BIEN 301
3
Saleh David Ramezani
COVER SHEET
Homework # 4
Chapter 2: Pressure Distribution in a Fluid
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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Saleh David Ramezani
1.0 PROBLEM 2.27
1.1 Given
Conduct an experiment to illustrate atmospheric pressure. Note: Do this over a
sink or you may get wet! Find a drinking glass with a very smooth, uniform rim at the
top. Fill the glass nearly full with water. Place a smooth, light, flat plate on top of the
glass such that the entire rim of the glass is covered. A glossy postcard works best. A
small index card or one flap of a greeting card best.
Figure 1.
Figure 1
Problem 2.27 Diagram
1.2 Find
a) Hold the card against the rim of the glass and turn the glass upside down.
Slowly release pressure on the card. Does the water fall out of the glass? Record your
experimental observations.
b) Find an expression for the pressure at points 1 and 2 in figure 1. Note that the
glass is now inverted, so the original top rim of the glass is at the bottom of the picture.
The weight of the card can be neglected.
c) Estimate the theoretical maximum glass height at which this experiment could
still work, such that the water would not fall out of the glass.
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1.3 Free Body Diagram, Simplifications, and Assumptions
P1
Δh=0.07 m
P2
Patm
Figure 2
Problem 2.27 Free Body Diagram
In this problem we are trying to find the hydrostatic pressure difference between
two points in a liquid. The points are at vertical distances from each other, on the top and
on at the bottom of the container.
1.4 Assumptions
We assume that the fluid is hydrostatic (not moving or tilting), water is pure (not
mixed with an external substance or liquid), and the temperature is constant.
1.5 Solution Calculations
a) Pressure at point 1:
Patm  P1   ( z 2  z1 )
101350 Pa  P1  
9790 N
(0  0.075m)
m3
P1  100616Pa  100.6kPa
b) Pressure at point 2:
Patm  P1   ( z 2  z1 )
101350 Pa  P2  
9790 N
(0)
m3
P2  101350Pa  101.4kPa
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Saleh David Ramezani
Patm  Ph   ( z 2  z1 )
101350 Pa  0  
9790 N
(0  h )
m3
h  10.35m
1.6 Computational Results
Pressure at P1, above the water.
P1  100616Pa  100.6kPa
Pressure at P2, at the bottom of the glass.
P1  101350Pa  101.4kPa
The theoretical maximum height at which this experiment could still work.
h  10.35m
1.7 Comparison of Analytical and Computational Results, with
Discussion
This experiment resembles a basic barometer with water as the liquid. In part (c)
in order to calculate the maximum height of the column, the pressure on the top of the
liquid was set equal to zero. This simply means that there would not be any force acting
on the top of the liquid in the negative y direction but the weight of the liquid.
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2.0 PROBLEM 2.34
2.1 Given
Sometimes monometer dimensions have a significant effect. In figure 2 containers
(a) and (b) are cylindrical and conditions are such that Pa= Pb.
D
L
D
SAE 30 oil
Water
H
h
d
Figure 3
Problem 2 Diagram
2.2 Find
Derive a formula for the pressure difference Pa-Pb when the oil water interface on
the right rises a distance Δh<h, for (a) d<<D and (b) d = 0.15D. What is the percentage
change in the value of Δp?
2.4 Assumptions
Hydrostatic pressure
Constant temperature
No mixing
2.5 Solution Calculations
In a regular condition (without increasing the oil water interface by Δh) we have:
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Saleh David Ramezani
Pa   water ( L  h)   oil (h  H )  Pb
If we add Δh to the oil water interface the hydrostatic relation becomes:
Pa   water ( L  h  h)   oil (h  H  h)  Pb
Pa  Pb  h( water   oil )
2.6 Computational Results
Pa  Pb  h( water   oil ) .
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Saleh David Ramezani
3.0 PROBLEM 2.47
3.1 Given
The cylindrical tank in Figure 3 is being filled with water at 20˚ C by a pump
developing an exit pressure of 175 kPa. At the instant shown, the air pressure is 110 kPa
and H=35 cm. the pump stops when it can no longer raise the water pressure.
Figure 4
Problem 3 Diagram
3.2 Find
For isothermal air compression, estimate H during this time.
3.4 Assumptions
The pump stops when it can no longer change the pressure; this means that at the
end of pumping, the outside and the inside (air) pressure should be the same. The process
is also Isothermal air compression, that means that T1=T2.
3.5 Solution Calculations
Pair 2  Pwater 2  Ppump
Pair 2  gH  PPump
Pair 2  9790 H  175000 Pa
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P1V1 P2V2

T1
T2
T1  T2 Isothermic 
P2 V1 Pair 2 Vair1



P1 V2 Pair1 Vair 2
Pair 2
R 2 (0.75m)
0.75m


2
110000 Pa R (0.75m  0.35m  H ) 1.1m  H
( 2)
From (1) and (2) we get:
0.75m(110000 Pa)
N
 9790 3 H  175000 Pa
1.1m  H
m
2
H  18.98H  11.24  0
Solving for H we get:
H  18.3635m
or
H  0.6118m
It’s clear that the first value of H (18.3635) is not a valid answer, therefore:
H  0.6118m
3.6 Computational Results
The pomp stops when the water height has been raised by 0.6118 m
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Saleh David Ramezani
4.0 PROBLEM 2.87
4.1 Given
The bottle of champagne (SG = 0.96) in Figure 4 is under pressure, as shown by
the mercury-manometer reading.
B
2m
Water
C
Figure 5
Problem 4 Diagram
4.2 Find
Compute the net force on the 2-in-radius hemispherical end cap at the bottom of
the bottle.
4.4 Assumptions
Hydrostatic fluid
Incompressible
Constant temperature
4.5 Solution Calculations
The horizontal component of water force is:
N 2m
FH  hCG A  (9790 3 )( )( 2m  3m)  58740 N
2
m
(3m)( 2m) 3
(
)(sin 90)
I xx sin 
12
YCP  

 0.33
2m
hCG Aproj
( )( 2m  3m)
2
So the horizontal force acts 0.3m below the center of gravity, or 0.667m above
point C.
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FV  Asec tionb  (9790
X CP 
3
Saleh David Ramezani
N  ( r  2 m) 2
)(
)(3m)  92268.6 N
4
m3
4 R 4  2m

 0.849
3
3
Similarly the vertical point acts 0.849m to the left of C.
M
C
 0 (2m) P  (58740 N )(0.667m)  (92268.6 N )(0.849m)  58757.8N
4.6 Computational Results
The horizontal force P required to hold the gate stationery is:
P  58.8kN
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Saleh David Ramezani
5.0 PROBLEM 2.64 (EXTRA CREDIT)
5.1 Given
Dam ABC in Figure 5 is 30 m wide into the paper and made of concrete (SG =
2.4).
Figure 6
Problem 4 Diagram
5.2 Find
Find the hydrostatic force on surface AB and its moment about C. Assuming no
seepage of water under the dam, could this force tip the dam over? How does your
argument change if there is seepage under the dam?
5.4 Assumptions
Enter the assumptions for this problem here.
5.5 Solution Calculations
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Saleh David Ramezani
Enter the calculations here.
The results of this analytical calculation can be seen below, in Section 4.6.
5.6 Computational Results
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Saleh David Ramezani
FE PROBLEMS (EXTRA CREDIT)
FE2.1
A gage attached to a pressurized nitrogen tank reads a gage pressure of 28 in of mercury.
If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank?
A.
B.
C.
D.
E.
95 kPa
99 kPa
101 kPa
194 kPa
203 kPa
FE2.2
On a sea-level standard day, a pressure gage, moored below the surface of the ocean (SG
= 1.025), reads an absolute pressure of 1.4 MPa. How deep is the instrument?
A.
B.
C.
D.
E.
4m
129 m
133 m
140 m
2080 m
FE2.3
In Figure 6, if the oil in region B has SG = 0.8 and the absolute pressure at point A is 1
atm, what is the absolute pressure at point B?
A.
B.
C.
D.
E.
5.6 kPa
10.9 kPa
107 kPa
112 kPa
157 kPa
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Saleh David Ramezani
BIOMEDICAL APPLICATION
Blood circulation and blood pressure in different parts of the body could be seen as a
biomedical application of this section. The blood pressure at two given points, adjacent to
each other, in a blood capillary should ideally be the same. Also we know that the
pressure gradient has a major role in exchange of waste materials in the kidneys. The
blood pressure inside a regular capillary is about 25 mmHg or 25 torr. However, the
blood pressure inside capillaries within the glomerulus (tiny little ball of blood
capillaries) is about 90 mmHg. This pressure gradient allows for the exchange of the
waste product.
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Saleh David Ramezani
COVER SHEET
Homework # 5
Chapter 2: Pressure Distribution in a Fluid
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 2.122
1.1 Given
A uniform block of steel (SG = 7.85) will “float” at a mercury water interface as
in Figure 1..
Figure 7
Problem 2.122 Diagram
1.2 Find
What is the ratio of the distance ‘a’ and ‘b’ for this condition?
1.4 Assumptions
The fluid is hydrostatic and incompressible.
Specific Gravity of water = 1.0
Specific Gravity of Mercury = 13.56
1.5 Solution Calculations
SG steel  steelVsteel  SG water  waterVwater  SG Mercury MercuryVMercury
7.85(a  b)  1.0a  13.56b
a 13.65  7.85

