KULIAH VIII - IX
MEKANIKA FLUIDA II
Nazaruddin Sinaga
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Entrance Length
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Shear stress and velocity distribution in pipe for
laminar flow
Typical velocity and shear distributions in turbulent flow near a
wall: (a) shear; (b) velocity.
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Solution of Pipe Flow Problems
• Single Path
– Find Dp for a given L, D, and Q
 Use energy equation directly
– Find L for a given Dp, D, and Q
 Use energy equation directly
Solution of Pipe Flow Problems
• Single Path (Continued)
– Find Q for a given Dp, L, and D
1. Manually iterate energy equation and friction factor
formula to find V (or Q), or
2. Directly solve, simultaneously, energy equation and
friction factor formula using (for example) Excel
– Find D for a given Dp, L, and Q
1. Manually iterate energy equation and friction factor
formula to find D, or
2. Directly solve, simultaneously, energy equation and
friction factor formula using (for example) Excel
Example 1
 Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The pipe
has 150-mm diameter, 500 m long, and the surface roughness is estimated
at 0.06 mm. Find the head loss and the pressure drop throughout the
length of the pipe.
Solution:
 From Table 1.3 (for water):  = 1000 kg/m3 and  =1.30x10-3 N.s/m2
V = Q/A and
A=R2
A = (0.15/2)2 = 0.01767 m2
V = Q/A =0.03/.0.01767 =1.7 m/s
Re = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000  turbulent flow
To find , use Moody Diagram with Re and relative roughness (k/D).
k/D = 0.06x10-3/0.15 = 4x10-4
From Moody diagram,   0.018
The head loss may be computed using the Darcy-Weisbach equation.
500x 1.7 2
L V2
hf  
 0.018x
 8.84m.
D 2g
0.15 x 2 x 9.81
The pressure drop along the pipe can be calculated using the relationship:
ΔP=ghf = 1000 x 9.81 x 8.84
ΔP = 8.67 x 104 Pa
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Example 2
 Determine the energy loss that will occur as 0.06 m3/s water flows
from a 40-mm pipe diameter into a 100-mm pipe diameter through
a sudden expansion.
Solution:
 The head loss through a sudden enlargement is given by;
2
V
hm  K a
2g
Va 
Q
0.06

 3.58 m / s
Aa  (0.04 / 2) 2
Da/Db = 40/100 = 0.4
From Table 6.3: K = 0.70
Thus, the head loss is
3.582
h Lm  0.70 x
 0.47m
2 x 9.81
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Example 3
 Calculate the head added by the pump when
the water system shown below carries a
discharge of 0.27 m3/s. If the efficiency of
the pump is 80%, calculate the power input
required by the pump to maintain the flow.
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Solution:
Applying Bernoulli equation between section 1 and 2
P1
V12
P2
V2 2
 z1 
 Hp 
 z2 
  H L12
g
2g
g
2g
(1)
P1 = P2 = Patm = 0 (atm) and V1=V2 0
Thus equation (1) reduces to:
H p  z 2  z1   H L12
(2)
HL1-2 = hf + hentrance + hbend + hexit
V2
 H L12 
2g
1000


 0.5  0.4  1
 0.015x
0.4


V2
 39.4
2g
From (2):
V2
H p  230 200 39.4
2x9.81
The velocity can be calculated using the
continuity equation:
V
Q
0.27

 2.15 m / s
2
A 0.4 / 2
Thus, the head added by the pump:
p 
Pin 
gQHp
gQH p
p
Hp = 39.3 m
Pin

1000 x9.81x 0.27 x39.3
0.8
Pin = 130.117 Watt ≈ 130 kW.
EGL & HGL for a Pipe System
•
Energy equation
V12 p1
V22 p2
1

 z1  hL   2

 z2
2g 
2g 
•
•
All terms are in dimension of length (head, or
energy per unit weight)
HGL – Hydraulic Grade Line
HGL 
•
p

EGL – Energy Grade Line
EGL  
•
•
z
V2 p
V2
  z  HGL  
2g 
2g
EGL=HGL when V=0 (reservoir surface, etc.)
EGL slopes in the direction of flow
EGL & HGL for a Pipe System
•
A pump causes an abrupt rise in EGL (and
HGL) since energy is introduced here
EGL & HGL for a Pipe System
• A turbine causes an
abrupt drop in EGL
(and HGL) as energy
is taken out
• Gradual expansion
increases turbine
efficiency
EGL & HGL for a Pipe System
•
•
When the flow passage changes diameter, the
velocity changes so that the distance between
the EGL and HGL changes
When the pressure becomes 0, the HGL
coincides with the system
EGL & HGL for a Pipe System
•
Abrupt expansion into reservoir causes a
complete loss of kinetic energy there
EGL & HGL for a Pipe System
•
When HGL falls below the pipe the pressure
is below atmospheric pressure
FLOW MEASUREMENT
• Direct Methods
– Examples: Accumulation in a Container; Positive
Displacement Flowmeter
• Restriction Flow Meters for Internal Flows
– Examples: Orifice Plate; Flow Nozzle; Venturi; Laminar
Flow Element
Definisi tekanan pada aliran di sekitar sayap
Flow Measurement
• Linear Flow Meters
– Examples: Float Meter
(Rotameter); Turbine;
Vortex; Electromagnetic;
Magnetic; Ultrasonic
Float-type variablearea flow meter
Flow Measurement
• Linear Flow Meters
– Examples: Float Meter (Rotameter); Turbine; Vortex;
Electromagnetic; Magnetic; Ultrasonic
Turbine flow meter
Flow Measurement
• Traversing Methods
– Examples: Pitot (or Pitot Static) Tube; Laser Doppler
Anemometer
The measured stagnation pressure cannot of itself be used
to determine the fluid velocity (airspeed in aviation).
However, Bernoulli's equation states:
Stagnation pressure = static pressure + dynamic
pressure
Which can also be written
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Solving that for velocity we get:
Note: The above equation applies only to incompressible fluid.
where:




V is fluid velocity;
pt is stagnation or total pressure;
ps is static pressure;
and ρ is fluid density.
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The value for the pressure drop p2 – p1 or Δp to Δh,
the reading on the manometer:
Δp = Δh(ρA-ρ)g
Where:
 ρA is the density of the fluid in the manometer
 Δh is the manometer reading
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EXTERNAL
INCOMPRESSIBLE
VISCOUS FLOW
Main Topics
•
•
•
•
•
The Boundary-Layer Concept
Boundary-Layer Thickness
Laminar Flat-Plate Boundary Layer: Exact Solution
Momentum Integral Equation
Use of the Momentum Equation for Flow with Zero
Pressure Gradient
• Pressure Gradients in Boundary-Layer Flow
• Drag
• Lift
The Boundary-Layer Concept
The Boundary-Layer Concept
Boundary Layer Thickness
Boundary Layer Thickness
• Disturbance Thickness, d where
Displacement Thickness, d*
Momentum Thickness, q
Boundary Layer Laws
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Governing Equations
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Boundary Conditions
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Equations are Coupled, Nonlinear, Partial Differential
Equations
• Blassius Solution:
– Transform to single, higher-order, nonlinear, ordinary
differential equation
Laminar Flat-Plate
Boundary Layer: Exact Solution
• Results of Numerical Analysis
Momentum Integral Equation
• Provides Approximate Alternative to
Exact (Blassius) Solution
Momentum Integral Equation
Equation is used to estimate the boundary-layer
thickness as a function of x:
1. Obtain a first approximation to the freestream velocity
distribution, U(x). The pressure in the boundary layer is
related to the freestream velocity, U(x), using the
Bernoulli equation
2. Assume a reasonable velocity-profile shape inside the
boundary layer
3. Derive an expression for tw using the results obtained
from item 2
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Simplify Momentum Integral Equation
(Item 1)
 The Momentum Integral Equation becomes
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Laminar Flow
– Example: Assume a Polynomial Velocity Profile (Item 2)
•
The wall shear stress tw is then (Item 3)
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Laminar Flow Results
(Polynomial Velocity Profile)
Compare to Exact (Blassius) results!
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
• Turbulent Flow
– Example: 1/7-Power Law Profile (Item 2)
Use of the Momentum Equation for Flow
with Zero Pressure Gradient
•
Turbulent Flow Results
(1/7-Power Law Profile)
Pressure Gradients in Boundary-Layer Flow
Drag
• Drag Coefficient
with
or
Drag
• Pure Friction Drag: Flat Plate Parallel to the
Flow
• Pure Pressure Drag: Flat Plate Perpendicular to
the Flow
• Friction and Pressure Drag: Flow over a Sphere
and Cylinder
• Streamlining
Drag
• Flow over a Flat Plate Parallel to the Flow: Friction
Drag
Boundary Layer can be 100% laminar,
partly laminar and partly turbulent, or
essentially 100% turbulent; hence
several different drag coefficients are
available
Drag
• Flow over a Flat Plate Parallel to the Flow: Friction
Drag (Continued)
Laminar BL:
Turbulent BL:
… plus others for transitional flow
Drag Coefficient
0.140
0.120
0.100
CD Laminar
0.080
CD Turbulen
0.060
0.040
0.020
0.000
1.E+00
1.E+01
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
1.E+08
1.E+09
Drag
• Flow over a Flat Plate Perpendicular to the
Flow: Pressure Drag
Drag coefficients are usually obtained empirically
Drag
• Flow over a Flat Plate Perpendicular to the
Flow: Pressure Drag (Continued)
Drag
• Flow over a Sphere and Cylinder: Friction and
Pressure Drag
Drag
• Flow over a Sphere and Cylinder: Friction and
Pressure Drag (Continued)
Streamlining
• Used to Reduce Wake and Pressure Drag
Lift
• Mostly applies to Airfoils
Note: Based on planform area Ap
Lift
• Examples: NACA 23015; NACA 662-215
Lift
• Induced Drag
Lift
• Induced Drag (Continued)
Reduction in Effective Angle of Attack:
Finite Wing Drag Coefficient:
Lift
• Induced Drag (Continued)
The End
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