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Introduction to Mechanical Engineering
• Chapter 1 The Mechanical Engineering Profession
• Chapter 2 Problem-Solving and Communication Skills
• Chapter 3 Forces in Structures and Machines
• Chapter 4 Materials and Stresses
• Chapter 5 Fluids Engineering
• Chapter 6 Thermal and Energy Systems
• Chapter 7 Motion and Power Transmission
• Chapter 8 Mechanical Design
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5-2
• Hydrodynamics • Aerodynamics • Steam – Water recycling • Biomedical Engineering • Buoyancy, drag and lift by fluid
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Properties of Fluids
1. Not equilibrium on shear stress
2. Liquid(incompressible)
3. Gas(compressible)
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5-5
Properties of Fluids
Continuous motion (flow) by shear stress
No-Slip Condition at microscopic level; only several molecules thick , adheres to solid surface
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Properties of Fluids
time)-thmass/(leng
; viscosityof units
fluid) (Newtonian
velocityflow ; v
friction stickness, of measure also
and flow tofluid a of resistance
);Viscosity(
kg/m.s
h
v
A
F
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P 0.01cP 1
.1.0)(P 1
cP P, ;ity for viscosunit special
sm
kgPoise
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Example 5.1 Machine Tool Guideways
• oil viscosity ; 240 cP • width of ways ; 8 cm • length of ways ; 40 cm • force ; 90 N • velocity ; 15 cm/s
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Example 5.1 Machine Tool Guideways
• oil viscosity ; 240 cP • width of ways ; 8 cm • length of ways ; 40 cm • force ; 90 N • velocity ; 15 cm/s • What is thickness of oil film?
mmx
smkg
smsmkgm
F
vA h
h
v
A
F
6.25 1056.2
)/.(90
)/ 15.0)(./ 24.0)()(4.0)(08.0)(2(
5
2
2
9
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Pressure and Buoyancy Force
fluid of volume; V
fluid ofdensity ;
gVw
ghPP o 1
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Buoyancy and pressure
ghpp
gAhApAp
balanceforcemequilibriu
Ahmmassliquid
01
01 0)(
,
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Units
• 1 pascal = 1 Pa(N/m2)
• psi = lb/in2
• psf = lb/ft2
• 1 atm = 1.013 x 105 Pa
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objectfluidB gVF
• Pressure and Buoyancy Force
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Example 5.2 Aircraft’s Fuel Capacity
• Fuel capacity 90,000 L • Density ; 840 kg/m3
• What is the weight of fuel?
kNs
mkgx
s
m
L
mL
m
kg
Vgmgw
6.741).
(10416.7
)81.9)(001.0)(000,90)(840(
2
5
2
3
3
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Example 5.3 Deep Submergence Rescue Vehicle
• Max. dive depth ; 5000 ft • What is water pressure in psi?
psiinx
ft
ft
lbx
fts
ftslugx
sft
slugx
fts
ft
ft
slug
ghppp
ghpp
2225) 1212
)((10204.3
)1
)(.
(10204.3).
(10204.3
)5000)(2.32)(99.1(
2
2
2
5
22
5
2
5
23
01
01
16
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Example 5.4 Great White Shark Attack
• 55 gal barrel weighing 35 lb • What force Shark to overcome ?
lblbs
ftslug
lbgal
ftgal
s
ft
ft
slugs
lbgVwFT
wTF
objectfluidB
B
2.436)(35).
(2.471
)(35)1337.0)(55)(2.32)(99.1(
)(35
0
2
3
23
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5.4 Laminar and Turbulent Fluid Flows
) ,viscosity,,(Re
number Reynolds
lengthsticcharacterispeeddensityvl
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Example 5.5 Reynolds Number
5000
)./(100.1
)01.0)(/5.0)(/1000(
pipe in the flowingfor water Re )2
10679.3
./5^108.1
)3^106.7)(/720)(3^/(2.1
6.7
/720:.
Re
number Reynolds
3
3
5
smkgx
msmmkgR
x
smkgx
mxsmmkgR
mmd
smvspeedWinchester
vl
e
e
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Example 5.5 Reynolds Number
7
3
3
3
10861021
1081026Re
sm/ 8 of speed and m 10 ofdiameter hull with Submarine 4)
63.17)./(26.0
)01.0)(/5.0)(/917(Re
... 30 3)
Re
number Reynolds
x.kg/(m.s) x .
m) m/s)()( kg/m
smkg
msmmkg
waterofinsteadoilSAE
vl
20
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5.5 Fluid Flows in Pipes
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In pipe, Not with any disturbances and low enough Re number(<2000), then flow is laminar
)2000Re of case ( 16
)2000Re of case ( ))(1(
2
max
2
max
inL
pdv
inR
rvv
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)2000Re (128
82
)2
(
2
)2000Re (2
1
;
/
4
max
2
max
2
max
max
casespecialL
pd
vd
v
d
vA
q
casespecialvv
Avq
rateflowVolumetric
txv
xAV
avg
avg
avg
Poiseuille’s law
23
)2000Re of case ( 16
2
max
inL
pdv
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constant2
2
2211
ghvp
vAvA
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Bernoulli’s equation
constant2
2
ghvp
1st; Work of pressure force 2nd; Kinetic energy of flow fluid 3rd; Gravitational potential energy of it
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Example 5.6 Automotive Fuel Line
1. A car speed ; 40 mph 2. With fuel economy rate; 28 miles/gallon 1. Fuel line inside dia.; 3/8 in.
8.467)./(101.6
.)/)(12/1.)()(8/3)(/10917.6)(/32.1(Re )
10917.6
).12
1(.)375.0(
4
)(10306.5
v
is rate flow line)
)(10306.5)1337.0)(10968.3(
10968.3)(3600
1
/28
/40
/28
/40
rate fuel
is rate flow c volumetri)
6
23
2
22
35
avg
35
34
4
sftslugx
inftinsftxftslugvlc
s
ftx
in
ftin
s
ftx
A
q
b
s
ftx
gal
ft
s
galx
s
galx
s
hx
galmi
hmi
galmi
hmispeedcarq
a
a) Volumetric flow of oil in ft3/s? b) Fuel avg. velocity in in./s? c) What Re number?