 0.8467
b
7.85  1
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Saleh David Ramezani
1.6 Computational Results
The ratio of the distances a and b for this condition is:
a
 0.8467
b
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Saleh David Ramezani
2.0 PROBLEM 2.146
2.1 Given
The tank in Figure 3 is filled with water and has a vent hole at point A. The tank
is 1 m wide into the paper. Inside the tank, a 10cm balloon filled with helium at 130 kPa,
is a tethered centrally by a string.
Figure 1
Figure 8
Problem 2.146 Diagram
2.2 Find
If the tank accelerates to the right at 5 m/s2 in rigid body motion, at what angle
will the balloon lean? Will it lean to the right or to the left?
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2.2 Assumptions
Constant Temperature
No mixing of gas and water
θ
ax
g=9.81
Figure 4
Problem 2.146 Free Body Diagram
2.5 Solution Calculations
  tan  1
ax
g
5
 tan  1 (
m
s2
)  27 
m
9.81
s2
2.6 Computational Results
2. 7 Conclusions and Applications to Biofluids
Summarize the conclusions that can be drawn here.
Recommend or briefly describe an application in biofluids where this knowledge
may prove useful or necessary in order to solve a more complex problem.
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3.0 PROBLEM 2.155
3.1 Given
The U tube in Figure 3 contains mercury at 20 degrees Celsius.
Figure 9
Problem 3 Diagram
3.2 Find
For what uniform rotation rate in r/min about axis C will the U-tube take the
configuration shown?
.
3.4 Assumptions
Enter the assumptions for this problem here.
3.5 Solution Calculations
Enter the calculations here.
ax  bx  cx  0
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(3.1)
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The results of this analytical calculation can be seen below, in Section 3.6.
3.6 Computational Results
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FE PROBLEMS (EXTRA CREDIT)
FE2.8
A 5-m diameter balloon contains helium at 125 kPa absolute and 15˚C, moored in sealevel standard air. If the gas constant of helium is 2077 m2/(s2.K) and balloon material
weight is neglected, what is the net lifting force of the balloon?
F. 67 N
G. 134 N
H. 522 N
I. 653 N
J. 787 N
FE2.9
A square wooden (SG = 0.6) rod, 5 cm by 5 cm by 10 m long, floats vertically in water at
20˚C when 6 kg of steel (SG = 7.84) are attached to one end. How high above the water
surface does the wooden end of the rod protrude?
F.
G.
H.
I.
J.
0.6 m
1.6 m
1.9 m
2.4 m
4.0 m
FE2.10
A floating body will be stable when its
F. Center of gravity is above its center of buoyancy
G. Center of buoyancy is below the waterline
H. Never
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Saleh David Ramezani
BIOFLUIDS APPLICATION
The application of buoyancy in human body could be as simple as float of platelets or
other molecular substances in the blood. Referring to problem 2.147, we could think of
the container as human skull, and the water as the cerebrospinal fluid within. When in
motion the substances that make up this fluid could act like the balloon in the water.
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Saleh David Ramezani
COVER SHEET
Homework # 6
Chapter 3: Integral Relations for a Control Volume
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
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Saleh David Ramezani
1.0 PROBLEM 3.4
1.1 Given
The following flow conditions are given:
a.
b.
c.
d.
e.
Flow near an automobile moving at 55 mi/h
Flow of the wind past a water tower
Flow in a pipe as the downstream valve is opened at a uniform rate
River flow over the spillway of a dam
Flow in the ocean beneath a series of uniform propagating surface waves.
1.2 Find
Discuss whether these flow rates are steady or unsteady. Also elaborate if each
case is ambiguous.
1.4 Assumptions
Objects such as water tower are fixed in position.
Moving objects such as the car have a constant velocity or acceleration.
Wind is blowing at a constant velocity/acceleration.
1.5 Solution Calculations
a. Steady- the automobile is moving at 55 mi/h (a constant velocity), breaking
through the air molecules that were fixed in position before this point.
b. Steady- according to our assumption, if the wind is flowing at a constant
velocity/acceleration past the fixed in position water tower, we have a steady
flow. Other wise, if wind is flowing at a varying acceleration the flow is unsteady.
c. Unsteady- the fluid behind the closed valve has a potential energy that will
accelerate the flow as the valve is opened.
d. Steady- unless the dam spillway has just been opened, in which case the flow be
instable and unsteady (Water’s potential energy behind the dam, that will increase
the flow rate and acceleration).
e. Steady- though waves do not have a steady movement, and are steadily tabulating,
the water beneath these waves are imperturbable and in a steady flow.
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2.0 PROBLEM 3.17
2.1 Given
Incompressible steady flow in the inlet between parallel plates in figure 3.1 is
uniform, u = U0 = 8 cm/s, while downstream the flow develops into the parabolic laminar
profile u = az(z0 – z), where a is a constant.
Figure 10.1
Problem 3.17 Diagram
2.2 Find
If z0 = 4 cm and the fluid is AE 30 oil at 20˚C, what is the value of umax in cm/s?
2.2 Assumptions
Incompressible flow
No flow through the sides
Steady flow
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2.5 Solution Calculations
Flowin  Flowout
Qin  Qout
w  width of the pathway in the paper
z0
U
z0
0
wdz   az ( z 0  z ) wdz
0
0
z 0 wU 0 
a
awz 0
6
3
6U 0
z0
2
6U 0 3U 0

16
8
We know that the maximum velocity would be at the furthest point of the
parabola in figure 3.1. In terms of z or height of the pathway this point is expected to be
in the center.
z
z max  0
2
u  u max  az ( z 0  z )
Substituting z with the zmax value and z0 with the given height value of 4 cm we
get:
a
z0
z
4
4
)( z 0  0 )  a( )( 4  )  4a
2
2
2
2
Substituting (1) in (2)
u  u max  a(
(2)
3U 0
2
cm
cm
U0  8
 u max  12
s
s
u max 
2.6 Computational Results
The maximum velocity between the plates is at the end of the parabola and the
value is 12 cm/s.
u max  12
Homework 3.doc
cm
s
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3.0 PROBLEM 3.30
3.1 Given
The V-shaped tank in Figure 3.2 has width b into the paper and is filled from the
inlet pipe to volume flow Q.
Figure 3.2
Problem 3.30 Diagram
3.2 Find
Derive expressions for (a) the rate of change dh/dt and (b) the time required for
the surface to rise from h1 to h2
3.4 Assumptions
Incompressible flow
Steady flow
Uniform angle with the base along the sides
3.5 Solution Calculations
w
20˚
h
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a)
h
tan 
b is the length of the container in the page
1
Volueme of the container : V  h.2 w.b
2
2
h .b
V 
tan 20
h 2 .b
d(
)
dV
dh
b
tan
20


(
)Q
dt
dt
dt tan 20
dh Q tan 20

dt
b
w
b)
dh Q tan 20

dt
b
b
 Q tan 20 dh   dt
H
2
b
t
dh
Q tan 20 H1
t
b
( H 2 )  ( H 1 )
Q tan 20
3.6 Computational Results
a)
dh Q tan 20

dt
b
b)
t
b
( H 2 )  ( H1 )
Q tan 20
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4.0 EXTRA PROBLEM 3.38
4.1 Given
An incompressible fluid in Figure 3.3 is being squeezed outward between two
large circular disks by the uniform downward motion V0 of the upper disk.
Figure 3.3
Problem 3.38 Diagram
4.2 Find
Assuming one-dimensional radial outflow, use the control volume to derive an
expression for V(r)
.
4.4 Assumptions
Steady flow of fluid off the sides
Incompressible flow
Uniform force applied on the discs
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4.5 Solution Calculations
Lest ' s say B  mass
dB dm

1
dm dm

d
d 
( Bsyst )    dv     (V .n)dA
dt
dt  CV
 CS
d
d
Conservation of mass  ( Bsyst )  (m syst )  0
dt
dt

d 
  dv     (V (r ).n)dA  0

dt  CV
 CS

2 r h
d
(
 .r.dhdrd ) 
dt 0 0 0
2 h
   .V .rdhd
0 0
d
( r 2 h)  V (r )2rh  0
dt
dh
r 2   V (r )2rh
dt
dh
r  2V (r )h
dt
dsplacement dh
Velocity 

 V0
time
dt
V0 r  2V (r )h
V (r )  
V0 r
2h
4.6 Computational Results
V (r )  
V0 r
2h
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3
Saleh David Ramezani
FE PROBLEMS (EXTRA CREDIT)
FE 3.1
In Figure 3.4 water exits from a nozzle into atmospheric pressure of 101 kPa. If
the flow rate is 160 gal/min, what is the average velocity at section 1?
a) 2.6 m/s
b) 0.81 m/s
c) 93 m/s
d) 23 m/s
e) 1.62 m/s
Figure 3.4
Problem FE 3.1 Diagram
1) Convert 160 gal/min to the SI unit
2) Q=0.10094 m3/s
3) To get the velocity with the m/s unit we would divide the flow rate by the
section area. Here Q/A=0.10094/0.003848  V=2.623 m/s
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Saleh David Ramezani
Biofluids Application
The Biomedical application of this section could be seen as flow rate
measurements of specific fluids (e.g. blood, oxygen, urine) within the human body. We
could also use the Reynolds transport theorem to track certain molecules within these
fluids. For example we may want to track and measure the behavior of certain injected
drugs in the blood stream.
Homework 3.doc
Page 33 of 123
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Saleh David Ramezani
Homework # 7
Chapter 3: Integral Relations for a Control Volume
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 3.85
1.1 Given
The thin-plate orifice in Figure 1 causes a large pressure drop. For 20˚C water
flow at 500 gal/min, with pipe D=10 cm and orifice d = 6 cm , p1-p2 ≈ kPa.
Figure 1
Problem 3.85 Diagram
1.2 Find
If the wall friction is negligible, estimate the force of the water on the orifice
plate.
1.2 Assumptions
Steady flow
Incompressible fluid
Frictionless flow
Flow along a single streamline
No shaft work between points 1 and 2
No heat transfer between points 1 and 2
1.5 Solution Calculations
d 

 F  dt   Vd    (m V )
 CV

i
i out
  (m iVi ) in
The volume integral term vanishes for steady flow:
Homework 3.doc
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BIEN 301
3
 F  (m V )   (m V )
 F  m V m V  ( A V )V
i
2
2
i out
1 1
i
i in
1 1
1
Saleh David Ramezani

