Pharos University
ME 259 Fluid Mechanics for
Electrical Students
Application of Bernoulli
Equation
1
• Euler’s equation of motion
• Bernoulli equation
2
INTRODUCTION
 The three equations commonly used in fluid mechanics are:
the mass, Bernoulli, and energy equations.
 The mass equation is an expression of the conservation of
mass principle.
 The Bernoulli equation: Conservation of kinetic, potential,
and flow energies ( viscous forces are negligible.)
3
Fluid Motion
• Two ways to describe
fluid motion
dx
– Lagrangian
V
i
dt
• Follow particles around
dy
dz
j k
dt
dt
– Eularian
V  ui  vj  wk
• Watch fluid pass by a
point or an entire region
– Flow pattern
• Streamlines – velocity
is tangent to them
4
STEADY AND UNSTEADY FLOW
Steady flow: the flow in which conditions at any
point do not change with time is called steady flow.
Then, P  0, V  0,   0, …etc.
t
t
t
• Unsteady flow: the flow in which conditions at any
point change with time, is called unsteady flow.
Then, P  0, V  0,   0, …etc.
t
t
t
5
UNIFORM AND NON-UNIFORM
• The flow in which the conditions at
all points are the same at the same
instant is uniform flow.
P
V

 0,
 0,
 0,
s
s
s
• The flow in which the conditions
vary from point to point at the
same instant is non-uniform flow.
P
V

 0,
 0,
 0,
s
s
s
6
ACCELERATION
• Acceleration = rate of change
of velocity
• Components:
– Normal – changing direction
– Tangential – changing speed


V  V ( s , t )et


de
 dV dV 
a

et  V t
dt
dt
dt
dV
V V
V

dt
s t

det V 
 en
dt
r
V V  V 2 

a  (V

)et 
en
s t
r
7
ACCELERATION
•
Cartesian coordinates
 


V  ui  vj  wk




a  ax i  a y j  az k
du u dx u dy u dz u u
u
u
u




. u  v w
dt x dt y dt z dt t
x
y
z
t
dv v dx v dy v dz v v
v
v
v
ay 



 . u  v w
dt x dt y dt z dt t
x
y
z
t
dw w dx w dy w dz w w
w
w
w
az 




. u
v
w
dt x dt y dt z dt t
x
y
z
t
ax 
•
•
•
•
•
In steady flow ∂u/∂t = 0 , local acceleration is zero.
In unsteady flow ∂u/∂t ≠ 0 ; local acceleration Occurs. Convective
Other terms u ∂u/∂x, v ∂u/∂y,.. are called convective
accelerations. Convective acceleration Occurs when the
velocity varies with position.
Uniform flow: convective acceleration = 0
Non-uniform flow: convective acceleration ≠0
Fluid Mechanics I
Local
8
Example
• Valve at C is opened
slowly
• The flow at B is non
uniform
• The flow at A is
uniform
Fluid Mechanics I
9
Laminar vs Turbulent Flow
• Laminar
• Turbulent
10
Flow Rate
• Volume rate of flow
– Constant velocity over
cross-section
Q  VA
– Variable velocity
Q   VdA
A
• Mass flow rate
m   VdA    VdA  Q
A
A
11
Flow Rate
•
Only x-direction component of
velocity (u) contributes to flow
through cross-section
Q   VdA   udA   V cosdA
A
A
A
or
Q   V  dA
A
or
Q V  A
12
CV Inflow & Outflow
 
Q V  A
Area vector always points
outward from CV
Qout  Qin  V2 A2  V1 A1
 
 
 V2 A2  V1  A1
 
 V  A
CS
13
Examples
Q  VA
• Discharge in a 25-cm pipe is
0.03 m3/s. What is the average
velocity?
V
• A pipe whose diameter is 8 cm
transports air with a temp. of
20oC and pressure of 200 kPa
abs. At 20 m/s. What is the
mass flow rate?
p
200000


