If the Syringe Volume Has Been Incorrectly Read Giving a Calculated System Volume

Learning Objectives

Past the end of this department, y'all volition exist able to:

• Calculate flow charge per unit.
• Ascertain units of volume.
• Describe incompressible fluids.
• Explain the consequences of the equation of continuity.

Period charge per unit Q is defined to be the book of fluid passing past some location through an area during a flow of time, as seen in Figure 1. In symbols, this tin can exist written as

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

where 5 is the volume and t is the elapsed fourth dimension. The SI unit for catamenia rate is grand³/southward, but a number of other units for Q are in common employ. For example, the eye of a resting adult pumps blood at a charge per unit of five.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters (ten^-3 m³ or 10³cmⁱⁱⁱ). In this text we shall use whatever metric units are most convenient for a given situation.

Figure i. Flow rate is the book of fluid per unit fourth dimension flowing past a signal through the area A. Here the shaded cylinder of fluid flows past point P in a uniform pipe in time t. The volume of the cylinder is Advertisement and the average velocity is [latex]\overline{v}=d/t\\[/latex] so that the flow rate is [latex]Q=\text{Advertizement}/t=A\overline{v}\\[/latex] .

Instance 1. Calculating Volume from Flow Rate: The Heart Pumps a Lot of Claret in a Lifetime

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

Strategy

Fourth dimension and flow rate Q are given, so the book V tin be calculated from the definition of menstruum charge per unit.

Solution

Solving Q=Five/t for volume gives

5 = Qt.

Substituting known values yields

[latex]\brainstorm{assortment}{lll}V& =& \left(\frac{5.00\text{ Fifty}}{\text{1 min}}\right)\left(\text{75}\text{y}\right)\left(\frac{i{\text{ thousand}}^{3}}{{\text{ten}}^{3}\text{ L}}\right)\left(5.26\times {\text{10}}^{v}\frac{\text{min}}{\text{y}}\correct)\\ \text{}& =& 2.0\times {\text{10}}^{5}{\text{chiliad}}^{3}\stop{array}\\[/latex].

Discussion

This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of h2o contained in a 6-lane 50-m lap pool.

Period rate and velocity are related, but quite unlike, concrete quantities. To make the distinction articulate, call back near the flow charge per unit of a river. The greater the velocity of the water, the greater the period rate of the river. Only period rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for case. The precise human relationship between menstruation rate Q and velocity [latex]\bar{five}\\[/latex] is

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

where A is the cross-sectional area and [latex]\bar{5}\\[/latex] is the average velocity. This equation seems logical enough. The 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 cantankerous-sectional area. Figure 1 illustrates how this relationship is obtained. The shaded cylinder has a volume

V = Ad,

which flows past the bespeak P in a time t. Dividing both sides of this relationship by t gives

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

We note that Q=5/t and the average speed is [latex]\overline{v}=d/t\\[/latex]. Thus the equation becomes [latex]Q=A\overline{5}\\[/latex]. Figure two shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any signal in the tube in a given fourth dimension to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic tin can be extended to say that the flow rate must be the same at all points along the piping. In particular, for points 1 and ii,

[latex]\brainstorm{cases}Q_{i} &=& Q_{two}\\ A_{i}v_{1} &=&A_{ii}v_{2} \cease{cases}\\[/latex]

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

Figure 2. When a tube narrows, the aforementioned volume occupies a greater length. For the same book to laissez passer points ane and two in a given time, the speed must exist greater at point 2. The process is exactly reversible. If the fluid flows in the contrary management, its speed will subtract when the tube widens. (Notation that the relative volumes of the two cylinders and the corresponding velocity vector arrows are not drawn to scale.)

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

Example 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 flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the h2o (a) in the hose and (b) in the nozzle.

Strategy

We tin can use the relationship between flow charge per unit and speed to find both velocities. We will use the subscript one for the hose and 2 for the nozzle.

