Contents

Flow Rate and Its Relation to Velocity

Flow rate $Q Q size 12{Q} {}$ is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in Figure 12.2. In symbols, this can be written as

Q = \frac{V}{t}, Q = \frac{V}{t}, size 12{Q= {  {V}  over  {t} } } {}

12.1

where $V V size 12{V} {}$ is the volume and $t t size 12{t} {}$ is the elapsed time.

The SI unit for flow rate is $m^{3} /s m^{3} /s size 12{m rSup { size 8{3} } "/s"} {}$ , but a number of other units for $Q Q size 12{Q} {}$ 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 1000 cubic centimeters ( $10^{- 3} 10^{- 3} size 12{"10" rSup { size 8{ - 3} } `m rSup { size 8{3} } } {}$ or $10^{3} 10^{3} size 12{"10" rSup { size 8{3} } `"cm" rSup { size 8{3} } } {}$ ). In this text we shall use whatever metric units are most convenient for a given situation.

The figure shows a fluid flowing through a cylindrical pipe open at both ends. A portion of the cylindrical pipe with the fluid is shaded for a length d. The velocity of the fluid in the shaded region is shown by v toward the right. The cross sections of the shaded cylinder are marked as A. This cylinder of fluid flows past a point P on the cylindrical pipe. The velocity v is equal to d over t.

Figure 12.2 Flow rate is the volume of fluid per unit time flowing past a point through the area

A A size 12{A} {}

. Here the shaded cylinder of fluid flows past point

P P size 12{P} {}

in a uniform pipe in time

t t size 12{t} {}

. The volume of the cylinder is

Ad Ad size 12{ ital "Ad"} {}

and the average velocity is

\bar{v} = d / t \bar{v} = d / t size 12{ {overline  {v}} =d/t} {}

so that the flow rate is

Q = Ad / t = A \bar{v} Q = Ad / t = A \bar{v} size 12{Q= ital "Ad"/t=A {overline  {v}} } {}

Example 12.1

Calculating Volume from Flow Rate: The Heart Pumps a Lot of Blood in a Lifetime

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

Strategy

Time and flow rate $Q Q size 12{Q} {}$ are given, and so the volume $V V size 12{V} {}$ can be calculated from the definition of flow rate.

Solution

Solving $Q = V / t Q = V / t size 12{Q=V/t} {}$ for volume gives

V = Qt . V = Qt . size 12{V= ital "Qt"} {}

12.2

Substituting known values yields

\begin{array}{l} V & = & (\frac{5.00 L}{1 min}) (75 y) (\frac{1 m^{3}}{10^{3} L}) (5 . 26 \times 10^{5} \frac{min}{y}) \\ = & 2 . 0 \times 10^{5} m^{3} . \end{array} \begin{array}{l} V & = & (\frac{5.00}{1 min}) (75 \\ = & 2 . 0 \times 10^{5} \end{array} alignl { stack {
            size 12{V= left ( {  {5 "." "00"" L"}  over  {"1 min"} }  right ) \( "75"" y" \)  left ( {  {1" m" rSup { size 8{3} } }  over  {"10" rSup { size 8{3} } " L"} }  right ) left (5 "." "26" times "10" rSup { size 8{5} }   {  {"min"}  over  {y} }  right )}  {} #
            "    "=2 "." 0 times "10" rSup { size 8{5} } " m" rSup { size 8{3} }  {}
            } } {}

12.3

Discussion

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

Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But 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 relationship between flow rate $Q Q size 12{Q} {}$ and velocity $\bar{v} \bar{v} size 12{ {overline {v}} } {}$ is

Q = A \bar{v}, Q = A \bar{v}, size 12{Q=A {overline  {v}} } {}

12.4

where $A A size 12{A} {}$ is the cross-sectional area and $\bar{v} \bar{v} size 12{ {overline {v}} } {}$ 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 average 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 area. Figure 12.2 illustrates how this relationship is obtained. The shaded cylinder has a volume

V = Ad, V = Ad, size 12{V= ital "Ad"} {}

12.5

which flows past the point $P P size 12{P} {}$ in a time $t t size 12{t} {}$ . Dividing both sides of this relationship by $t t size 12{t} {}$ gives

\frac{V}{t} = \frac{Ad}{t} . \frac{V}{t} = \frac{Ad}{t} . size 12{ {  {V}  over  {t} } = {  { ital "Ad"}  over  {t} } } {}

12.6

We note that $Q = V / t Q = V / t size 12{Q=V/t} {}$ and the average speed is $\bar{v} = d / t \bar{v} = d / t size 12{ {overline {v}} =d/t} {}$ . Thus the equation becomes $Q = A \bar{v} Q = A \bar{v} size 12{Q=A {overline {v}} } {}$ .

