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Chapter 3.
Fluid Dynamics


24

Chapter T
hree


F
F
l
l
u
u
i
i
d
d


D
D
y
y
n
n
a
a
m
m
i
i
c
c
s
s




(
(










)
)




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 l
arge 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, whi
le 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
















(3.1)

where m is the mass of the obje
ct 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 stati
onary, 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















(3.2)



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 liqui
d

(see Fig. 3.1)
,
its weight (
mg
) will exert on the
bottom
surface of the imaginary cylinder.




T
herefore



Medical Physics


25


On the other hand, a container at the earth’s surface


has the pressure of atmosphere. So






(3.3)


Pascal’s principle
:


W
hen a change in pressure is applied to an enclosed

fluid, the change is transmitted undiminished
ly

to


every point in the fluid and to walls of the container.
















According to Pascal’s prin
ciple, the pressure P
1

at one end should be equal to P
2

at
another end

shown in Fig. 3.2
.
So we have


3.2.
2

Buoyancy (
浮力
) and Archimede’s principle


Buoyancy is a fami liar phenomenon; a body immersed in water seems to have less weight
than when immersed
in air, a body whose average densit y is less t han t hat of t he fluid in
which it is immersed can float in t hat liquid. Examples are t he human body in water, or a
helium
-
fi lled balloon in air.


Archi mede’s pri nciple

states t hat
when a body is immersed in a f
luid, t he f luid exert s an
upward f orce on t he body equal t o t he weight of t he f luid t hat is displaced by t he body
. Or in
other word, we can say t hat
a

body immersed in a f luid i s buoyed up by a f orce equal t o t he
weight of f luid which t he body has displace
d.
T
he buoyancy can be worked out by




w
here
V

is t he volume of fluid displaced by t he body,

is t he densit y of t he fluid and
g

is t he
gravitat ional accelerat ion. (
A st ory describes about how Archimede
t o
g
e
t hi s principle. It is
s
aid t hat t he king in his count ry asked him t o f ind a met hod t o dist inguish whet her
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 th
e street to tell the good
news to his king
.)

h

P
0



A

AA
A


F
F


Fig. 3.1

F
1

F
F
2
2


P
1

P
2

Fig. 3.2 Hydraulic pressure

Chapter 3.
Fluid Dynamics


26



§
3.3 Hydrodynamics (
流体动力学
)


3.3.1 Steady flow and ideal fluid



1
.

Stead
y

flow



Streamline (
流线


In the motion of fluid,

the speed of particles in the fluid may be

different. At any given moment, we can

draw some lines. The tangent direction of

every point on the line ha
s the same direction

and speed.
the lines

are

call
ed


streamlines

(see Fig. 3.3)

(It is also

defined that a curve
whose tangent,

at any point, is in the direction

of the fluid velocity at that point).





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


A
1



v
1
, A
2



v
2
,
…,
A
n



v
n



During a time interval
D
t, 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




=

×

A
i
(
v


t
)


Choosing two cross
-
sections, the amount of fluid through A
1

should be equal to that
through A
2
. So we have


1

A
1

(v
1


t) =

2

A
2

(v
2


t)

A

B

C

v
v
A

v
v
B

v
v
c

Fig. 3.3 at different point on the streamline,
the velocity does not change at any time. The
sha
pe of streamline doesn’t change.
v
v
A

… do
not change
.

Medical Physics


27


1

=

2

for the same and incompressible fluid. This result gives A
1

v
1

= A
2

v
2
. Since A
1
and
A
2

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: (v
2

= A
1
v
1
/A
2
)

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 toward
s

the walls. In this case, we must interpret the
velocity in the continuity equation as average velocity.


Example 3.1

in a normal restin
g result, the heart pumps blood into the aorta (
主动脉
) at an
average volume flow rate (
Av
) of about 9.0
×
10
-
5

m
3
/s. Calculate the average speed of the
blood in an aorta of inside diameter 1.40 cm

Solution:

known A
1
v
1
= 9.00
×
10
-
5

m
3
/s

A
2

=

r
2
= 3.14 (7.0
×
10
-
3
)
2


v
2

= A
1
v
1
/A
2

= 0.58 m/s


3.3.3 Work
-
energ
y and Bernoulli’s equation

The work
-
energy principle may be applied to the ide
al 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 P
1
, A
1
,


and v
1

respectively
and in point 2, P
2
, A
2

and v
2
.



P = F/A



F
1
= P
1
A
1
,

F
2
= P
2
A
2

The work done by F
1

and F
2

in the time period
is given below:


W
1

= F
1
∙ (v
1

t) = P
1
A
1
v
1

t = P
1

V
1


W
2

=

F
2
∙ (v
2

t) =


P
2
A
2
v
2

t =


P
2

V
2

Where
W
2

is negative because the liquid moves in a direction opposite to the ap
plied force.






















