MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Chapter 3:
Finite control volume analysis
Contents
3.1.
Objectives ..................................................................................................................4
3.2.
Discharge and mean velocity .....................................................................................4
3.2.1.
Discharge ...........................................................................................................4
3.2.1.1.
Definition ...................................................................................................4
3.2.1.2.
Volume flow rate .......................................................................................4
3.2.1.3.
Mass flow rate............................................................................................4
3.2.2.
Mean velocity.....................................................................................................5
3.2.2.1.
Velocity profiles.........................................................................................5
3.2.2.2.
Mean velocity.............................................................................................5
3.3.
Continuity of flow......................................................................................................8
3.3.1.
Conservation of mass.........................................................................................8
3.3.2.
Continuity equation for steady flow ................................................................10
3.4.
Momentum equation and its applications ................................................................12
3.4.1.
Momentum in a flowing fluid ..........................................................................12
3.4.2.
Momentum equation for two- and three-dimensional flow along a stream line
13
3.4.3.
Force exerted by a jet striking a flat plate........................................................14
3.4.3.1.
Stationary flat plate ..................................................................................14
3.4.3.1.1. Case: Inclined plate...............................................................................14
3.4.3.1.2. Case: Vertical plate (θ = 90°) ...............................................................15
3.4.3.2.
Hinged plate .............................................................................................18
3.4.3.3.
Moving plate ............................................................................................21
3.4.3.3.1. Force exerted by the jet on the moving plate........................................22
3.4.3.3.2. Work done per second by the jet on the plate.......................................22
3.4.3.3.3. Efficiency of transmission of the jet .....................................................23
3.4.3.3.4. Case: vertical plate (θ = 90°) ................................................................24
3.4.4.
Force due to the deflection of a jet by a curved vane ......................................26
3.4.4.1.
Stationary curved vane.............................................................................26
3.4.4.1.1. Case: Jet strikes the curved vane at the centre......................................26
3.4.4.1.2. Case: Jet strikes the curved vane at one end tangentially .....................28
3.4.4.2.
Moving curved vane ................................................................................33
3.4.4.2.1. Case: Jet strikes the curved vane at the centre......................................33
3.4.4.2.2. Case: Jet strikes the curved vane at one end .........................................36
3.5.
Energy of a flowing fluid.........................................................................................42
3.5.1.
Conservation of energy ....................................................................................42
3.5.1.1.
First law of thermodynamics ...................................................................42
3.5.1.2.
Forms of energy .......................................................................................42
3.5.2.
Energy of the fluid ...........................................................................................43
3.5.2.1.
Kinetic energy..........................................................................................43
3.5.2.1.1. Definition ..............................................................................................43
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 1
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.5.2.1.2. Kinetic energy per unit weight..............................................................43
3.5.2.1.3. Kinetic energy per unit volume.............................................................43
3.5.2.2.
Potential energy .......................................................................................44
3.5.2.2.1. Definition ..............................................................................................44
3.5.2.2.2. Potential energy per unit weight ...........................................................44
3.5.2.2.3. Potential energy per unit volume ..........................................................44
3.5.2.3.
Internal energy .........................................................................................44
3.5.2.3.1. Definition ..............................................................................................44
3.5.2.4.
Pressure energy ........................................................................................45
3.5.2.4.1. Definition ..............................................................................................45
3.5.2.4.2. Pressure energy per unit weight............................................................45
3.5.3.
External energy (shaft work) ...........................................................................45
3.5.3.1.
Definition .................................................................................................45
3.5.4.
Heat exhange between the fluid and the surroundings ....................................46
3.5.5.
Steady flow energy equation............................................................................46
3.5.5.1.
General equation ......................................................................................46
3.5.5.1.1. Mathematical form................................................................................47
3.5.5.1.2. Validity of the equation ........................................................................47
3.5.5.2.
Bernoulli's equation .................................................................................47
3.5.5.2.1. Background ...........................................................................................47
3.5.5.2.2. Bernoulli's equation for frictionless incompressible flow ....................48
3.5.5.2.3. General expression of Bernoulli's equation for incompressible flow ...50
3.5.5.2.4. Validity of the equation ........................................................................51
3.5.5.3.
Euler's equation........................................................................................55
3.5.5.3.1. Background ...........................................................................................55
3.5.5.3.2. Mathematical form................................................................................56
3.5.5.3.3. Notes .....................................................................................................56
3.5.6.
Power in fluid flow ..........................................................................................57
3.5.6.1.
Definition .................................................................................................57
3.5.7.
Application of Bernoulli's equation .................................................................58
3.5.7.1.
Pitot tube ..................................................................................................58
3.5.7.1.1. Background ...........................................................................................58
3.5.7.1.2. Description............................................................................................58
3.5.7.1.3. Stagnation pressure ...............................................................................59
3.5.7.1.4. Measurement of flow velocity in a pitot tube .......................................59
3.5.7.1.5. Other arrangements of pitot tube ..........................................................60
3.5.7.1.6. Energy line and hydraulic grade line ....................................................61
3.5.7.1.6.1. Bernoulli's equation for steady, frictionless, incompressible flow 61
3.5.7.1.6.2. Energy line .....................................................................................61
3.5.7.1.6.3. Hydraulic grade line.......................................................................61
3.5.7.1.6.4. Representation of energy line and hydraulic grade line.................61
3.5.7.2.
Venturi meter ...........................................................................................62
3.5.7.2.1. Background ...........................................................................................62
3.5.7.2.2. Description............................................................................................62
3.5.7.2.3. Measurement of flow rate in a venturi meter........................................63
3.5.7.3.
Orifice ......................................................................................................69
3.5.7.3.1. Velocity of the jet..................................................................................69
3.5.7.3.2. Discharge ..............................................................................................70
3.5.7.3.3. Vena contracta ......................................................................................70
3.5.7.3.4. Relation between Cd, Cv, and Cc ...........................................................71
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 2
MEC441 Fluid Mechanics 1
3.6.
Chapter 3: Finite control volume analysis
Problems ..................................................................................................................74
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 3
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.1. Objectives
After completing this chapter students should be able to:
ƒ
3.2. Discharge and mean velocity
3.2.1. Discharge
3.2.1.1. Definition
ƒ
It is a total quantity of fluid flowing in unit time past any particular crosssection of a stream.
ƒ
It can be measured either in terms of mass (mass flow rate) or in terms of
volume (volume flow rate).
3.2.1.2. Volume flow rate
ƒ
It is a volume of flowing fluid per unit time.
ƒ
Unit: unit of volume/unit of time, m³/s.
ƒ
Mathematical expression:
Q = ∫ udA
:
u
velocity of flowing fluid at a point in the cross-section
normal to the section
dA :
•
element of area of the point
If v is the mean velocity of the flowing fluid across the section
Q = Av
A : cros-section of the stream
v
: mean velocity of the flow passing through the section
3.2.1.3. Mass flow rate
ƒ
It is a mass of flowing fluid per unit time.
ƒ
Unit: unit of volume/unit of time, kg/s.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 4
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Mathematical expression:
i
m = ρAv = ρQ
ρ
: mass density of the flowing fluid
A : cross-section of the stream
v
: mean velocity of the flow passing through the section
3.2.2. Mean velocity
3.2.2.1. Velocity profiles
3.2.2.2. Mean velocity
v=
Q
A
Example No. 3 - 1
Air flows between two parallel plates 80 mm apart. The following velocities were
obtained from experiment.
Distance from lower plate
(mm)
0
10 20 30 40 50 60 70 80
Velocity (m/s)
0
23 28 31 32 29 22 14
0
Determine the velocity distribution curve and the mean velocity.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 5
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
ƒ
Velocity distribution curve
ƒ
Discharge
Q = ∫ udA
dA = width of the plate × dy = bdy
⇒ Q = b ∫ udy
∫ udy
= area enclosed by the curve
= a1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a8
By assuming that each area is a triangle or trapezium,
a1 =
a2 =
a3 =
a4 =
( 0 + 23) × (10 − 0 )
= 115
2
( 23 + 28 ) × ( 20 − 10 )
2
= 255
( 28 + 31) × ( 30 − 20 )
= 295
2
( 31 + 32 ) × ( 40 − 30 )
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
2
= 315
Page 6
MEC441 Fluid Mechanics 1
a5 =
a6 =
a7 =
a8 =
•
Chapter 3: Finite control volume analysis
( 32 + 29 ) × ( 50 − 40 )
2
( 29 + 22 ) × ( 60 − 50 )
2
= 305
= 255
( 22 + 14 ) × ( 70 − 60 )
= 180
2
(14 + 0 ) × ( 80 − 70 )
2
= 70
Total area enclosed by the curve
∫ udy = 1790
•
Discharge
Q = 1790 b
ƒ
Cross-sectional area of the stream
A = 80 b
ƒ
Mean velocity
v=
Q
A
=
1790 b
= 22.4 m/s
80 b
Example No. 3 - 2
If 70 litres of oil can be discharge from a tank within 50 seconds, determine the
flow rate in m³/s. If the oil is discharged using a pipe of 50 mm diameter, find the
average velocity of the flow.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 7
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
ƒ
Flow rate
Q=
ƒ
volume discharged
discharged time
=
70 ×10 −3 m 3
= 1.4 × 10 −3 m 3 /s
50 s
Average velocity
Q = vA ⇔
v=
Q
1.4 × 10 −3 m 3 /s
=
= 0.713 m/s
A π× (50 ×10 -3 ) 2
m2
4
3.3. Continuity of flow
3.3.1. Conservation of mass
ƒ
Formulated by Lavoisier1
Matter cannot be created nor
destroyed.
1
Antoine Laurent Lavoisier
born Aug. 26, 1743
died May 8, 1794
French nobleman prominent in the histories of chemistry, finance, biology, and economics.
The "father of modern chemistry," he stated the first version of the Law of
conservation of matter, recognized and named oxygen (1778), disproved
the phlogiston theory, introduced the Metric system, invented the first
periodic table including 33 elements, and helped to reform chemical
nomenclature. He was also an investor and administrator of the "Ferme
Générale," a private tax collection company; chairman of the board of the
Discount Bank (later the Banque de France); and a powerful member of a
number of other aristocratic administrative councils.
Having also served as a leading financier and public administrator before
the French Revolution, he was guillotined at the age of 51 with other
financiers during the revolutionary terror.
Sources: http://en.wikipedia.org/wiki/Antoine_Lavoisier
http://www.britannica.com/eb/article-9106472
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 8
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
Note:
This statement does not apply to nuclear reaction, where matter can be
transformed into energy.
ƒ
Conservation of mass applied to a flowing fluid
A system is composed of the same
quantity of matter at all times.
 Increase of mass


