pharos university me 259 fluid mechanics ii review of previous fluid mechanics dr. shibl
TRANSCRIPT
Pharos University
ME 259 Fluid Mechanics II
Review of Previous Fluid Mechanics
Dr. Shibl
Fluid Properties:Liquid or Gas
• Liquids are:– Incompressible, V ≠ f(P)– Viscous (high viscosity)– Viscosity decreases with temperature
• Gases are:– Compressible, V = f(P)– Low viscosity– Viscosity increases with temperature
Equations for Fluid Property
• Circular Area: • Weight: w = m*g Newton• Density: = m/V Kg/m3
• Specific Weight: = w/V N/m3
• Specific gravity: SG=/water
Area = /4*D2
Viscosity• Dynamic Viscosity
= Shear Stress/Slope of velocity profile
• Kinematic Viscosity cS (centistokes) or m2/Sec.
yv
AF
/
/
Slope = v/y
v
y
F
cP (centipoise) or Pa-sec
Pressure and Elevation
• Change in pressure in homogeneous liquid at rest due to a change in elevationP = h
Where,
P = change in pressure, kPa = specific weight, N/m3
h = change in elevation, m
Pressure-Elevation Relationship
• Valid for homogeneous fluids at rest (static)
Free Surface Free Surface
P1
P2
P1 > P2
P2 = Patm + gh
Example: Manometer
• Calculate pressure (psig) or kPa (gage) at Point A. Open end is at atmospheric pressure.
A
Hg: SG = 13.54
Water
0.4 m
0.15 m
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
P = *h
Patm
d
Vertical wall
Flow Classification • Classification of Fluid Dynamics
Apr 19, 2023 10/27
Inviscidµ = 0
Viscous
Laminar
Turbulent
Compressible Incompressibleϱ = constant
Internal External
Definitions
• Volume (Volumetric) Flow Rate– Q = Cross Sectional Area*Average Velocity
of the fluid– Q = A*v
• Weight Flow Rate– W = *Q
• Mass Flow Rate– M = *Q
Volumev
Q = Volume/Unit timeQ = Area*Distance/Unit Time
Key Principles in Fluid Flow
• Continuity for any fluid (gas or liquid)– Mass flow rate In = Mass Flow Rate out
– M1 = M2
– 1*A1*v1 = 2*A2*v2
• Continuity for liquids– Q1 = Q2
– A1*v1 = A2*v2
M1 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
• Two points along the same pipe: E1 = E2
• Bernoulli’s Equation:
Pz
v
g
Pz
v
g1
112
22
22
2 2
g
vwzw
PwE
2
2
g
wvwz
wp
g
wvwz
wp
22
22
22
21
11
Conservation of Energy
Pz
v
gh h h
Pz
v
gA R L1
112
22
22
2 2
17
1
211
2
222 z
2g
VPz
2g
VP
gg
Bernoulli’s Equation
18
zρg
P
2g
V2
piezometric head
kinetic head
1
211
2
222 z
2g
VPz
2g
VP
gg
19
Flow through a contraction
constant2g
Vz
ρg
P 2
position
head Total Energy
Piezometric Head
Kinetic Head
1 2
20
position
Energy Grade Line
Piezometric Head
Kinetic Head
head
Hydraulic Grade Line
21
Bernoulli’s Equation• No shaft work• Steady state• Constant temperature• Incompressible
• No heat transfer• No shear work (frictionless)• Single uniform inlet and
single uniform outlet
22
Frictional Effects
• Pipes are NOT frictionless• Add a loss due to friction to Bernoulli’s eq.
L2
222
1
211 hz
2g
V
ρg
Pz
2g
V
ρg
P
Head loss due to friction
23
position
head EGL (ideal)
EGL (actual)
HGL (actual)
hL
zρg
P
2g
V2
24
Friction Losses
fmL hhh
2g
VKh
2
mm
hL = head losses due to friction
2g
V
D
Lfh
2
f
Fittings (valves, elbows, etc)
Pipe friction
fmL ΔPΔPΔP
2
ρVKΔP
2
mm
2
ρV
D
LfΔP
2
f
25
Minor LossesKm - minor loss coefficient
2
ρVKΔP
2
mm fittings
mm KK
26
Pipe Friction Losses
2
ρV
D
LfΔP
2
fDarcy-Weisbach equation
Kinetic pressure
Length/diameter ratio
Darcy friction factor- pipe roughness (Table 6.1)- Reynolds number
Reynolds Number
• Dimensionless• Ratio of inertial forces to viscous forces• Used to characterize the flow regime
27
ν
VL
μ
ρVLRe
Reynolds Number
• Describes if the flow is:– Laminar - smooth and steady– Turbulent - agitated, irregular
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
28
Osborne Reynolds Tests
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Laminar
Turbulent
29
Friction Factor
dRe
64f
•For circular pipe Re ≈ 2300 for transition•L = Diameter of pipe
•Laminar flow
•Turbulent flow
1/2d
1/2 fRe
2.51
3.7d
ε2.0log
f
1 Colebrook Equation
30
Piping Systems
• Three examples of piping systems1. Pipes in series
321BA hhhh 321 QQQ
32
33
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.
A B
D = 2 cmL = 5 m
D = 4 cmL = 5 m
z
= 1000 kg/m3
= 0.001 kg/(m·s) = 0.05 mm
Centrifugal PumpsCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
PL2
222
1
211 hhz
2g
V
ρg
Pz
2g
V
ρg
P
Head increase over pump
2g
V
D
Lf
2g
VKzz
2g
V
2g
V
ρg
P
ρg
Ph
22
m12
21
2212
p
2
ρV
D
Lf
2
ρVKVV
2
ρzzρgPPΔP
22
m2
1221212P
Fixed system pressure
Variable system pressure
2Vf
2Qf
2BQAP system
Nature of Dimensional Analysis
Example: Drag on a Sphere
Drag depends on FOUR parameters:sphere size (D); speed (V); fluid density (); 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, , , 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 …
systemP
Q
Static head
System Curve
pumpP
Q
Pump Curve
2
ρV
D
Lf
2
ρVKVV
2
ρzzρgPPΔP
22
m2
1221212P
Pump pressureSystem pressure
P
Q
Operating point
P
Q
P
Q
2BQAP system
changing
P
Q
2BQAP system
changing
Pump Power
• Recall that )( 12 PPvdPvvdPw p
ppp PQwmW
p
pp
PQW
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 /d = 0.001. Compute the pump horsepower required