 ( A2V2 )V2
From conservation of mass we have:
m 1  m 2  m
Since the inlet density, and area are the same as those of the output V1=V2. and
we’ll have:
F  0
Now if we write out the sum of forces exerted on the system (which in this
problem are along the x-axes):
F
x
 p1 A1  p 2 A2  Fplate   F  0
Fplate  ( p1  p 2 ) A
Fplate  (145000 Pa)( )(
0.1m 2
)
2
Fplate  1138.83 N
1.6 Computational Results
The force acting on the plate is thus,
F plate  1138.83 N
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Saleh David Ramezani
2.0 PROBLEM 3.138
2.1 Given
Students in the fluid mechanics laboratory at Penn State use a very simple device
to measure the viscosity of water as a function of temperature. The viscometer shown in
Figure 2, consists of a tank , a long vertical capillary tube, a graduated cylinder, a
thermometer, and a stopwatch. Because the tube has such a small diameter, the flow
remains laminar. Because the tube is so long, entrance losses are negligible. It will be
shown in Chapter 6 that the laminar head loss through a long pipe is given by hf-laminar =
(32μLV)(ρgd2), where V is the average speed through the pipe.
Figure 2
Problem 3.138 Diagram
2.2 Find
(a) In a given experiment, diameter d, length L, and water level height H are
known, and value flow rate Q is measured with the stopwatch and graduated cylinder.
The temperature of the water is also measured. The water density at this temperature is
obtained by weighting a known volume of water. Write an expression for the viscosity of
the water as a function of these variables.
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(b) Here are some actual data from an experiment: T=16.5˚C, ρ=998.7 kg/m3,
d=0.041 in, Q=0.310 mL/s, L=36.1 in, and H=0.153 m. Calculate the viscosity of theater
in kg/(m.s) based on these experimental data.
(c) Compare the experimental result with the published value of μ at this
temperature, and report a percentage error.
(d) Compute the percentage error in the calculation of μ that would occur if a
student forgot to include the kinetic energy flux correction factor in part (b) (compare
results with and without inclusion of kinetic energy flux correction factor). Explain the
importance (or lack of importance) of kinetic energy flux correction factor in a problem
such as this.
2.2 Assumptions
Steady flow
Incompressible fluid
Flow along a single streamline
No shaft work between points 1 and 2
No heat transfer between points 1 and 2
2.5 Solution Calculations
a) The energy equation is:
 p V 2

 p V 2

 
 z    
 z   h friction  h pump  hturbine
2g
2g

 in  
 out
Since both inlet and outlet surfaces are exposed to atmospheric pressure, the
pressure terms cancel out. Also the inlet velocity is equal to zero, therefore:


 z   h friction  h pump  hturbine
 2g
 out
z in    2V
2
 V 2

( H  L)   2
 0   h friction  0  0
 2g
 out
32 LV
h friction 
gd 2
h friction  H  L 

 2V 2
2g

32 LV
gd 2

 ( H  L)( 2 g )   2V 2 gd 2 

64 gLV



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Saleh David Ramezani
We can write Q in term of V:
 d 2 
V
Q  AV  
 4 
 4Q 
V  2
 d 
Therefore the viscosity becomes:
2


 4Q  
2

(
H

L
)(
2
g
)



gd



2
2
 d  
 



 4Q 


64 gL 2 


 d 


Simplify to get:

gd 4
128LQ
( H  L) 
 2 Q
16L
b)
T  16.5  C
  998.7
kg
m3
0.0254m
 0.001041m
1in
mL
1L
1m 3
Q  0.310


 3.1  10 7 m 3
s 1000mL 1000 L
0.0254m
L  36.1in 
 0.9169m
1in
H  0.153m
d  0.041in 
And for a fully developed laminar pipe flow α = 2.0. We solve for μ:
Homework 3.doc
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BIEN 301


3
gd 4
128 LQ
( H  L) 
Saleh David Ramezani
2
16L
kg
m
kg
4
 9.81 2  0.001041m 
2  998.7 3  3.1  10 7 m 3
3
m
s
m
(0.153m  0.917 m) 
16    0.917 m
128  0.917m  3.1  10 7 m 3
  998.7
  0.001045
N .s
m2
c)
From Table A.1 the published value of μ at 10˚ C and 20˚ C are respectively
N .s
N .s
1.307  10 3 2 and 1.003  10 3 2 . With a simple linear interpolation at 16.5˚ C we
m
m
get:
0.001003  16.5
N .s
20  16.5
  Theoretical (16.5)  0.001109 2

m
20  10
0.001003  0.001307
The percent error equation, thus, yields:
Percent error 
 exp erimental   theoretical
 theoretical
 100 
0.001045  0.0011094
0.0011094
 100  5.8%
d)
If the kinetic energy coefficient factor was left out:

kg
m
kg
4
 9.81 2  0.001041m 
998.7 3  3.1  10 7 m 3
3
m
s
m
(0.153m  0.917 m) 
7
3
16    0.917 m
128  0.917 m  3.1  10 m
  998.7
  0.001056
N .s
m2
Which is very close to the number calculated in part b) where the kinetic energy
coefficient factor was taken in account.
Homework 3.doc
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Saleh David Ramezani
2.6 Computational Results
a)

gd 4
128LQ
( H  L) 
2
16L
b)
  0.001045
kg
m.s
c)
Percent Error  5.8%
d)
  0.001056
Homework 3.doc
N .s
m2
Page 41 of 123
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Saleh David Ramezani
3.0 EXTRA PROBLEM 3.132
3.1 Given
Consider a turbine extracting energy from a penstock in a dam, as in Figure 3. for
turbulent pipe flow (Chapter 6), the friction head loss is approximately hf=CQ2 where the
constant C depends on penstock dimensions and the properties of water.
Figure 3
Problem 3.132 Diagram
3.2 Find
Show that, for a given penstock geometry and variable river flow Q, the
maximum turbine power possible in this case is Pmax=2ρgHQ/3 and occurs when the flow
rate is Q  H /(3C ) .
3.4 Assumptions
Steady flow
Incompressible fluid
Flow along a single streamline
3.5 Solution Calculations
3.6 Computational Results
Homework 3.doc
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BIEN 301
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Saleh David Ramezani
FE PROBLEMS (EXTRA CREDIT)
FE 3.9
Water flowing in a 2mooth 6-cm diameter pipe enters a venture contraction with a
throat diameter of 3 cm. Upstream pressure is 120 kPa. If cavitation occurs in the throat
at a flow rate of 155 gal/min, what is the estimated fluid vapor pressure, assuming deal
frictionless flow?
f) 6 kPa
g) 12 kPa
h) 24 kPa
i) 31 kPa
j) 52 kPa
Homework 3.doc
Page 43 of 123
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BIEN 301
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Saleh David Ramezani
BIOFLUIDS APPLICATION
The Biomedical application of this section could be seen as steady flow measurements of
a fluid such as blood down the veins. How long it takes for the blood to travel a certain
distance, how might body temperature and friction of the veins affect this rate, are
examples of the energy equation applications. The z terms in the energy equation is
another useful variable that would allow us to compare the mentioned characteristics at
different elevations, and in respect with one another.
Homework 3.doc
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BIEN 301
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Saleh David Ramezani
COVER SHEET
Homework # 8
Chapter 3: Integral Relations for a Control Volume
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 3.161
1.1 Given
A necked-down section in a pipe flow, called a venturi, develops a low throat
pressure that can aspirate fluid upward from a reservoir, as in Figure 1.
Figure 1
Problem 3.161 Diagram
1.2 Find
Using Bernoulli’s equation with no losses, derive an expression for the velocity
V1 that is just sufficient to bring reservoir fluid into the throat.
1.4 Assumptions
Steady flow
Incompressible fluid
Frictionless flow
Flow along a single streamline
No shaft work between points 1 and 2
No heat transfer between points 1 and 2
1.5 Solution Calculations
The general form of Bernoulli equation for steady frictionless
incompressible flow along a streamline is:
p1 1 2
p 1 2
 V1  gz1  2  V2  gz2
 2
 2
Homework 3.doc
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3
Saleh David Ramezani
Now if:
p1  p atm  gh
p 2  p atm
And:
z1 = z2 = 0
The equation becomes:
patm  gh

1 2 p
1 2
 V1  atm  V2
2

2
Since we have a steady flow:
Qin  Qout
V1 A1  V2 A2
 D12 
 D2 2 




V1 
  V2  4 
4




V1D1  V2 D2
2
2
2
D 
V2  V1  1 
 D2 
The equation thus becomes,
p atm  gh

V1 
D2
2
p
1
1 2 D
 V1  atm  V1  1
2

2  D2
2 gh
D2  D1
4
4
or V1 
 D2
2
2
2 gh
D2  D1
4



4
1.6 Computational Results
V1 
D2
2
2 gh
D2  D1
Homework 3.doc
4
4
or V1 
 D2
2
2 gh
D2  D1
4
4
Page 47 of 123
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Saleh David Ramezani
2.0 PROBLEM 3.172
2.1 Given
The 35˚ C water flow in Figure 2 discharges to sea level standard atmosphere.
Figure 2
Problem 3.172 Diagram
2.2 Find
a) Neglecting losses, for what nozzle diameter D will cavitation begin to occur?
b) To avoid cavitation, should you increase or decrease D from this critical value?
2.2 Assumptions
Steady flow
Incompressible fluid
Frictionless flow
Flow along a single streamline
No shaft work between points 1 and 2
No heat transfer between points 1 and 2
2.5 Solution Calculations
Cavitation occurs when water pressure is lowered to a point where water starts to
boil. The saturation points of water at various temperatures are available in
thermodynamic property Table A.5. For easiness I used the exact value from the
thermodynamic textbook table. Thus, the saturation pressure point at 35˚ C was found to
be 5628 Pa.
We start out by writing the Bernoulli equation from point 1 to 3.
p
1 2
1 2
 V1  gz1  3  V3  gz 3
 2
 2
p1
Homework 3.doc
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Saleh David Ramezani
Where:
p1  5628Pa
p 2  p a  101325Pa
kg
m3
z1  z 2  0 or z  0
  998.7
For a steady flow we also have:
Qin  Qout
V1 A1  V3 A3
 D1 2
V1 
 4
 D 2