 Mass of fluid entering   Mass of fluid leaving  

=
 + of fluid in the control 

 per unit time
  per unit time
 
 volume per unit time 
•
For steady flow
 Mass of fluid entering   Mass of fluid leaving 

=

 per unit time
  per unit time

Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 9
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.3.2. Continuity equation for steady flow
ƒ
General expression for continuity equation
i
i
m1 = m 2 = constant ⇔ ρ1 A1 v1 = ρ 2 A2 v 2 = constant
In a steady flow, the mass flow rate
through any section of the flow is
constant.
ƒ
Case: incompressible flow
ρ1 = ρ 2 = constant ⇔
A1 v1 = A2 v 2 = constant
⇔ Q1 = Q 2 = constant
In a steady incompressible flow, the
volume flow rate through any section
of the flow is constant.
Example No. 3 - 3
Consider a piping system as shown. Velocity of water entering the pipe AB is 4.5
m/s and velocity of water leaving pipe CD is 1.5 m/s. Determine the velocity of
flow in pipes BC and CE, and the total discharge.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 10
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
Velocity of flow in pipes BC and CE
ƒ
Continuity equation
Q1 = Q 2 = Q3 + Q 4
⇔
⇔
A1 v1 = A2 v 2 = A3 v3 + A4 v 4
πd12
4
v1 =
πd 2 2
4
v2 =
πd 3 2
4
v3 +
πd 4 2
4
v4
⇔ d12 v1 = d 2 2 v 2 = d 3 2 v3 + d 4 2 v 4
⇔ ( 0.05 ) 2 × 4.5 = ( 0.075 ) 2 × v 2 = ( 0.071) 2 × 1.5 + ( 0.03) 2 × v 4
v2 =
⇔
v4 =
( 0.05 ) 2
( 0.075 ) 2
× 4.5 = 2 m/s
( 0.05 ) 2 × 4.5 − ( 0.071) 2 × 1.5
( 0.03) 2
= 4.1 m/s
Total discharge
Q total = Q1 = Q 2 = Q3 + Q 4
⇔ Q total = Q1 = A1 v1 =
⇔
πd12
4
v1
π× ( 0.05 ) 2
× 4.5 = 8.836 × 10 −3 m 3 /s
Q total =
4
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 11
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.4. Momentum equation and its applications
3.4.1. Momentum in a flowing fluid
ƒ
Momentum of a particle or object is defined as the product of its mass and its
velocity:
Momentum = mv
•
Change of velocity (in magnitude and/or in direction) will change the
momentum.
ƒ
Consider a control volume
ƒ
Newton's second law of motion
The resultant force acting on a fluid
mass is simply the product of its mass
and its acceleration.
F = ma
∆v
∆t
m∆v
⇒ F=
∆t
a=
ƒ
Impulse-momentum equation
F ∆t = m∆v = m(v2 − v1 )
Impulse of force acting on a fluid mass
in a short interval of time is equal to
the change of momentum in the
direction of force.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 12
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Momentum equation
•
Also called law of conservation of momentum or momentum principle.
F=
i
m
∆v = m(v2 − v1 )
∆t
The net force acting on a fluid mass is
equal to the rate of change in
momentum (change in momentum per
unit time).
By Newton's third law (ActionReaction), the fluid will exert an equal
force but in opposite direction on the
surroundings (solid boundary or body
of fluid producing the change of
velocity).
•
Force exerted by the fluid on the surroundings
i
F = m(v1 − v2 )
ƒ
Notes:
Force, impulse and momentum are vectors.
3.4.2. Momentum equation for two- and three-dimensional flow
along a stream line
ƒ
It is obtained by decomposing the momentum equation into x, y, and z
direction.
i
Fx = m(v2 x − v1x )
i
F = m(v2 − v1 ) ⇔
i
Fy = m(v2 y − v1 y )
i
Fz = m(v2 z − v1z )
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 13
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Force exerted by the fluid on the surroundings
i
Fx = m(v1x − v2 x )
i
F = m(v1 − v2 ) ⇔
i
Fy = m(v1 y − v2 y )
i
Fz = m(v1z − v2 z )
ƒ
Notes:
For two-dimensional flow, two of the three equations above will be used.
3.4.3. Force exerted by a jet striking a flat plate
3.4.3.1. Stationary flat plate
3.4.3.1.1.
Case: Inclined plate
ƒ
Schematic diagram
ƒ
Let x be the horizontal axis, positive to the right.
ƒ
Let y be the vertical axis, positive upward.
ƒ
Let n be the normal axis to the plate.
i
F = m ( v1 − v 2 ) ⇔
i
Fx = m ( v1 x − v 2 x )
i
F y = m ( v1 y − v 2 y )
The force must be normal to the plate. It means both components of force (in x
and y directions) will be present. To simplify, let's consider the normal direction to
the plate.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 14
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Force component in normal direction
i
Fn = m ( v1 n − v 2 n )
i
m
: mass flow rate of the jet
= ρA jet v jet
v1 n
: velocity of the jet striking the plate in normal direction
= v jet sin θ
v2 n
: velocity of the jet after impact in normal direction
=0
ƒ
⇒
Fn = ρA jet v jet ( v jet sin θ − 0 )
⇔
Fn = ρA jet v jet 2 sin θ
Force component in x- and y-direction
Fx = Fn sin θ = ρA jet v jet 2 sin 2 θ
F y = Fn cos θ = ρA jet v jet 2 sin θ cos θ
3.4.3.1.2.
ƒ
Case: Vertical plate (θ = 90°)
Schematic diagram
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 15
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Force exerted by the fluid on the plate
Fx = ρA jet v jet 2
Fy = 0
Example No. 3 - 4
A flat plate is struck normally by a jet of water 50 mm in diameter. If the discharge
is 0.0353 m³/s, calculate the force on the plate when it is stationary.
Solution
ƒ
Schematic diagram
ƒ
Force exerted by the fluid on the plate
F = ρA jet v jet 2
A jet
: area of the jet
=
v jet
: velocity of the jet
=
⇒
π× ( 50 × 10 −3 ) 2
= 1.96 × 10 −3 m 2
4
Q
A jet
=
0.0353 m 3 /s
1.96 × 10 −3 m 2
= 17.98 m/s
F = 1000 ×1.96 × 10 −3 × (17.98 ) 2 = 633.5 N
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 16
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Example No. 3 - 5
A jet of water of diameter 75 mm moving with a velocity of 25 m/s strikes a fixed
plate in such a way that the angle between the jet and the plate is 60°. Find the
force exerted by the jet on the plate:
a. in the direction normal to the plate.
b. in the direction of the jet.
Solution
ƒ
Schematic diagram
ƒ
Force exerted by the jet on the plate in the direction normal to the plate
Fn = ρA jet v jet 2 sin θ
A jet
: cross-sectional area of the jet
=
⇒
ƒ
π× ( 75 × 10 −3 ) 2
= 4.42 × 10 −3 m 2
4
Fn = 1000 × ( 4.42 ×10 −3 ) × 25 2 × sin 60 = 2391.2 N
Force exerted by the jet on the plate in the direction of the jet
Fx = Fn sin θ = 2391.2 × sin 60 = 2070.9 N
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 17
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.4.3.2. Hinged plate
ƒ
Schematic diagram
ƒ
Forces acting on the plate
•
Normal force due to water jet (see stationary flat plate)
Fn = ρA jet v jet 2 sin θ
•
ƒ
Weight of the plate, W
Equilibrium of the plate
•
Moment about the hinge (point O)
∑M
= 0 ⇔ d A' × Fn − d Gx × W = 0
(
)
d 
⇔  A  × ρA jet v jet 2 sin θ − ( d G cos θ ) × W = 0
 sin θ 
⇔ d A ρA jet v jet 2 − d G W cos θ = 0
d A ρA jet v jet 2
⇔
cos θ =
or
d A ρA jet v jet
θ = cos −1 
dGW

Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010


dGW
2 



Page 18
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Example No. 3 - 6
A rectangular plate weighing 60 N is supported vertically by a hinge on the top
horizontal edge. The centre of gravity of the plate is 100 mm from the hinge. A
horizontal jet of water 20 mm diameter whose axis is 150 mm below the hinge
impinges normally on the plate with a velocity of 5 m/s. Find:
a. the horizontal force applied at centre of gravity to maintain the plate in its
vertical position.
b. the corresponding velocity of the jet if the plate is deflected through 30° and
the same force continues to act at the plate.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 19
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
Horizontal force applied at centre of gravity to maintain the plate in its vertical position
ƒ
Forces acting on the plate
•
Force exerted by the jet on the plate in the direction normal to the plate
Fx = ρA jet v jet 2
A jet
: cross-sectional area of the jet
=
π× ( 20 × 10 −3 ) 2
= 3.14 × 10 −4 m 2
4
⇒ Fx = 1000 × ( 3.14 ×10 −4 ) × 5 2 = 7.85 N
ƒ
•
Weight of the plate, W = 60 N
•
Horizontal force applied to the plate to maintain the plate in its vertical
position, FC
Equilibrium of the plate
•
Moment about the hinge
∑M
= 0 ⇔ (150 × 10 −3 ) × Fx − (100 × 10 −3 ) × FC + 0 × W = 0
⇔ 0.15 × 7.85 = 0.1 FC
⇔
FC = 11.78 N
Corresponding velocity of the jet if the plate is deflected through 30° and the same force
continues to act at the plate
ƒ
Forces acting on the plate
•
Force exerted by the jet on the plate in the direction normal to the plate
Fn = ρA jet v jet 2 sin θ
A jet
: cross-sectional area of the jet
= 3.14 × 10 −4 m 2
θ
: angle between the jet and the plate
= 90 − 30 = 60
⇒ Fn = 1000 × ( 3.14 × 10 −4 ) × v jet 2 × sin 60 = 0.27 v jet 2
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 20
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
•
Weight of the plate, W = 60 N
•
Horizontal force applied to the plate to maintain the plate in its vertical
position, FC = 11.78 N
Equilibrium of the plate
•
Moment about the hinge
∑M
=0
⇔
0.15
cos 30
× Fn − ( 0.1× cos 30 ) × FC − ( 0.1× sin 30 ) × W = 0
⇔ 0.17 × 0.27 v jet 2 − 0.087 ×11.78 − 0.05 × 60 = 0
⇔ 0.047 v jet 2 = 4.025
⇔
v jet =
4.025
= 9.28 m/s
0.047
3.4.3.3. Moving plate
ƒ
Schematic diagram
ƒ
Let:
•
vjet : absolute velocity of the jet
•
u : velocity of the plate
•
θ : angle between the jet and the plate
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 21
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Velocity diagram
v jet = u + vr
⇔ vr = v jet − u
The component vr = (vjet - u) is the relative velocity of the jet with respect to the
plate. It is the velocity with which the jet strikes the plate. It may be thought
that the jet strikes a stationary inclined flat plate at a velocity of vr = (vjet - u).
3.4.3.3.1.
Force exerted by the jet on the moving plate
As in the case of stationary inclined flat plate, using the same method, we
obtain:
Fn = ρAjet vr 2 sin θ = ρAjet (v jet − u )2 sin θ
where Fn
ƒ
: normal force exterted by the jet on the moving plate
ρ
Ajet
: mass density of the fluid
: cross-sectional area of the jet
vr
v jet
: relative velocity of the jet with respect to the plate
: initial velocity of the jet
u
θ
: velocity of the plate
: angle between the jet and the plate
Force components in horizontal and vertical direction
Fx = Fn sin θ = ρAjet vr 2 sin 2 θ = ρAjet (v jet − u )2 sin 2 θ
Fy = Fn cos θ = ρAjet vr 2 sin θ cos θ = ρAjet (v jet − u ) 2 sin θ cos θ
ƒ
Notes:
If the plate moves toward the jet, the relative velocity will become vr = (v + u).
Hence, the normal force will be:
Fn = ρAjet vr 2 sin θ = ρAjet (v jet + u )2 sin θ
3.4.3.3.2.
Work done per second by the jet on the plate
It may be thought as power transmitted by the jet to the plate.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 22
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Work done by the jet on the plate
W = Fn idisplacement of the plate
⇔ W = ( Fn sin θ) × ∆x
⇔ W = Fx × ∆x
•
ƒ
Unit: N.m
Work done per second by the jet on the plate
Work done by the jet on the plate
elapsed time
W Fx × ∆x
⇔ P=
=
∆t
∆t
P=
⇔
•
ƒ
P = Fxu = ρAjet vr 2u sin 2 θ = ρAjet (v jet − u )2 u sin 2 θ
Unit: N.m/s or watt (W)
Notes:
•
3.4.3.3.3.
In the case of stationary plate, the work done is zero, because there is no
displacement of the plate.
Efficiency of transmission of the jet
If the plate is moving because of the jet strikes on it, it may be necessary to
calculate the efficiency of transmission of the jet.
ƒ
Initial power of the jet
Pjet = Kinetic energy of the jet per second
⇔
⇔
⇔
ƒ
1 mv 2
jet
2
1 m
= × × v jet 2
∆t
2 ∆t
i
1
1
Pjet = × m× v jet 2 = × ρAjet v jet × v jet 2
2
2
1
Pjet = ρAjet v jet 3
2
Pjet =
Efficiency of transmission
η=
⇔
ρAjet (v jet − u ) 2 u sin 2 θ
Power transmitted
=
1 ρA v 3
Initial power of the jet
jet jet
2
η=
2(v jet − u )2 u sin 2 θ
v jet 3
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 23
MEC441 Fluid Mechanics 1
3.4.3.3.4.
ƒ
Chapter 3: Finite control volume analysis
Case: vertical plate (θ = 90°)
Force exerted by the jet on the moving plate
Fx = Fn = ρAjet vr 2 = ρAjet (v jet − u )2
Fy = 0
ƒ
Work done per second by the jet on the plate
P = Fxu = ρAjet vr 2u = ρAjet (v jet − u )2 u
ƒ
Efficiency of transmission
η=
2(v jet − u )2 u
v jet 3
Example No. 3 - 7
A flat plate is struck normally by a jet of water 50 mm in diameter. If the discharge
is 0.0353 m³/s, calculate the force on the plate, the work done per second and
the efficiency when it moves in the same direction as the jet at 6 m/s.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 24
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
ƒ
Schematic diagram
ƒ
Force on the plate
Fx = Fn = ρA jet ( v jet − u ) 2
A jet
: diameter of the jet
=
v jet
: velocity of the jet
=
⇒
ƒ
π× ( 50 × 10 −3 ) 2
= 1.96 ×10 −3 m 2
4
Q
0.0353 m 3 /s
=
= 17.98 m/s
A jet 1.96 × 10 −3 m 2
Fx = 1000 × 1.96 × 10 −3 (17.98 − 6 ) 2 = 281.3 N
Work done per second
P = Fx u = 281.3 × 6 = 1687.8 N.m/s
ƒ
Efficiency
•
Initial power of the jet
1
1
Pjet = ρA jet v jet 3 = ×1000 × (1.96 × 10 −3 ) × (17.98 ) 3 = 5696.3 N.m/s
2
2
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 25
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
Efficiency
1687.8
η= P =
= 0.2963 = 29.63%
Pjet 5696.3
3.4.4. Force due to the deflection of a jet by a curved vane
3.4.4.1. Stationary curved vane
3.4.4.1.1.
Case: Jet strikes the curved vane at the centre
ƒ
Schematic diagram
ƒ
Observation
•
The jet is deflected by the curved vane.
•
The jet leaves the vane in the tangential direction of the curved vane at
outlet at an angle β which is the angle of the vane at outlet.
•
If there is no loss (no loss due to impact of the jet, and the vane is
smooth), the jet will leave the vane with the same velocity as the coming
jet.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 26
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Momentum equation in x-direction
i
Fx = m ( v1 x − v 2 x )
Fx
i
: force exerted by the jet on the vane in x -direction
m
: mass flow rate of the jet
= ρA jet v jet
v1 x
: velocity of the coming jet in x -direction
= v jet
v2 x
: velocity of the jet leaving the vane in x -direction
= − v jet cos β
⇒
Fx = ρA jet v jet ( v jet + v jet cos β )
⇔
Fx = ρA jet v jet 2 (1 + cos β )
Example No. 3 - 8
A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a fixed
symmetrical curved plate at the centre. Find the force exerted by the jet of water
in the direction of the jet, if the jet is deflected through an angle of 120° at the
outlet of the curved plate.
Solution
ƒ
Schematic diagram
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 27
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Force exerted by the jet in the direction of the jet
Fx = ρA jet v jet 2 (1 + cos β )
π× ( 50 × 10 −3 ) 2
A jet =
= 1.963 ×10 −3 m 2
4
⇒ Fx = 1000 × (1.963 × 10 −3 ) × ( 40 ) 2 × (1 + cos(180 − 120 ))
⇔
3.4.4.1.2.
Fx = 4712.4 N
Case: Jet strikes the curved vane at one end tangentially
ƒ
Schematic diagram
ƒ
Observation
•
The jet strikes the vane at an angle α which is the angle of the vane at
inlet.
•
The jet leaves the vane at an angle β which is the angle of the vane at
outlet.
•
If there is no loss (no loss due to impact of the jet, and the vane is
smooth), the jet will leave the vane with the same velocity as the coming
jet.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 28
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Momentum equation in x-direction
i
Fx = m(v1x − v2 x )
Fx
i
: force exerted by the jet on the vane in x-direction
m
: mass flow rate of the jet
= ρAjet v jet
v1x
: velocity of the coming jet in x-direction
= v jet cos α
v2 x : velocity of the jet leaving the vane in x-direction
= −v jet cos β
•
⇒
Fx = ρAjet v jet (v jet cos α + v jet cos β)
⇔
Fx = ρAjet v jet 2 (cos α + cos β)
If the vane is symmetrical (α = β)
Fx = 2ρAjet v jet 2 cos α
ƒ
Momentum equation in y-direction
i
Fy = m(v1 y − v2 y )
Fy
: force exerted by the jet on the vane in y -direction
v1 y
: velocity of the coming jet in y -direction
= v jet sin α
v2 y : velocity of the jet leaving the vane in y -direction
= v jet sin β
•
⇒
Fy = ρAjet v jet (v jet sin α − v jet sin β)
⇔
Fy = ρAjet v jet 2 (sin α − sin β)
If the vane is symmetrical (α = β)
Fy = 0
ƒ
Magnitude and direction of resultant force
•
Resultant force
F = Fx 2 + Fy 2
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 29
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
Direction angle (with horizontal direction)
tan θ =
Fy
Fx
⇒
θ = tan −1
Fy
Fx
Example No. 3 - 9
Calculate the magnitude and direction of the vertical, horizontal and total force
components exerted on the stationary blade by a 50 mm jet of water moving at
15 m/s as shown. Given: α = 30°, β = 45°.
Solution
ƒ
Horizontal force
Fx = ρA jet v jet 2 (cos α + cos β )
A jet =
π× ( 50 ×10 −3 ) 2
= 1.96 × 10 −3 m 2
4
⇒ Fx = 1000 × (1.96 ×10 −3 ) × (15 ) 2 × (cos 30 + cos 45 )
⇔
Fx = 693.75 N
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
[ →]
Page 30
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Vertical force
F y = ρA jet v jet 2 (sin α − sin β )
ƒ
⇔
F y = 1000 × (1.96 × 10 −3 ) × (15 ) 2 × (sin 30 − sin 45 )
⇔
F y = −91.33 N
↓ 
Total force
•
Magnitude
F=
•
Fx 2 + F y 2 =
( 693.75 ) 2 + ( −91.33) 2 = 699.74 N
Direction angle with horizontal
tan θ =
Fy
Fx
=
−91.33
= −0.1316 ⇒
693.75
θ = −7.5
Example No. 3 - 10
A jet of water from a nozzle is deflected through 60° from its original direction by
a curved plate which it enters tangentially without shock with a velocity of 30 m/s
and leaves with a mean velocity of 25 m/s. If the discharge from the nozzle is 0.8
