Download Chapter 3 Fluid dynamics

Chapter 3. Fluid Dynamics
Chapter Three
Fluid Dynamics
(流体动力学)
Fluid is the general name for gases and liquids. It has no definite size and shape unless it
is hold in a container.
The difference between solid and fluid is that the molecules (or atoms) in liquids can
move in a large area or even from one place to another and those in solids can only vibrate in
a very small range.
The distinction between gases and liquids is that they differ to a great degree in the
compressibilities ( 可 压 缩 性 ). Gases may be compressed with ease, while liquids are
practically incompressible.
§3.1 Density
You have little trouble walking upstream against a 5 km/h current of air (气流). Fighting the
same flow waist-deep of water requires considerable effort, while opposing a similar current
in a mudflow （ 泥 浆 流 ） is beyond your capacity. The distinction is largely due to
DENSITY (密度).
The density  of a homogeneous (均匀的）substance is defined as the mass of the
material per unit volume
m

(3.1)
V
where m is the mass of the object and V is its volume (体积). Units of density are one
kilogram per cubic meter (1 kg m-3).
§3.2 Hydrostatics (流体静力学)
Hydrostatics is the study of fluids at rest. Liquids flow under the action of unbalanced
forces, so that if an amount of liquid is stationary, or static, the net force acting on the amount
is zero. That is why it is named as hydrostatics.
3.2.1 Pressure and Pascal’s principle
Pressure （压强） is defined as the magnitude of the force per unit area. It is found that the
hydrostatic pressure P at some point in liquid is
F
P   gh
(3.2)
A
 is the density of fluid, h is the height from the top of liquid, and g is the gravitational
acceleration.
Derivation of this formula needs to consider a cylindrical volume of liquid (see Fig. 3.1),
its weight (mg) will exert on the bottom surface of the imaginary cylinder.
F  mg  Vg   (hA) g
Therefore
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F
 gh
A
On the other hand, a container at the earth’s surface
has the pressure of atmosphere. So
(3.3)
Pabs  P0  gh
P
P0
A
AAA
Pascal’s principle:
h
When a change in pressure is applied to an enclosed
fluid, the change is transmitted undiminishedly to
every point in the fluid and to walls of the container.
F2
F1
P1
P2
F
Fig. 3.1
Fig. 3.2 Hydraulic pressure
According to Pascal’s principle, the pressure P1 at one end should be equal to P2 at
another end shown in Fig. 3.2. So we have
F1 F2
A