Figure 12.3 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 point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2,

\begin{matrix} Q_{1} = Q_{2} \\ A_{1} {\bar{v}}_{1} = A_{2} {\bar{v}}_{2} \end{matrix}} . \begin{matrix} Q_{1} = Q_{2} \\ A_{1} {\bar{v}}_{1} = A_{2} {\bar{v}}_{2} \end{matrix}} . size 12{ left none  matrix {
Q rSub { size 8{1} } =Q rSub { size 8{2} }  {} ##
A rSub { size 8{1} }  {overline  {v rSub { size 8{1} } }} =A rSub { size 8{2} }  {overline  {v rSub { size 8{2} } }}
}  right rbrace  "." } {}

12.7

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 end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.

The figure shows a cylindrical tube broad at the left and narrow at the right. The fluid is shown to flow through the cylindrical tube toward right along the axis of the tube. A shaded area is marked on the broader cylinder on the left. A cross section is marked on it as A one. A point one is marked on this cross section. The velocity of the fluid through the shaded area on narrow tube is marked by v one as an arrow toward right. Another shaded area is marked on the narrow cylindrical on the right. The shaded area on narrow tube is longer than the one on broader tube to show that when a tube narrows, the same volume occupies a greater length. A cross section is marked on the narrow cylindrical tube as A two. A point two is marked on this cross section. The velocity of fluid through the shaded area on narrow tube is marked v two toward right. The arrow depicting v two is longer than for v one showing v two to be greater in value than v one.

Figure 12.3 When a tube narrows, the same volume occupies a greater length. For the same volume to pass 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 decrease when the tube widens. (Note 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. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion.

Example 12.2

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 water (a) in the hose and (b) in the nozzle.

Strategy

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

Solution for (a)

First, we solve $Q = A \bar{v} Q = A \bar{v} size 12{Q=A {overline {v}} } {}$ for $v_{1} v_{1} size 12{v rSub { size 8{1} } } {}$ and note that the cross-sectional area is $A = {πr}^{2} A = {πr}^{2} size 12{A=πr rSup { size 8{2} } } {}$ , yielding

{\bar{v}}_{1} = \frac{Q}{A_{1}} = \frac{Q}{{πr}_{1}^{2}} . {\bar{v}}_{1} = \frac{Q}{A_{1}} = \frac{Q}{{πr}_{1}^{2}} . size 12{ {overline  {v rSub { size 8{1} } }} = {  {Q}  over  {A rSub { size 8{1} } } } = {  {Q}  over  {πr rSub { size 8{1}   rSup { size 8{2} } } } } } {}

12.8

Substituting known values and making appropriate unit conversions yields

{\bar{v}}_{1} = \frac{(0.500 L/s) (10^{- 3} m^{3} / L)}{π (9 . 00 \times 10^{- 3} m)^{2}} = 1 . 96 m/s . {\bar{v}}_{1} = \frac{(0.500}{π (9 . 00 \times 10^{- 3}} = 1 . 96 . size 12{ {overline  {v rSub { size 8{1} } }} = {  { \( 0 "." "500"" L/s" \)  \( "10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } /L \) }  over  {π \( 9 "." "00" times "10" rSup { size 8{ - 3} } " m" \)  rSup { size 8{2} } } } =1 "." "96"" m/s"} {}

12.9

Solution for (b)

We could repeat this calculation to find the speed in the nozzle ${\bar{v}}_{2} {\bar{v}}_{2} size 12{ {overline {v rSub { size 8{2} } }} } {}$ , but we will use the equation of continuity to give a somewhat different insight. Using the equation which states

A_{1} {\bar{v}}_{1} = A_{2} {\bar{v}}_{2}, A_{1} {\bar{v}}_{1} = A_{2} {\bar{v}}_{2}, size 12{A rSub { size 8{1} }  {overline  {v rSub { size 8{1} } }} =A rSub { size 8{2} }  {overline  {v rSub { size 8{2} } }} } {}

12.10

solving for ${\bar{v}}_{2} {\bar{v}}_{2} size 12{ {overline {v rSub { size 8{2} } }} } {}$ and substituting ${πr}^{2} {πr}^{2} size 12{πr rSup { size 8{2} } } {}$ for the cross-sectional area yields

{\bar{v}}_{2} = \frac{A_{1}}{A_{2}} {\bar{v}}_{1} = \frac{{πr}_{1}^{2}}{{πr}_{2}^{2}} {\bar{v}}_{1} = \frac{r_{1^{2}}}{r_{2^{2}}} {\bar{v}}_{1} . {\bar{v}}_{2} = \frac{A_{1}}{A_{2}} {\bar{v}}_{1} = \frac{{πr}_{1}^{2}}{{πr}_{2}^{2}} {\bar{v}}_{1} = \frac{r_{1^{2}}}{r_{2^{2}}} {\bar{v}}_{1} . size 12{ {overline  {v rSub { size 8{2} } }} = {  {A rSub { size 8{1} } }  over  {A rSub { size 8{2} } } }  {overline  {v rSub { size 8{1} } }} = {  {πr rSub { size 8{1}   rSup { size 8{2} } } }  over  {πr rSub { size 8{2}   rSup { size 8{2} } } } }  {overline  {v rSub { size 8{1} } }} = {  {r rSub { size 8{1}   rSup { size 8{2} } } }  over  {r rSub { size 8{2}   rSup { size 8{2} } } } }  {overline  {v rSub { size 8{1} } }} } {}