kg/s, calculate the magnitude and direction of the resultant force on the vane, if
the vane is stationary.
Solution
ƒ
Schematic diagram
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 31
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Horizontal force
i
i
Fx = m ( v1 x − v 2 x ) = m ( v1 − v 2 cos 60 )
⇒ Fx = 0.8 × ( 30 − 25 × cos 60 )
⇔
ƒ
Fx = 14 N
[ →]
Vertical force
i
i
F y = m ( v1 y − v 2 y ) = m ( 0 − v 2 sin 60 )
⇒ F y = 0.8 × ( −25 × sin 60 )
⇔
ƒ
F y = −17.32 N
↓ 
Total force
•
Magnitude
F=
•
Fx 2 + F y 2 =
(14 ) 2 + ( −17.32 ) 2 = 22.27 N
Direction angle with horizontal
tan θ =
Fy
Fx
=
−17.32
= −1.237 ⇒
14
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
θ = −51.05
Page 32
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.4.4.2. Moving curved vane
3.4.4.2.1.
Case: Jet strikes the curved vane at the centre
ƒ
Schematic diagram
ƒ
Velocity diagram
v = u + vr
⇔ vr = v − u
The component vr = (vjet - u) is the relative velocity of the jet with respect to the
vane. It is the velocity with which the jet strikes the vane. It may be thought
that the jet strikes a stationary curved vane at a velocity of vr = (vjet - u).
ƒ
Force exerted by the jet on the moving vane in x-direction (direction of the jet)
As in the case of stationary curved vane, using the same method, we obtain:
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 33
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Fx = ρAjet (v jet − u ) 2 (1 + cos β)
Fx
: force exerted by the jet on the vane in x-direction
v jet − u : relative velocity of the jet with respect to the vane
ƒ
v jet
= vr
: velocity of the jet striking the vane
u
: velocity of the vane
Work done per second by the jet on the vane (power transmitted by the jet to
the vane)
•
Work done by the jet on the vane
W = F idisplacement of the vane
⇔ W = Fx × ∆x
•
ƒ
Unit: N.m
Work done per second by the jet on the vane
Work done by the jet on the vane
elapsed time
W Fx × ∆x
⇔ P=
=
∆t
∆t
P=
⇔
•
ƒ
P = Fxu = ρAjet vr 2u (1 + cos β) = ρAjet (v jet − u )2 u (1 + cos β)
Unit: N.m/s or watt (W)
Efficiency of transmission of the jet
•
Initial power of the jet
1
Pjet = Kinetic energy of the jet per second = ρAjet v jet 3
2
•
Efficiency of transmission
ρAjet (v jet − u )2 u (1 + cos β)
Power transmitted
η=
=
1 ρA v 3
Initial power of the jet
jet jet
2
⇔
η=
2(v jet − u ) 2 u (1 + cos β)
v jet 3
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 34
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Example No. 3 - 11
A jet of water of diameter 7.5 cm strikes a curved plate at its centre with a velocity
of 20 m/s. The curved plate is moving with a velocity of 8 m/s in the direction of
the jet. The jet is deflected through an angle of 165°. Assuming the plate to be
smooth, find:
a. the force exerted on the plate in the direction of the jet.
b. the power of the jet.
c. the efficiency of the jet.
Solution
ƒ
Schematic diagram
ƒ
Force exerted by the jet in the direction of the jet
Fx = ρA jet ( v jet − u ) 2 (1 + cos β )
A jet =
π× ( 7.5 × 10 −2 ) 2
= 0.004417 m 2
4
⇒ Fx = 1000 × 0.004417 × ( 20 − 8) 2 × (1 + cos(180 − 165 ))
⇔
ƒ
Fx = 1250.38 N
Power of the jet
P = Fx u = 1250.38 × 8 = 10003.04 W = 10 kW
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 35
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Efficiency of the jet
η=
⇔
3.4.4.2.2.
Power transmitted
10003.04
=
Initial power of the jet 1 ρA v 3
jet jet
2
η=
10003.04
1
× 1000 × 0.004417 × 20 3
2
= 0.564 = 56.4%
Case: Jet strikes the curved vane at one end
ƒ
Schematic diagram
ƒ
Velocity triangle at inlet
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 36
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
Notation:
v1 : velocity of the jet at inlet
v1 =
vw1 : whirl velocity (parallel to u1 )
v f 1 : flow velocity (perpendicular to u1 )
α1 : angle of of the jet (with u1) at inlet
v1 =
vw1 = v1 cos α1
v f 1 = v1 sin α1
u1 : velocity of the vane at inlet
vr1 : relative velocity of the jet (with respect to the vane) at inlet
β1 : angle of the relative velocity of the jet (with u1) at inlet. It is also angle
of the vane at inlet.
v1 =
ƒ
vw1 = u1 + vr1 cos β1
v f 1 = vr1 sin β1
Velocity triangle at outlet
•
Notation:
v2 : velocity of the jet leaving the vane at outlet
v2 =
vw2 : whirl velocity (parallel to u2 )
v f 2 : flow velocity (perpendicular to u2 )
α2 : angle of of the leaving jet (with u2) at outlet
v2 =
vw2 = v2 cos α 2
v f 2 = v2 sin α 2
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 37
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
u2 : velocity of the vane at outlet
vr2 : relative velocity of the leaving jet (with respect to the vane) at outlet
β2 : angle of the relative velocity of the leaving jet (with u2) at outlet. It is
also angle of the vane at outlet.
v2 =
ƒ
vw2 = u2 + vr 2 cos β2
v f 2 = vr 2 sin β2
Force exerted by the jet on the vane in the direction of motion of the vane
In this case, the direction of motion of the vane is as indicated by the velocity
u, i.e. in x-direction.
i
Fx = m(v1x − v2 x )
i
m
: mass flow rate of the jet striking the vane
= ρAjet vr1
v1x
: velocity of the jet striking the vane at inlet in x -direction
= vr1 cos β1 = vw1 − u1
v2 x : velocity of the jet leaving the vane at outlet in x-direction
= vr 2 cos β2 = −(u2 − vw2 )
⇒
Fx = ρAjet vr1 [ vr1 cos β1 − vr 2 cos β2 ]
or
⇔
ƒ
Fx = ρAjet vr1 [ (vw1 − u1 ) + (u2 − vw2 )]
Work done per second by the jet on the vane (power transmitted by the jet to
the vane)
•
Work done by the jet on the vane at inlet
W = F idisplacement of the vane at inlet
⇔ W = Fx × ∆x1
•
ƒ
Unit: N.m
Work done per second by the jet on the vane at inlet
Work done by the jet on the vane at inlet
elapsed time
W Fx × ∆x1
⇔ P=
=
∆t
∆t
P=
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 38
MEC441 Fluid Mechanics 1
⇔
Chapter 3: Finite control volume analysis
P = Fxu1 = ρAjet vr1 (vr1 cos β1 − vr 2 cos β2 )u1
or
⇔
•
ƒ
P = Fxu1 = ρAjet vr1 [ (vw1 − u1 ) + (u2 − vw2 ) ] u1
Unit: N.m/s or watt (W)
Notes:
•
To calculate the work done by the jet on the vane at outlet, replace u1 with
u2.
•
u1 is different than u2 can be found in Francis2 turbine or in centrifugal
pump.
Francis turbine
2
James Bicheno Francis
born May 18, 1815
died Sept. 18, 1892
British-American hydraulic engineer and inventor of Francis turbine that was used for low-pressure
installations.
In 1833 Francis went to the United States and was hired by the engineer G.W.
Whistler to help construct the Stonington (Conn.) Railway. In Lowell he joined
the Proprietors of the Locks and Canals on the Merrimack River as a
draftsman and at age 22 became chief engineer of the company.
In his 40 years of managing the company's waterpower interests and acting
as a consulting waterpower engineer to factories, he contributed greatly to the
rise of Lowell as an industrial centre.
He also investigated timber preservation, the testing and design of cast-iron
girders, and fire protection systems. In addition to the Francis turbine, he is
known for his formulas for the flow of water over weirs and many other
hydraulic studies. Francis wrote more than 200 technical papers and, although
unschooled, was considered one of the foremost civil engineers of his time.
Source: Encyclopædia Britannica
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 39
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
u1 is the same as u2 can be found in Pelton3 wheel.
Pelton wheel
3
Lester Allen Pelton
born 1829
died March 14, 1908
US engineer who developed a highly efficient water turbine used to drive both
mechanical devices and hydroelectric power turbines using large heads of water.
The Pelton wheel remains the only hydraulic turbine of the impulse type in
common use today.
From Ohio, Pelton joined the California gold rush at the age of 20. He observed
the water wheels used at the mines to power machinery, and came up with
improvements. By 1879 he had tested a prototype at the University of California.
A patent was granted 1889, and he later sold the rights to the Pelton Water
Wheel Company of San Francisco.
The energy to drive these wheels was supplied by powerful jets of water which struck the base of the
wheel on hemispherical cups. Pelton's discovery was that the wheel rotated more rapidly, and hence
developed more power, with the jet striking at the inside edge of the cups, rather than the centre; he
built a wheel with split cups.
By the time of his death, Pelton wheels developing thousands of horsepower in hydroelectric schemes
at efficiencies of more than 90% were in operation.
Source: http://www.cartage.org.lb/en/themes/Biographies/MainBiographies/P/Pelton/1.html
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 40
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Example No. 3 - 12
Calculate the force exerted by the jet (50 mm diameter) on the moving blade as
shown.
Solution
ƒ
Velocity triangle at inlet
ƒ
Velocity triangle at outlet
ƒ
Force exerted by the jet in the direction of the jet
Fx = ρA jet v r 1 ( v r 1 cos β1 − v r 2 cos β 2 )
A jet =
π× ( 50 × 10 −3 ) 2
= 1.96 ×10 −3 m 2
4
β1 = 0
β 2 = 180
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 41
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
v r 1 = v r 2 (we assume no loss, smooth vane surface)
v r 1 = v1 − u = 45 − 30 = 15 m/s
⇒
Fx = 2 ρA jet v r 12 = 2 × 1000 × (1.96 × 10 −3 ) × 15 = 882 N
3.5. Energy of a flowing fluid
3.5.1. Conservation of energy
3.5.1.1. First law of thermodynamics
For any mass system, the net energy
supplied to the system equals the
increase of energy of the system plus
the energy leaving the system.
Ein = Eout + ∆E
3.5.1.2. Forms of energy
ƒ
Energy of the fluid
•
ƒ
Mechanical energy of the fluid
o
Kinetic energy
o
Potential energy
•
Internal energy of the fluid
•
Pressure energy or flow energy or flow work
External energy (shaft work)
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 42
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
•
Work taken from the fluid
•
Work supplied to the fluid
Heat exchange between the fluid and the surroundings
3.5.2. Energy of the fluid
3.5.2.1. Kinetic energy
3.5.2.1.1.
ƒ
Definition
It is due to the velocity of the mass of fluid.
KE = 1 mv 2
2
m : mass of fluid, kg
v
ƒ
Unit : Joule or N.m
3.5.2.1.2.
ƒ
: mean velocity of the flow, m/s
Kinetic energy per unit weight
It is also called velocity head.
hv =
KE v 2
=
mg 2 g
g : gravitational acceleration, m/s 2
ƒ
Unit : m
3.5.2.1.3.
ƒ
Kinetic energy per unit volume
It is also called dynamic pressure.
p dynamic =
KE
1
= ρv 2
volume 2
ρ : mass density of fluid, kg/m 3
ƒ
Unit : N/m² or Pa
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 43
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.5.2.2. Potential energy