  V3  3

 4


V1 D1  V2 D2
2




2
2
D 
V1  V3  3 
 D1 
0.0254m
D1  1in 
 0.0254m
1in
 D3 
V1  V3 

 0.0254m 
2
The equation becomes:
2
5628 Pa 1   D3  
101325Pa 1 2
 V3 
 V 0
   0 
kg 2   0.0254m  
kg 2 3
998.7 3
998.7 3
m
m
0.351625
V3 
D3  0.000645
From example 3.21 we learned that the discharge velocity equals the speed that a
frictionless particle would attain if it fell freely from point 1 to point 2. Suggested by
Evangelista Torricelli I, discharge velocity is independent of the fluid density, a
characteristic of gravity-driven flow. This velocity is approximately equal to
Vdischarg e  2 gh .
Knowing this, we can solve for D3:
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Vdischarg e  V3  2 gh  2  9.81
5.99 
Saleh David Ramezani
m
0.3048m
m
 6 ft 
 5.99
2
1 ft
s
s
0.351625
D3  0.000645
D3  0.004091m 
1 ft
 0.1342 ft
0.3048m
D3  0.1342 ft
Cavitation is defined as “The production of voids in a liquid system due to
extreme reduction of internal pressure” (www.trident-itw.com/glossary.asp). So to avoid
cavitation we would need to increase the pressure inside our system. From the Bernoulli
equation, it’s evident that an increase in velocity will result in an internal pressure
reduction. Since here the outlet diameter is inversely related to the outlet velocity:
0.351625
a decrease in diameter will increase the velocity. So to avoid
V3 
D3  0.000645
cavitation we would want to decrease the outlet diameter.
2.6 Computational Results
a)
D3  0.1342 ft
b)
To avoid cavitation, we would decrease the diameter.
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Saleh David Ramezani
3.0 EXTRA PROBLEM 3.180
3.1 Given
The large tank of incompressible liquid in Figure 3 is at rest when, at t=0. the
valve is opened to the atmosphere.
Figure 3
Problem 3.180 Diagram
3.2 Find
Assuming h ≈ constant (negligible velocities and accelerations in the tank). Use
the unsteady frictionless Bernoulli equation to derive and solve a differential equation for
V(t) in the pipe.
3.4 Assumptions
Unsteady flow
Incompressible fluid
Frictionless flow
Flow along a single streamline
No shaft work between points 1 and 2
No heat transfer between points 1 and 2
3.5 Solution Calculations
We start by writing the unsteady Bernoulli equation

2
1
2 dp
V
1
ds  
 V22  V12  g z 2  z1   0
1 
t
2
Homework 3.doc


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Since both inlet and outlet surfaces are exposed to the atmospheric pressure and
there is no variation of density with pressure, P1=P2 and thus the pressure integral will
cancel out. We also have:
z1  h
z2  0
V1  0
V2  V (t )
If we rewrite the equation we’ll have:

2
1
V
1
ds  V 2  gh  0
t
2
Where ds is the change in length and could be rewritten as dL, the integral
becomes:
V
1
dL  V 2  gh  0
0 t
2
V
1 2
L  V  gh  0
t
2

L
This is a first order non-linear differential equation. I tried to solve the equation
but I was unsuccessful. This could have been a Bernoulli differential equation if it was in
the standard form of:
V
 P(t )V  f (t )V n
t
Which would have been easier to solve. But this is not the case since the first order V is
missing in the equation.
3.6 Computational Results
The answer of the following differential equation:
V
1
L  V 2  gh  0
t
2
Homework 3.doc
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FE PROBLEMS (EXTRA CREDIT)
FE3.10
Water flowing in a smooth 6-cm diameter pipe enters a venture contraction with a
throat diameter of 4 cm. upstream pressure is 120 kPa. If the pressure in the throat is 50
kPa. What is the flow rate, assuming ideal frictionless flow?
k) 7.5 gal/min
l) 236 gal/min
m) 263 gal/min
n) 745 gal/min
o) 1053 gal/min
This problem is worked out exactly like example 3.23 on page 189. We start by
writing the Bernoulli equation where z1=z2. We then solve V2 in terms of V1 and
pressure difference. We finally find Q by substituting the values into the equation.
Homework 3.doc
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BIOFLUIDS APPLICATION
The Biomedical application of this section could be seen as steady or unsteady
flow rate measurements of specific fluids (e.g. blood, oxygen, urine) within the human
body. Let’s say that we want to find out the time it takes for a drug or a substance to
reach a certain point in a vein. If we know the start point’s diameter and the end point’s
diameter, as well as the initial velocity, using Bernoulli’s equation we would be able to
calculate the time it would take for that substance to reach the second point. Of course
other factors such as blood density and pressure are involved in such problems.
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BIEN 301
3
Saleh David Ramezani
Homework # 9
Chapter 4: Differential Relations for Fluid Flow
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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6-feb.-16
BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 4.9
1.1 Given
An idealized incompressible flow has the proposed three-dimensional velocity
distribution:
V  4 xy2 i  f ( y) j  zy 2 k
1.2 Find
Find the appropriate form of the function f(y) that satisfies the continuity relation.
1.3 Assumptions
Incompressible
Density and velocity are continuum functions
Steady or Unsteady
Viscous or frictionless
Compressible or Incompressible
1.4 Solution Calculations
We start by writing the continuity equation:
 


 ( u )  ( v)  ( w)  0
t x
y
z
Since the flow is incompressible, the density changes are negligible. Therefore,

 0 regardless of steadiness or unsteadiness of the flow. Ignoring the density term in
t
above equation we will get:
u v w


0
x y z
Homework 3.doc
Page 56 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
If u  4xy 2 , v  f ( y ) , and w  zy 2 :
 (4 xy2 )  ( f ( y ))  ( zy 2 )


0
x
y
z
df ( y )
4 y2 
 y2  0
dy
df ( y )
 3 y 2  df ( y )  3 y 2 dy
dy
 df ( y)    3 y dy
2
f ( y)   y 3  C
1.5 Computational Results
f ( y)   y 3  C
Homework 3.doc
Page 57 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
2.0 PROBLEM 4.38
2.1 Given
The approximate velocity profile in Problem P.3.18 and Figure 1 for steady
laminar flow through a duct, was suggested as

y 2  z 2 
u  u max 1  2 1  2 
 b  h 
With v=w=0, it satisfied the no-slip condition and gave a reasonable volume flow
estimate (which was the point of problem P.3.18).
Figure 1
Problem 4.38 Diagram
2.2 Find
Show, however, that it does not satisfy the x-momentum Navier Stokes equation
for duct flow with constant pressure gradient p / x  0 . For extra credit, explain briefly
how the actual exact solution to this problem is obtained (see, for example, Ref. 5, pp.
120-121).
2.3 Assumptions
Incompressible
Constant density and viscosity
Steady flow
u and v independent of time.
Homework 3.doc
Page 58 of 123
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BIEN 301
3
Saleh David Ramezani
1.4 Solution Calculations
We start by writing the x-momentum of the Navier-Stoke equation:
  2u  2u  2u 
p
du
   2  2  2   
x
dt
y
z 
 x
The problem states that the pressure gradient is a constant smaller than zero.
g x 
p
C 0
x
  2u
 2u
 2u 
du
g x    2  2  2   
C
dt
y
z 
 x
Where
du u
u
u
u
and since v  w  0 , the equation becomes:

u
v
w
dt t
x
y
z
  2u
 2u
 2u 
u
u


g x    2  2  2      u   C
x 
y
z 
 t
 x

y 2  z 2 
Substituting u  u max 1  2 1  2  into the equation we will have:
 b  h 
 2

y 2 
z2 
2

y 2 
z 2 
g x    2 u max 1  2 1  2   2 u max 1  2 1  2    C
 b  h  z
 b  h  
 y
u 
 2u
 u
Note that the    u  , and 2 terms were canceled out since there were no
x 
x
 t
x and t variables in the velocity profile equation u. If we continue with the
differentiations we’ll have:
g x  
 y2 y2z2
2

u
 2 2
max  
2
y 2
b h
 b

 z2 y2z2
2
   2 u max   2  2 2
z
b h

 h

 2
2
2z 2 
2y2





u


max 
2
2
b 2 h 2 
b2h2
 b
 h
 2( z 2  y 2  b 2  h 2 ) 
  C
g x  u max 
2 2
b
h


g x  u max  
Homework 3.doc
Page 59 of 123

  C


  C

6-feb.-16
BIEN 301
3
Saleh David Ramezani
Though b and h, the width and height on the left hand side are constant, the y and
z variables are changing. Therefore,
 2( z 2  y 2  b 2  h 2 ) 
  C
2 2
b
h


g x  u max 
2.5 Computational Results
 2( z 2  y 2  b 2  h 2 ) 
  C
b2h2


g x  u max 
Homework 3.doc
Page 60 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
3.0 PROBLEM 4.41
3.1 Given
As mentioned in section 4.11, the velocity profile for laminar flow between two
plates, as in Figure 2, is
u
4u max y(h  y)
h2
vw0
Figure 2
Problem 4.41 Diagram
3.2 Find
If the wall temperature is Tw at both walls, use the incompressible flow
energy equation (4.75) to solve for the temperature distribution T(y) between the walls
for steady flow.
3.3 Assumptions
Incompressible
Steady flow
Constant density and viscosity
3.4 Solution Calculations
The temperature dependent energy equation is:
Homework 3.doc
Page 61 of 123
6-feb.-16
BIEN 301
c p
3
Saleh David Ramezani
dT
 k 2T  
dt
Where  is given by:
2
2
2
2
2
  u  2
 v 
 w   v u   w v   vu w  
   2   2   2       
   

 
 z   x y   y z   z x  
  x 
 y 
But since v  w  0 , the v and w containing partial derivatives will cancel out and
we’ll have:
  u  2  u  2  u  2 
   2        
  x   y   z  
There are also no x and z terms in the velocity equation, therefore:
 u 
    
 y 
2
We know that  2T is defined as:
 2T 
 2T  2T  2T


x 2 y 2 z 2
Since T is a linear function of y (T=T(y)) the x and z containing partial derivatives
would vanish:
 2T
 T 2
y
2
Rearranging the energy formula we’ll get:
 u 
dT
 2T
c p
 k 2    
dt
y
 y 
2
From equation 4.54 we can expand the
dT
term to get:
dt
dT T
T
T
T

u
v
w
dt
dt
dx
dy
dz
Homework 3.doc
Page 62 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
Again since T is a linear function of only y, and v=w=0, all partial derivatives
including the time dependent derivative will cancel out and we’ll have:
dT
0
dt
Rearranging the energy equation again we’ll have:
 u 
 2T
k 2     
y
 y 
2
Now that we have a simpler form of the energy equation we can substitute the
velocity equation for u; and to find T, we would futher integrate both sides of the
resulting equation:
  4u max y (h  y )  
 

