Download Fluids and Fluid Mechanics Fluids in motion – Dynamics Equation of

Fluids and Fluid Mechanics
Fluids in motion – Dynamics
Equation of Continuity
After having worked on fluids at rest we turn to a moving fluid. To describe a moving fluid we develop
two equations that govern the motion of the fluid through some medium, like a pipe. These two equations are the
equation of continuity (which says that the fluid flow rate is continuous) and the second is Bernoulli’s equation
(which is a statement of conservation of energy.) Consider the pipe in Figure #1 with varying cross-sectional area.
At point 1 let the cross-sectional area be A1 and at point 2 let the cross sectional area of the pipe be A2 . Fluid
enters point 1 from the left moving at velocity v1 . What we’d like to determine, is with what velocity will it leave
point 2. To do this we need to determine the mass flow rate is a constant throughout the pipe. That is, the mass of
fluid per unit time entering at point 1 and the mass of fluid per unit time leaving point 2. Providing that there are
no leaks in the system (the pipe) the mass of fluid entering the pipe at point 1 (per unit time) has to equal the mass
of fluid leaving the pipe at point 2 (per unit time.) To quantify this we write
the density of the fluid,
Δm1 Δm2
. Thus we can write using
=
Δt
Δt
Δm1 ρ1ΔV1
for the rate of change of the mass of fluid flowing past point 1 and similarly
=
Δt
Δt
Δm2 ρ 2 ΔV2
for the mass of fluid flowing past point 2. ΔV is the volume of fluid passing any point in space and
=
Δt
Δt
equating the two expressions gives
Δm1 Δm2
ρ ΔV ρ ΔV
ρ A Δl ρ A Δl
=
→ 1 1= 2 2→ 1 1 1= 2 2 2.
Δt
Δt
Δt
Δt
Δt
Δt
Figure #1: Diagram showing the relevant quantities used to
determine the equation of continuity. D. Giancoli, Physics:
Principles with Applications, 7th Ed., Prentice Hall.
The density of the fluid at point 1 and point 2 could be different allowing for the fact that the fluid may be
compressible. If the fluid is incompressible, as we will assume for everything that we will do, then ρ1 = ρ 2 and we
identify
Δl
ρ A Δl ρ A Δl
= v as the speed of the fluid at any point. Therefore, we have 1 1 1 = 2 2 2 → A1v1 = A2 v2 , which
Δt
Δt
Δt
we call the equation of continuity. The equation of continuity gives us a way of determining the velocity of the
fluid moving at point 2. By the equitation of continuity, the amount of fluid moving past point 1 per unit time is
the same as that moving past point 2 per unit time. This implies that ρ1V1 = ρ 2V2 and since the fluid is
incompressible the volume of fluid in (per unit time) at point 1 is equal to the volume of fluid out (per unit time) at
point 2. This is an alternate version of the equation of continuity and says the same thing.
Bernoulli’s Equation: Conservation of Energy in a Moving Fluid
Having the equation of continuity, we are now in a position to describe the moving fluid. Let’s assume that
we have a pipe as shown in Figure #2. The left end of the pipe at point 1 is at a height y1 above the ground and the
fluid enters point 1 with a speed v1 through the pipe with cross-sectional area A1 . The fluid at point 2 leaves with a
speed v2 through a section of pipe with cross-sectional area A2 and point 2 is at a vertical height y2 above the
ground. We’d like, again, to determine the speed of the fluid when it leaves point 2. To do this we need to get the
fluid moving into point 1. This takes work and the work done to move the fluid into point 1 is given by some
applied force and how far into the pipe you need to move the mass of fluid. Thus, W1 = F1Δl1 = P1 A1Δl1 , where P1 is
the pressure of the fluid at point 1 and Δl1 is the distance it takes to move a mass of fluid m1 past point 1. To keep
the fluid moving at point 2, we have to push on the fluid ahead of the incoming fluid that is moving toward point 2
from the left. The fluid immediately to the right of point 2 offers resistance against the incoming fluid. This takes
energy out of the fluid and costs work. The work done against the incoming fluid is W2 = −F2 Δl2 = −P2 A2 Δl2 ,
where P2 is the pressure of the fluid at point 2 and Δl2 is the distance it takes to move a mass of fluid m2 past point
2. The net work done is equal to the change in energy of the system. Thus,
ΔEsystem = W1 + W2 = P1 A1Δl1 − P2 A2 Δl2 = ΔK T + ΔK R + ΔU g + ΔU s . We assume that there is no change in spring
potential energy and that the fluid is not rotating.