3.5.2.2.1.
ƒ
Definition
It is due to the mass of fluid being at a height above the datum level and acted
upon by gravity.
PE = mgz
m : mass of fluid, kg
g
: gravitational acceleration, m/s 2
z
: elevation above the datum level, m
ƒ
Unit : Joule or N.m
ƒ
Notes :
We are usually interested only in difference of elevation, therefore the datum
plane can be placed anywhere by considerations of convenience.
3.5.2.2.2.
ƒ
Potential energy per unit weight
It is also called potential head or elevation head.
hz =
ƒ
Unit : m
3.5.2.2.3.
ƒ
PE
=z
mg
Potential energy per unit volume
It is also called hydrostatic pressure.
p hydrostatic =
PE = ρgz
volume
ρ : mass density of fluid, kg/m 3
ƒ
Unit : N/m² or Pa
3.5.2.3. Internal energy
3.5.2.3.1.
ƒ
Definition
It is due to the activity of the molecules of the fluid forming the mass (motion
of molecules and forces of attraction between them).
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 44
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
ƒ
See Thermodynamics course for more detailed information.
ƒ
Notation :
•
Internal energy per unit mass: e (unit : Joule/kg or N.m/kg)
•
Internal energy per unit weight: e/g (unit : m)
3.5.2.4. Pressure energy
3.5.2.4.1.
ƒ
Definition
It is due to the pressure force exerted to the cross-sectional area of the
control volume which makes the fluid moves a short distance s.
pressure energy = pAs = p × volume of displacement
p : pressure exerted on the fluid, N/m 2
A : cross-sectional area where the fluid enters the control volume, m 2
s
: displacement of the fluid, m
ƒ
It is also called flow energy or flow work
ƒ
Unit : Joule or N.m
3.5.2.4.2.
ƒ
Pressure energy per unit weight
It is also called pressure head.
hp =
pressure energy p × volume
p
=
=
mg
mg
ρg
p : static pressure, N/m 2
ρ : mass density of fluid, kg/m 3
g : gravitational acceleration, m/s 2
ƒ
Unit : m
3.5.3. External energy (shaft work)
3.5.3.1. Definition
ƒ
It is additional (external) work supplied into (or taken from) the flowing fluid.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 45
MEC441 Fluid Mechanics 1
ƒ
ƒ
Chapter 3: Finite control volume analysis
Example:
•
Pump : gives energy to the flowing fluid.
•
Turbine : takes energy from the fluid.
Notes :
•
If the work is supplied to the system, the direction is from surroundings into
the system.
•
If the work is taken from the system, the direction is from the system to
surroundings.
3.5.4. Heat exhange between the fluid and the surroundings
ƒ
It can be heat supplied to the system or taken from the system.
ƒ
Example of heat supplied to the system:
•
ƒ
Example of heat taken from the system:
•
ƒ
The pipe is heated when the fluid is flowing inside.
Condensor in the refrigeration cycle.
Notes :
•
If the heat is supplied to the system, the direction is from surroundings into
the system.
•
If the heat is generated by the system (or taken from the system), the
direction is from the system to the surrounding.
3.5.5. Steady flow energy equation
3.5.5.1. General equation
ƒ
Schematic diagram
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 46
MEC441 Fluid Mechanics 1
ƒ
Let:
•
q : the net heat supplied to the fluid per unit mass (sum of all heat given to
the fluid minus sum of all heat taken from the fluid).
•
w : the net work done by the fluid per unit mass (sum of all works taken
from the fluid minus sum of all works given to the fluid).
3.5.5.1.1.
ƒ
Chapter 3: Finite control volume analysis
Mathematical form
It is derived from conservation of energy.
p1 e1 v12
p
e
v 2
q
w
+ +
+ z1 + = 2 + 2 + 2 + z2 +
g ρ2 g g 2 g
g
ρ1g g 2 g
or
p1
v2
+ e1 + 1 + z1g + q =
ρ1
2
p1
+ e1 = H1
ρ1
p2
v 2
+ e2 + 2 + z2 g + w
ρ2
2
: enthalpy of fluid at section 1
p2
+ e2 = H 2 : enthalpy of fluid at section 2
ρ2
⇒
3.5.5.1.2.
v2
v 2
H1 + 1 + z1g + q = H 2 + 2 + z2 g + w
2
2
Validity of the equation
ƒ
Steady flow
ƒ
All fluids: liquids, vapour or gases (real or ideal fluids)
ƒ
Continuous flow
3.5.5.2. Bernoulli's equation
3.5.5.2.1.
Background
Derived from momentum equation by Daniel Bernoulli4.
4
Daniel Bernoulli
born Feb. 8, 1700, Groningen , Netherlands
died Mar. 17, 1782, Basel, Switzerland
Dutch-born Swiss mathematician, son of Johann Bernoulli (1667-1748) and the nephew of Jacques
Bernoulli both important mathematicians.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 47
MEC441 Fluid Mechanics 1
3.5.5.2.2.
ƒ
Chapter 3: Finite control volume analysis
Bernoulli's equation for frictionless incompressible flow
Schematic diagram
At the age of 13 he was sent to Basel University to study philosophy
and logic. He obtained his baccalaureate examinations in 1715 and
went on to obtain his master's degree in 1716. During the time he
studied philosophy at Basel, he was learning the methods of the
calculus from his father and his older brother Nicolaus(II) Bernoulli.
He studied medicine at Heidelberg in 1718 and Strasbourg in 1719. He
returned to Basel in 1720 to complete his doctorate in medicine.
Having failed to obtain an academic post in Basel, he went to Venice to
study practical medicine. While in Venice he worked on mathematics
and his first mathematical work was published in 1724. This consisted
of four separate parts being four topics that had attracted his interest
while in Venice.
In 1725, he and his brother Nikolaus (1695-1726) were invited to work at the St. Petersburg Academy
of Sciences. These were the most fruitful years of Daniel Bernoulli's life, because of his collaboration
with Leonhard Euler, who came to St. Petersburg in 1727.
In 1733 Daniel returned to Basel, where as a professor he taught botany and anatomy and later
physiology and physics. His most important work was in the field of hydrodynamics (fluid mechanics).
Bernoulli's Law was published in "Hydrodynamica" (1738), his most important work. In this book he
also gave a theoretical explanation of the pressure of a gas on the walls of a container. Assuming that
the gas comprised a large number of small particles, moving randomly and in straight lines at high
velocity, he suggested that the pressure was caused by the impact of the particles on the wall. He is
therefore considered one of the founders of the kinetic theory of gases.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 48
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Bernoulli's equation expressed in head (energy per unit weight)
v 2
p
v 2
+ 1 + z1 = 2 + 2 + z 2
ρg 2 g
ρg 2 g
p1
⇔
p v2
+
+ z = constant = H
ρg 2 g
For steady incompressible flow of a
frictionless fluid along a streamline,
the total energy per unit weight (total
head) remains constant from point to
point although its division between the
three forms of energy may vary.
•
Terms used
o
Pressure head (or pressure energy)
o
Velocity head (or kinetic energy)
o
Potential head (or potential energy)
o
Total head
It is the constant in the Bernoulli's equation (or sum of pressure head,
velocity head and potential head).
o
Piezometric head
It is the sum of the pressure head and the potential head.
Piezometric head =
ƒ
p
+z
ρg
Bernoulli's equation expressed in energy per unit volume
p + 1 ρv 2 + ρgz = constant = pT
2
•
Terms used
o
Static pressure
It is the pressure measured by moving along with the fluid, thus being
"statics" relative to the moving fluid. Another way to measure the static
pressure is to connect a piezometer to the pipe wall.
Static pressure = p
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 49
MEC441 Fluid Mechanics 1
o
Chapter 3: Finite control volume analysis
Dynamic pressure
It is the pressure related to the velocity head.
1
Dynamic pressure = ρv 2
2
o
Hydrostatic pressure
It is the pressure related to the potential head or elevation head.
Hydrostatic pressure = ρgz
o
Total pressure
It is the pressure related to the total head (or sum of static pressure,
dynamic pressure and hydrostatic pressure).
pT = ρgH = p1 + 1 ρv 2 + ρgz
2
o
Piezometric pressure
It is the sum of the static pressure and the hydrostatic pressure.
Piezometric pressure = p + ρgz
3.5.5.2.3.
ƒ
General expression of Bernoulli's equation for incompressible flow
Schematic diagram
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 50
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
Bernoulli's equation expressed in head (energy per unit weight)
p1
ρg
+
v12
2g
+ z1 + q =
p2
ρg
+
v2 2
2g
+ z2 + w + h
q
: head supplied to the flowing fluid between 1 and 2 (ex: pump)
w
: work done per unit weight between 1 and 2 (ex: turbine)
h
: loss of head during the process from 1 to 2
For steady incompressible flow along
a streamline, the sum of heads entering
the system is equal to the sum of heads
leaving the system.
3.5.5.2.4.
Validity of the equation
ƒ
Steady flow
ƒ
Incompressible flow
•
This form of Bernoulli's equation is mainly used for liquids.
ƒ
Continuous flow
ƒ
Irrotational flow
Example No. 3 - 13
Gasoline (density 800 kg/m³) is flowing upward through a vertical pipeline which
tapers from 30 cm to 15 cm diameter. A gasoline mercury differential manometer
is connected between the pipe section to measure the rate of flow. The distance
between the manometer tapping is 1 meter and gauge reading is 50 cm of
mercury. Find:
a. the differential gauge reading in terms of gasoline head.
b. Using Bernoulli's equation, find the flow rate.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 51
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
ƒ
Pressure gauge
p1 − p 2 = ( p1 − p A ) + ( p A − p B ) + ( p B − p 2 )
= − ρ gasoline g ( z1 − z A ) − ρ Hg g ( z A − z B ) − ρ gasoline g ( z B − z 2 )
= − ρ gasoline g ( z1 − z A + z B − z 2 ) − ρ Hg g ( z A − z B )
= − ρ gasoline g ( ( z1 − z 2 ) + ( z B − z A ) ) + ρ Hg g ( z B − z A )
= −800 × 9.81× ( ( −1) + 0.5 ) + 13600 × 9.81× ( 0.5 )
= 70632 N/m 2
ƒ
Pressure gauge in terms of gasoline head
h=
ƒ
p1 − p 2
70632
=
= 9 m of gasoline
ρ gasoline g 800 × 9.81
Flow rate
•
Bernoulli's equation
p1
ρ gasoline g
+
v12
2g
+ z1 =
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
p2
ρ gasoline g
+
v2 2
2g
+ z2
Page 52
MEC441 Fluid Mechanics 1
⇔
Chapter 3: Finite control volume analysis
p1 − p 2
v 2 2 v1 2
+ z1 − z 2 =
−
ρ gasoline g
2g 2g
v 2 − v12
⇔ 9 −1 = 2
2g
⇔ v 2 2 − v12 = 8 × 2 × 9.81 = 156.96
•
Continuity equation
A1 v1 = A2 v 2
⇔
πd12
4
× v1 =
πd 2 2
4
× v2
d 2
( 0.3) 2
⇔ v 2 = 1 × v1 =
× v1 = 4 v1
( 0.15 ) 2
d22
•
Velocity at 1
16 v1 2 − v1 2 = 156.96
⇔ 15 v12 = 156.96
⇔ v1 = 3.235 m/s
•
Flow rate
Q = A1 v1 =
πd12
4
× v1 =
π× ( 0.3) 2
× 3.235 = 0.23 m 3 /s
4
Notes:
ƒ
From difference of pressure between 1 and 2, we have:
p1 − p 2 = − ρ gasoline g ( ( z1 − z 2 ) + ( z B − z A ) ) + ρ Hg g ( z B − z A )
⇔
ρ Hg
p1 − p 2
( zB − zA )
= − ( z1 − z 2 ) − ( z B − z A ) +
ρ gasoline g
ρ gasoline
⇔
 ρ Hg