 2.378 kg / m 3
RT 287 * 293
m  VA
Q
Q
0.03


 0.611m / s
A  d 2  (0.25) 2
4
4

 2.378 * 20 * (0.08) 2  0.239 kg / s
4
14
Example: The velocity distribution in a circular duct is
r
v ( r )  Vo, (1  )
R
where r is the radial location in the duct, R is the duct radius,
and Vo is the velocity on the axis. Find the ratio of the mean
velocity to the velocity on the axis.
•
Find:
V
Vo
r
v(r )  Vo (1  )
R
R
Q   VdA   Vo (1  r / R)2rdr
A
0
2
3
R
r
r
R2 R2
 2Vo (  )  2Vo (  )
2 3R 0
2
3
1
 Vo R 2
3
1

Vo R 2
V
Q
1

3 2

Vo AVo
R Vo
3
15
Example: Air (ρ =1.2 kg/m^3) enters the duct shown:
•
•
•

Find: Q, V , m
V/10=y/.5  V=20y
dA=1*dy
0.5
0.5
0
0
Q  2  VdA  2  20 ydy
2 0.5
y
 40
2
 5 m3 / s
0
Q
5
  5m / s
A 1
m
  Q  1.2 * 5  6 kg / s
V 
16
Example: In this flow passage the discharge is varying with time according to the
t
following expression: Q  Qo  Q1 . At time t=0.5 s, it is known that at section Ato
A the velocity gradient in the S direction
is +2m/s per meter. Given that Qo, Q1,
and to are constants with values of 0.985 m^3/s, 0.5 m^3/s, and 1 s, respectively,
and assuming that one-dimensional flow, answer the following questions for time
t=0.5 s.
a. What is the velocity a A-A?
b. What is the local acceleration at A-A?
c. What is the convective acceleration at A-A?
Q  Qo  Q1
t
 0.985  0.5t
to
V
 2m/ s
s
V
Q

A
Qo  Q1

2
t
to

0.985  0.5(0.5)

 3.4743 m / s
2
d
(0.5)
4
4
V  (Q / A)
 Q1
 0.5
aL 



 2.55 m / s 2
 2

t
t
d to
(0.5) 2 (1)
4
4
V
aC  V
 3.743 * 2  7.49 m / s 2
Fluid Mechanics I
s
17
Example



2
Given : V  3ti  xzj  ty k

Find : Acceleration, a
u  3t ; v  xz; w  ty 2
u
u
u
u
u  v w
 0(3t )  0( xz )  0(ty 2 )  3  3
x
y
z
t
v
v
v
v
 u  v  w   z (3t )  0( xz )  x(ty 2 )  0  3 zt  xy 2t
x
y
z
t
w
w
w
w

u
v
w
 0(3t )  2ty( xz )  0(ty 2 )  y 2  2 xyzt  y 2
x
y
z
t
ax 
ay
az





2 
2 
a  a x i  a y j  a z k  3i  (3tz  txy ) j  (2 xyzt  y )k
18
Systems, Control Volume, and Control Surface
System (sys)
A fluid system: contains the
same fluid particles.
Mass does not cross the
system boundaries.
Thus the mass of the system
is constant.
Control Volume (C.V)
A control volume is a selected
volumetric region in space. It’s
shape and position may change
with time.(Open System)
Control Surface (C.S)
The surface enclosing the
control volume is called
the control surface.
19
Systems, Control Volume, and Control Surface (continued )
Consider the tank shown, assume:
•
the control volume is defined by the tank
walls and the top of the liquid.
•
The control surface that encloses the control
volume is designated by the dashed line.
•
The liquid in the tank at time t is elected as
the system and is indicated by the solid line.
At this instant in time, the system completely
occupies the control volume and is contained
by the control surface. Thus, at this time:
M sys (t )  M cv (t ) .........(1)
After a time  t some liquid
has flowed out of the control
volume to the right. The
amount that flowed out is:
.
M out  m out  t
During the same period
some liquid has entered the
control volume from the left,
the amount being:
.
M in  m in  t
20
Systems, Control Volume, and Control Surface (continued )
Now the system has been deformed as shown in Fig. (b).
Part of the system is the liquid that has flowed out across
the control surface.
The system remaining in the control volume has been
deformed by the mass that has flowed in across the
control surface.
Also, the height of the control volume has changed to
accommodate the net flow into the tank.
The mass of the system at time t + Δt can be determined
by taking the mass in the control volume, subtracting the
mass that entered, and adding the mass that left.
M sys (t  t )  M cv (t  t )  M out  M in ......(2)
Subtracting (1) from (2), we have:
M sys (t  t )  M sys (t )  M cv (t  t )  M cv (t )  M out  M in
.
.
M sys (t  t )  M sys (t )  M cv (t  t )  M cv (t )  (mout  min )t
Dividing by Δt and taking the limit as Δt 0 yields
dM sys
dt
.
dM cv .