Figure #2: Diagram showing the relevant quantities used to
determine the Bernoulli’s equation. D. Giancoli, Physics:
Principles with Applications, 7th Ed., Prentice Hall.
Thus, we have ΔEsystem = P1 A1Δl1 − P2 A2 Δl2 = ΔK T + ΔU g =
the volume of the fluid at any point, we have P1V − P2V =
(
(
1
2
1
2
m2 v22 − 12 m1v12 ) + ( m2 gy2 − m1gy1 ) . Noting that is AΔl
m2 v22 − 12 m1v12 ) + ( m2 gy2 − m1gy1 ) , where by the
equation of continuity the volume of fluid in at point 1 equals the volume of fluid out at point 2, namely
A1Δl1 = A2 Δl2 → V1 = V2 = V . Now we divide both sides by the volume of fluid and note that the density of the
fluid is simply the ratio of the mass of the fluid divided by its volume. Further, we assume again that the fluid is
incompressible so that ρ1 = ρ 2 = ρ . We have P1 − P2 =
(
1
2
ρ v22 − 12 ρ v12 ) + ( ρ gy2 − ρ gy1 ) , or
ΔE = 0 = ( P2 − P1 ) + ( 12 ρ v22 − 12 ρ v12 ) + ( ρ gy2 − ρ gy1 ) , in the absence of any external forces. This is called
Bernoulli’s equation and it’s a statement of conservation of energy in a moving fluid. The first term on the right is
the difference in pressure between points 1 and 2. The second term is the difference in the kinetic energy per unit
volume of fluid and the third is the difference in gravitational potential energy per unit volume of fluid. So, we
identify the volume kinetic energy as
K
= 12 ρ v 2 and the volume gravitational potential energy as
Volume
Ug
= ρ gy .
Volume
In fact, since Bernoulli’s equation is a statement of conservation of energy, it should apply to any situation.
How about a fluid at rest? Consider Figure #3 below in which the segment of fluid is at rest with respect to the
surrounding fluid. Thus v2 = v1 = 0 . The bottom of the cylinder of fluid is at a depth y2 below the surface while
the top of the cylinder is at depth y1 below the surface, so that y2 − y1 = h . Therefore from Bernoulli’s equation the
pressure at the cylinder’s bottom P2 is: ΔE = 0 = ( P2 − P1 ) + ( ρ gy2 − ρ gy1 ) → P2 = P1 + ρ g ( y1 − y2 ) = P1 + ρ gh . This
is exactly the result we obtained when we looked the forces on a static fluid, namely that the pressure increases
linearly with increasing depth.
Figure #3: Variation of pressure with
depth. D. Giancoli, Physics: Principles
with Applications, 7th Ed., Prentice Hall.
Now let’s apply this to a few example problems.
Example #1: Fluid exiting a syringe
The body of a syringe is held horizontally and has a cross sectional area Abody = 2.5 × 10 −5 m 2 while the needle has a
cross-sectional area of Aneedle = 1.0 × 10 −8 m 2 . Suppose that a force of Fapplied = 2N is applied to the plunger on the
body side of the syringe, what is the speed of the fluid as it leaves the needle’s tip? Assume that the fluid inside
has a density of ρ fluid = 1050 mkg3 and that in the absence of the applied force, the pressure on all sides of the syringe
is 1Atm = 1.01× 10 5
N
m2
.
