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PHAROS UNIVERSITYME 259
FLUID MECHANICS FOR ELECTRICAL STUDENTS
Basic Equations for a Control Volume
Main TopicsFlow ClassificationBasic Laws for a SystemRelation of System Derivatives
to the Control Volume FormulationConservation of MassBernoulli Equation
Flow Classification Classification of Fluid Dynamics
Inviscidµ = 0
Viscous
Laminar
Turbulent
Compressible Incompressibleϱ = constant
Internal External
Basic Laws for a System
Conservation of Mass
Relation of System Derivatives to the Control Volume Formulation
Extensive and Intensive Properties
Relation of System Derivatives to the Control Volume Formulation
Reynolds Transport Theorem
• Continuity Equation
• Bernoulli’s Equation
• Momentum Equation
• Energy Equation
FLUID FLOW
Basic Laws Conservation of mass: dM/dt=0 for
system∂/∂t ∫ϱ d 𐐏 +∫ ϱ ⊽. d Ᾱ=0 for control
vol.
Newton’s 2nd law Σ F = ma Σ F = ∂/∂t ∫ ⊽ ϱ d 𐐏 + ∫ ⊽ ϱ ⊽. d Ᾱ
First Law of Thermo: Q - W = dE/dt
Q-w=∂/∂t∫e ϱ d𐐏+∫(p/ ϱ+0.5V2+gz)ϱ⊽. dᾹ
Relation of System Derivatives to the Control Volume Formulation
Interpreting the Scalar Product
Conservation of Mass
Basic Law, and Transport Theorem
Conservation of Mass
Conservation of Mass
Basic Law for a System
Conservation of Mass
Incompressible Fluids
Steady, Compressible Flow
DefinitionsVolume (Volumetric) Flow Rate• Q = Cross Sectional Area*Average
Velocity of the fluid• Q = A*v cms
Weight Flow Rate• W = *Q N/s
Mass Flow Rate• M = *Q kg/s
Volume vQ = Volume/Unit time
Q = Area*Distance/Unit Time
Flow in non-circular sectionsFlow rate is determined by:• Q = A*v
Where,
A = Net flow area
v = average velocity Example:Dlarge, i = 0.5 mDsmall, o = 0.25 mDsmall, i = 0.2 mVsmall, i = 1 m/secVlarge, i = 1 m/secFind Qlarge and Qsmall
Continuity Equation
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
17/27
Equation of continuity
1 1 2 2(volume flow rate) constantvR Av A v
Volume flow rate has units m3/s
Mass flow rate has units kg/s
Units and Conversion Factors
Q: m3/secM: kg/sec,Volume Flow Rate:• 1 L/min = 0.06 m3/h• 1 m3/sec = 60,000 L/min• 1 gal/min = 3.785 L/min
Example #1If d1 and d2 are 50 mm and 100 mm,
respectively, and water at 70° C is flowing at 8 m/sec in section 1, determine: v2, Q, W, M.
1 2d1d2
v1 v2
© Pritchard
Example # 2
If d1, d2 and d3 are 10 cm, 20 cm, and 50 cm, determine Q and the velocities, v2 and v3 if v1 = 1 m/sec.
V1 = 1 m/sec
d1 = 10 cm d2 = 20 cm d3 = 50 cm
Example # 3
Determine the required size standard Schedule 40 steel pipe to carry 192 m3/hr with a maximum velocity of 6.0 m/sec.
Example # 4The tank is being filled with water by two 1-D inlets. Air is at the top of the tank. The water height is h. (a) Find an expression for the change in water height dh/dt. (b) Compute dh/dt if D1 = 1 cm, D2 = 3 cm, V1 = 3 m/s, V2 = 2 m/s and At = 2 m2.
Tank Area At
a
w
h
12
Example # 5
Consider the entrance region of a circular pipe for laminar flow. What is mean velocity of the fluid.
© PritchardApr 19, 2023 24/27
Assume:
the flow of fluids is laminar (not turbulent) or steady flow
- the fluid has no viscosity (no friction).
Ideal Fluids in Motion:Continuity & Bernoulli’s equation
A fluid element traces out a streamline as it moves. The velocity vector of the element is tangent to the streamline at every point.
The steady flow of a fluid around an air foil, as revealed by a dye tracer that was injected into the fluid upstream of the airfoil
Conservation of EnergyBernoulli’s Equation
Energy cannot be created or destroyed, just transformed
Three forms of energy in fluid system:• Potential• Kinetic• Flow energy
Potential Energy
Due to the elevation of the fluid element
Where,
w = weight of fluid element
z = elevation with respect to a reference level
zwPE
Kinetic Energy
Due to the velocity of the fluid element
Where,
v = average velocity of the fluid element
g
vwKE
2
2
Flow Energy
Flow work or pressure energyAmount of energy necessary to move a fluid
element across a certain section against pressure
Where,p = pressure on the fluid element
P
wPE
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
Heads
Pessure H ead
P r _
z E leva tion H ead _
v
gVelocity H ead
2
2 _
Pz
v
gTo ta l H ead
2
2_
Assumption: No energy is added or lost
Assumption: Energy level remains constant
Restrictions on Bernoullis’ Equation
Valid only for incompressible fluidsNo energy is added or removed by pumps,
brakes, valves, etc.No heat transfer from or to liquidNo energy lost due to friction
Application of Bernoulli’s Equation
Write Bernoulli’s equation in the direction of flow,
Label diagramSimplify equation by canceling terms that
are zero, or equal on both sides of the equation
Solve equation and find desired result(s)
Example
A hose carries water at a flow rate of 0.01 m3/sec. The hose has an internal diameter of 12 mm, and the gauge pressure at faucet is 100 kPa. Determine the pressure at the end of the hose Z = 10 m
Torricelli’s Theorem
For a liquid flowing from a tank or reservoir with constant fluid elevation, the velocity through the orifice is given by:
where, h is the difference in elevation between the orifice and the top of the tank
Example: If h = 3.00 m, compute v2
v gh2 2
h
Take Home ExperimentA reservoir of water has the surface at 310m above the outlet nozzle
of a pipe with diameter 15mm. What is the a) velocity, b) the discharge out of the nozzle and c) mass flow rate.
Water Velocity = (2gh)0.5
= (2x 9.81 x 310)0.5 = 78 m/s
Individual Experiment Pipe Flow: Ideal flow Assumption and Energy Equation
The aim is to study Continuity equation and Bernoulli equation as will as pressure losses due to viscous ( frictional) effects in fluid flows through pipes
Flow meter
Differential Pressure Gauge- measure ΔP
LValve
H
Reservoir
PipeD
Schematic of experimental Apparatus
• Pipes with different Diameter and Length will be used later for the experiments to study Energy Equation and pressure losses