Fig. 3.4 derivation of Bernoulli

s equation


F
2

v
1


t

v
2


t

F
1
=
P
1
A
1

h
1

h
2

Chapter 3.
Fluid Dynamics


28




As the ideal fluid is incompressible, the amount of fluid running through A
1

must go through
A
2

as well. Therefore,

V
1
=

V
2

=

V, so the total work done by F
1

and F
2

is

W

= W
1

+ W
2

= (P
1

P
2
)

V

Since the mass of the liquid in

V should be



V, the change in kinetic energy is




K
E



m v
2
2


½

m v
1
2


= ½



V(v
2
2


v
1
2
)

The change in potential energy is





P
E

=

m g h
2




m g h
1





=

r
V g (h
2

h
1
)

The total change of th
e kinetic and potential energy should be equal to the work done by F
1

and F
2





W =

K
E

+

P
E


Substituting W,

K
E

and

P
E

in
to

the above equation, we have



(P
1

P
2
)

V = ½



V(v
2
2


v
1
2
) +



V g (h
2

h
1
)
P
1
+ ½


v
1
2

+


g h
1

= P
2

+ ½


v
2
2

+


g
h
2









(3.8)

As point 1 and point two are chosen freely, we have

P

+ ½

v
2

+


g h = constant









(3.9)


This is know
n

as Bernoulli’s equation. Now we have a look at two special cases:


1.

Hydrostatics
(v
1

= v
2

= 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 h
2









(3.9a)

as the point 1 and 2 are chosen freely, so





P +


g h
=constant










(3.10)


from (
3.9a)




P
1

= P
2

+


g (h
2

h
1
)


comparing it with (3.3), the P
2

should be the atmospheric pressure P
0

and (h
2

h
1
) should be

h
.


2.
Hydrodynamics

(h
1

= h
2

= 0)


In many cases, fluid flows in a horizontal pipe, that means h
1

= h
2
. Bernoulli’s equation
bec
omes


P
1

+ ½


v
1
2

= P
2

+ ½


v
2
2

So we have



P + ½


v
2

= constant











(3.11)

This explains that w
hen 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
)



Medical Physics


29

Pitot tube’s principle

In Fig.3.6, the tube
t
1

is a straight tube,
t
2

is a bend tube. At the two tube’s ends (c & d), they

are in the same level, i. e. h
c

= h
d
. So Bernoulli’s equation becomes

P
c

+
½


v
c
2

=
P
d

+
½


v
d
2

As d is the dead end of t
ube
t
2
, v
d
= 0. So















P
d

= P
c

+
½

v
2



(3.11
a
)

w
here
v

=
v
c

.Therefore, dynamic pressure

can be changed to static pressure in a dead

end. Of course, if we know P
c

& P
d
, the speed


v can be worked out. Since P
e

= P
a

= P
0
,

using the fo
rmula (3.3) for static pressure,

we have



P
b

= P
e

+


g h

From the fig. 3.6, we know that


P
d



P
c

= P
b



P
e

=


g h





½


v
2
=


g h




v = (2 g h)
½

So the speed of fluid can be calculated by

measured
h.





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
th
e
position
c
,

the fluid velocity is the full stream velocity

v
. So
using the equation
(3.11a) , we have





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






h

d

c

v

Fig. 3.
6

The Pitot tube

½

v
2

Fig. 3.
5

Principle of kinemometer

t
1

t
2

h

d

a

e

C

Chapter 3.
Fluid Dynamics


30

S
o we could calculate the velocity of the fluid b
y knowing the height
h,















(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 do
es 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.



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 t
wo 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





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 t
he connected two
layers and velocity gradient.

x

x+

x

v

v+

v

Fig. 3.
7

The laminar
flow


x


v

x

x+

x

v+

v

v

Fig. 3.
8

Velocity varies as x

Medical Physics


31


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 particl
es 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 r
adius 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.



R
e

=

v r /














(3.13)

R
e

is called Reynold number. Experiments show that:



R
e



1000

laminar flow


1000


R
e



1500

unsteady flow



R
e



1500

turbulent flow

In biological transport system,
R
e

is too small to create turbulent flow, but if a blood vessel
(
脉管
) is bent, turbulent flow will occur when
R
e

is small. So in all bent and br
anches
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

point
s of a cross
-
section (
横截面
). The

outmost layer of fluid clings to (
粘附
) the

walls of the tube and its velocity is zero.

The tube’s walls exert a backward drag (
拽力
)

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 a
nd decrease to zero at the wall
s
.


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 c
ylinder.
T
he force on the left end is
P
1

r
2
,

and that on the right end
P
2

r
2
,

the net force on the imaginary tube is thus




Since the fluid in the pipe does not accelerate, this force must balance the viscous retarding
force at the

surface of this cylindrical fluid.
T
his 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
P
1

l


P
2

Chapter 3.
Fluid Dynamics


32

should be
dv/dr
. Note that the area
A

of the cylinder
relating

to the reta
rding force is 2

r
l
.
T
hus
the viscous force is


E
quating this to the net force due to pressure on the ends and rearranging, we find that

.