Solution: We apply Bernoulli’s equation to the situation. Defining position 2 to be the needle and position 1 to be
(
) (
the body of the needle we have: ΔE = 0 = Pneedle − Pbody +
1
2
)
2
2
ρ fluid vneedle
− 12 ρ body vbody
. We solve this for the kinetic
2
2
= Pbody − Pneedle + 12 ρ body vbody
energy (per unit volume) of the fluid in the needle and we have 12 ρ fluid vneedle
. The
pressure applied at the body of the needle is due to atmospheric pressure and the force applied over the crosssectional area of the body of the needle. Solving for the speed of the fluid in the needle we have
⎡⎛ F
⎤
⎞
2 ⎢⎜ applied A + Patm ⎟ − Patm ⎥
⎛
⎞
P
−
P
⎝
⎠
body
⎦ + v 2 = 2Fapplied + v 2 .
2
2
vneedle
= 2 ⎜ body needle ⎟ + vbody
= ⎣
body
body
ρ fluid
ρ fluid
Abody ρ fluid
⎝
⎠
To solve for the speed of the fluid exiting the needle, we need to know how fast the fluid is moving in the body of
the needle. Here we will use the equation of continuity, which says that A1v1 = A2 v2 → Abody vbody = Aneedlevneedle .
⎛A
⎞
Therefore solving for the velocity of the fluid in the body of the needle we have vbody = ⎜ needle ⎟ vneedle . Inserting
⎝ Abody ⎠
this into our equation we can solve for the velocity of the exiting fluid. Substituting we have
2
2
needle
v
2Fapplied
2Fapplied ⎛ Aneedle ⎞ 2
2
=
+ vbody
=
+
vneedle
Abody ρ fluid
Abody ρ fluid ⎜⎝ Abody ⎟⎠
2
⎛
⎞
2Fapplied
Aneedle
2
vneedle
1−
=
2
⎜
⎟
Abody ⎠ Abody ρ fluid
⎝
vneedle =
2
⎞
2Fapplied ⎛
Aneedle
1−
2
⎜
Abody ρ fluid ⎝
Abody ⎟⎠
−1
⎛ (1.0 × 10 −8 m 2 )2 ⎞
2 × 2N
=
⎜ 1−
⎟
2.5 × 10 −5 m 2 × 1050 mkg3 ⎜⎝ ( 2.5 × 10 −5 m 2 )2 ⎟⎠
−1
vneedle = 12.3 ms
Comment: Notice Aneedle << Abody that so that when we formed the ratio, the ratio was so small compared to one that
−1
⎛
A2 ⎞
we could have just ignored the term ⎜ 1− needle
2
⎟ .
Abody
⎝
⎠
Example #2: Lift on an airplane wing
Consider the airplane shown below. What is the aerodynamic lift produced on the wing if the wing has a 60m 2
surface area and the velocity of the airflow across the top and bottom of the wing are 340 ms and 290 ms
respectively? Assume that the wing is thin enough so that the airflow over and under the wing does not have an
appreciable change in height. And in fact, one could argue that the front and back edge of the wing are the same
height, so what ever increase in gravitational potential the air gets going over the front edge it loses when it comes
back down to the back edge, effectively experiencing no net change in height.
Ptop
Pbottom
Figure #4: Photograph of one of the U.S. Air
Force Thunderbird aircraft. The red arrows
show the pressures on the upper and lower
surface of the wings. The difference in
pressure gives rise to lift. Photo: S. LaBrake
Solution: We start with Bernoulli’s equation and use the stated assumption.