 F2  2 F1
A1 A2
A1
3.2.2 Buoyancy (浮力) and Archimede’s principle
Buoyancy is a familiar phenomenon; a body immersed in water seems to have less weight
than when immersed in air, a body whose average density is less than that of the fluid in
which it is immersed can float in that liquid. Examples are the human body in water, or a
helium-filled balloon in air.
Archimede’s principle states that when a body is immersed in a fluid, the fluid exerts an
upward force on the body equal to the weight of the fluid that is displaced by the body. Or in
other word, we can say that a body immersed in a fluid is buoyed up by a force equal to the
weight of fluid which the body has displaced. The buoyancy can be worked out by
FB  mg  Vg
where V is the volume of fluid displaced by the body,  is the density of the fluid and g is the
gravitational acceleration. (A story describes about how Archimede to get his principle. It is
said that the king in his country asked him to find a method to distinguish whether his crown
is made of pure gold or not. He worked out this problem while he was taking his bath. He was
so excited that he did not realize that he was naked when he ran on the street to tell the good
news to his king.)
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Chapter 3. Fluid Dynamics
§3.3 Hydrodynamics (流体动力学)
vA
vc
C
3.3.1 Steady flow and ideal fluid
B
1. Steady flow
• Streamline (流线): In the motion of fluid,
the speed of particles in the fluid may be
vB
different. At any given moment, we can
draw some lines. The tangent direction of
every point on the line has the same direction
and speed. the lines are called
streamlines (see Fig. 3.3) (It is also
Fig. 3.3 at different point on the streamline,
defined that a curve whose tangent,
the velocity does not change at any time. The
at any point, is in the direction
shape of streamline doesn’t change. vA … do
of the fluid velocity at that point).
not change.
A
• Steady flow: If the speed of every point on the streamline does not change at different
moment, this kind of fluid is called steady flow or streamline flow.
• No streamlines can be crossed!
2. Ideal fluid
In order to make the question simpler, we use an ideal model to replace the real fluid. It
is called ideal fluid. The ideal fluid has the following properties:
• non-viscous (no internal friction)
• incompressible
• Moving in a streamline motion
3.3.2 Continuity equation
• Now we need a method for calculating the velocity change that occurs when a fluid
flows in a pipe of variable cross section.
• Consider an ideal fluid in streamline flow in a pipe of variable cross-section (横截面).
•As the pipe has different cross-sections in different places, do the different places
have the same flow speed?
•Consider the ideal fluid flowing in a pipe (see figure 3.4)
• Suppose that the magnitude of velocity at all points on the cross-section A has the
same speed v
A1  v1, A2  v2, …, An  vn
• During a time interval Dt, the amount of fluid flowing through any cross-section
should be exactly the same because the ideal fluid is incompressible.
The amount of fluid = density of fluid × volume in t
= × Ai (v t)
•Choosing two cross-sections, the amount of fluid through A1 should be equal to that
through A2. So we have
1 A1 (v1 t) = 2 A2 (v2 t)
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1 = 2 for the same and incompressible fluid. This result gives A1 v1 = A2 v2. Since A1 and
A2 are any two cross-sections, this means
A v = constant
This is called continuity equation. From this equation, we have this result: the velocity of
fluid is inversely proportional to the cross-sectional area: (v2 = A1v1/A2)
A
v2  1 v1
A2
Thus the fluid moves faster at narrow places and slower at wider places. Av is called volume
flow rate as it is the volume of liquid per unit time flowing through.
blood is a reasonably incompressible fluid that moves through the blood vessels in a
streamline flow. However, blood is decidedly viscous and this property causes a velocity that
is high in the center in the artery but less towards the walls. In this case, we must interpret the
velocity in the continuity equation as average velocity.
Example 3.1 in a normal resting result, the heart pumps blood into the aorta (主动脉) at an
average volume flow rate (Av) of about 9.0×10-5 m3/s. Calculate the average speed of the
blood in an aorta of inside diameter 1.40 cm
Solution: known A1v1= 9.00×10-5 m3/s
A2 = r2 = 3.14 (7.0 ×10-3)2 ∴v2 = A1v1/A2 = 0.58 m/s
3.3.3 Work-energy and Bernoulli’s equation
The work-energy principle may be applied to the ideal fluid flowing in a pipe. (Fig. 3.4)
Consider the flow for a time interval t. In point 1, the pressure, cross-section and speed are
assumed to be P1, A1, and v1 respectively and in point 2, P2 , A2 and v2.
∵ P = F/A
∴ F1= P1A1, F2 = P2A2
The work done by F1 and F2 in the time period is given below:
W1 = F1· (v1t) = P1A1 v1t = P1 V1
W2 = －F2· (v2t) = － P2A2 v2t = －P2 V2
Where W2 is negative because the liquid moves in a direction opposite to the applied force.
v2 t
F1=P1A1
F2
v1 t
h2
h1
Fig. 3.4 derivation of Bernoulli’s equation
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Chapter 3. Fluid Dynamics
As the ideal fluid is incompressible, the amount of fluid running through A1 must go through
A2 as well. Therefore, V1 = V2 = V, so the total work done by F1 and F2 is
W = W1 + W2 = (P1－P2) V
Since the mass of the liquid in V should be  V, the change in kinetic energy is
KE =½ m v22 －½ m v12 = ½  V(v22 － v12)
The change in potential energy is
PE = m g h2 － m g h1
= rV g (h2 －h1)
The total change of the kinetic and potential energy should be equal to the work done by F1
and F2
∴
W = KE + PE
Substituting W, KE and PE into the above equation, we have
(P1－P2) V = ½  V(v22 － v12) +  V g (h2 －h1)P1 + ½  v12 +  g h1
= P2 + ½  v22 +  g h2
(3.8)
As point 1 and point two are chosen freely, we have
P + ½ v2 +  g h = constant
(3.9)
This is known as Bernoulli’s equation. Now we have a look at two special cases:
1. Hydrostatics (v1 = v2 = 0)
It is interesting to note that Equation (3.3) is a special case of (3.8) when the speed of
fluid is zero. In this case, the equation (3.8) becomes
P 1 +  g h 1 = P 2 +  g h2
(3.9a)
as the point 1 and 2 are chosen freely, so
P +  g h =constant
(3.10)
from (3.9a)
P1 = P2 +  g (h2 －h1)
comparing it with (3.3), the P2 should be the atmospheric pressure P0 and (h2 －h1) should be
h.
2. Hydrodynamics (h1 = h2 = 0)
In many cases, fluid flows in a horizontal pipe, that means h1 = h2. Bernoulli’s equation
becomes
P1 + ½  v12 = P2 + ½  v22
So we have
P + ½  v2 = constant
(3.11)
This explains that when fluid flows in a horizontal pipe, bigger pressure occurs in low
velocity position and small pressure happens in high velocity position.
3.3.4 Applications of Bernoulli’s equation
1. Kinemometer (流速计)
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Pitot tube’s principle
In Fig.3.6, the tube t1 is a straight tube, t2 is a bend tube. At the two tube’s ends (c & d), they
are in the same level, i. e. hc = hd. So Bernoulli’s equation becomes
Pc + ½  vc2 = Pd + ½  vd2
As d is the dead end of tube t2, vd= 0. So
t1
t2
e
½v
h
2
a
d
C
Fig. 3.5 Principle of kinemometer
Pd = Pc + ½ v2
(3.11a)
where v = vc .Therefore, dynamic pressure
can be changed to static pressure in a dead
end. Of course, if we know Pc & Pd, the speed
v can be worked out. Since Pe = Pa = P0,
using the formula (3.3) for static pressure,
we have
Pb = Pe +  g h
From the fig. 3.6, we know that
Pd － Pc = Pb － Pe =  g h
∴ ½  v2 =  g h
∴ v = (2 g h)½
So the speed of fluid can be calculated by
measured h.
v
c
d
h
Fig. 3.6 The Pitot tube
2. The Pitot tube
At the position d in Fig 3.6, the fluid velocity is zero since a dead area arises at the inlet to the
tube. At the position c, the fluid velocity is the full stream velocity v. So using the equation
(3.11a) , we have
1
Pd  Pc  v 2
2