p1 − p 2
+ ( z1 − z 2 ) = ( z B − z A ) 
− 1
 ρ gasoline

ρ gasoline g


Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 53
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
From Bernoulli's equation, we have:
p1 − p 2
v 2 v 2 v 2 − v12
+ z1 − z 2 = 2 − 1 = 2
2g 2g
2g
ρ gasoline g
ƒ
Combining both equation, we obtain:
 ρ Hg

v 2 2 − v1 2
= ( zB − zA ) 
− 1
 ρ gasoline

2g


 ρ Hg

⇔ v 2 2 − v12 = 2 g ( z B − z A ) 
− 1
 ρ gasoline



ƒ
Continuity equation
A1 v1 = A2 v 2
A
⇔ v 2 = 1 v1 = mv1
A2
A
m= 1
A2
ƒ
: area ratio
Substituting continuity equation into previous equation, we have:
 ρ Hg

( mv1 ) 2 − v12 = 2 g ( z B − z A ) 
− 1
 ρ gasoline



 ρ Hg

⇔ ( m 2 − 1) v12 = 2 g ( z B − z A ) 
− 1
 ρ gasoline



⇒ v1 =
ƒ

2 g ( z B − z A )  ρ Hg
− 1


( m 2 − 1)  ρ gasoline

Volume flow rate
Q = A1 v1 = A1
ƒ

2 g ( z B − z A )  ρ Hg
− 1


( m 2 − 1)  ρ gasoline

Remarks:
From the formula obtained, we observed that to calculate the flow rate, we do
not need to know the position of section 1 and section 2 (z1 and z2). By
knowing the difference of level of the fluid in the manometer (zB - zA), we can
calculate the flow rate.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 54
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
In general form:
Q = A1
2 g ( z B − z A )  ρ mano

− 1


( m 2 − 1)  ρ
ρ mano
: density of manometric fluid
ρ
: density of flowing fluid
A1
: inlet area of the flowing fluid
m
: area ratio, m =
zB − zA
: difference of level in the manometer
A1
A2
3.5.5.3. Euler's equation
3.5.5.3.1.
Background
Derived from momentum equation by Leonhard Euler5.
5
Leonhard Euler
born Apr. 15, 1707, Basel, Switzerland
died Sept. 18, 1783, St Petersburg, Russia
Swiss mathematician, made important contributions to practically every area of pure and applied
mathematics.
He entered the University in 1720, at the age of 14, first to obtain a general education before going on
to more advanced studies. Johann Bernoulli soon discovered Euler's great potential for mathematics
in private tuition that Euler himself engineered. Euler's own account given in his unpublished
autobiographical writings.
In 1723 he completed his Master's degree in philosophy having
compared and contrasted the philosophical ideas of Descartes and
Newton. He began his study of theology in the autumn of 1723, following
his father's wishes.
Euler obtained his father's consent to change to mathematics after
Johann Bernoulli had used his persuasion. He completed his studies at
the University of Basel in 1726. He had studied many mathematical
works during his time in Basel. By 1726 Euler had already a paper in
print, a short article on isochronous curves in a resisting medium. In
1727 he published another article on reciprocal trajectories and
submitted an entry for the 1727 Grand Prize of the Paris Academy on the
best arrangement of masts on a ship.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 55
MEC441 Fluid Mechanics 1
3.5.5.3.2.
Chapter 3: Finite control volume analysis
Mathematical form
∂p
+ vdv + gdz = 0
ρ
3.5.5.3.3.
ƒ
Notes
Bernoulli's equation for frictionless incompressible flow can be easily obtained
by integrating Euler's equation.
∫
∂p
+ vdv + ∫ gdz = constant
ρ ∫
ρ = constant (incompressible flow)
⇒
p v2
+ + gz = constant
ρ 2
or
⇒
ƒ
p v2
+
+ z = constant
ρg 2 g
Bernoulli's equation for frictionless compressible flow can be obtained from
Euler's equation by knowing the equation of state which relates the pressure,
the density and the temperature.
•
Isothermal process (T = constant)
o
Equation of state
p
= RT = constant = C1 (because T is constant)
ρ
p
⇔ ρ=
C1
In April 1727 Euler left Basel for St Petersburg where, later, he joined the St. Petersburg Academy of
Science. Through the requests of Daniel Bernoulli and Jakob Hermann, Euler was appointed to the
mathematical-physical division of the Academy.
Euler served as a medical lieutenant in the Russian navy from 1727 to 1730. In St Petersburg he lived
with Daniel Bernoulli. He became professor of physics at the academy in 1730 and, since this allowed
him to became a full member of the Academy, he was able to give up his Russian navy post.
When Daniel Bernoulli who held the senior chair in mathematics at the Academy left St Petersburg to
return to Basel in 1733 it was Euler who was appointed to this senior chair of mathematics.
Euler became totally blind in 1771 but loss of sight did not lessen his output, which included writings
on optics, algebra, and the theory of the Moon's motion. Even after his death, his articles continued to
appear in the press for nearly 50 more years.
The basic laws of ideal, frictionless fluids were given mathematical form by Leonhard Euler in 1755.
Euler based his work in part on earlier work by Daniel and Jacques Bernoulli.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 56
MEC441 Fluid Mechanics 1
⇒
o
∂p
∂p
p
= C1 ∫
= C1 ln p = ln p
ρ
ρ
p
Bernoulli's equation for compressible flow undergoing isothermal
process
⇒
•
∫
Chapter 3: Finite control volume analysis
p
v2
ln p + + gz = constant
ρ
2
Adiabatic process (no heat exchange)
o
Adiabatic relation (obtained from equation of state)
p
ργ
1
= constant = C2
 p γ
⇒ ρ=

 C2 
1
1
− 1 +1  p
∂p
∂p
1
⇒ ∫
= C2 γ ∫ 1 = C2 γ
p γ =
 γ
1
ρ
ρ
− +1
pγ
γ
1
∂p
γ pγ
⇔ ∫
=
p
ρ γ −1 ρ
o
γ−1
γ
=
1
 γ  γ  γ−γ 1
 
p
  γ −1 
γ p
γ −1 ρ
Bernoulli's equation for compressible flow undergoing adiabatic
process
⇒
γ p v2
+ + gz = constant
γ −1 ρ 2
3.5.6. Power in fluid flow
3.5.6.1. Definition
ƒ
It is energy of the flowing fluid per unit time.
Power =
⇔
ƒ
energy energy weight
mg
volume
=
×
= head ×
= head × ρg ×
time
weight time
t
t
P = ρgQ × head
Notes:
•
The head used here can be any head; it depends which power we want to
calculate.
•
For example:
o
To calculate the supplied power of a flowing fluid, use total head H.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 57
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
o
To calculate the power to maintain the flow against friction loss, use
head loss due to friction.
o
To calculate the power of a fluid jet, use velocity head.
3.5.7. Application of Bernoulli's equation
3.5.7.1. Pitot tube
3.5.7.1.1.
Background
Named after Henri Pitot6.
3.5.7.1.2.
Description
It is a device that consists of a tube having a short right-angled bend which is
placed vertically in a moving body of fluid with the mouth of the bent part directed
upstream and that is used with a manometer to measure the velocity of fluid flow.
6
Henri Pitot
born May 3, 1695, Aramon, France
died Dec. 27, 1771, Aramon, France
French hydraulic engineer and inventor of the pitot tube, which measures flow velocity .
Beginning his career as a mathematician and astronomer, Pitot won election to the Academy of
Sciences in 1724. He became interested in the problem of flow of water in rivers and canals and
discovered that much contemporary theory was erroneous—for example, the idea that the velocity of
flowing water increased with depth. He devised a tube, with an opening facing the flow, that provided
a convenient and reasonably accurate measurement of flow velocity and that has found wide
application ever since (e.g., in anemometers for measuring wind speed). Appointed chief engineer for
Languedoc, he performed a variety of maintenance and construction works on canals, bridges, and
drainage projects. His major work was construction of an aqueduct for the city of Montpellier (1753 86), including a stone-arch Roman-type section a kilometer (more than 1/2 mile) in length.
Source: Encyclopædia Britannica
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 58
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
The principle used in this devise is that if the velocity of flow at a point becomes
zero (stagnation point), the pressure at that point will be increased due to the
conversion of the kinetic energy into pressure energy (stagnation pressure).
The flow velocity is then determined by measuring the rise of liquid in the pitot
tube.
3.5.7.1.3.
Stagnation pressure
It is the pressure at a point in a flow where the velocity is zero. In this case, point
2 is a stagnation point, thus the pressure at point 2 is stagnation pressure.
ƒ
Bernoulli's equation at points 1 and 2
p1 v12
+
+ z1 =
ρg 2 g
z1 = z2
v2 = 0
⇒
p2 v22
+
+ z2
ρg 2 g
(horizontal flow)
(stagnation point)
p2 = p1 + 1 ρv12
2
The pressure at stagnation point (p2) is greater than the static pressure (p1) by
an amount of dynamic pressure.
3.5.7.1.4.
ƒ
Measurement of flow velocity in a pitot tube
Rearranging the stagnation pressure equation, we have:
 p2 − p1 
 p − p1 
v1 = 2  2