 mout  min .......(3)
dt
The equation relates the rate of change of the mass of the
system to the rate of change of mass in the control volume
plus the net outflow (efflux) across the control surface
21
Systems, Control Volume, and Control Surface (continued )
By definition, the mass of the system is constant
so
dM sys
0
Lagrangian statement
dt
And Eq. (3) becomes
.
dM cv .
 m out  m in  0
dt
The corresponding Eulerian statement and can be
written as:
.
dM cv .
 m in  m out
dt
This equation states that there is an increasing mass in the
control volume if there is a net mass influx through the
control surface and decreasing mass if there is a net mass
efflux. This is identified as the continuity equation.
22
Systems
•
Laws of Mechanics
– Written for systems
– System = arbitrary quantity of
mass of fixed identity
– Fixed quantity of mass, m
• Conservation of Mass
– Mass is conserved and
does not change
dm
0
dt
• Momentum
– If surroundings
exert force on
system, mass
will accelerate

 d (mV )
F
dt
• Energy
– If heat is added to
system or work is
done by system,
energy will change
dE dQ dW


dt
dt
dt
23
Control Volumes
• Solid Mechanics
– Follow the system, determine
what happens to it
• Fluid Mechanics
– Consider the behavior in a
specific region or Control
Volume
• Convert System approach to
CV approach
– Look at specific regions, rather
than specific masses
• Reynolds Transport Theorem
– Relates time derivative of
system properties to rate of
change of property in CV
B   bdm   bd
CV
CV
 mass, momentum, energy (extensive)
dB
dm
 amount of B per unit mass (intensive)
b
24
CV Inflow & Outflow
bm  B
 



Bnet  Bout  Bin  bm out  bm in   bV  A   bm
CS
CS
M sys ,t  t  M CV ,t  t  M out  M in
Bsys ,t  t  BCV ,t  t  Bout  Bin
25
Continuity Equation
•
In the case of the continuity equation, the extensive property in the control volume
equation is the mass of the system, Msys, and the corresponding intensive variable,
b, is the mass per unit mass, or
b
•
M sys
1
Substituting b equal to unity in the control volume equation yields the general
form of the continuity equation.
dM sys
dt
•
M sys
 
d
   d    V  dA
cs
dt cv
This is sometimes called the integral form of the continuity equation.
26
Continuity Equation
•
•
•
The term on the left is the rate of change of the mass of the system. However, by
definition, the mass of a system is constant. Therefore the left-hand side of the
equation is zero, and the equation can be written as
 
d
cs  V  dA   dt cv  d
This is the general form of the continuity equation. It states that the net rate of the
outflow of mass from the control volume is equal to the rate of decrease of mass
within the control volume.
The continuity equation involving flow streams having a uniform velocity across
the flow section is given as
 
d

V

A


 d


cv
dt
cs
27
Example: at a certain time rate of
rising is 0.1 cm/s:
• Continuity equation
 
d
0
V  A
 d   
dt CV
CS
.
dM cv .
 m in  m out  0
dt
d
( Atank h)  Vin Ain  Vout Aout
dt
dh
 Atank
 Vin Ain  Vout Aout
dt
 0.1* 0.1x10  2  Vin (0.0025)  2 g *1(0.0025)

Vin  4.47 m / s
28
Example: Both pistons are moving to the left, but piston A has a speed
twice as great as that of piston B. Then the water level in the tank is: a)
rising, b) not moving up or down, c) falling.
 