2
 T
h2



k 2   

y
y




k
 2T
 4u h 8u max y 
    max


2
2
y
h2 
 h
2
2
 4u max
 2T
h  2 y 2
k 2 
2
y
h
 4u max 2
 2T
k 2 
h  4y2  4y
2
y
h

k
 4u max 2
 2T

h  4y2  4y
2
2
y
h

k
16 u max y 2 16 u max y
 2T



4
u


max
y 2
h2
h2


k
 2T 4 u max

y 2
h2
T 4 u max
k

y
h2
 4 y  4 y
2

 h 2 dy
 2 4y3

 2 y 
 h 2 y  C 
3


Integrating again to get T, we’ll have:
Homework 3.doc
Page 63 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
 
T 4u max   2 4 y 3

 h 2 y  C1 dy 
   2 y 
2
y
3
h
 
 

4u max  2 y 3 y 4 h 2 y 2

T


 C1 y   C 2
2
3
2
h
 3

Not sure what boundary conditions are, but that at y = 0 and y = h, T=Tw.
k
3.5 Computational Results
T
4umax
h2
Homework 3.doc
 2 y3 y 4 h2 y 2


 
 C1 y   C2
3
2
 3

Page 64 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
4.0 EXTRA PROBLEM 4.17
4.1 Given
A reasonable approximation for the two-dimensional incompressible laminar
boundary layer on the flat surface in Figure 3 is
 2y y2 
u  U 
 2  for y   where   C x , C  constant
  
Figure 3
Problem 4.17 Diagram
4.2 Find
(a) Assuming a no-slip condition at the wall, find an expression for the velocity
component v(x, y) for y   .
(b) Then find the maximum value for v at the station x=1 m, for the particular
case of airflow, where U = 3 m/s and  =1.1 cm
.
4.3 Assumptions
4.4 Solution Calculations
4.5 Computational Results
Homework 3.doc
Page 65 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
FE PROBLEMS (EXTRA CREDIT)
FE 4.1
Given the steady, incompressible velocity distribution V  3xi  Cyj  0k , where
C is a constant, if conservation of mass is satisfied, the value of C should be
a) 3
b) 3/2
c) 0
d) -3/2
e) -3
Homework 3.doc
Page 66 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
BIOFLUIDS APPLICATION
The Biomedical application of this section could be seen as the steady or non
steady flow rate measurements of specific fluids (e.g. blood, oxygen, urine) within the
human body. For example we can measure the flow rate of the high pressure oxygenated
blood that’s leaving the heart in multiple directions and at the same time. The differential
form of the energy equation is another useful equation that contains the Temperature
variable within. This equation should ideally allow us to study the effect of temperature
on specific flow rates.
Homework 3.doc
Page 67 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
Homework # 10
Chapter 4: Differential Relations for Fluid Flow
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
Page 68 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 4.59
1.1 Given
Consider the two-dimensional incompressible velocity potential
  xy  x 2  y 2
1.2 Find
(a) Is it true that  2  0 , and if so, what does this mean?
(b) If it exists, find the stream function  ( x, y ) of this flow.
(c) Find the equation of the streamline that passes through (x, y) = (2, 1).
1.3 Assumptions
Incompressible
Two-dimensional flow
1.4 Solution Calculations
a)
 2 
 2  2  2


x 2 y 2 z 2
 2  2

 22  0
x 2 y 2
Yes it’s true. This means that the flow is irrotational and the incompressible
continuity equation checks.
 2 
b)


u  x  y  2 x

v    x  2 y

y


u  y  y  2 x

v     x  2 y

x
y2
 
 2 xy  f ( x)
2
Homework 3.doc
(1)
(2)
(3)
Page 69 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
If we differentiate (3) with respect to x and compare with (2), we’ll get:

 2 y  f ' ( x)  2 y  x
x
x2
f ( x)  
C
2

f ' ( x)   x
And therefore,
 
y2
x2
 2 xy 
C
2
2
c)
Substituting (2, 1) for (x, y) in the stream function and choosing a value for C,
here C=0, we’ll have:
y2
x2
 2 xy  
C
2
2
1
22
 (2,1)   2(2)(1)   0
2
2
5
 (2,1) 
2
According to example 4.7 this is a constant value equal to . Therefore the
equation of the line passing through the point (2, 1) would become:
 ( x, y ) 
 ( x, y ) 
Homework 3.doc
y2
x2 5
 2 xy  

2
2 2
Page 70 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
2.0 PROBLEM 4.81
2.1 Given
Figure 1 is given:
Figure 1
Problem 4.81 Diagram
2.2 Find
(a) Modify the analysis of Figure 1 to find the velocity uθ when the inner cylinder
is fixed and the outer cylinder rotates at angular velocity Ω0.
(b) May this solution be added to equation below to represent the flow caused
when both inner and outer cylinders rotate? Explain your conclusion.
v   i ri
r0 / r  r / r0
r0 / ri  ri / r0
2.3 Assumptions
Incompressible
Constant density and viscosity
Circular symmetry
No axial motion or end effect (vz = 0)
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
2.4 Solution Calculations
For the conditions of this problem, almost like the one on page 276, all terms of
the continuity equation are zero but the last term (for the same reasons stated on page
276)
1 d  dv  v
r

r dr  dr  r 2
The left hand side of this equation is a laplacian in cylindrical coordinate with the
z terms equal to zero. This linear second order differential equation can be solved to get:
 2 v 
C2
r
v  C1 r 
The constants are found by the no-slip condition at the inner and outer cylinders:
C2
ri
Inner, at r = ri
v  0  C1ri 
Outer, at r= ro
v   o ro  C1r0 
C2
r0
Solve for C1 and C2:
C1 
  0 r0
2
ri  ro
2
2
ri  o ro
2
C2 
ri  ro
2
2
2
If we substitute C1 and C2 in the solved differential equation, we’ll get:
v 
  o ro
2
ri  ro
2
2
ri  o ro
2
r
r
2
  o ro r 2  ri  o ro
r
2
2
i

2
ro r 2  ri ro
2
r
v   o ro
2

r 2  ri
r
i
Homework 3.doc
2
 ro r
2
i
2
2
 ro r
2
v   o

 ro r
2
i
2
v 
2
2
2

 ro r
2
Page 72 of 123
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BIEN 301
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Saleh David Ramezani
2.5 Computational Results
r 2  ri
2
2
ri  ro r
2
v  o ro
2
Homework 3.doc


Page 73 of 123
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BIEN 301
3
Saleh David Ramezani
FE PROBLEMS (EXTRA CREDIT)
FE 4.3
Given the steady, incompressible velocity distribution V=3xi+Cyj+0k, where C is
a constant, the shear stress  xy at the point (x, y, z) is given by
f) 3μ
g) (3x+Cy)μ
h) 0
i) Cμ
j) (3+C)μ
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Saleh David Ramezani
BIOFLUIDS APPLICATION
The Biomedical application of this section could be seen as the steady or non
steady flow rate measurements of specific fluids (e.g. blood, oxygen, urine) within the
human body. For example we can measure the flow rate of the high pressure oxygenated
blood that’s leaving the heart in multiple directions and at the same time. The differential
form of the energy equation is another useful equation that contains the Temperature
variable within. This equation should ideally allow us to study the effect of temperature
on specific flow rates.
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Homework # 11
Chapter 6: Viscous Flow in Ducts
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 6.7
1.1 Given
Cola, approximated as pure water at 20˚ C, is to fill an 8-oz container (1 U.S gal
=128 fl oz) through a 5-mm-diamter tube.
No Sketch
1.2 Find
a) Estimate the minimum filling time if the tube flow is to remain laminar.
b) For what cola (water) temperature would this minimum time be 1 min?
1.3 Assumptions
Incompressible
Fully developed
Steady flow
1.4 Solution Calculations
a)
We start by writing the Reynolds equation:
Re 
Vd

From example 6.1 we accept the transition Reynolds number for flow in a circular
pipe to be around 2300.
Re  2300
kg
  998 3
m
  1.003  10 3
N .s
kg
or
2
m.s
m
d  0.005m
We solve for Velocity:
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kg
 2300
m
m.s
 0.4623
kg
s
998 3  0.005m
m
To find the flow rate:
Q
4Q
V 
A  d 2
 Re
V

d
Q
1.003  10 3
 d 2 V
4


4
 0.005m 2  0.4623
Q
m

s
 9.08  10 6
m3
s
This much cola passes through the tube every second. To find the time it takes for
8 oz of the Cola to pass through the tube we would divide the amount by the flow
rate:
1gal
3.785  10 4 m 3
8 fl  oz 

v
2.366  10  4 m 3
128 fl  oz
1gal
t 

 26 s
3
3
Q
6 m
6 m
9.08  10
9.08  10
s
s
b)
To fill an 8 fl-oz of cola container with cola at one minute:
v  2.366  10 4 m 3
t  60 s
3
v 2.366  10  4 m 3
6 m
Q 
 3.943  10
t
60 s
s
Q  VA
 Re
V 
d
3
 Re  d 2 
  2300   (0.005m) 2
6 m