 = 2g 
 ρ 
 ρg 
ƒ
Difference of pressure between points 2 and 1
•
Pressure at point 1 (static pressure)
pA − p1 = −ρg ( zA − z1 ) = −ρgh1
⇔
p1 = pA + ρgh1 = patm + ρgh1
h1 : pressure head at point 1
•
Pressure at point 2 (stagnation pressure)
pB − p2 = −ρg ( zB − z2 ) = −ρgh2
⇔
p2 = pB + ρgh2 = patm + ρgh2
h2 : pressure head at point 2
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 59
MEC441 Fluid Mechanics 1
•
Chapter 3: Finite control volume analysis
Difference of pressure between points 2 and 1
p2 − p1 = patm + ρgh2 − patm − ρgh1 = ρg (h2 − h1 )
⇔
ƒ
p2 − p1
= h2 − h1
ρg
Substituting in velocity equation
v1 = 2 g (h2 − h1 ) = 2 gh
h = h2 − h1 (difference of liquid levels
in piezometer and pitot tube)
ƒ
Notes:
The flow velocity calculated is the theoretical value. To obtain the actual
value, the loss during the process must be taken into consideration by
introducing a coefficient of velocity (Cv) to the previous equation.
v1 = Cv 2 gh
3.5.7.1.5.
Other arrangements of pitot tube
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 60
MEC441 Fluid Mechanics 1
3.5.7.1.6.
3.5.7.1.6.1.
Chapter 3: Finite control volume analysis
Energy line and hydraulic grade line
Bernoulli's equation for steady, frictionless, incompressible flow
p v2
+
+ z = constant = H
ρg 2 g
3.5.7.1.6.2.
Energy line
ƒ
It is the line that represents the total head available to the fluid.
ƒ
It can be obtained by measuring the stagnation pressure using Pitot tube.
3.5.7.1.6.3.
Hydraulic grade line
ƒ
It is the line that represents piezometric head.
ƒ
It can be obtained by measuring the elevation of piezometer.
3.5.7.1.6.4.
Representation of energy line and hydraulic grade line
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 61
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.5.7.2. Venturi meter
3.5.7.2.1.
Background
Named after Giovanni Battista Venturi7.
3.5.7.2.2.
7
Description
Giovanni Battista Venturi
born 1746, near Reggio, Reggio nell'Emilia, Italy
died 1822
Italian physicist
Ordained a priest (1769), he was appointed professor of geometry and
philosophy at the University of Modena (1773), and later became
professor of physics. His research concentrated on the flow of fluids, and
he kept in close touch with the work of Bernoulli and Euler in fluid
mechanics. He is remembered for his discovery of the Venturi effect, the
decrease in the pressure of a fluid in a pipe where the diameter has been
reduced by a gradual taper. The effect has many applications, such as in
the carburetor and fluid-flow measuring instruments.
Source: http://www.allbiographies.com/biography-GiovanniBattistaVenturi33350.html
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 62
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
It is a device used for measuring the rate of flow in a pipeline. It consists of:
ƒ
A short converging part
ƒ
Throat
ƒ
Diverging part.
The flow rate is determined based on Bernoulli's equation by measuring the
difference of pressure at two points: before and at the throat.
3.5.7.2.3.
ƒ
Measurement of flow rate in a venturi meter
Bernoulli's equation at sections 1 and 2
p1 v12
p
v 2
+
+ z1 = 2 + 2 + z2
ρg 2 g
ρg 2 g
⇔
v22 − v12 p1 − p2
=
+ ( z1 − z2 )
ρg
2g
 p − p2

⇔ v22 − v12 = 2 g  1
+ ( z1 − z2 ) 
 ρg

ƒ
Continuity equation
A1v1 = A2v2
A
⇔ v2 = 1 v1 = mv1
A2
A
m= 1
A2
ƒ
: area ratio
Substituting in Bernoulli's equation
 p − p2

⇒ m 2v12 − v12 = 2 g  1
+ ( z1 − z2 ) 
 ρg

 p − p2

⇔ (m 2 − 1)v12 = 2 g  1
+ ( z1 − z2 ) 
 ρg

⇔ v1 =
ƒ
 p − p2

+ ( z1 − z2 ) 
2g  1
 ρg

m2 − 1
1
Volume flow rate
Q = A1v1 =
 p − p2

2g  1
+ ( z1 − z2 ) 
 ρg

m2 − 1
A1
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 63
MEC441 Fluid Mechanics 1
ƒ
Chapter 3: Finite control volume analysis
The indication from the manometer can be used to simplify the above
equation
p1 − p2
= ( p1 − pA ) + ( pA − pB ) + ( pB − p2 )
= −ρg ( z1 − zA ) − ρmano g ( zA − zB ) − ρg ( zB − z2 )
= −ρg ( ( z1 − zA ) + ( zB − z2 ) ) − ρmano g ( zA − zB )
= −ρg ( ( z1 − z2 ) + ( zB − zA ) ) + ρmano g ( zB − zA )
= −ρg ( ( z1 − z2 ) + h ) + ρmano gh
⇔
⇔
ƒ
Substituting in volume flow rate equation
Q=
ƒ
ρ
ρ

p1 − p2
= −( z1 − z2 ) − h + mano h = ( z1 − z2 ) + h  mano − 1
ρg
ρ
 ρ

ρ

p1 − p2
+ ( z1 − z2 ) = h  mano − 1
ρg
 ρ

ρ

2 gh  mano − 1
 ρ

m2 − 1
A1
Notes:
•
The final equation for flow rate is independent of z1 and z2. It is valid for
any position of the venturi meter (inclined, horizontal or vertical).
•
The value of h is positive if B is higher than A, otherwise h is negative.
•
The volume flow rate calculated is the theoretical value. To obtain the
actual value, the loss during the process must be taken into consideration
by introducing a coefficient of dicharge (Cd) to the previous equation.
Qactual = Cd
 p − p2

2g  1
+ ( z1 − z2 
 ρg

m2 − 1
A1
or
Qactual = Cd
ρ

2 gh  mano − 1
 ρ

m2 − 1
A1
Example No. 3 - 14
A 50 cm-diameter pipeline, to carry oil of specific gravity 0.8 is laid on a slope of
30° with the horizontal. The oil is flowing from the bottom towards the top of the
slope. A venturi meter, throat diameter 20 cm, is inserted in the line to measure
the flow; the distance between the mouth and the throat of the meter being 600
cm. If a U-tube gauge containing mercury of specific gravity 13.6, shows 50 cm
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 64
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
difference level, compute the flow in the pipe in litres/s. The coefficient of the
meter is unity.
Solution
 ρ mano

− 1
2 gh 
 ρ

m 2 −1
A1
Qactual = C d
Cd = 1
A1 =
π× ( 0.5 ) 2
= 0.1963 m 2
4
A2 =
π× ( 0.2 ) 2
= 0.0314 m 2
4
A
m = 1 = 6.25
A2
h = 0.5 m
ρ mano = 13.6 × 1000 = 13600 kg/m 3
ρ = 0.8 × 1000 = 800 kg/m 3
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 65
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
0.1963
⇒ Qactual = 1×
⇔
( 6.25 ) 2 − 1
13600 
− 1
2 × 9.81× 0.5 × 
 800

Qactual = 0.4 m 3 /s = 400 litres/s
Example No. 3 - 15
A 10 cm-diameter pipeline carrying water is laid on a slope of 30° with the
horizontal. The water is flowing from the bottom towards the top of the slope
giving a discharge of 0.02 m³/s. A venturi meter is fitted to the pipe. If a U-tube
gauge containing mercury of specific gravity 13.6, shows 60 cm difference level,
calculate the throat diameter. The coefficient of discharge of the venturi meter is
0.95.
Solution
Qactual = C d
A1
m 2 −1
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
 ρ mano

− 1
2 gh 
 ρ

Page 66
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
C d = 0.95
π× ( 0.1) 2
A1 =
= 7.85 × 10 −3 m 2
4
A2 =
m=
πd 2 2
A1
A2
4
=
d1 2
d2
2
=
( 0.1) 2
d2
2
=
0.01
d22
h = 0.6 m
ρ mano = 13.6 ×1000 = 13600 kg/m 3
ρ = 1000 kg/m 3
⇒ 20 × 10 −3 = 0.95 ×
7.85 × 10 −3
2
 0.01 

 −1
d 2 
 2 
13600 
2 × 9.81× 0.6 × 
− 1
 1000

Solving this equation, we'll obtain:
⇒
d 2 = 0.046 m = 46 mm
Example No. 3 - 16
A vertical venturi meter measures the flow of oil of specific gravity 0.82 and has
an entrance of 125 mm diameter and a throat of 50 mm diameter. There are
pressure gauges at the entrance and at the throat, which is 300 mm above the
entrance. If the coefficient for the meter is 0.92, find the flow in m³/s when the
pressure difference is 27.5 kN/m².
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 67
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Solution
ƒ
Schematic diagram
ƒ
Volume flow rate
 p − p2

+ ( z1 − z2 ) 
2g  1
 ρg

m2 − 1
Cd = 0.92
Q = Cd
A1
A1 =
π× (0.125)2
= 0.01227 m 2
4
π× (0.05)2
= 0.00196 m 2
4
A1
= 6.25
m=
A2
A2 =
p1 − p2 = 27.5 × 103 N/m 2
z1 − z2 = −0.3 m
ρ = 0.82 ×1000 = 820 kg/m3
⇒ Q = 0.92 ×
⇔
 27.5 × 103

− 0.3
2 × 9.81× 
(6.25) 2 − 1
 820 × 9.81

0.01227
Q = 0.0143 m3 /s
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 68
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.5.7.3. Orifice
3.5.7.3.1.
ƒ
Velocity of the jet
Bernoulli's equation for A and B
pA vA 2
p
v 2
+
+ zA = B + B + zB
ρg 2 g
ρg 2 g
pA = pB
(atmospheric pressure)
vA ≈ 0
(large reservoir)
z A − zB = H
⇒
ƒ
8
vB2
=H
2g
⇒
vB = 2 gH
Torricelli's theorem8
Evangelista Torricelli
born Oct. 15, 1608, Faenza, Romagna (now Italy)
died Oct. 25, 1647, Florence, Tuscany (now Italy)
Italian physicist and mathematician who invented the barometer and
whose work in geometry aided in the eventual development of integral
calculus. Inspired by Galileo's writings, he wrote a treatise on mechanics,
De Motu (“Concerning Movement”), which impressed Galileo. In 1641
Torricelli was invited to Florence, where he served the elderly astronomer
as secretary and assistant during the last three months of Galileo's life.
Torricelli was then appointed to succeed him as professor of mathematics
at the Florentine Academy.
Two years later, pursuing a suggestion by Galileo, he filled a glass tube 4
feet (1.2 m) long with mercury and inverted the tube into a dish. He
observed that some of the mercury did not flow out and that the space
above the mercury in the tube was a vacuum.
Torricelli became the first man to create a sustained vacuum. After much observation, he concluded
that the variation of the height of the mercury from day to day was caused by changes in atmospheric
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 69
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
The velocity of the issuing jet is
proportional to the square root of the
head producing flow.
ƒ
Notes:
•
The velocity of the jet obtained above is the theoretical value, because we
did not consider any loss during the process.
•
To obtain the actual value, we must multiply the theoretical value obtained
with a coefficient of velocity, Cv.
( vB )actual = Cv × ( vB ) theoretical = Cv
3.5.7.3.2.
ƒ
2 gH
Discharge
Theoretical discharge
Q = Aorifice vB = Aorifice 2 gH
ƒ
Actual discharge
( Q )actual = Cd × ( Q )theoretical = Cd Aorifice
Cd
3.5.7.3.3.
2 gH
: coefficient of discharge
Vena contracta
ƒ
From the right figure it can be seen that the actual area of the jet is less than
the area of the orifice.
ƒ
The diameter of the jet is reduced from point C until it reaches point B where
the diameter becomes constant.
ƒ
The section through B is called vena contracta.
ƒ
Actual area of the jet
pressure. He never published his findings, however, because he was too deeply involved in the study
of pure mathematics - including calculations of the cycloid, a geometric curve described by a point on
the rim of a turning wheel. In his Opera Geometrica (1644; "Geometric Works"), Torricelli included his
findings on fluid motion and projectile motion.
Source: Encyclopædia Britannica
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 70
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
Ajet = Cc × Aorifice
Cc : coefficient of contraction
3.5.7.3.4.
Relation between Cd, Cv, and Cc
Actual discharge = Actual area of the jet at B × Actual velocity of the jet at B
= Cc Aorifice × Cv 2 gH
Cd Aorifice 2 gH
⇔
Cd = Cc × Cv
Example No. 3 - 17
a. A water jet with a discharge of 1800 litres per minute strikes a fixed vertical
flat plate. The force exerted to the plate is 0.21 kN. Calculate the velocity of
the jet.
b. The jet is produced from a 100 mm diameter orifice under a constant head of
3 m. Determine the theoretical velocity and theoretical discharge of the jet.
c. Obtain the coefficient of velocity and the coefficient of discharge of the jet.
d. Calculate the coefficient of contraction of the jet.
Solution
ƒ
Velocity of the jet
•
Force exerted by the jet on the plate
Fx = ρA jet v jet 2
ƒ
⇔
Fx = ρQv jet
⇔
v jet =
Fx
ρQ
=
0.21× 10 3
= 7 m/s
 1800 × 10 −3 
1000 × 

