3.3 Flow
Rate and Its Relation to Velocity
Summary:
·
Calculate flow rate.
·
Define units of volume.
·
Describe incompressible fluids.
·
Explain the consequences of the equation of continuity.
Flow rate 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 1. In symbols, this can
be written as
Q=Vt,
where V is
the volume and t is the elapsed time.
The
SI unit for flow rate is m3/s,
but a number of other units for 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 (10−3m3 or 103cm3). In this text we shall
use whatever metric units are most convenient for a given situation.
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Figure 1: 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 point P in a uniform pipe in time t. The volume of the cylinder is Ad and the average velocity is v−=d/t so that the flow rate is Q=Ad/t=Av−.
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Example 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 are given, and so the volume V can
be calculated from the definition of flow rate.
Solution
Solving Q=V/t for
volume gives
V=Qt.
Substituting known values yields
V==(5.00L1 min)(75y)(1m3103L)(5.26×105miny)2.0×105m3.
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 and velocity v− is
Q=Av−,
where A is
the cross-sectional area and 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 1 illustrates how
this relationship is obtained. The shaded cylinder has a volume
V=Ad,
which
flows past the point P in a time t. Dividing both sides of
this relationship by t gives
Vt=Adt.
We
note that Q=V/t and
the average speed is v−=d/t.
Thus the equation becomes Q=Av−.
Figure 2 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,
Q1=Q2A1v−1=A2v−2}.
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.
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Figure 2: 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.)
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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 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=Av− for v1 and note that the
cross-sectional area is A=πr2, yielding
v−1=QA1=Qπr21.
Substituting known values and making appropriate
unit conversions yields
v−1=(0.500L/s)(10−3m3/L)π(9.00×10−3m)2=1.96m/s.
Solution for (b)
We
could repeat this calculation to find the speed in the nozzle v−2, but we will use the
equation of continuity to give a somewhat different insight. Using the equation
which states
A1v−1=A2v−2,
solving
for v−2 and
substituting πr2 for the
cross-sectional area yields
v−2=A1A2v−1=πr21πr22v−1=r12r22v−1.
Substituting known values,
v−2=(0.900cm)2(0.250cm)21.96m/s=25.5 m/s.
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
n1A1v−1=n2A2v−2,
where n1 and n2 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 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 8.0μm, calculate the number
of capillaries in the blood circulatory system.
Strategy
We
can use Q=Av− 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=Av− or v−=Qπr2 for a cylindrical
vessel.
Substituting the known values (converted to
units of meters and seconds) gives
v−=(5.0L/min)(10−3m3/L)(1min/60s)π(0.010 m)2=0.27m/s.
Solution for (b)
Using n1A1v−1=n2A2v−1, assigning the
subscript 1 to the aorta and 2 to the capillaries, and solving for n2 (the number of capillaries)
gives n2=n1A1v−1A2v−2. Converting all
quantities to units of meters and seconds and substituting into the equation
above gives
n2=(1)(π)(10×10−3m)2(0.27 m/s)(π)(4.0×10−6m)2(0.33×10−3m/s)=5.0×109capillaries.
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 mm3, or about 200×106 per 1 kg of
muscle. For 20 kg of muscle, this amounts to about 4×109 capillaries.
Section Summary
·
Flow rate Q is defined to be the volume V flowing
past a point in time t, or Q=Vt where V is
volume and t is
time.
·
The SI unit of volume is m3.
·
Another common unit is the liter (L), which is 10−3m3.
·
Flow rate and velocity are related by Q=Av− where A is
the cross-sectional area of the flow and v− is its average
velocity.
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