q=A\bar{5}}[/latex]or[latex]\boldsymbol{\bar{v}=\frac{Q}{\pi{r}^2}}[/latex]for a cylindrical vessel.

Substituting the known values (converted to units of meters and seconds) gives

[latex]\boldsymbol{\bar{v}\:=}[/latex][latex]\boldsymbol{\frac{(5.0\textbf{ L/min})(ten^{-3}\textbf{ m}^3\textbf{/L})(one\textbf{ min/}60\textbf{ s})}{\pi(0.010\textbf{ m})^2}}[/latex][latex]\boldsymbol{=\:0.27\textbf{ m/due south.}}[/latex]

Solution for (b)

Using[latex]\boldsymbol{n_1A_1\bar{v}_1=n_2A_2\bar{v}_1},[/latex]assigning the subscript i to the aorta and 2 to the capillaries, and solving for[latex]\boldsymbol{n_2}[/latex](the number of capillaries) gives[latex]\boldsymbol{n_2=\frac{n_1A_1\bar{v}_1}{A_2\bar{five}_2}}.[/latex]Converting all quantities to units of meters and seconds and substituting into the equation in a higher place gives

[latex]\boldsymbol{n_2\:=}[/latex][latex]\boldsymbol{\frac{(1)(\pi)(10\times10^{-three}\textbf{ 1000})^2(0.27\textbf{ yard/southward})}{(\pi)(iv.0\times10^{-six}\textbf{ grand})^2(0.33\times10^{-three}\textbf{ m/s})}}[/latex][latex]\boldsymbol{=\:5.0\times10^9\textbf{ capillaries}}.[/latex]

Word

Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the meaning increase in the full cantankerous-exclusive area at the capillaries. This depression speed is to allow sufficient time for effective commutation to occur although it is as of import for the flow not to become stationary in social club to avoid the possibility of clotting. Does this large number of capillaries in the torso seem reasonable? In active muscle, one finds about 200 capillaries per[latex]\boldsymbol{\textbf{mm}^iii},[/latex]or about[latex]\boldsymbol{200\times10^half-dozen}[/latex]per i kg of muscle. For 20 kg of muscle, this amounts to about[latex]\boldsymbol{iv\times10^9}[/latex]capillaries.

Department Summary

• Flow rate[latex]\boldsymbol{Q}[/latex]is defined to be the volume[latex]\boldsymbol{V}[/latex]flowing by a bespeak in time[latex]\boldsymbol{t},[/latex]or[latex]\boldsymbol{Q=\frac{5}{t}}[/latex]where[latex]\boldsymbol{V}[/latex]is volume and[latex]\boldsymbol{t}[/latex]is time.
• The SI unit of volume is[latex]\boldsymbol{\textbf{m}^three}.[/latex]
• Another common unit is the liter (L), which is[latex]\boldsymbol{10^{-3}\textbf{ m}^3}.[/latex]
• Flow rate and velocity are related by[latex]\boldsymbol{Q=A\bar{v}}[/latex]where[latex]\boldsymbol{A}[/latex]is the cross-exclusive area of the menstruum and[latex]\boldsymbol{\bar{5}}[/latex]is its average velocity.
• For incompressible fluids, flow charge per unit at various points is constant. That is,

  [latex]\begin{array}{c} \boldsymbol{Q_1=Q_2} \\ \boldsymbol{A_1\bar{v}_1=A_2\bar{five}_2} \\ \boldsymbol{n_1A_1\bar{v}_1=n_2A_2\bar{5}_2.} \cease{array}[/latex][latex]\rbrace[/latex]

Conceptual Questions

1: What is the divergence between period rate and fluid velocity? How are they related?

2: Many figures in the text bear witness streamlines. Explicate why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the human relationship between fluid velocity and the cross-sectional area through which it flows.)

iii: Identify some substances that are incompressible and some that are not.

Problems & Exercises

ane: What is the average menstruation rate in[latex]\boldsymbol{\textbf{cm}^3\textbf{/s}}[/latex]of gasoline to the engine of a machine traveling at 100 km/h if it averages x.0 km/Fifty?

2: The heart of a resting adult pumps blood at a charge per unit of 5.00 L/min. (a) Convert this to[latex]\boldsymbol{\textbf{cm}^iii\textbf{/s}}.[/latex](b) What is this rate in[latex]\boldsymbol{\textbf{m}^three\textbf{/south}}?[/latex]

three: Blood is pumped from the heart at a charge per unit of v.0 L/min into the aorta (of radius ane.0 cm). Determine the speed of blood through the aorta.

4: Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s. Make up one's mind the flow charge per unit and the volume that passes through the artery in a menses of 30 south.

5: The Huka Falls on the Waikato River is one of New Zealand'southward well-nigh visited natural tourist attractions (see Figure 3). On average the river has a menstruum rate of about 300,000 50/due south. At the gorge, the river narrows to 20 m wide and averages 20 grand deep. (a) What is the boilerplate 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 sixty thou and its depth increases to an average of 40 m?