60


Theoretical velocity of the jet
(v jet ) theo = 2 gH = 2 × 9.81× 3 = 7.67 m/s
ƒ
Theoretical discharge of the jet
(Q ) theo = Aorifice (v jet ) theo =
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
π× (100 ×10−3 ) 2
× 7.67 = 0.06 m3/s
4
Page 71
MEC441 Fluid Mechanics 1
ƒ
Coefficient of velocity of the jet
Cv =
ƒ
v jet
(v jet ) theo
=
7
= 0.913
7.67
Coefficient of discharge of the jet
Cd =
ƒ
Chapter 3: Finite control volume analysis
Q jet
(Q jet ) theo
=
1800×10−3
60
0.06
= 0.5
Coefficient of contraction of the jet
C
0.5
Cc = d =
= 0.548
Cv 0.913
Example No. 3 - 18
A tank has an orifice in its side, located 25 cm below the water surface in the
tank. The free jet coming out from the orifice reaches the floor which is 15.4 cm
below the orifice at a horizontal distance of 37.5 cm. Calculate the coefficient of
velocity of the jet.
Solution
ƒ
Time taken for the jet to reach a horizontal distance x
x = v jet t
⇔ t=
x
v jet
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 72
MEC441 Fluid Mechanics 1
ƒ
Vertical distance that the jet reaches within t seconds
1
2
y = gt
ƒ
Chapter 3: Finite control volume analysis
2
 x
y = g
 v jet

1
2
⇒
2
Velocity of jet (from Bernoulli's equation)
v jet = 2 gH
ƒ




or
( v jet )actual = Cv
2 gH
Substituting in equation of y

y = 1 g
2 C
 v
⇒
Cv =
2
2

 = x
2 gH 
4 Cv 2 H
x
37.5
2 15.4 × 25
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
⇒ Cv =
x
2
yH
= 0.956
Page 73
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
3.6. Problems
1.
Oil flows through a pipeline which contracts from 450 mm diameter at A to
300 mm at B then forks, one branch being 150 mm diameter discharging at
C and the other branch 225 mm diameter discharging at D. If the velocity at
A is 1.8 m/s and the velocity at D is 3.6 m/s, calculate:
a) the discharge at C and D.
b) the velocities at B and C.
2.
Pipe AB branches into two pipes C and D as shown. The pipe has a
diameter of 45 cm at A, 30 cm at B, 20 cm at C and 15 cm at D. If the
velocity at A is 2 m/s, determine:
a) the discharge at A.
b) the velocity at B and the velocity at D, given that the velocity at C is 4
m/s.
3.
The diameter for pipe A is 20 mm, the diameter for pipe B is 10 mm with a
velocity of 0.3 m/s. Pipe C is 15 mm in diameter with velocity of 0.6 m/s.
Find the initial velocity (v), and the flow rate (Q) at the entrance of pipe A.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 74
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
4.
A fluid of constant density flows at the rate of 15 litres/sec. along a pipe AB
of 100 mm diameter. This pipe branches at B into two pipes BC and BD,
each of 25 mm diameter and a third pipe BE of 50 mm diameter. The flow
rates are such that the flow through BC is three times the flow rate through
BE and the velocity through BD is 4 m/s. Find the flow rates in the three
branches BC, BD and BE and the velocities in pipes AB, BC and BE.
5.
A closed tank of fixed volume is used for the continuous mixing of two
liquids which enter at A and B, and are discharged completely mixed at C.
The diameter of inlet pipe A is 150 mm and the liquid flows at the rate of
0.056 m³/s and has a specific gravity of 0.93 and inlet pipe B is 100 mm
diameter and the liquid (sp. gr. = 0.87) flow rate is 0.03 m³/s. The diameter
of outlet pipe C is 175 mm. Determine the mass flow rate, velocity and
specific gravity of the mixture discharged.
6.
A jet (5 cm diameter) discharges 30 litres/s of water perpendicular to a flat
plate. Calculate:
a) the force on the plate when it is stationary.
b) the force on the plate, the work done per second and the efficiency when
it moves at 5 m/s.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 75
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
7.
A jet of water which is flowing freely in the atmosphere is deflected
horizontally by a 90° curved vane as shown. The water jet has a diameter of
20 mm and velocity of 5 m/s. Find the resultant force acting on the vane.
8.
A jet of water 5 cm in diameter discharges 0.03 m³/s. Calculate the force
required to move the plate towards the jet with a velocity of 5 m/s. The jet
strikes the plate perpendicularly.
9.
A square plate mass 12.7 kg of uniform thickness and 300 mm edge is hung
so that it can swing freely about its upper horizontal edge. A horizontal jet 20
mm in diameter strikes the plate with a velocity of 15 m/s and the centerline
of the jet is 150 mm below the upper edge of the plate so that when the
plate is vertical the jet strikes the plate normally at its center. Find:
a) the force that must be applied at the lower edge of the plate to keep it
vertical.
b) the inclination to the vertical at which the plate will assume under action
of the jet to swing freely.
10. A jet of water 25 mm diameter strikes a 800 N flat plate normally with a
velocity of 30 m/s at 150 mm below the upper edge. The plate is hinged at
the upper edge. Its centre of gravity is 100 mm from the hinge.
a) What is the angle that the plate will form with the vertical axis.
b) What force should be applied, 100 mm below the axis of the jet, in order
to keep the plate vertical?
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 76
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
11. A jet of water of 100 mm diameter is moving at 36 m/s and is deflected by a
vane moving at 15 m/s in a direction at 30° to the direction of the jet. The
water leaves the vane with no velocity in the direction of motion of the vane.
Determine:
a) draw the sketch of the vane. Indicate all velocity components and their
angle.
b) the inlet and outlet angles of the vane assuming no shock at entry or exit.
c) the force on the vane in the direction of motion.
Take the velocity of the water at outlet relative to the vane to be 0.85 of the
relative velocity at entry.
12. Define the following terms:
a)
b)
c)
d)
Potential head.
Pressure head.
Velocity head.
Total head for a liquid in motion.
13. What are the forms of energy to consider in analyzing fluid flow in pipeline
problem? Describe each of them.
14. State Bernoulli's theorem for a liquid.
15. A vertical venturi meter has the inlet and throat diameter of 150 mm and 75
mm respectively. The throat is 225 mm above the inlet and Cd = 0.96. Petrol
of specific gravity 0.78 flows up through the meter at a rate of 0.029 m³/s.
Find the pressure difference between inlet and the throat.
16. A vertical venturi meter measures the flow of oil of specific gravity 0.82 and
has an entrance of 125 mm diameter and a throat of 50 mm diameter. There
are pressure gauges at the entrance and at the throat, which is 300 mm
above the entrance. If the coefficient for the meter is 0.92, find the flow in
m³/s when the pressure difference is 27.5 kN/m².
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 77
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
17. Using Bernoulli's equation and continuity of flow, show that the theoretical
discharge Q through an inclined venturi meter may be expressed as:
Q=
w

2 gx  m − 1
2
w


m −1
where
A1
A1
A2
m
: area ratio,
wm
: specific weight of mercury
w
: specific weight of the discharged liquid.
18. A venturi meter has a main diameter of 65 mm and a throat diameter of 26
mm. When measuring the flow of a liquid of density 898 kg/m³, the reading
of a mercury differential-pressure gauge was 71 mm. Calculate the flow
through meter in m³/h if the coefficient of the meter is 0.97 and the specific
gravity of mercury is 13.6.
19. A venturi meter is fitted to a 100 mm-diameter horizontal pipe. Water flows
in the pipe giving a discharge of 20 litres/s. A differential U-tube mercury (sp.
gr. 13.6) manometer placed between the entrance of the venturi meter and
the throat shows a difference of level of 0.6 m as shown. If the coefficient of
discharge is 0.95, find the throat diameter of the venturi meter.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 78
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
20. Water is flowing from section 1 to 2. At section 1, which is 25 mm in
diameter, the gauge pressure is 345 kPa and the velocity of flow is 3.0 m/s.
At section 2, which is 50 mm in diameter, is 2.0 m above section 1.
Assuming there are no energy losses in the system, calculate pressure p2.
21. A pipe, as shown, carries water and taper uniformly from a diameter of 0.1
m at A to 0.2 m at B over a length of 2 m. Pressure gauges are installed at
A, B and C (which is mid-point of AB). If the pipe centerline slopes upwards
from A to B at an angle of 30° and the pressure at A and B are 2.0 and 2.3
bar respectively, determine the flow through the pipe and pressure at point
C.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 79
MEC441 Fluid Mechanics 1
Chapter 3: Finite control volume analysis
22. A 100 mm diameter orifice dicharges 36 litres per second of water under a
constant head of 2.6 m. A flat plate held just downstream from the orifice
requires a force of 240 N to resist the impact of the jet. Determine:
a) the coefficient of velocity.
b) the coefficient of discharge.
c) the coefficient of contraction.
Assoc. Prof. Dr. Wirachman Wisnoe
1/13/2010
Page 80