ΔE = 0 = ( P2 − P1 ) + ( 12 ρ v22 − 12 ρ v12 ) + ( ρ gy2 − ρ gy1 ) = ( P2 − P1 ) + ( 12 ρ v22 − 12 ρ v12 ) . We notice that since energy is
conserved, where the velocity is higher, by Bernoulli’s equation, the pressure must be lower at that point in order
to keep the energy a fixed quantity. Thus since the velocity of the air is lower on the bottom of the wing, the
pressure must be higher there than on top of the wing and these pressures are shown in figure #4 above. Next we
define position 2 as the bottom of the wing and position 1 the top of the wing. We have (using the density of air as
ρ air = 1.3 mkg3 )
(
) (
2
2
ρ vbottom
− 12 ρ vtop
)
2
2
Pbottom − Ptop = 12 ρ air vtop
− 12 ρ air vbottom
=
1
2
0 = Pbottom − Ptop +
1
2
(1.3 )(( 340 ) − ( 290 ) ) = 20475
kg
m3
m 2
s
m 2
s
N
m2
Now, the pressure is related to the force that’s applied divided by an area. We define the pressure on the bottom
and top of the wings to be
F
Fbottom
and top respectively. The difference in forces between the top and bottom of
Awing
Awing
the wing define the lift Flift = Fbottom − Ftop , which is perpendicular to the wing’s surface. Here since the pressure on
the wing’s bottom is greater than on the wing’s top, the net force is directed vertically upward and the force of lift
acts vertically upward to raise the plane up in the air. Thus we have
Pbottom − Ptop = 20475 mN2 =
(
F
Fbottom Ftop
−
= lift . We can solve for the force of lift and we get
Awing Awing Awing
)
Flift = Awing Pbottom − Ptop = 60m 2 × 20475 mN2 = 1.23 × 10 6 N .
Comment: If the lift force is larger than the weight of the airplane then the airplane accelerates upward at some
rate, call it a plane , and the airplane rises. If the weight of the airplane is greater than the lift (say by the airplane
slowing down its forward velocity) then the plane falls at −a plane . If the weight of the airplane and the lift force are
equal the plane neither rises nor falls but travels forward at a constant velocity. This is provided of course that the
thrust produced by the engines pushing the plane forward equals the drag on the airplane produced by air friction.
Example #3: An old West water tower
Consider a water tower, like one might find in the old West, to be circular with a top that has a cross-sectional area
Atop and the sides a depth d . Further let the water tower be open to air on the top. A Wild West outlaw shoots a
hole in the side of the water tower at a depth h < d below the surface and this hole has a cross-sectional area Aside .
The water flows out of the side of the water tower through the small hole. What is the speed of the water flowing
out of the small hole on the side of the water tower if h = 5m ?
Solution: We apply Bernoulli’s equation and define position 2 to be the hole on the side of the tank and position 1
to be the top of the tank as shown in figure #6 below.
v2
v1
Figure #6: A tank of water, open to the air on
top, with water exiting through a side hole a
depth h below the top. This is a model of the
“water tower”. D. Giancoli, Physics:
Principles with Applications, 7th Ed., Prentice
Hall.
(
) (
We have ΔE = 0 = Pside − Ptop +
1
2
) (
)
2
2
ρ vside
− 12 ρ vtop
+ ρ gyside − ρ gytop . Next we note that the top is open to air and
that the water flowing out of the hole on the side is flowing into air. Thus the pressure on the top and side of the
2
tank is due to air and Pside = Ptop = Pair . Returning to Bernoulli’s equation we have solving for vside
:
(
)
2
2
vside
= vtop
+ −2gyside + 2gytop . Defining the zero of the gravitational potential energy to be at ytop , we can solve
2
2
− 2g ( −h ) = vtop
+ 2gh . Again, we need to relate the speed of the water moving across
for vside . Thus vside = vtop
the top surface of the water tower to the speed of the water coming out of the side of the tower. We return to the
⎛A
⎞
equation of continuity and we get A1v1 = A2 v2 → Asidevside = Atop vtop → vtop = ⎜ side A ⎟ vside . Assuming that
⎝
top ⎠
Atop >> Aside we have
Aside
⎛A
⎞
<< 1 and vtop = ⎜ side A ⎟ vside ~ 0 . So, the speed of the water exiting the hole on the side
⎝
top ⎠
Atop
2
+ 2gh ~ 2gh = 2 × 9.8 sm2 × 5m = 9.9 ms .
of the water tower is approximately vside = vtop
Comment: Notice that the density of the fluid disappeared here. This is exactly analogous to dropping a mass m
from rest and letting it fall through a height h . The velocity of the falling mass does not depend on the mass, but
only the height through which it fell.