is the density of the fluid. Because of the additional pressure, the surfaces of the fluid
in the U-tube with density  are different and with a height difference h. But in such a
condition, we should consider the weight of the fluid in flow. Therefore the additional
pressure caused in this case is
Pd  Pc  (     ) gh
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Chapter 3. Fluid Dynamics
So we could calculate the velocity of the fluid by knowing the height h,
2(     ) gh
v

(3.12)
3.3.5 The flow of viscous fluid.
The real fluid has internal friction and it is viscous. It is different from the ideal fluid. It
wastes energy when it flows since internal friction does work in opposite direction. So
Bernoulli’s equation can not explain the law of viscous fluid. Some new concept are required
to be put forward.
1. Laminar flow (层流)
Put some transparent glycerol in a tube and then put in some colored glycerol. Let them flow.
From the colored glycerol’s shape change, we can see the flow velocity of glycerol is
different in different places.
•The speed is smaller near the wall of the tube and flow velocity is the biggest in the center of
the tube. This means that viscous fluid flows in different layers. We call it Laminar Flow.
• when fluid flows in laminar flow, there is internal friction or viscous force between two
close layers.•the internal friction is caused by the interaction forces between molecules. The
internal friction is bigger in liquids than in gases.
x
2. Viscosity
In laminar flow, the fluids
are imagined as being
made up of very thin
liquid layers stacked (堆叠)
parallel to the surface
across which flow occurs.
x ---- the distance changed
between two layers
v --- the velocity changed
between two layers
The velocity change per
unit length is the slope the
function v(x) at point x. it
is also called velocity gradient
in x-direction, expressed as
v
x+x
v+v
Fig. 3.7 The laminar flow
v+ v
v
v
x
x
v dv
slope  lim