12.11

Substituting known values,

{\bar{v}}_{2} = \frac{(0.900 cm)^{2}}{(0.250 cm)^{2}} 1.96 m/s = 25 . 5 m/s . {\bar{v}}_{2} = \frac{(0.900}{(0.250} 1.96 . 5 m/s . size 12{ {overline  {v rSub { size 8{2} } }} = {  { \( 0 "." "900"" cm" \)  rSup { size 8{2} } }  over  { \( 0 "." "250"" cm" \)  rSup { size 8{2} } } } 1 "." "96"" m/s"="25" "." "5 m/s"} {}

12.12

Discussion

A speed of 1.96 m/s is about right for water 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 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 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 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

n_{1} A_{1} {\bar{v}}_{1} = n_{2} A_{2} {\bar{v}}_{2}, n_{1} A_{1} {\bar{v}}_{1} = n_{2} A_{2} {\bar{v}}_{2}, size 12{n rSub { size 8{1} } A rSub { size 8{1} }  {overline  {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} }  {overline  {v rSub { size 8{2} } }} } {}

12.13

where $n_{1} n_{1} size 12{n rSub { size 8{1} } } {}$ and $n_{2} n_{2} size 12{n rSub { size 8{2} } } {}$ are the number of branches in each of the sections along the tube.

Example 12.3

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 body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is $8.0 μ m 8.0$ , calculate the number of capillaries in the blood circulatory system.

Strategy

We can use $Q = A \bar{v} Q = A \bar{v} size 12{Q=A {overline {v}} } {}$ to calculate the speed of flow in the aorta and then use 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 $Q = A \bar{v} Q = A \bar{v} size 12{Q=A {overline {v}} } {}$ or $\bar{v} = \frac{Q}{{πr}^{2}} \bar{v} = \frac{Q}{{πr}^{2}} size 12{ {overline {v}} = { {Q} over {πr rSup { size 8{2} } } } } {}$ for a cylindrical vessel.

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

\bar{v} = \frac{(5.0 L/min) (10^{- 3} m^{3} /L) (1 min/ 60 s)}{π {(0 . 010 m)}^{2}} = 0 . 27 m/s . \bar{v} = \frac{(5.0) (10^{- 3}) (1)}{π {(0 . 010 m)}^{2}} = 0 . 27 . size 12{ { bar  {v}}= {  { left (5 "." 0`"L/min" right ) left ("10" rSup { size 8{ - 3} } `m rSup { size 8{3} } "/L" right ) left (1`"min/""60"`s right )}  over  {π left (0 "." "010 m" right ) rSup { size 8{2} } } } =0 "." "27"`"m/s"} {}

12.14

Solution for (b)

Using $n_{1} A_{1} {\bar{v}}_{1} = n_{2} A_{2} {\bar{v}}_{1} n_{1} A_{1} {\bar{v}}_{1} = n_{2} A_{2} {\bar{v}}_{1} size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}$ , assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for $n_{2} n_{2} size 12{n rSub { size 8{2} } } {}$ (the number of capillaries) gives $n_{2} = \frac{n_{1} A_{1} {\bar{v}}_{1}}{A_{2} {\bar{v}}_{2}} n_{2} = \frac{n_{1} A_{1} {\bar{v}}_{1}}{A_{2} {\bar{v}}_{2}}$ . Converting all quantities to units of meters and seconds and substituting into the equation above gives

n_{2} = \frac{(1) (π) {(10 \times 10^{- 3} m)}^{2} (0.27 m/s)}{(π) {(4.0 \times 10^{- 6} m)}^{2} (0.33 \times 10^{- 3} m/s)} = 5.0 \times 10^{9} capillaries . n_{2} = \frac{(1) (π) {(10 \times 10^{- 3})}^{2} (0.27 m/s)}{(π) {(4.0 \times 10^{- 6})}^{2} (0.33 \times 10^{- 3})} = 5.0 \times 10^{9} . size 12{n rSub { size 8{2} } = {  { left (1 right ) left (π right ) left ("10" times "10" rSup { size 8{ - 3} } " m" right ) rSup { size 8{2} }  left (0 "." "27"" m/s" right )}  over  { left (π right ) left (4 "." 0 times "10" rSup { size 8{ - 6} } " m" right ) rSup { size 8{2} }  left (0 "." "33" times "10" rSup { size 8{ - 3} } " m/s" right )} } =5 "." 0 times "10" rSup { size 8{9} } " capillaries"} {}

12.15

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 total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per ${mm}^{3} {mm}^{3} size 12{"mm" rSup { size 8{3} } } {}$ , or about $200 \times 10^{6} 200 \times 10^{6} size 12{"200" times "10" rSup { size 8{6} } } {}$ per 1 kg of muscle. For 20 kg of muscle, this amounts to about $4 \times 10^{9} 4 \times 10^{9} size 12{4 times "10" rSup { size 8{9} } } {}$ capillaries.