Example #4: Fluid flow in arteries and veins
Suppose that the aorta has a radius of about raorta = 1.25cm and that the typical blood
velocity is around 30 cms with an average density of ρ = 1050 mkg3 . A simplified model
of the human circulatory system is shown in Figure #7. The human circulatory system
is a closed system, so the flow rate of blood out of the heart has to be the same as the
flow rate of blood coming back to the heart.
Questions and Solutions:
a. What is the total flow rate of blood through the aorta?
r
The total flow rate is determined from the equation of continuity. We have
3
2
2
Q = Aaorta vaorta = π raorta
vaorta = π ( 0.0125m ) × 0.3 ms = 1.5 × 10 −4 ms
(
)
b. What is the average blood velocity in the major arteries if the total crosssectional area of the major arteries is 20cm 2 ?
Figure #7: Simplified
model of the human
circulatory system. D.
Giancoli, Physics:
Principles with
Applications, 7th Ed.,
Prentice Hall.
First we convert 20cm 2 into square meters. We have
1m
20cm 2 × ( 100cm
)2 = 2 × 10 −3 m 2 . Then we use the equation of continuity to
determine the blood velocity in the major arteries. From the equation of
3
Q
1.5 × 10 −4 ms
continuity we have Q = Aaorta vaorta = Aarteries varteries → varteries =
=
= 0.075 ms = 7.5 cms .
Aarteries
2 × 10 −3 m 2
c. On the assumption that all the blood in the circulatory system goes through the capillaries, what is the total
cross sectional area of the capillaries if the average velocity of the blood in the capillaries is 0.03 cms ?
Again we use the equation of continuity. We have for the total cross-sectional area of all of the capillaries
3
Q
1.5 × 10 −4 ms
Q = Aarteries varteries = Acapillaries vcapilaies → Acapillaries =
=
= 0.5m 2 .
−4 m
vcapilaries
3 × 10 s
rr
d. If a typical capillary has a cross sectional area of Acapillary = 3 × 10 −11 m 2 , about how many capillaries are
there in the human body?
Acapillaries
0.5m 2
The approximate number of capillaries in the human body is # =
=
= 1.7 × 1010 or
−11
2
Acapillary 3 × 10 m
roughly 17 billion.
e. If a capillary has an average length of l = 0.75mm what is the average time that a red blood cell spends in a
capillary?
Assuming that the blood velocity is constant in the capillaries then the average time a red blood cell spends
l
l
0.75 × 10 −3 m
in a capillary is vcapilary = → t =
=
= 2.5s .
t
vcapilary
3 × 10 −4 ms
f. What are the kinetic energy per unit volume for blood in the aorta, the major arteries, and the capillaries?
2
1
K
2 mv
=
= 12 ρ v 2 .
The kinetic energy per unit volume of blood in the major systems is given by
Volume
V
Evaluating this for each system we find:
Aorta:
K
2
= 12 ρ v 2 = 12 × 1050 mkg2 ( 0.3 ms ) = 47.3 mJ3 .
Volume
Arteries:
K
2
= 12 ρ v 2 = 12 × 1050 mkg2 ( 0.075 ms ) = 2.95 mJ3 .
Volume
Capillaries:
K
= 12 ρ v 2 = 12 × 1050 mkg2 ( 3 × 10 −4
Volume
m
s
)
2
= 4.73 × 10 −5
J
m3
.
Example #5: The Heart as a mechanical pump
The human heart can be modeled as a mechanical pump. The aorta is a large artery that carries oxygenated blood
away from the heart to various organs in the body. For an individual at rest, the blood ( ρ blood = 1050 mkg3 ) in the aorta
of radius raorta = 1.25cm flows at a rate of 5 ×10−4
m3
min
. In what follows, we will be talking about power. Power is
the rate at which work is done or the rate at which energy is transferred into or out of a system by an external force.
Thus we can define the work done on an object as P =
equivalently Watts, where 1W = 1 Js .
