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