T
his shows that the velocity changes more and more rapidly as we go fr
om the center
(r = 0)
to the pipe wall (r =
r
0
). The negative sign must be introduced because v decreases as r
increases.
I
ntegrating on both sides, we get




T
he velocity at the position with the distance
r

from

the center of the tube

can be calculated
and is given by













(3.14)

w
here
r
0

is the radius of the tube and
r

is dista
nce from the center of the tube to a point in

the
tube and
l

is the length of the tube
.

F
rom 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 (
P
1
-
P
2
)/
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 pip
e 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

end of this element in a time

t

is
v dA

t
, where v is the

velocity at t
he radius r and
dA

is the white area, equal to

2

rdr
.

Taking the expression for v from Eq.(3.14)
,

we obtain





T
he volume flowing across the entire section is obtained by integrating over all elements from
the center to the wall o
f the pipe.
S
o on the left hand side, the volume is from
dQ

to

Q

for a
time
dt

and the radius is from 0 to
r
0
, and we have




T
he
total volume of
flow
per unit time
,

Q
/

t,

called
volume flow rate, denoted by
Q
,

is
given by



,











(3.1
5
)


r

d
r

Fig. 3.10 cross
-
section of
the pipe


Medical Physics


33

w
here

p
=
P
1


P
2

is
the difference of the pressures at the two inlets of the
pipe

and R =
8

l/
(
p

r
0
4
).

Equation (3.1
5
) is called Poiseuille’s law. It is similar to the Ohm’s law. So
R

is
called
flow resistance

(流阻)
.

T
he 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 dia
meter
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.

I
f fluid flows several
tubes

one by one, we have



R
total

= R
1

+ R
2

+ … + R
n









(3.1
6
)

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












(3.1
7
)

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 fricti
on 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 r
0













(3.17)

Where
v

and
r
0

are the speed and radius of sphere


respectively
,


is the coefficient of
viscosi
ty.

This relation is called Stoke’s law.


3. Terminal velocity


Assuming that a ball falls in liquid due to gravity. The force
s

on the ball
are

gravity,
buoyancy and the resistant force given by Stokes law.


gravity =
m

g

= 4/3


r
0
3


g

Buoyancy =
4/3



r
0
3


g

Stoke’s force =
6




v r
0


The total force on this ball is


F
total

= gravity + buoyancy + Stoke’
s force



= 4/3


r
0
3
(


-


)
g


6




v r
0

When
F
total

is greater than zero, the ball will accelerate

downwards and the speed of ball will incre
ase
.
T
he inceasing

speed

will lead the resistance force to increase

and

will

decrease the total force. At some time later, the total force

acted on the ball will be reduced to zero. It is ease to obtain

the balance velocity that





(3.18)







Fig. 10
all forces on the ball

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

.

mg

Buoyancy

Stoke’s
force

v

Chapter 3.
Fluid Dynamics


34


Problems


1.

Water flows in

a pipe. The volume flow rate (Q) of water is 0.24 m
3

s
-
1
. We choose two
points A and B in the

pipe. The cross
-
sections are 2
00 cm
2

at point A and 1
00 cm
2

at
point B. The pressure at point A is 2.5


10
5

Pa and point B is 1.5 meters higher than
point A. S
uppose 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 m
3
,



S
A

=
2
00 cm
2

= 2


10
-
2

m
2
,

, h
A

= 0



S
B

= 1
00 cm
2

= 1


10
-
2

m
2
,

h
B

= 1.5 m



(2) Find:
v
A
,
v
B

and
P
B






using
,








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 tw
o 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

m
3

s
-
1
, 3.00


10
-
3

Pa

s
and 1.20


10
3

kg

m
-
3
.

hint:


R=
1.30


10
-
2

m
, L = 0.4 m, Q =
1.00


10
-
4

m
3

s
-
1
,

=
3.00


10
-
3

Pa

s
,






=
1.20


10
3

kg

m
-
3
.






3.

Water stands at a depth H in a large open tank whose side walls are v
ertical

(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
secon
d hole
be cut so that the
stream emerging from it would have the same range?


4.

A cylindrical vessel, open at the top,

is 20
cm

high and 10
cm

in diameter.

A circular hole
whose cross
-
sec
tional area is
1
cm
2

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 140
cm
3

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 hori
zontal pipeline the veloc
ity is
a

m

s
-
1

and the gauge pressure is
1.0

10
4
Pa above
at
mospheric.

Find the gauge pressure

at a second point in the
pipe
line
.

Note that
the cross
-
sectional area at the second point

is one
-
half that at the fi
r
st and

t
he
liquid in the pipe is water.

6.
Water in an enclosed tank is subjected to a gauge pressure of 2

10
4
Pa, applied by
compressed air introduced into the top of the tank. There is a small hole in the side of the
tank 5m be
low the level of the water. Calculate the speed with which water escapes from
this hole.

7.

A tall vessel, with 5 h
oles drilled in a vertical line
,

is filled with water and maintained at a
constant level
X

by placing it un
d
er a tap.

Water issue
s horizontal
ly from the 5 holes
.

Suppo
s
e that the height between

any two holes,

the
spacing from the top of water to the
highe
s
t ho
l
e 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?

Medical Physics


35


(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
-
3
and 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?