Download ME 215.3 Fluid Mechanics

Pharos University
ME 259 Fluid Mechanics II
Review of Previous Fluid Mechanics
Dr. Shibl
Fluid Properties:
Liquid or Gas
• Liquids are:
– Incompressible, DV ≠ f(DP)
– Viscous (high viscosity)
– Viscosity decreases with temperature
• Gases are:
– Compressible, DV = f(DP)
– Low viscosity
– Viscosity increases with temperature
Equations for Fluid Property
•
•
•
•
•
Circular Area: Area = p/4*D2
Weight: w = m*g Newton
Density: r = m/V Kg/m3
Specific Weight: g = w/V N/m3
Specific gravity: SG=r/rwater
Viscosity
• Dynamic Viscosity
 = Shear Stress/Slope of velocity profile
F/A
v/ y
cP (centipoise) or Pa-sec
n
F
y
v
Slope = v/y
• Kinematic Viscosity   
m2/Sec.
r
cS (centistokes) or
Pressure and Elevation
• Change in pressure in homogeneous
liquid at rest due to a change in elevation
DP = gh
Where,
DP = change in pressure, kPa
g = specific weight, N/m3
h = change in elevation, m
Pressure-Elevation Relationship
• Valid for homogeneous fluids at rest
(static)
P2 = Patm + rgh
Free Surface
Free Surface
P2
P1
P1 > P2
Example: Manometer
• Calculate pressure
(psig) or kPa (gage)
at Point A. Open end
is at atmospheric
pressure.
A
0.15 m
Water
0.4 m
Hg: SG = 13.54
Forces due to Static Fluids
• Pressure =Force/Area (definition)
• Force = Pressure*Area
• Example:
– If a cylinder has an internal diameter of 50
mm and operates at a pressure of 20 bar,
calculate the force on the ends of the cylinder.
Force-Pressure: Rectangular Walls
Patm
Vertical wall
d
DP = g*h
Flow Classification
• Classification of Fluid Dynamics
Laminar
Inviscid
µ=0
Viscous
Turbulent
Compressible
Incompressible
ϱ = constant
22-May-17
Internal
External
10/27
Definitions
• Volume (Volumetric) Flow Rate
– Q = Cross Sectional Area*Average Velocity
of the fluid
v
Q = Volume/Unit time
Volume
Q = Area*Distance/Unit Time
– Q = A*v
• Weight Flow Rate
– W = g*Q
• Mass Flow Rate
– M = r*Q
Key Principles in Fluid Flow
• Continuity for any fluid (gas or liquid)
– Mass flow rate In = Mass Flow Rate out
– M1 = M2
M1
– r1*A1*v1 = r2*A2*v2
• Continuity for liquids
– Q1 = Q2
– A1*v1 = A2*v2
M2
Newton’s Laws
• Newton’s laws are relations between motions of bodies
and the forces acting on them.
– First law: a body at rest remains at rest, and a body in motion
remains in motion at the same velocity in a straight path when
the net force acting on it is zero.
– Second law: the acceleration of a body is proportional to the net
force acting on it and is inversely proportional to its mass.
– Third law: when a body exerts a force on a second body, the
second body exerts an equal and opposite force on the first.
Momentum Equation
• Steady Flow
• Average velocities
• Approximate momentum flow rate
Total Energy and Conservation
of Energy Principle
• E = FE + PE + KE
v2
E  w  w z  w
g
2g
P
• Two points along the same pipe:
E1 = E2
• Bernoulli’s Equation:
wv12 wp2
wv22
 wz1 

 wz2 
g
2g
g
2g
wp1
v12
P2
v22
 z1 

 z2 
g
2g g
2g
P1
Conservation of Energy
v12
P2
v22
 z1 
 hA  hR  hL 
 z2 
g
2g
g
2g
P1
Bernoulli’s Equation
 P 2 V2
  P1 V1






z


z
2
1
 rg 2g
  rg 2g


 

2
2
17
 P 2 V2 2
  P1 V12






z


z
2
1
 rg 2g
  rg 2g


 

P
 z  piezometric head
ρg
V2
 kinetic head
2g
18
Flow through a contraction
1
2
P
V2 
  z 
  constant
2g 
 ρg
head
Total Energy
Kinetic Head
Piezometric Head
position
19
head
Energy Grade Line
Kinetic Head
Piezometric Head
Hydraulic Grade Line
position
20
Bernoulli’s Equation
• No heat transfer
• No shear work (frictionless)
• Single uniform inlet and
single uniform outlet
•
•
•
•
No shaft work
Steady state
Constant temperature
Incompressible
21
Frictional Effects
• Pipes are NOT frictionless
• Add a loss due to friction to Bernoulli’s eq.
2
2
P1 V1
P 2 V2

 z1 

 z2  h L
ρg 2g
ρg 2g
Head loss due to friction
22
head
EGL (ideal)
hL
EGL (actual)
2
V
2g
P
z
ρg
HGL (actual)
position
23
Friction Losses
hL = head losses due to friction
h L  hm  hf
 V2 

h m  K m 
 2g 
L  V2 

h f  f 
D  2g 
ΔPL  ΔPm  ΔPf
2


ρV
Fittings (valves,

ΔPm  K m 
elbows, etc)
 2 
Pipe friction
L  ρV
ΔPf  f 
D 2
2



24
Minor Losses
 ρV 2 

ΔPm  K m 
 2 
Km - minor loss coefficient
Km 
K
m
fittings
25
Pipe Friction Losses
L  ρV
ΔPf  f 
D 2
2