x 0 x
dx
x+ x
Fig. 3.8 Velocity varies as x
Experiments show that on one hand if the velocity gradient is large, the frictional force F is
large as well; on the other hand, the greater the area of connected two layers, the larger the
frictional force. Therefore, the friction force F is proportional to the area of the connected two
layers and velocity gradient.
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F =  A dv/dx
(3.12)
This equation is called Newton’s law of viscosity. Where  is the coefficient of viscosity and
determined by the nature and temperature of the fluids. It is a measure (量度) of the frictional
force in liquids, hence, their resistance (阻力) to flow.
3. Turbulent flow (湍流) and Reynold number
When the velocity of fluid is bigger than some value, the flow is neither steady flow nor
laminar flow and it is very irregular. The particles of fluid in out-layer is involved into the
inner layer, moving in little eddies (旋涡). This flow is called turbulent flow.
The turbulent flow is not only caused by the large velocity, but also density  of fluid, the
coefficient  of viscosity and the radius of a tube. A parameter (参数), called Reynold
number, is given by Reynold to determine whether the fluid is in laminar flow, turbulent flow
or unsteady flow.
Re =  v r / 
(3.13)
Re is called Reynold number. Experiments show that:
Re ﹤ 1000
laminar flow
1000 ﹤ Re ﹤ 1500
unsteady flow
Re ﹥ 1500
turbulent flow
In biological transport system, Re is too small to create turbulent flow, but if a blood vessel
(脉管) is bent, turbulent flow will occur when Re is small. So in all bent and branches
turbulent flow often occurs, such as in the heart, arteries and bronchi in the human body.
3.3.6 Poiseuille’s law and stoke’s law
1. Poiseuille’s law
When a viscous fluid flows in a tube,
the flow velocity is different at different
points of a cross-section (横截面). The
outmost layer of fluid clings to (粘附) the
l
walls of the tube and its velocity is zero.
P
P2
The tube’s walls exert a backward drag (拽力) 1
on this layer and so on. Fig. 3.9 shows that
if the real flow flows from left to right in a
horizontal tube, the pressure will be higher
Fig. 3.9 the flow of a real fluid
at the left hand side of the tube and lower at
the other end. If the velocity of the fluid is not too high, the flow is still laminar flow, with a
velocity that is greatest at the center of the tube and decrease to zero at the walls.
Let us consider the variation of velocity with radius foe a cylindrical pipe of inner radius
R. an imaginary liquid cylinder coaxial with the pipe, of radius r, and length L, shown in Fig.
3.9 for the smaller cylinder. The force on the left end is P1r2, and that on the right end P2r2,
the net force on the imaginary tube is thus
F  ( P1  P2 )r 2
Since the fluid in the pipe does not accelerate, this force must balance the viscous retarding
force at the surface of this cylindrical fluid. This force is given by Newton’s law of viscosity
(3.12), but since the velocity varies with the distance r from the center, the velocity gradient
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Chapter 3. Fluid Dynamics
should be dv/dr. Note that the area A of the cylinder relating to the retarding force is 2rl.
Thus the viscous force is
dv
F  2rl
dr
Equating this to the net force due to pressure on the ends and rearranging, we find that
dv
( P  P2 )r
 1
.
dr
2l
This shows that the velocity changes more and more rapidly as we go from the center (r = 0)
to the pipe wall (r = r0). The negative sign must be introduced because v decreases as r
increases. Integrating on both sides, we get
r
0
P1  P2 0
  dv 
rdr
2l r
v
The velocity at the position with the distance r from the center of the tube can be calculated
and is given by
P  P2 2