Solution for (a)

First, we solve [latex]Q=A\overline{v}\\[/latex] for five ₁ and annotation that the cross-sectional area is A=πr ², yielding

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

Substituting known values and making appropriate unit conversions yields

[latex]\bar{v}_{i}=\frac{\left(0.500\text{ L/s}\correct)\left(10^{-3}\text{ thou}^{iii}\text{L}\right)}{\pi \left(9.00\times 10^{-iii}\text{ chiliad}\right)^{two}}=one.96\text{ m/s}\\[/latex].

Solution for (b)

We could echo this calculation to discover the speed in the nozzle [latex]\bar{5}_{2}\\[/latex], only we will use the equation of continuity to give a somewhat different insight. Using the equation which states

[latex]{A}_{one}{\overline{v}}_{1}={A}_{2}{\overline{v}}_{2}\\[/latex],

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

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

Substituting known values,

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

Word

A speed of 1.96 thousand/s is about right for h2o emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the menses to a narrower tube.

The solution to the last function of the case shows that speed is inversely proportional to the foursquare of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for instance, past 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 heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but information technology 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]{northward}_{ane}{A}_{i}{\overline{v}}_{ane}={n}_{2}{A}_{2}{\overline{v}}_{2}\\[/latex],

where n _one and n ₂ are the number of branches in each of the sections along the tube.

Example 3. Calculating Flow 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 circulate around the torso. (a) Calculate the average speed of the blood in the aorta if the catamenia rate is v.0 L/min. The aorta has a radius of x mm. (b) Blood likewise flows through smaller blood vessels known every bit capillaries. When the charge per unit of blood catamenia in the aorta is five.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average bore of a capillary is 8.0 μ m , calculate the number of capillaries in the claret circulatory system.

Strategy

We can utilize [latex]Q=A\overline{5}\\[/latex] to calculate the speed of flow in the aorta then use the general form of the equation of continuity to summate the number of capillaries as all of the other variables are known.

Solution for (a)

The flow charge per unit is given by [latex]Q=A\overline{v}\\[/latex] or [latex]\overline{v}=\frac{Q}{{\pi r}^{two}}\\[/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{ L/min}\right)\left(10^{-three}{\text{ m}}^{3}\text{/L}\right)\left(1\text{ min/}60\text{s}\right)}{\pi {\left(0.010\text{ k}\correct)}^{ii}}=0.27\text{ m/s}\\[/latex].

Solution for (b)

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

[latex]{n}_{2}=\frac{\left(ane\right)\left(\pi \correct){\left(\text{10}\times {\text{10}}^{-3}\text{m}\right)}^{2}\left(0.27 \text{ m/southward}\correct)}{\left(pi \right){\left(4.0\times {\text{10}}^{-6}\text{m}\correct)}^{2}\left(0.33\times {\text{10}}^{-iii}\text{thousand/s}\correct)}=5.0\times {\text{10}}^{nine}\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 increment in the total cantankerous-sectional area at the capillaries. This depression speed is to allow sufficient time for effective exchange to occur although it is equally important for the menstruation not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the trunk seem reasonable? In agile muscle, ane finds most 200 capillaries per mm³, or nearly 200 × 10^{half dozen} per 1 kg of muscle. For 20 kg of musculus, this amounts to near 4 × 10⁹ capillaries.

Section Summary

• Menses charge per unitQ is defined to be the volume Vflowing past a point in fourth dimension t, or [latex]Q=\frac{V}{t}\\[/latex] where V is volume and t is time.
• The SI unit of book is g³.
• Some other common unit of measurement is the liter (L), which is x^-3 m³.
• Flow rate and velocity are related by [latex]Q=A\overline{five}\\[/latex] where A is the cross-sectional expanse of the menstruum and[latex]\overline{v}\\[/latex] is its average velocity.
• For incompressible fluids, menstruum rate at various points is constant. That is,

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

Conceptual Questions

ane. What is the departure between menstruum rate and fluid velocity? How are they related?

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

iii. Identify some substances that are incompressible and some that are non.