Darcy-Weisbach equation
Kinetic pressure
Length/diameter ratio
Darcy friction factor
- pipe roughness (Table 6.1)
- Reynolds number
26
Reynolds Number
• Dimensionless
• Ratio of inertial forces to viscous forces
• Used to characterize the flow regime
ρVL VL
Re 

μ
ν
27
Reynolds Number
• Describes if the flow is:
– Laminar - smooth and steady
– Turbulent - agitated, irregular
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28
Osborne Reynolds Tests
Laminar
Turbulent
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
29
Friction Factor
•For circular pipe Re ≈ 2300 for transition
•L = Diameter of pipe
•Laminar flow
64
f
Re d
•Turbulent flow
ε


1
2.51
 d
 Colebrook


2.0log
1/2
1/2 

Equation
f
3.7 Re d f


30
Piping Systems
•
Three examples of piping systems
1. Pipes in series
Q1  Q2  Q3
h A  B  h1  h 2  h 3
32
Home work
• Two reservoirs are connected by a pipe as
shown. The volume flow rate in pipe A is
2.2 L/s. Find the difference in elevation between
the two surfaces.
r = 1000 kg/m3
 = 0.001 kg/(m·s)
e = 0.05 mm
Dz
A
B
D = 2 cm
L=5m
D = 4 cm
L=5m
33
Centrifugal Pumps
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2
2
P1 V1
P 2 V2

 z1 

 z2  h L  h P
ρg 2g
ρg 2g
Head increase over pump
 V2 
 P 2 P1   V2 2 V12 
L  V2 
  z 2  z1   K m 
  f 

h p      



D  2g 
 ρg ρg   2g 2g 
 2g 
2
2




ρ 2
ρV
L
ρV
2
  f 

ΔPP  P2  P1   ρgz 2  z1   V2  V1  K m 
2
D 2 
 2 

Fixed system
pressure
DPsy stem  A  BQ

Variable system
pressure
2
 
 f Q 
f V
2
2
Nature of Dimensional Analysis
Example: Drag on a Sphere
 Drag depends on FOUR parameters:
sphere size (D); speed (V); fluid density (r); fluid viscosity ()
 Difficult to know how to set up experiments to determine
dependencies
 Difficult to know how to present results (four graphs?)
Nature of Dimensional Analysis
Example: Drag on a Sphere
 Only one dependent and one independent variable
 Easy to set up experiments to determine dependency
 Easy to present results (one graph)
Buckingham Pi Theorem
• Step 1:
List all the dimensional parameters
involved
Let n be the number of parameters
Example: For drag on a sphere, F, V, D,
r, , and n = 5
Buckingham Pi Theorem
• Step 5
Set up dimensional equations, combining the
parameters selected in Step 4 with each of the other
parameters in turn, to form dimensionless groups
There will be n – m equations
Example: For drag on a sphere
Dimensional Analysis and
Similarity
• Geometric Similarity - the model must be the
same shape as the prototype. Each dimension
must be scaled by the same factor.
• Kinematic Similarity - velocity as any point
in the model must be proportional
• Dynamic Similarity - all forces in the model
flow scale by a constant factor to
corresponding forces in the prototype flow.
• Complete Similarity is achieved only if all 3
conditions are met.
Flow Similarity and Model
Studies
• Example: Drag on a Sphere
For dynamic similarity …
… then …
Flow Similarity and Model
Studies
• Scaling with Multiple Dependent Parameters
Example: Centrifugal Pump
(Negligible Viscous Effects)
If …
… then …
DPsystem
System Curve
Static head
Q
DPpump
Pump Curve
Q
2
2




ρ 2
ρV
L
ρV
2
  f 

ΔPP  P2  P1   ρgz 2  z1   V2  V1  K m 
2
D 2 
 2 


System pressure
Pump pressure
DP
Operating point
Q
DP
Q
DP
changing
DPsy stem  A  BQ2
Q
DP
DPsy stem  A  BQ
changing
Q
2
Pump Power
• Recall that
w p   vdP  v  dP  v (P2  P1 )
 w p  QDPp
W p  m
W p 
QDPp
p
Power provided to fluid
Power required by pump
Home Work
Water is pumped between two reservoirs at 0.2 ft3/s (5.6 L/s) through
400 ft (124m) of 2-in (50mm) -diameter pipe and several minor losses,
as shown. The roughness ratio is e/d = 0.001.
Compute the pump horsepower required