Q


 3.943  10
d  4    0.005m
4
s

m2
 4.366  10 7

s


This is the kinetic viscosity equation  v   . From appendix A.1, and with a


linear interpolation, we find the corresponding temperature value to be:
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4.75  10 7  4.366  10 7
70  T

7
7
70  60
4.75  10  4.14  10

T  64 C
1.5 Computational Results
a)
The minimum filling time:
t  26s
b)
For the filling time to be exactly 1 min, the temperature needs to be:
T  64  C
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2.0 PROBLEM 6.56
2.1 Given
Consider a horizontal 4-ft-diamater galvanized iron pipe simulating the Alaskan
pipeline. The oil flow is 70 million U.S. gallon per day, at a density of 910 kg/m3 and
viscosity of 0.01 kg/(m.s) (see figure A.1 for SAE 30 oil at 100˚ C). Each pump along the
line raises the oil pressure to 8 MPa, which then drops, due to head loss, to 400 kPa at the
entrance to the next pump.
No Sketch
2.2 Find
Estimate
(a) the appropriate distance between pumping stations and
(b) the power required if the pus are 88 percent efficient.
2.3 Assumptions
Incompressible
Fully developed
Steady flow
1.4 Solution Calculations
We would first need to find the Reynolds number. We start by writing each
individual term in the equation:
Re 
Vd

Diameter of the pipe is given in the problem as:
d  4 ft 
0.3048m
 1.2192m
1 ft
kg
m3
kg
  0.01
m.s
Q
V 
A
  910
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BIEN 301
3
gal 0.003785m 3 1day
1hr
1 min
m3




 3.067
day
1gal
24hr 60 min
60 s
s
Q  7  10 7
A
d 2
4

Saleh David Ramezani
 1.2192 2
4
 1.167 m 2
3
m
s  2.627 m
V 
s
1.167 m 2
3.067
And finally:
Vd
Re 


910
kg
m
 2.627  1.2192m
3
s
m
 291458
kg
0.01
m.s
To find the distance between the pumping stations we would use equation 6.10:
hf  f
LV2
d 2g
We would first find the value of f, the friction factor. Equation 6.49 gives f as:
1.11

  

1
6.9  d  
 1.8 log 


Re d  3.7  
f


 
From table 6.1 the recommended roughness value of the galvanized iron pipe was
found to be:
  0.15mm  0.00015m
Using this info we can find the roughness ratio ε/d to be:

d

0.00015m
 0.000123
1.2192m
Friction factor equation becomes:
1.11
 6.9
 0.000123  
 1.8 log 

 
f
 291458  3.7  
1
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Solving for f we’ll get:
f  0.01548
The friction head equation 6.8 gives:
h f  z 
p
g
Assuming the pipe level to be constant, Δz = 0. The length equation becomes:
p
LV2
 f
g
d 2g
2pd 2  800000 Pa  400000 Pa   1.2192m
L

 190605.6m
2
fV 2
kg 
m
0.01548  910 3   2.627 
s
m 
b)
Similar to example 4.12, the power required is the product of flow rate and
pressure drop, therefore:
Power  Qp  3.067


m3
 8  10 6  4  10 5  23309200W
s
If the pumps are 88% efficient:
Power  23309200W  0.88  20512096W  20MW
2.5 Computational Results
a)
190605.6m
b)
Power  20MW
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3.0 PROBLEM 6.108
3.1 Given
The water pump in Figure 1 maintains a pressure of 6.5 psig at point 1. There is a
filter, a half-open disk value, and two regular screwed elbows. There are 80 ft of 4-in
diameter commercial steel pipe.
Figure 1
Problem 6.108 Diagram
3.2 Find
(a) If the flow rate is 0.4 ft3/s, what is the loss coefficient of the filter?
(b) If the disk valve is wide open and Kfilter=7, what is the resulting flow rate?
3.3 Assumptions
Incompressible
Room Temperature
Steady flow
Fully developed
3.4 Solution Calculations
a)
We start by writing the energy equation:
p

p1 V12
V2

 z1   2  2  z 2   htot
g 2 g
 g 2 g

2
V  fL

htot 
 K

2g  d

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p
 V 2  fL
p1 V12
V2


 z1   2  2  z 2  
 K

g 2 g

 g 2 g
 2g  d
And:
 K K valve  K filter  K elbows  K exit
The equation becomes:
p
 V 2  fL
p1 V12
V2


 z1   2  2  z 2   1 
 K valve  K filter  K elbows  K exit 
g 2 g

 g 2 g
 2g  d
Given in the problem:
6.5lb 144in 2
lb

 936 2
2
2
in
1 ft
ft
1 ft
ft
9.81m
g

 32.18 2
2
0.3048m
s
s
z1  0
p1 
z 2  9 ft
L  80 ft
d  0.4in 
1 ft 1
 ft
12in 3
Assuming that the pump is placed at 50˚F, from table A.1 we find density to be:
  1.94
slug
ft 3
To find Velocity at point 1:
V1 
Q
A
Q  0.4
ft 3
s
2
1 
  ft 
2
d
 2
3 
A


ft
4
4
36
ft 3
0. 4
s  4.584 ft
V1 
 2
s
ft
36
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Saleh David Ramezani
The outlet velocity is almost negligible (existence of a vertical tank), and the
pressure at this point is independent of the pump pressure. Therefore:
V2  0
p2  0
Table 6.5 gives the resistance coefficients of a 4 in diameter, regular elbow, at a
90˚ angle to be:
K elbow  0.64
And since there are two elbows the coefficient doubles:
K elbow  2  0.64  1.28
For a half opened disk valve, we find the average loss coefficient from Figure
6.18b to be:
K valve  2.9
And for a submerged exit K exit  1.0 regardless of the pipe’s shape.
To find f, we would first need to calculate the Reynolds’s number:
Re 
Vd


lb
ft 1
 4.58  ft
3
s 3
ft
 108488.4
lb.s
2.73  10 5 2
ft
1.940
And for commercial steel, from table 6.1, we find the roughness value to be:
  0.00015 ft
The roughness ratio ε/d to be:

d

0.00015 ft
 0.00045
1
ft
3
Using this information and Equation 6.49, the friction factor f is found to be:
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1.11

  
6
.
9
 1.8 log 
  d  

Re
 3.7  
f
 d 
 
1
f  0.019645
Plugging in the obtained values in the energy equation we would have:
p
 V 2  fL
p1 V12
V2


 z1   2  2  z 2   1 
 K valve  K filter  K elbows  K exit 
g 2 g

 g 2 g
 2g  d
2
2
lb
ft 
ft 



936 3
 4.58 
 4.58 2  

0
.
019645

80
ft
ft
s
s
  9 ft  
 
 
 3  K filter  1.28  1.0 
1
ft 
ft  




slug 
ft 
ft
2 32.18 2  

1.94 3  32.18 2  2 32.18 2 
3


s
s




ft
s




15.32 ft  9 ft  0.3259K filter  5.3389
K filter  14.05
b)
If the disk valve is wide open, according to table 6.18:
K valve  0
Writing out the energy equation one more time, and substituting all known
variables but velocity we’ll have:




f
80
ft


 9 ft 
 3  7  1.28  1.0 

ft 
ft   1



slug 
ft 
ft
2 32.18 2  
1.94 3  32.18 2  2 32.18 2 


s 
s  3


ft 
s 

936
lb
ft 3
V1 2
V1 2
Assuming the same roughness ratio ε/d:

0.00015 ft
 0.00045
1
d
ft
3
And writing Reynolds number as a function of velocity:

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3
Re 
Vd


Saleh David Ramezani
lb
1
 V1  ft
3
3
ft
lb.s
2.73  10 5 2
ft
1.940
Using these information and Equation 6.49, the friction factor f becomes
1.11

  
6
.
9
 1.8 log 
  d  

Re d  3.7  
f

 

1








1.11 

1
6.9
 0.00045  
 1.8 log 


lb
1   3.7  

f
  1.940 ft 3  V1  3 ft 




  2.73  10 5 lb.s 


2
 

ft



f 
1.6364
  V1  6.4561

 10.006 
ln 
V1

 
2
Plugging f back into the energy equation and solving for V1:
936
lb
ft 3

slug 
ft 
1.94 3  32.18 2 
ft 
s 


V1 2
ft 

2 32.18 2 
s 

1.6364


80 ft


2
   V1  6.4561


 10.006 
 ln 

2
V1
V1 

 

