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ME 304 FLUID MECHANICS II
Prof. Dr. Haşmet Türkoğlu
Çankaya University
Faculty of Engineering
Mechanical Engineering Department
Spring, 2017
du
 yx   dy
 yx
 du 
 k  
 dy 
n
du
 yx  k dy
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du
du

dy
dy
2
The Fundamental Laws
Experience have shown that all fluid motion analysis must be consistent with the
following fundamental laws of nature.
The law of conservation of mass: Mass can be neither created nor destroyed. It can
only be transported or stored.
Newton’s three laws of motion:
- A mass remains in a state of equilibrium, that is, at rest or moving at constant
velocity, unless acted on by an unbalanced force.
- The rate of change of linear momentum of mass is equal to the net force acting on
the mass.
- Any force action has a force reaction equal in magnitude and opposite in direction.
The first law of thermodynamics (law of conservation of energy) Energy, like mass, can
be neither created nor destroyed. Energy can be transported, changed in form, or
stored.
The second law of thermodynamics: The entropy of the universe must increase or, in
the ideal case, remain constant in all natural processes.
The state of postulate (law of property relations): The various properties of a fluid are
related. If a certain minimum number (usually two) of fluid’s properties are specified,
the remainder of the properties can be determined.
Differential versus Integral Formulation
We must now consider the level of detail of the resulting flow analysis. We must
choose between a detailed point by point description and a global or lumped
description.
When a point by point (local) description is desired, fundamental laws are applied to
an infinitesimal control volume. The result will be a set of differential equations with
the fluid velocity and pressure as dependent variables and the location (x, y, z) and
time as independent variables. Solution of these differential equations, together with
boundary conditions, will be two functions V(x, y, z, t), and P(x, y, z, t) that can tell us
the velocity and pressure at every point.
When global information such as flow rate, force and temperature change between
inlet and outlet is desired, the fundamental laws are applied to a finite control volume.
The result will be a set of integral equations.
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FLUID STATICS
Basic Equation of fluid statics

 grad P  g  0
p  p0  gh
dp
  g  
dz
HYDROSTATIC FORCE ACTING ON A PLANE SUBMERGED SURFACE
Integral Method 
FR    pdA
A
yFR   ypdA  y 
A
xFR   xpdA  x 
A

FR  FR   pdA
1
FR
 ypdA
A
1
xpdA
FR A

 Magnitude of FR
A

 Direction of FR is normal and toward the surface
Algebraic method
FR  ghc
Pressure Prism Method



FR  k  dP   P k
P
x'  xR 
x 
I xyc
yc A
 xc
y'  yR 
1
1
1
xPdA

x

ghdA

FR A
 P A
P
I xc
 yc
yc A
 xd  X G
P
and
y 
1
1
1
yPdA

y

ghdA

FR A
 P A
P
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 yd  YG
P
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BASIC EQUATIONS FOR A FINITE CONTROL VOLUME (Equations In Integral Form)
 

 N 


d



V
 dA
Reynolds Transport Equation: 




t

t

 System
C
CS
 

d   V  dA  0
t C
CS
Equation of Conservation of Mass (Continuity Equation):
Linear Momentum Equation:
 

  
 
F  FS  FB 
V

d


V
 V  dA
t C
CS
Bernoulli equation subject to restrictions:
Vs2
Bernoulli Equation:

 gz  C 1. Steady flow
 2
2. No friction
p
3. Incompressible flow
4. Flow along a streamline
Moment of Momentum Equation:

 

   
  
r  Fs   r  gd Tshaft 
 r  Vd   r  VV  dA

t
C
C
CS
Euler Turbine Equation:
Energy Equation:



Tshaft  r2Vt 2  r1Vt1 m
 





Q  Ws  Wshear  Wother 
ed   e  pv V  dA

t C
CS
2
2
 p1
  p2

V
V
1
2
Extended Bernouilli Equation: 






z




z
1
1
2
2   hf
 g

2g
2

  g

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HEAD LOSS (PRESSURE DROP)
The head loss (pressure loss) in closed conduits can be dived into two part:
1)
Major (friction) loss: Losses due to viscous effects on the duct wall.
2)
Minor (local) losses): Losses due to the flow through valves, tees, elbows and
other non-constant cross-sectional area portions of the system.
Major Head Loss (Pressure Drop)
L V 2
P  f
d 2
LV2
hf  f
d 2g
For laminar flow, friction factor,
For turbulent flow, friction factor,
hf 
P
g
64
f 
Re
f  f Re,  s / d 
Friction factors for turbulent flows are given in charts (Mood Diagram) or as
correlations.
Minor (Local) Head Loss
The minor head losses in variable area parts are proportional to the velocity head of the
fluid, i.e.
V2
hf  k
2g
Minor head losses for valves, fittings and bends can be calculated using the equivalent
length technique, which may be given by the following equation:
Le V 2
hf  f
d 2g
Total loss = (Major loss) + (Minor loss)
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INTRODUCTION TO DIFFERENTIAL ANALYSIS OF FLUID MOTION (Chapter 5)
In course Fluid Mechanics I, we developed the basic equations in integral form for a
finite control volume. The integral equations are particularly useful when we are
interested in the gross behavior of a flow and its effect on various devices. However,
the integral approach does not enable us to obtain detailed point by point data of
the flow field.
To obtain this detailed knowledge, we must apply the equations of fluid motion in
differential form.
In this chapter, we will derive fundamental equations in differential form and
apply this equations to simple flow problems.
EQUATION OF CONSERVATION OF MASS (CONTINUITY EQUATION)
The application of the principle of conservation of mass to fluid flow yields an
equation which is referred as the continuity equation. We shall derive the
differential equation for conservation of mass in rectangular and in cylindrical
coordinates.
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Rectangular Coordinate System
The differential form of the continuity equation may be obtained by applying the
principle of conservation of mass to an infinitesimal control volume. The sizes of the
control volume are dx, dy, and dz. We consider that, at the center, O, of the control
volume, the density is  and the velocity is


 
V  u ı  vj  wk
For the control voule, equation of consevation of mass in integral form is
  

V
CS  dA  t CV dV  0


 V  dA
To evaluate the first term
in this equation, we must evaluate the mass
CS
flow rate over each face of the control volume.
To be completed in class
The values of the mass fluxes at each of six faces of the control volume may be
obtained by using a Taylor series expansion of the density and velocity components
about point O. For example, at the right face,
 x  dx
2
2
   dx     1  dx 
       2    
 x  2  x  2!  2 
2
Neglecting higher order terms, we can write
and similarly,
   dx
 x  dx     
2
 x  2
 u  dx
u x  dx  u   
2
 x  2
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Corresponding terms at the left face are
 dx
   dx 
 x dx          
2
x 2
 x  2 
u dx
 u  dx 
u x dx  u       u 
x 2
2
 x  2 
To be completed in class
Table. Mass flux through the control surface of a rectangular differential control volume
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The net rate of mass flux out through control surface is
   u v w 

V
CS  dA  x  y  z dxdydz
The rate of change of mass inside the control volume is given by



d
V

dxdydz

t CV
t
Therefore, the continuity equation in rectangular coordinate is
u v w 



0
x
y
z
t
Since the vector operator, , in rectangular coordinates, is given by
    
 ı  j k
x
y
z
 
   V 
0
t
The continuity equation may be simplified for two special cases.
1. For an incompressible flow, the density is constant, the
continuity equation becomes,

 V  0

2. For a steady flow, the partial derivatives with respect to time are
zero, that is _________.
Then, ……………………………….
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Example: For a 2-D flow in the xy plane, the velocity component in the y direction is
given by
v  y2  x2  2 y
a)
b)
Determine a possible velocity component in the x direction for steady flow of an
incompressible fluid. How many possible x components are there?
Is the determined velocity component in the x-direction also valid for unsteady
flow of an incompressible fluid?
To be completed in class
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Example: A compressible flow field is described by



V  axi  bxyj e  kt
Determine the rate of change of the density at point x=3 m, y=2 m and z=2 m for
t=0.
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Derivation of Continuity Equation Cylindrical Coordinate System
In cylindrical coordinates, a suitable differential control volume is shown
in the figure. The density at center, O, is  and the velocity there is




V  vr er  v e  vz ez
Figure. Differential control volume in cylindrical coordinates.
 