d
0
• Select a CV that moves up
 d   V  A
dt CV
CS
and down with the water
surface
• Continuity Equation
CS
0
d
 d   2VB AA  VB AB
dt CV
AA  4 32 ; AB  4 6 2 ; AA  14 AB
dh
0  A  2VB ( 14 AB )  VB AB
dt
dh
A  12 VB AB  0  surface is rising
dt
VB
VA=2VB
h
Fluid Mechanics I
29
Euler Equation
•
Fluid element accelerating in l
direction & acted on by pressure
and weight forces only (no
friction)
Newton’s 2nd Law
•
F
l
 Mal
pA  ( p  p )A  W sin    lAal
W  lA
p  ( p  p )  l sin    lal
dp
dz
 g
  al
dl
dl
a
d p
 (  z)  l
dl 
g

  g
30
Example 1:
•
•
Given: Steady flow. Liquid is
decelerating at a rate of 0.3g.
Find: Pressure gradient in flow direction
in terms of specific weight.
a
d p
 (  z)  l
dl 
g
dp

dz
  al  
dl
g
dl
Flow l
30o

  ( 0.3g )   sin 30o
g
  (0.3  0.5)
dp
 0.2
dl
31
Example 2:
•
•
vertical
Given:  = 10 kN/m3, pB-pA=12 kPa.
Find: Direction of fluid acceleration.
A
d p
a
 (  z)  z
dz 
g
1 dp dz
az   g (
 )
 dz dz
p  pB
az   g ( A
 1)
1m
B

 12,000
 1)
10,000
a z  g (1.2  1)  0
az   g (
(accelerat ion is up)
32
Example 3:
• Given: Steady flow. Velocity varies
linearly with distance through the
nozzle.
• Find: Pressure gradient ½-way
through the nozzle
a
d p
(  z)  x
dx 
g
dp
dV
  a x    (V
)
dx
dx
V1/2=(80+30)/2 ft/s = 55 ft/s

dV/dx = (80-30) ft/s /1 ft = 50 ft/s/ft
 ( 1.94 slugs / ft 3 ) * (55 ft / s ) * (50 ft / s / ft )
 5,355lbf / ft 2 / ft
33
Bernoulli Equation
d p
1
 (  z )  at
ds 
g
1 dV
 V
g ds
d  V 2 

ds  2 g 
d  p
V 2 
z
0


ds  
2g 
V2
z
 Constant

2g
p
• Consider steady flow
along streamline
• s is along streamline,
and t is tangent to
streamline
p

 z  Piezometric head
V2
Velocity( dynamic) head
2g
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
34
The Bernoulli equation states that
the sum of the kinetic, potential, and
flow energies of a fluid particle is
constant along a streamline during
steady flow.
An alternative form of the Bernoulli
equation is expressed in terms of heads as:
The sum of the pressure, velocity, and
elevation heads is constant along a
streamline.
35
Example 4:
• Given: Velocity in outlet pipe from
reservoir is 6 m/s and h = 15 m.
• Find: Pressure at A.
• Solution: Bernoulli equation
Point 1
V12 p A
V A2
 z1 

 zA 

2g

2g
p1
0
pA
V A2
h

0

2g 
2g
0
pA
pA
Point A
V A2
18
  ( h  )  9810(15 
)
2g
9.81
 129.2 kPa
36
Example 5:
• Given: D=30 in, d=1 in, h=4 ft
• Find: VA
• Solution: Bernoulli equation
V12 p A
V A2
 z1 