 9 ft 
 3  7  1.28  1.0 

1
ft


ft

2 32.18 2  
3
s 






We would solve for V1, then using Q=AV equation we would find the flow rate
(and hopefully we won’t have to do a problem like this on the test!)
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Saleh David Ramezani
3.5 Computational Results
K filter  14.05
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4.0 PROBLEM 6.116
4.1 Given
For the series-parallel system of Figure 2, all pipes are 8-cm-diamter asphalted
cast iron,
Figure 2
Problem 6.116 Diagram
4.2 Find
if the total pressure drop p1-p2 = 750 kPa, find the resulting flow rate Q m3/h for
water at 20˚C. Neglect minor losses.
4.3 Assumptions
4.4 Solution Calculations
4.5 Computational Results
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5.0 EXTRA PROBLEM 6.146
5.1 Given
A pipe connecting two reservoirs, as in Figure 3, contains a thin plate orifice.
Figure 3
Problem 6.146 Diagram
5.2 Find
For waster flow at 20˚C, estimate (a) the volume flow through the pipe and (b) the
pressure drop across the orifice plate.
5.3 Assumptions
5.4 Solution Calculations
5.5 Computational Results
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FE PROBLEMS (EXTRA CREDIT)
FE 6.1
In flow through a straight, smooth pipe, the diameter Reynolds number for
transition to turbulence is generally taken to be
k) 1500
l) 2300
m) 4000
n) 250,000
o) 500,000
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FE 6.5
Minor losses through valves, fittings, bends, contractions, and the like are
commonly modeled as proportional to:
a)
b)
c)
d)
e)
Total head
Static head
Velocity head
Pressure drop
Velocity
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Saleh David Ramezani
BIOFLUIDS APPLICATION
The biomedical application of this chapter once again has to do with flow rate of blood
through out the body. The entire, or small sections of the cardiovascular system could be
thought of as a network of ducts with a fluid in flow in different directions. Through
experiments we could also develop resistance or loss coefficients for different cases (e.g.
a fully or semi coagulated blood vessel vs. a normal vessels) that would increase the
accuracy of energy equation and flow rate calculations.
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Homework # 12
Chapter 5: Dimensional Analysis and Similarity
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
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1. Define “Dimensional Analysis”, and describe how this is important for our work
in fluid mechanics.
Answer: Experiments that might result in tables of output, or even multiple
volumes of tables, might be reduced to as a single set of curves – or even a single
curve – when suitably nondimensionalized. The technique for doing this is called
dimensional analysis.
Most practical fluid flow problems are too complex, both geometrically and
physically, to be solved analytically. That’s why we would test them by
experiments.
2. Describe the Principle of Dimensional Homogeneity (PDH)
Answer: If an equation truly expresses a proper relationship between variables in
a physical process, it will be dimensionally homogeneous; that is, each of its
additive terms will have the same dimensions. This is the Principle of
Dimensional Homogeneity.
3. Explain the Pi theorem
Answer: If a physical process satisfies the PDH and involves n dimensional
variables, it can be reduced to a relation between only k dimensionless variables
or πs. The reduction j=n-k equals the maximum number of variables that do not
form a pi among themselves and is always less than or equal to the number of
dimensions describing the variables.
4. Explain why nondimensionalization is important for some types of analyses.
Answer: instead of using the pie theorem to find the dimensionless parameters,
we could amend basic equations of flow to reveal basic dimensionless parameters.
Using this technique we would be able to decide when those parameters are
negligible.
5. Explain the following terms:
a. Geometric Similarity
Answer: a model and prototype are geometrically similar if and only if all
body dimensions in all three coordinates have the same linear scale ratio.
b. Kinematic Similarity
Answer: the motions of two systems are kinematically similar if homologous
particles lie at homologous points at homologous times.
Homework 3.doc
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BIEN 301
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Saleh David Ramezani
c. Dynamic Similarity
Answer: dynamic similarity exists when the model and the prototype have the
same length scale ratio, time scale ratio, and force scale (or mass scale) ratio.
6. In as much detail as possible, describe the steps you would consider in designing
a test stand to study a scaled-up version of a blood flow obstruction to determine
wall shear stresses and stagnation points at the obstruction.
Answer: the scaled up version of our blood flow obstruction must be dynamically
similar to the actual case. Reduce all dimensions by the same ratio
7. Which of the items above would be most important?
Answer: making the model with the same length scale ratio
8. How would you go about ensuring the validity of your scaled-up experiment?
Answer: Since blood is an incompressible fluid make sure that the Reynolds,
Froude, and Webber numbers are correspondingly equal to the actual flow model.
9. Why would doing this scaled-up test be useful in the first place?
Answer: we would get more accurate and realistic measurements.
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3
Saleh David Ramezani
BIOFLUIDS APPLICATION
Since organs of human body are characterized by physically and geometrically complex
shapes, it’s often difficult to solve biofluids problems analytically. We could get help
from dimensional analysis techniques and experiments to produce tables that can help us
with those measurements.
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Saleh David Ramezani
Homework # 13
Chapter 7: Flow Passed Immersed Bodies
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
Page 98 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 7.1
1.1 Given
For flow at 20 m/s past a thin flat plate:
No Sketch
1.2 Find
Estimate the distance x from the leading edge at which the boundary layer
thickness will be either 1 mm or 10 cm for:
(a) Air
(b) Water at 20˚ C and 1 atm.
1.3 Assumptions
Incompressible
Steady flow
1.4 Solution Calculations
a)
From table A.2 for air at 20˚C:
  1.2
kg
m3
  1.8  10 5
kg
m.s
If we try the laminar flow for 1mm thickness:

5.0
5.0

1/ 2
1/ 2
x Re x
 Vx 


  
 2 25

x 2 Vx

x
 V

25
2
Homework 3.doc
0.001m 2  1.2 kg  20 m
m.s
25  1.8  10 5
kg
m.s
s  0.053 m
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Saleh David Ramezani
Calculating the Reynolds number and comparing it with the local Reynolds
numbers on page 453:
kg
m
1.2
 20  0.053m
Vx
m.s
s
Re x 

 71111.1 1 Checks. Laminar Flow

5 kg
1.8  10
m.s
For the 10 cm thick boundary layer we try the turbulent equation:

x

0.16
0.16

1/ 7
1/ 7
Re x
 Vx 


  
1/ 6
8.483  V 
1/ 6
x
 1/ 6
 7/6
kg
m

7/6
8.483  1.2
 20   0.1m 
s
 m.s

 6.063m
1/ 6
kg


5
1.8  10

m.s 

Calculating the Reynolds number and comparing it with the local Reynolds
numbers on page 453:
Re x 
b)
Vx


1.2
kg
m
 20  6.063m
m.s
s
 8083917.63 Checks. Turbulent Flow
5 kg
1.8  10
m.s
From table A.1 for water at 20˚C:
  998
kg
m3
kg
m.s
If we try the laminar flow for 1mm thickness:
  1.003  10 3
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BIEN 301

3
Saleh David Ramezani
5.0
5.0

1/ 2
1/ 2
x Re x
 Vx 


  
 2 25

x 2 Vx

x
 V

25
2
0.001m 2  998 kg  20 m
m.s
s  .0.796m
kg
m.s
Calculating the Reynolds number and comparing it with the local Reynolds
numbers on page 453:
Re x 
25  1.003  10 3
Vx


998
kg
m
 20  0.796m
m.s
s
 15840876.18 Does NOT Check!
3 kg
1.003  10
m.s
So we try the equation for turbulent flow:

x

0.16
0.16

1/ 7
1/ 7
Re x
 Vx 


  
1/ 6
8.483  V 
1/ 6
x
 1/ 6
 7/6
kg
m

7/6
8.483   998
 20   0.001m 
m.s
s


 0.0446m
1/ 6

3 kg 
1.003  10

m.s 

Calculating the Reynolds number and comparing it with the local Reynolds
numbers on page 453:
Vx
Re x 


Homework 3.doc
998
kg
m
 20  0.446m
m.s
s
 878745.05 Checks. Turbulent Flow
3 kg
1.003  10
m.s
Page 101 of 123
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Saleh David Ramezani
For the 10 cm thick boundary layer we try the turbulent equation again:

x

0.16
0.16

1/ 7
1/ 7
Re x
 Vx 


  
1/ 6
8.483  V 
1/ 6
x
 1/ 6
 7/6
kg
m

7/6
8.483   998
 20   0.1m 
m.s
s


 9.513m
1/ 6

3 kg 
1.003  10

m.s 

Calculating the Reynolds number and comparing it with the local Reynolds
numbers on page 453:
Vx
Re x 


Homework 3.doc
998
kg
m
 20  9.513m
m.s
s
 189319881.9 Checks. Turbulent Flow
3 kg
1.003  10
m.s
Page 102 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
2.0 PROBLEM 7.52
2.1 Given
a
 6 
Clift et al. give the formula F  
 4   Vb for the drag of a
b
 5 
prolate spheroid in creeping motion, as shown in Figure 1. The half thickness b is 4 mm.
if the Fluid is SAE 50W oil at 20˚ C:
Figure 1
Problem 7.52 Diagram
2.2 Find
(a) Check the Reb < 1
(b) Estimate the spheroid length if the drag is 0.02 N.
2.3 Assumptions
Incompressible
Constant density and viscosity
Steady flow
2.4 Solution Calculations
From Table A.3 for SAE 50W oil we have:
Homework 3.doc
Page 103 of 123
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BIEN 301
3
Saleh David Ramezani
kg
m3
kg
  0.86
m.s
  902
Calculating the Reynolds number for the half thickness b:
Vb
Re b 


902
kg
m
 0.2  0.004m
m.s
s
 0.839 < 1 Checks
kg
0.86
m.s
Given the drag force, F=0.02N, and the Clift formula we can easily estimate the
spheroid length a:
a
 6 
F 
 4   Vb
b
 5 
5F
5  0.02 N
a
 4b 
 4  0.004m  0.01484m
kg
m
6V
6  0.86
 0.2
m.s
s
Spheroid length= 2a  0.0297m
2.5 Computational Results
a)
Re b  0.839  1
b)
Spheroid Length= 0.0297m
Homework 3.doc
Page 104 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
3.0 PROBLEM 7.90
3.1 Given
In the great hurricane of 1938, winds of 85 mi/h blew over a boxcar in
Providence, Rhode Island. The boxcar was 10 ft high, 40 ft long, and 6 ft wide, with a 3 ft
clearance above tracks 4.8 ft apart.
40ft
V
10ft
6ft
3ft
4.8ft
Figure 2
Problem 7.90 Diagram
3.2 Find
What wind speed would topple a boxcar weighing 40,000 lbf?
3.3 Assumptions
Incompressible
Steady flow
3.4 Solution Calculations
Homework 3.doc
Page 105 of 123
6-feb.-16
BIEN 301
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Saleh David Ramezani
Figure 3
Problem 7.90 Free Body Diagram
Assuming 68˚F at 1 atm, from Table A.2 we have:
  2.34  10 3
slug
ft 3
From Table 7.3 we can estimate CD=1.2
Using Equation 7.62 we can then solve for Drag force:
Drag Force  F 
C D V A