v 1
r0  r 2 
(3.14)
4l
where r0 is the radius of the tube and r is distance from the center of the tube to a point in the
tube and l is the length of the tube. From this formula, we could find that the maximum
velocity is located at the center of the pipe and its value is proportional to the square the pipe
radius and is also proportional to the pressure change per unit length (P1-P2)/l, called the
pressure gradient. Thus the velocity decreases from a maximum value at the center to zero at
the wall of the pipe. On the other hand, when the length of the pipe is longer enough and the
different pressure between the two ends of the pipe is not too large, the velocity at the far end
could be very small.
In order to make the flow speed constant in the tube, a force to counteract （抵消，中
和）the internal friction is needed and this comes from the different pressure at the two inlets
of the tube.
Considering the thin-walled element of the water cylinder
in the pipe in Fig. 3.10. The volume of fluid dQ crossing the
dr
r
end of this element in a time t is v dA t, where v is the
velocity at the radius r and dA is the white area, equal to
2rdr. Taking the expression for v from Eq.(3.14),
we obtain
Fig. 3.10 cross-section of the pipe
P1  P2 2
(r0  r 2 )2rdrt
4l
The volume flowing across the entire section is obtained by integrating over all elements from
the center to the wall of the pipe. So on the left hand side, the volume is from dQ to Q for a
time dt and the radius is from 0 to r0, and we have
r
r04 ( P1  P2 )
P  P2 0 2
2
Q  1
(
r

r
)
2

rdr

t

t
0
4l 0
8l
dQ 
The total volume of flow per unit time, Q/t, called volume flow rate, denoted by Q, is
given by
r04 P P
,
(3.15)
Q

8L
R
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where p = P1 – P2 is the difference of the pressures at the two inlets of the pipe and R =
8l/(p r04). Equation (3.15) is called Poiseuille’s law. It is similar to the Ohm’s law. So R is
called flow resistance （流阻）. The volume flow rate is inversely proportional to the flow
resistance as we expected, proportional to the pressure gradient along the pipe and it varies as
the fourth power of the radius. For example, if the radius is halved, the flow rate is reduced by
a factor of 16. This relation is familiar to physicians in connection with the selection of
needles for hypodermic syringes (皮下注射管). Needle size is much more important than
thumb pressure in determining the flow rate from the needle; doubling the needle diameter
has the same effect as increasing the thumb force sixteen-fold. Similarly, blood flow in
arteries and veins can controlled over a wide range by relatively small changes in diameter, an
important temperature-control mechanism in warm-blood animals.
If fluid flows several tubes one by one, we have
Rtotal = R1 + R2 + … + Rn
(3.16)
It is just like the total resistance in series electrical circuits （串联电路）. If fluids flow
several tubes at the same time, the total R can be calculated as
1
1
1
1