 V  dA , we must consider the mass flux through each of
To evaluate
CS
the six faces of the control surface. The properties at each of the six
faces of the control surface are obtained from Taylor series expansion
about point O.
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Table. Mass flux through the control surface of a cylindrical differential control volume
The net rate of mass flux out through the control surface is given by
  
v v
v 

V
CS  dA  vr  r r r    r z z  drddz
The rate of change of mass inside the control volume is given by



dV

rdrddz

t CV
t
In cylindrical coordinates the continuity equation becomes
vr v
vz

vr  r

r
r
0
r

z
t
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Dividing by r gives
vr
r

vr 1 v vz 



0
r
r 
z
t
or
1 (rvr ) 1 ( v ) ( vz ) 



0
r r
r 
z
t
In cylindrical coordinates the vector operator  is given by

  1  
 
er 
e  ez
r
r 
z
Then the continuity equation can be written in vector notation as
 
  V 
0
t


e
 
er 

 Note : r  e and   er 


The continuity equation may be simplified for two special cases:
1. For an incompressible flow, the density is constant, i.e.,
2. For a steady flow,
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Derivation of Continuity Equation in Cylindrical Coordinate sytem Using Vector Form
of the Equation
To be completed in class
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Example: Consider one-dimensional radial flow in the r plane, characterized by vr =
f(r) and v = 0. Determine the conditions on f(r) required for incompressible flow.
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STREAM FUNCTION FOR TWO-DIMENSIONAL INCOMPRESSIBLE FLOW
For a two-dimensional flow in the xy plane of the Cartesian coordinate systems,
the continuity equation for an incompressible fluid reduces to
u v

0
x y
If a continuous function  ( x, y, t ) , called stream function, is defined such that
u

y
and
v

x
Then continuity equation is satisfied exactly, since
u v  2  2
 

0
x y xy yx
Streamlines are tangent to the direction of flow at every point in the flow field.
Thus, if dr is an element of length along a streamline, the equation of streamline is
given by

 

 

V  dr  0  (u ı  vj )  (dx ı  dyj )  (udy  vdx)k
Substituting for the velocity components of u and v, in terms of the stream
function 
udy  vdx  0
(A)
At a certain instant of time, t0, the stream function may be expressed as
   ( x, y, t0 ) . At this instant, the streamfunction


dx 
dy  0
x
y
(B)
Comparing equations (A) and (B), we see that along instantaneous streamline
 = constant. In the flow field, 2-1, depends only on the end points of
integration, since the differential equation of  is exact.
d 


dx 
dy
x
y
d  0
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Now, consider the two-dimensional flow
of an incompressible fluid between two
instantaneous streamlines, as shown in
the Figure. The volumetric flow rate
across areas AB, BC, DE, and DF must be
equal, since there can be no flow across
a streamline.
For a unit depth, the flow rate across AB is
Along AB, x = constant and
Q
y2
y1
d 
x1
y1
y1
2

dy  d   2  1
1
y
Along BC, y = constant and d 
x2
y2

dy
y

dy . Therefore,
y
For a unit depth, the flow rate across BC is Q 
Q  
y2
Q   udy  

x2
x1
vdx  
x2
x1

dx
x

dx . Therefore,
x
1

dx   d   2  1
2
x
Thus, the volumetric flow rate per unit depth between any two streamlines, can be
expressed as the difference between constant values of  defining the two
streamlines.
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In r plane of the cylindrical coordinate system, the incompressible continuity
equation reduces to
rvr v

0
r

The streamfunction  (r, ,t) then is defined such that
vr 
1 
r 
v  

r
Example: Consider the stream function given by  = xy. Find the corresponding
velocity components and show that they satisfy the differential continuity equation.
Then sketch a few streamlines and suggest any practical applications of the resulting
flow field.
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