2
2
1.2  2.34  10 3
slug
 V 2  40 ft  10 ft
3
ft
 0.5616V 2 lbf
2
Summing moments around the right leg as shown in Figure 3, we’ll have:
 M  0.5616V
2
 8 ft  40000lbf  2.4 ft  0
V 2  21367.52
V  146
ft
s
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
3.5 Computational Results
V  146
ft
s
Homework 3.doc
Page 107 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
BIOFLUIDS APPLICATION
Platelets have a key role in maintaining the haemostatic system by the formation of blood
clots at the site of injury. However, blood clots can also be formed in the blood stream
restricting blood flow to critical areas of the body causing serious consequences such as
stroke, cardiac ischemia, and myocardial infarction. Blood clots could be seen as an
immersed body that blocks the flow of blood in those veins.
Homework 3.doc
Page 108 of 123
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BIEN 301
3
Saleh David Ramezani
Homework # 14
Chapter 8: Potential Flow and Computational Fluid Dynamics
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
Page 109 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 8.71
1.1 Given
Figure 1 shows the streamline and potential lines of flow over a thin-plate weir as
computed by the complex potential method.
Figure 1
Problem 8.71 Diagram
Figure 2
Problem 8.71 Diagram
1.2 Find
Compare qualitatively with Figure 2. State the proper boundary conditions at all
boundaries. The velocity potential has equally spaced values. Why do the flow-net
“squares” become similar in the overflow jet?
1.3 Assumptions
Incompressible flow
Frictionless flow
Neglected gravity
Steady flow
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
1.4 Solution Calculations
The stream function at the bottom and weir wall of both diagrams is zero. If we
draw the mesh model for both cases, since the weir and bottom wall make 90
degrees angle (square mesh), the flow rate in those areas will almost be zero.
Homework 3.doc
Page 111 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
2.0 PROBLEM C8.1
2.1 Given
Did you know that you can solve simple fluid mechanics problems with
Microsoft Excel? The successive relaxation technique for solving the Laplace equation
for potential flow problems is easily set up on a spreadsheet, since the stream functionat
each interior cell is simply the average of its four neigbours.
Figure 3
Problem C8.1 Diagram
2.2 Find
As and example, solve for the irrotational potential flow through a contraction, as
given in Figure 3.
2.3 Assumptions
Square mesh model
2.4 Solution Calculations
5.0000
4.0000
3.0000
2.0000
1.0000
0.0000
5.0000
3.9091
2.8400
1.8230
0.8873
0.0000
Homework 3.doc
5.0000
3.7960
2.6278
1.5646
0.7260
0.0000
5.0000
3.6470
2.3101
1.0817
0.4519
0.0000
5.0000
3.4817
1.8838
0.0000
0.0000
0.0000
5.0000
3.3958
1.7435
0.0000
Page 112 of 123
5.0000
3.3579
1.6943
0.0000
5.0000
3.3417
1.6757
0.0000
5.0000
3.3330
1.6670
0.0000
6-feb.-16
BIEN 301
3
Saleh David Ramezani
BIOFLUIDS APPLICATION
Not all blood vessels or passage ways in our body have a uniform thickness or shape. In
fact most of them are characterized by complicated geometries. The most obvious
biomedical application of this problem is the numerical analysis of velocities or flow
rates through those more complex shaped passage ways.
Homework 3.doc
Page 113 of 123
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BIEN 301
3
Saleh David Ramezani
Homework # 15
Chapter 9: Compressible Flow
Submitted by:
Saleh David Ramezani
On:
6-Feb-16
HONOR CODE STATEMENT
On my honor, I promise that I have not received inappropriate assistance on this assignment.
Inappropriate assistance for homework: Copying off another person’s paper, copying information from
the solution of homework from previous homework, and any sort of computer file sharing.
Inappropriate assistance on pop quizzes and exams: All work must be your own (no looking at other
people’s paper, no talking, no cheat sheets, and no use of electronic information.
Inappropriate assistance on projects: Refer to the guidelines on the strain gauge project and the graduate
student final project in syllabus.
For a complete set of the honor code rules and regulations applicable to this course, consult the
Louisiana Tech University Honor Code at:
http://www.latech.edu/students/judicial-affairs.shtml
_____________________
Student Signature / Date
Homework 3.doc
Page 114 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
1.0 PROBLEM 9.11
1.1 Given
AT 300˚C and 1 atm:
No Sketch
1.2 Find
Estimate the speed of sound of
(a) Nitrogen
(b) Hydrogen
(c) Helium
(d) Steam
(e) 238UF6 (k ≈ 1.06)
1.3 Assumptions
Ideal gas
Constant temperature
Constant pressure
1.4 Solution Calculations
Using equation 9.16 and table A.4 in the back of the book, for:
a) Nitrogen
a  kRT  1.4  297
m2
m
 300  273K   488.11
2
s
s K
b) Hydrogen
a  kRT  1.41  4124
m2
m
 300  273K   1825.35
2
s
s K
c) Helium
Homework 3.doc
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6-feb.-16
BIEN 301
3
a  kRT  1.66  2077
Saleh David Ramezani
m2
m
 300  273K   1405.56
2
s
s K
d) Steam
m2
m
a  kRT  1.33  461 2  300  273K   592.72
s
s K
e) 238UF6 (k ≈ 1.06)
R is not given in the table, we use equation 9.3 and the molecular weight to solve
for R, and then speed of sound:

8314
m2

 23.62 2
M gas 238  6  19 
s .K
R gas 
a  kRT  1.06  23.62
m2
m
 300  273K   119.77
2
s
s K
1.5 Computational Results
a) a  488.11
m
s
b) a  1825.35
m
s
c) a  1405.56
m
s
d) a  592.72
m
s
e) a  119.77
m
s
Homework 3.doc
Page 116 of 123
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BIEN 301
3
Saleh David Ramezani
2.0 PROBLEM 9.43
2.1 Given
Air flows isentropically through a duct with T0=300˚C. At two sections
with identical areas of 25cm2, the pressures are p1=120 kPa and p2=60 kPa.
No Sketch
2.2 Find
Determine:
(a) The mass flow
(b) The throat area
(c) Ma2
2.3 Assumptions
2.4 Solution Calculations
2.5 Computational Results
Homework 3.doc
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BIEN 301
3
Saleh David Ramezani
3.0 PROBLEM 9.78
3.1 Given
The orientation of a hole can make a difference. Consider holes A and B in Figure
1, which are identical but reversed.
Figure 1
Problem 9.78 Diagram
3.2 Find
For the given air properties on either side, Compute the mass flow through each
hole and explain why they are different.
3.3 Assumptions
Ideal gas
Constant temperature
Constant pressure
3.4 Solution Calculations
Since the entrance area in duct A is smaller than the exit area the throat is
considered to be at critical or sonic condition and therefore duct A is said to be
choked. Equation 9.46.b can be used to find the maximum mass flow rate which
occurs at the choking condition:
m 
0.6847 p0 A
RT0
Homework 3.doc

0.6847  150000Pa  0.00002m 2
287  293
Page 118 of 123
 0.00708
kg
s
6-feb.-16
BIEN 301
3
Saleh David Ramezani
Duct B is not chocked because the entrance area is larger than the exit area and it
doesn’t create a critical or sonic condition. To find the mass flow rate in this case,
we would use equation 9.46a
m  AV
None of the variables are known. To solve for density we would first need to find
the exit temperature. We use equation 9.26 for this purpose:
p 100kPa

 0.667
p 0 150kPa
We use table B.1 to find the Mach number:
0.6690  0.6560 0.78  0.8

0.6690  0.667 0.78  Ma
Ma  0.783
T0
k 1
 1
Ma 2
T
2
293
1.4  1
0.7832
 1
T
2
T  261K

p

RT
kg
100000 Pa
 1.33 3
2
m
m
287 2  261K
s .K
Knowing temperature we can also solve for speed of sound which will help us
find the velocity using the Mach number:
m2
m
a  kRT  1.4  287 2  261K  323.84
s
s .K
V  Ma  a  0.783  323.84
m  1.33
Homework 3.doc
m
m
 253.56
s
s
kg
m
kg
 0.00002m 2  253.56  0.0067
3
s
s
m
Page 119 of 123
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Saleh David Ramezani
3.5 Computational Results
m a  0.00708
m b  0.0068
Homework 3.doc
kg
s
kg
s
Page 120 of 123
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Saleh David Ramezani
4.0 PROBLEM 9.88
4.1 Given
Ai enters a 5- by 5-cm square duct at V1=900 m/s and T1= 300 K. The friction
factor is 0.02.
No Sketch
4.2 Find
(a) For what length duct will the flow exactly decelerate to Ma=1.0?
(b) If the duct length is 2 m, will there be a normal shock in the duct?
(c) If so, at what Mach number will it occur?
4.3 Assumptions
Ideal gas
Constant temperature
Constant pressure
4.4 Solution Calculations
4.5 Computational Results
Homework 3.doc
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3
Saleh David Ramezani
5.0 EXTRA PROBLEM 9.114
5.1 Given
W have simplified things here by separating friction (Sec. 9.7) from heat addition
(Sec. 9.8). Actually they occur together, and their effects must be evaluated
simultaneously.
No Sketch
5.2 Find
Show that, for flow with friction and heat transfer in a constant diameter pipe, the
continuity, momentum, and energy equations may be combined into the flowing
differential equation for Mach number changes:


dMa 2 1  kMa 2 dQ kMa 2 2  k  1Ma 2 fdx


D
Ma 2
1  Ma 2 c p T
2 1  Ma 2


Where dQ is the heat added. A complete derivation, including many additional
combined effects such as area change and mass addition, is given in Chapter 8 of
Ref. 5.
5.3 Assumptions
Incompressible
Constant density and viscosity
Circular symmetry
No axial motion or end effect (vz = 0)
5.4 Solution Calculations
5.5 Computational Results
Homework 3.doc
Page 122 of 123
6-feb.-16
BIEN 301
3
Saleh David Ramezani
BIOFLUIDS APPLICATION
The only application that I can think of is the flow measure of oxygen and other
compressible gases present in the air in and out of the lungs.
Homework 3.doc
Page 123 of 123
6-feb.-16