Issues & Exercises

i. What is the boilerplate flow charge per unit in cm^three/s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L?

two. The heart of a resting developed pumps claret at a rate of 5.00 50/min. (a) Convert this to cmⁱⁱⁱ/s . (b) What is this rate in m³/s ?

iii. Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius ane.0 cm). Determine the speed of claret through the aorta.

4. Blood is flowing through an artery of radius 2 mm at a rate of xl cm/s. Determine the flow rate and the volume that passes through the artery in a period of thirty s.

v. The Huka Falls on the Waikato River is i of New Zealand's most visited natural tourist attractions (see Figure 3). On average the river has a flow rate of well-nigh 300,000 50/s. At the gorge, the river narrows to 20 m wide and averages 20 m 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 information technology widens to 60 thousand and its depth increases to an average of 40 thousand?

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

6. A major artery with a cross-exclusive area of 1.00 cm² branches into eighteen smaller arteries, each with an average cross-exclusive area of 0.400 cm². By what factor is the average velocity of the blood reduced when information technology passes into these branches?

7. (a) As blood passes through the capillary bed in an organ, the capillaries join to form venules (modest veins). If the blood speed increases by a cistron of iv.00 and the total cantankerous-exclusive area of the venules is 10.0 cm², what is the total cross-sectional area of the capillaries feeding these venules? (b) How many capillaries are involved if their boilerplate diameter is x.0 μ m?

8. The human apportionment system has approximately ane × x⁹ capillary vessels. Each vessel has a diameter of about viii μ thousand. Assuming cardiac output is 5 L/min, determine the average velocity of blood flow through each capillary vessel.

9. (a) Estimate the fourth dimension it would accept to fill a private pond puddle with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at 5000 chiliad³/south, into information technology?

10. The menstruation charge per unit of blood through a ii.00 × 10^-6-radius capillary is 3.80 × 10⁹. (a) What is the speed of the blood period? (This small speed allows fourth dimension for diffusion of materials to and from the blood.) (b) Assuming all the blood in the trunk passes through capillaries, how many of them must there exist to bear a total flow of 90.0 cm³/southward? (The big number obtained is an overestimate, but it is however reasonable.)

11. (a) What is the fluid speed in a burn hose with a ix.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per 2nd? (c) Would your answers exist different if table salt water replaced the fresh water in the fire hose?

12. The principal uptake air duct of a forced air gas heater is 0.300 one thousand in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the firm'southward interior every fifteen min? The within book of the house is equivalent to a rectangular solid 13.0 thousand wide by 20.0 thousand long by ii.75 m high.

xiii. Water is moving at a velocity of 2.00 grand/s through a hose with an internal diameter of one.60 cm. (a) What is the menstruation rate in liters per 2nd? (b) The fluid velocity in this hose's nozzle is 15.0 grand/s. What is the nozzle'southward inside diameter?

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

15. Water emerges straight downwards from a faucet with a one.80-cm diameter at a speed of 0.500 m/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/southward? (b) What is the diameter of the stream 0.200 m below the faucet? Neglect any effects due to surface tension.

sixteen. Unreasonable ResultsA mountain stream is 10.0 m broad and averages 2.00 chiliad in depth. During the spring runoff, the flow in the stream reaches 100,000 m³/s. (a) What is the average velocity of the stream under these weather? (b) What is unreasonable about this velocity? (c) What is unreasonable or inconsistent nigh the premises?

Glossary

flow rate:

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

liter:

a unit of measurement of volume, equal to x⁻ⁱⁱⁱ m³

Selected Solutions to Problems & Exercises

1. ii.78 cm³/southward

3. 27 cm/s

v. (a) 0.75 yard/south (b) 0.13 k/s

7. (a) xl.0 cm² (b) 5 . 09 × 10⁷

nine. (a) 22 h (b) 0.016 s

eleven. (a) 12.6 thou/due south (b) 0.0800 chiliad^three/southward (c) No, independent of density.

xiii. (a) 0.402 L/southward (b) 0.584 cm

15. (a) 128 cm³/due south (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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