(3.17)
Rtotal R1 R2
Rn
It is like the resistance in parallel electrical circuits (并联电路).
2. Stokes’ law
If an object moves at constant velocity in viscous fluid, there is a layer of fluid on the
surface of the object. There is internal friction between fluid-layer of object and other fluid
layers. If the object is a sphere (ball), its resistant force is given by
F = 6   v r0
(3.17)
Where v and r0 are the speed and radius of sphere respectively,  is the coefficient of
viscosity. This relation is called Stoke’s law.
3. Terminal velocity
Assuming that a ball falls in liquid due to gravity. The forces on the ball are gravity,
buoyancy and the resistant force given by Stokes law.
gravity = m g = 4/3  r03  g
Buoyancy = 4/3  r03  g
Buoyancy
Stoke’s force = 6   v r0
Stoke’s
The total force on this ball is
Ftotal = gravity + buoyancy + Stoke’s force
force
3
= 4/3  r0 ( - ) g– 6   v r0
When Ftotal is greater than zero, the ball will accelerate
downwards and the speed of ball will increase. The inceasing
v
speed will lead the resistance force to increase and will
decrease the total force. At some time later, the total force
mg
acted on the ball will be reduced to zero. It is ease to obtain
the balance velocity that
2r 2 g     
(3.18)
Fig. 10 all forces on the ball
v 0
9
The velocity is called terminal velocity. From this equation,
if we want to increase the terminal velocity, we could
increase the radius and density of sphere and decrease .
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Chapter 3. Fluid Dynamics
Problems
1. Water flows in a pipe. The volume flow rate (Q) of water is 0.24 m3s-1. We choose two
points A and B in the pipe. The cross-sections are 200 cm2 at point A and 100 cm2 at
point B. The pressure at point A is 2.5  105 Pa and point B is 1.5 meters higher than
point A. Suppose that water is an ideal fluid. Find the flow speed at A and B and the
pressure at B.
hint:
(1). Known as:
volume flow rate Q = 0.24 m3,   1000kg  m 3
SA = 200 cm2 = 2  10-2 m2, PA  2  10 5 Pa , hA = 0
SB = 100 cm2 = 1  10-2 m2, hB = 1.5 m
(2) Find: vA, vB and PB
using S A v A  S B vB  Q , 
2. The radius of an adult’s artery is 1.30  10-2 m. If the blood flows for 0.4 meters, what are
the flow resistance and the difference of blood pressure between the two ends of the 0.40
meter artery; Find the Renold number and describe its type of blood flow. Suppose that
the volume flow rate, the coefficient of the viscosity and the density of blood are known,
given respectively as 1.00  10-4 m3s-1, 3.00  10-3 Pas and 1.20  103 kgm-3.
hint:
R= 1.30  10-2 m, L = 0.4 m, Q = 1.00  10-4 m3s-1, = 3.00  10-3 Pas,
 = 1.20  103 kgm-3.
R f  8L R 4 ,
Re  vR 
3. Water stands at a depth H in a large open tank whose side walls are vertical (Fig.13-16).
A hole is made in one of the walls at a depth h below the water surface.
a) At what distance R from the foot of the wall does the emerging stream of water strike
the floor?
b) At what height above the bottom of the tank could a second hole be cut so that the
stream emerging from it would have the same range?
4. A cylindrical vessel, open at the top, is 20cm high and 10cm in diameter. A circular hole
whose cross-sectional area is 1cm2 is cut in the center of the bottom of the vessel. Water
flows into the vessel from a tube above it at the rate of 140cm3s-1.How high will the
water in the vessel rise? When the water in the tank reaches to a height H and stop pouring
water into it, how long does it take to drain the water completely out from the tank?
5. At a certain point in a horizontal pipeline the velocity is a ms-1 and the gauge pressure is
1.0104Pa above atmospheric. Find the gauge pressure at a second point in the pipeline.
Note that the cross-sectional area at the second point is one-half that at the first and the
liquid in the pipe is water.
6. Water in an enclosed tank is subjected to a gauge pressure of 2104Pa, applied by
compressed air introduced into the top of the tank. There is a small hole in the side of the
tank 5m below the level of the water. Calculate the speed with which water escapes from
this hole.
7. A tall vessel, with 5 holes drilled in a vertical line, is filled with water and maintained at a
constant level X by placing it under a tap. Water issues horizontally from the 5 holes.
Suppose that the height between any two holes, the spacing from the top of water to the
highest hole and the distance from the lowest hole to the bottom of water are all h. Find
(1) At which hole will water be projected outwards with maximum velocity?
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Medical Physics
(2) From which hole will the water attain maximum range?(Neglect air resistance.)
8. The densities of steel and of glycerine are 8.5gcm-3and 1.32gcm-3, respectively, and the
viscosity of glycerine is 8.3 poise.
a) With what velocity is a steel ball 1mm in radius falling in tank of glycerine at an instant
when its acceleration is one-half that of a freely falling body?
b) What is the terminal velocity of the ball?
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