Figure 3. The Huka Falls in Taupo, New Zealand, demonstrate flow charge per unit. (credit: RaviGogna, Flickr)

6: A major artery with a cantankerous-sectional surface area of[latex]\boldsymbol{1.00\textbf{ cm}^2}[/latex]branches into xviii smaller arteries, each with an boilerplate cross-sectional surface area of[latex]\boldsymbol{0.400\textbf{ cm}^two}.[/latex]By what factor is the average velocity of the blood reduced when it passes into these branches?

seven: (a) Equally claret passes through the capillary bed in an organ, the capillaries bring together to form venules (minor veins). If the blood speed increases past a factor of iv.00 and the total cantankerous-sectional area of the venules is[latex]\boldsymbol{x.0\textbf{ cm}^2},[/latex]what is the full cross-sectional area of the capillaries feeding these venules? (b) How many capillaries are involved if their boilerplate diameter is[latex]\boldsymbol{x.0\:\mu\textbf{grand}}?[/latex]

viii: The homo circulation arrangement has approximately[latex]\boldsymbol{1\times10^nine}[/latex]capillary vessels. Each vessel has a diameter of nearly[latex]\boldsymbol{eight\:\mu\textbf{m}}.[/latex]Assuming cardiac output is 5 L/min, make up one's mind the average velocity of blood flow through each capillary vessel.

9: (a) Estimate the fourth dimension information technology would take to fill a private swimming puddle with a capacity of 80,000 50 using a garden hose delivering 60 L/min. (b) How long would it take to fill up if you could divert a moderate size river, flowing at[latex]\boldsymbol{5000\textbf{ m}^three\textbf{/s}},[/latex]into information technology?

10: The menstruation rate of blood through a[latex]\boldsymbol{2.00\times10^{-6}\textbf{ -m}}[/latex]-radius capillary is[latex]\boldsymbol{3.80\times10^nine\textbf{ cm}^3\textbf{/s}}.[/latex](a) What is the speed of the blood menstruation? (This small speed allows time for diffusion of materials to and from the blood.) (b) Bold all the blood in the trunk passes through capillaries, how many of them must there be to carry a full flow of[latex]\boldsymbol{90.0\textbf{ cm}^3\textbf{/due south}}?[/latex](The large number obtained is an overestimate, simply it is even so reasonable.)

11: (a) What is the fluid speed in a burn hose with a nine.00-cm diameter carrying 80.0 Fifty of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be unlike if common salt water replaced the fresh water in the burn down hose?

12: The main uptake air duct of a forced air gas heater is 0.300 thousand in bore. What is the average speed of air in the duct if information technology carries a volume equal to that of the business firm'due south interior every 15 min? The inside book of the house is equivalent to a rectangular solid thirteen.0 m wide by 20.0 1000 long by 2.75 m high.

13: Water is moving at a velocity of ii.00 m/s through a hose with an internal diameter of 1.lx cm. (a) What is the menses charge per unit in liters per second? (b) The fluid velocity in this hose's nozzle is fifteen.0 yard/southward. What is the nozzle'south inside diameter?

xiv: Prove that the speed of an incompressible fluid through a constriction, such as in a Venturi tube, increases by a cistron equal to the square of the cistron by which the bore decreases. (The converse applies for catamenia out of a constriction into a larger-bore region.)

15: Water emerges straight down from a faucet with a 1.lxxx-cm diameter at a speed of 0.500 m/southward. (Because of the construction of the faucet, in that location is no variation in speed across the stream.) (a) What is the flow rate in[latex]\boldsymbol{\textbf{ cm}^3\textbf{/s}}?[/latex](b) What is the bore of the stream 0.200 chiliad below the faucet? Fail any effects due to surface tension.

16: Unreasonable Results

A mountain stream is ten.0 g broad and averages 2.00 m in depth. During the spring runoff, the menses in the stream reaches[latex]\boldsymbol{100,000\textbf{ m}^3\textbf{/s}}.[/latex](a) What is the average velocity of the stream under these conditions? (b) What is unreasonable nigh this velocity? (c) What is unreasonable or inconsistent about the premises?

Glossary

period charge per unit

abbreviated Q, information technology is the volume V that flows past a particular point during a fourth dimension t, or Q = V/t

liter

a unit of volume, equal to 10⁻³ mⁱⁱⁱ

Solutions

Problems & Exercises

1:

[latex]\boldsymbol{2.78\textbf{ cm}^three\textbf{/s}}[/latex]

iii:

27 cm/s

5:

(a) 0.75 thou/s

(b) 0.thirteen k/s

7:

(a)[latex]\boldsymbol{forty.0\textbf{ cm}^2}[/latex]

(b)[latex]\boldsymbol{v.09\times10^7}[/latex]

nine:

11:

(a) 12.half dozen k/s

(b)[latex]\boldsymbol{0.0800\textbf{ m}^3\textbf{/s}}[/latex]

(c) No, contained of density.

13:

(a) 0.402 Fifty/south

(b) 0.584 cm

15:

(a)[latex]\boldsymbol{127\textbf{ cm}^3\textbf{/southward}}[/latex]

(b) 0.890 cm

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Source: https://pressbooks.bccampus.ca/collegephysics/chapter/flow-rate-and-its-relation-to-velocity/

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