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Drag forces and Viscosity
dragoftcoefficienC
speedrelativev
objectofareafrontalA
densitysfluid
CAvF
D
DD
:
:
:
':
2
1 2
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Nearly constant
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Drag forces and Viscosity
pois) , ty(kg/m.s viscosisfluid' ;μ
diameter ssphere' ; d
)1Re ; spherefor case Special( 3 dvFD
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Example 5.7 Golf Ball in Flight
1. 1.68 in. dia. golf ball speed 70 mph 2. Drag force? A) smooth ball B) CD=0.27
lbxs
ftslugxF
insteadCWithB
lbxs
ftslugx
s
ftftx
ft
slugxF
Csohighveryxvl
CdecidingfornumberknowToA
CAvF
D
D
D
D
D
DD
2
2
2
2
2
2
222
3
3
4
2
10104.5.
1010.5
, 27.0 )
10452.9.
10452.9
)5.0()7.102)(10538.1)(1033.2(2
1
)apply , (10813.8Re
Re )
2
1
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34
)1(Re )(18
)6/( ,6/
,3
0
2
33
fluidobject
objectobject
objectfluidBD
DB
ρρgd
v
ρdmdV
mgWgVρπμdv , FF
-wFFbalanceforce
Terminal Velocity
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Drag and Lift forces in fluids
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Wind tunnel
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equation sBernoulli'
constant2
2
ghvp
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constant2
2
ghvp
Lift force by pressure difference on the airfoil
bottombottom ppthenvv
vp
toptop
2
, If
constant2
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lift oft coefficien ;
2
1 2
L
LL
C
CAvF
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1. Buoyancy in fluids
2. Drag forces in fluids
3. Lift forces in fluids
objectfluidB gVF
)1(Re 3 dvFD
2
1 2
DD CAvF
LLCAvF 2
2
1
vlRe
46
Summary
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Problem 4.36: (a) A luxury sports car has a frontal area of 22.4 ft2 and a 0.29 coefficient of drag at 60 mi/hr. What is the drag force on the vehicle at this speed? (b) A sport utility vehicle has CD = 0.45 at 60 mi/hr, and the slightly larger frontal area of 29.1 ft2. What is the drag force in this case?
DD CAvF 2
2
1
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s
ft
s
hr
mi
ft
hr
miv 8810778.2528060 4
lbs
ftftftslugF
D6.5829.0884.22/1033.2
2
12
233
lbs
ftftftslugF
D11845.0881.29/1033.2
2
12
233
Approach: Apply Equation using the density of air
listed in Table 4.3. Convert velocity to consistent units when calculating
the drag force. Solution:
a) Luxury sports car Velocity:
b) Sport utility vehicle
33 /1033.2 ftslug
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Problem 4.28: A steel storage tank is filled with gasoline. The tank has partially corroded on its inside, and small particles of rust have contaminated the fuel. The rust particles are spherical, have diameter 25 µm, and density 5.3 g/cm3. (a) What is the terminal velocity of the particles as they fall through the gasoline? (b) How long would it take the particles to fall 5 m and settle out of the tank?
• Approach: Apply terminal velocity Equation to find the terminal velocity using the density(680 kg/m3) and viscosity(2.9x10-4 kg/m.s) of gasoline listed in Table 4.3. Calculate the Reynolds number to confirm that the velocity is low enough that the equation was correctly applied.
3
3
35300001.01003.5
m
kg
g
kg
m
cm
cm
gsphere
)/(0054.0
6805300/109.218
1025/81.9)(
18 334
2622
sm
m
kg
m
kg
smkg
msmgdv gp
Solution: a) Terminal velocity
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which is dimensionless. Since the numerical value of above Re is less than one, the velocity was properly calculated.
31.0/109.2
1025/0054.0/680Re
4
63
smkg
msmmkgvd
Check Reynolds number,
b) Fall time
min4.15926/0054.0
5 s
sm
m
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Problem 4.39: Submarines dive by opening vents which allow air to escape from ballast tanks, and water to flow in and fill them. In addition, diving planes located at the bow are angled downward to help push the boat below the surface. Calculate the diving force produced by a 20 ft2 hydroplane that is inclined by 3o as the boat cruises at 15 knots (1 knot = 1.152 mi/hr).
Approach: Apply Equation (4.19) using consistent units for velocity. From Figure 4.33
at an angle of attack of 3o, CL = 0.32. Use the density of sea water(1.99 slut/ft3)
s
ft
s
hr
mi
ft
kt
hrmiktv
Velocity
3.2510778.25280/
1516.115 4
lbs
ftft
ft
slugCAvL
Downward
L 408832.03.25201099.12
1
2
1
force diving
2
2
3
32
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Reference
• Wickert J., Lewis K., “An Introduction to Mechanical Engineering”, 3rd Edition, 2013.
• www.slideshare.com
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