 zA 

2g

2g
p1
Point 1
Point A
0 0
V A2
h
 0

2g 
2g
0
V A  2 gh
 16 ft / s
37
Static, Stagnation, Dynamic, and Total Pressure: Bernoulli Equation
Dynamic
Pressure
Static
Pressure
Hydrostatic
Pressure
Static Pressure: moves along the fluid “static” to the motion.
p1  h
Dynamic Pressure: due to the mean flow going to forced stagnation.
Hydrostatic Pressure: potential energy due to elevation changes.
Following a streamline:
p2 
1
1
V2 2  z2  p1  V12  z1
2
2
0
0, no elevation
p2  p1 
1
V12
2
V1   H  h
Note:
Follow a Streamline from point 1 to 2
0, no elevation
p2  H
“Total Pressure = Dynamic Pressure + Static Pressure”
H>h
In this way we obtain a measurement of the centerline flow with piezometer tube.
38
• Bernoulli equation: The sum of flow, kinetic, and potential
energies of a fluid particle along a streamline is constant.
• Each term in this equation has pressure units, and thus
each term represents Energy per unit volune
• P is the static pressure.
• ϱV2/2 is the dynamic pressure.
• ϱ gz is the hydrostatic pressure
• The sum of the static, dynamic, and hydrostatic pressures
is called the total pressure. Bernoulli equation states that
the total pressure along a streamline is constant.
• The sum of the static and dynamic pressures is called the
stagnation pressure, and it is expressed as
39
• The static, dynamic, and
stagnation pressures are
shown.
• When static and stagnation
pressures are measured at a
specified location, the fluid
velocity at that location can
be calculated from
40
Stagnation Tube
V12 p2
V22
 z1 

 z2 

2g

2g
p1
V12 p2


 2g 
2
V12  ( p2  p1 )
p1


2

( (l  d )  d )
V1  2 gl
41
Stagnation Tube in a Pipe
H
p

z
V2
2g
V2
2g
p

Pipe
2
Flow
1
z
z0
42
Pitot Tube
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
p1 V12 p2 V22



 2g  2g
V2  2 g[(
p1

 z1 )  (
p1

 z1 )
V  2 g ( h1  h2 )
43
Example – Venturi Tube
•
•
•
Given: Water 20oC, V1=2 m/s, p1=50 kPa,
D=6 cm, d=3 cm
Find: p2 and p3
Solution: Continuity Eq.
V1 A1  V2 A2
A
D
V2  V1 1  V1  
A2
d 
•
2
V2 p
V2
 z1  1  2  z2  2

2g

2g
p1
 p1 

2

2
(V12  V22 )
[1  D / d 4 ]V12
1000
[1  6 / 34 ]22 Pa
2
p2  120 kPa
 150,000 
D
d
2
1
Bernoulli Eq.
p2  p1 
D
3
Nozzle: velocity
increases, pressure
decreases
Diffuser: velocity
decreases, pressure
increases
Similarly for 2  3, or 1  3
p3  150 kPa
Pressure drop is fully recovered, since we
assumed no frictional losses
Knowing the pressure drop 1  2 and
d/D, we can calculate the velocity and
flow rate
Fluid Mechanics I
V2 
2( p1  p2 )
 [1  d / D 4 ]
44
Ex
•
•
•
Given: Velocity in circular duct = 30
m/s, air density = 1.2 kg/m3.
Find: Pressure change between circular
and square section.
Solution: Continuity equation
Vc Ac  Vs As
30(

D 2 )  Vs D 2
4
Vs  m / s
•
Bernoulli equation
pc
Vc2
 zc 


2g
pc  p s 
Air conditioning (~ 60 oF)
Vs2
 zs 

2g
ps

2
pc  ps  N / m
2
(Vs2  Vc2 )
45
Ex
•
•
•
Given:  = 1000 kg/m3 V1= 30 m/s,
and A2/A1=0.5, m=20000 N/m3
Find: h
Solution: Continuity equation
V1 A1  V2 A2
V2  V1
•
A1
m/ s
A2
Bernoulli equation
V2 p
V2
 z1  1  2  z2  2

2g 
2g
p1
p1  p2 

2
(V22
Heating (~ 170 oF)
•
Manometer equation
 V12 )
p1  p2  h( m   air )
p1  p2 
h  m
 N / m2
46
Pitot Tube Application
V
1
z1-z2
p1  ( z1  z2 ) k  l k  y Hg  (l  y ) k  p2
p1  p2  y ( Hg   k )  ( z1  z2 ) k
2
p1  p2
k
l
 z1  z2 
y ( Hg   k )
k
h1  h2  y ( Hg /  k  1)
V  2 gy ( Hg /  k  1)  m / s
y
47