ΔE ΔW FΔx
=
=
= Fv in units of Joules per second or
Δt
Δt
Δt
Questions and Solutions:
a. With every beat, the heart does work moving the blood into the aorta. The heart does work at a rate of
0.5 Js . Derive an expression for the energy per unit volume of blood associated with the blood flow into the
aorta?
We start with the expression for power and we see that the question ask us to determine an expression for
the energy per unit volume. Thus we will take our equation for power and multiply it by a factor of one.
This will generate the energy per unit volume we seek. In addition we generate an extra term that, the
volume of blood flow per unit time. This is simply the flow rate, which we will call Q . Thus we have
P=
ΔE
P
ΔE Volume
ΔE
Volume
ΔE
= .
×
=
×
=
× Q and the energy per unit volume is
Volume Q
Δt Volume Volume
Δt
Volume
Using the numbers given in the problem we can calculate a value for this quantity, namely
ΔE
P
0.5 J
= = 5×10−4 ms 3 = 60000 mJ3 = 60000 mN2 .
Volume Q
60 s
b. The energy per unit volume (or kinetic energy per unit volume) of a moving fluid corresponds to a
difference in pressure between two different points in space. Suppose you have the condition called
atherosclerosis. Atherosclerosis is a disease in which plaque builds up inside walls of your arteries.
Arteries are blood vessels that carry oxygen-rich blood to your heart and other parts of your body. Plaque is
made up of fat, cholesterol, calcium, and other substances found in the blood. Over time, plaque hardens
and narrows your arteries and constricts blood flow. This limits the flow of oxygen-rich blood to your
organs and other parts of your body. Atherosclerosis can lead to serious problems, including heart attack,
stroke, or even death.). Suppose that you had the condition of atherosclerosis, what would be the radius of
the opening that remains and what percent of the aorta would be blocked? Assume that the buildup on the
walls of the aorta is uniform so that the opening that remains is circular and that the patient is lying
horizontal. The situation is shown in Figure #8 below.
FromHearttoBody
Blockage
Blockage
Figure #8 Cartoon model of the human
aorta with a blockage for a person lying
horizontally.
In order to answer the question of how much of the aorta is blocked, we first calculate the speed of the
blood in the unblocked portion of the aorta. We can calculate this speed from the flow rate and the crosssectional area of the aorta. From the information given in the problem, we have the flow rate where
−4
Q = Aaorta vaorta and thus vaorta
3
5.4×10 m
Q
60 s
=
=
= 0.017 ms .Next we turn to Bernoulli’s equation and
Aaorta π ( 0.0125m )2
for a person who is lying horizontally, the difference in pressure between the unblocked and blocked
portion of the aorta is P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2 . Defining position 1 to be in the unblocked
2
2
= Pblockage + 12 ρ vblockage
aorta and position 2 in the blockage, we have Paorta + 12 ρ vaorta
. Solving for the speed of
the blood flow in the blockage we get
vblockage =
(
)
2 60000 mN2
2
2
2
Paorta − Pblockage + vaorta
=
+ ( 0.017 ms ) = 10.1 ms . Having now calculated the
kg
ρ
1050 m 3
(
)
speed of the blood in the blockage and since the flow rate is continuous, we can determine the area of the
blockage. From the area we can determine the radius of the blockage and what percent is blocked. The
area of the blockage can be determined from the flow rate. The flow rate and area of the blockage are
Q = Ablockagevblockage → Ablockage =
Q
vblockage
=
5.4×10 −4 m 3
60 s
m
s
10.7
= 8.41× 10 −7 m 2 . Therefore the radius of the blockage,
assuming that the blockage is circular is given by
2
Ablockage = π rblockage
→ rblockage =
Ablockage
8.41× 10 −7 m 2
=
= 0.0005m = 0.05cm .
π
π
⎡r − r ⎤
⎡1.25cm − 0.05cm ⎤
Lastly the percent of the aorta that is blocked is % = ⎢ i f ⎥ ×100% = ⎢
⎥⎦ ×100% = 96% .
⎣
1.25cm
⎣ ri ⎦