Lecture 4
Final Version
Contents
• Mass Conservation – Example 3
• Bernoulli Equation - Derivation
What Did We Do In Last Lecture?
Control Volume / Control Surface
Classification of Forces
• Body Force vs Surface Force
• Static Force vs Dynamic Force
Reynolds Number
Re =
UL
ν
• Non-dimensional number
• Ratio of inertia forces and viscous forces
What Did We Do In Last Lecture?
Streamlines
/
Stream Tubes
Laminar / Turbulent Flow
What Did We Do In Last Lecture?
Other Non-dimensional Numbers
Conservation of Mass /Continuity Equation
∑ρ Q = 0
EXAMPLE
CLASSES
Bring along any
questions you may have …
Example 3:
• A cylindrical open-top tank with height 1.3 m and a
radius of 0.5m is filled with water. Water drains from a
hole with diameter 0.04m near bottom. How long does
it take until water level has dropped to a height of 0.4
m. The average outflow velocity is given by...
V = 2gh
h0 = 1.3 m
h2 = 0.4 m
rtank = 0.5 m
Solution:
• Conservation of mass for a control volume CV
requires...
•
•
min − mout
dm cv
=
dt
Continued ...
•
However, here no mass enters the CV ...
•
•
min = 0
•
mout
dm cv
=
dt
(1)
Mass flow rate out is ...
•
mout = (ρ V A jet )out
•
With given outflow velocity V one gets...
•
mout = (ρ 2 g h A jet )out
(2)
with
2
A jet = π r jet
•
Mass of water in tank at any time is ...
mcv = ρ Vol. = ρ Atan k h
(3)
with
2
Atank = π rtank
•
Now substitute (2) and (3) into (1). This gives ...
Continued ...
•
After substituting in we have ...
− ρ 2 g h A jet =
d ( ρ Atank h )
dt
With areas of tank and jet
2
A jet = π r jet
−ρ 2 g hπ
2
Atank = π rtank
2
r jet
(
2
d ρ π rtank
h
=
dt
Get constants out of the d(...)/dt
2
2
− ρ 2 g h π r jet
= ρ π rtank
2
2
− 2 g h rjet
= rtank
dh
dt
dh
dt
)
Continued ...
•
Rearranging last equation from previous slide gives ...
2
rtank
dt = − 2
r jet
2
rtank
dt = − 2
r jet
•
dh
2gh
1 dh
2g h
Integrate ...
t
h2
2
rtank
dt = − 2
r jet
0
h
∫
∫
1 dh
2g h
0
On right-hand side get constants in
front of integral
t
∫ dt = −
0
2
rtank
2
r jet
1
2g
h2
∫
h0
dh
h
Continued ...
•
Carry out integration of integral on previosu slide ...
t
∫
dt = −
0
2
rtank
2
r jet
2
rtank
t=− 2
r jet
t=
•
2
rtank
2
r jet
1
2g
1
2g
h2
∫
h0
⎡
⎢⎣2 h
1
dh
h
2⎤
h2
⎥⎦
h
0
1 ⎞
2 ⎛ 12
⎜⎜ h0 − h2 2 ⎟⎟
2g⎝
⎠
Substituting in given values yields ...
(0.5 m )2
2
( 1.3 m − 0.4 m )
t=
2
2
(0.02 m ) 2 ⋅ 9.81 m/s
= 143.3 s = 2 mins 23.3 s
Note: We neglected viscosity. Thus, in reality actual
time for tank to drain will be somewhat longer.
Derivation of Bernoulli Equation
Introductory Remarks
•
•
Previously considered pressures in static fluids (i.e.
hydrostatic pressure).
Learnt that:
“The pressure at any two points at the same level
in a body of fluid AT REST will be the same ”
•
Will now look at fluids in motion. Will find that
pressure at same level in a body of fluid also depends
on flow velocity.
What we will find is ...
• Velocity decreases, satisfy mass
•
conservation
Velocity decrease accompanied
by pressure increase.
High Flow Velocity
Low Pressure
Low Flow Velocity
High Pressure
Continued...
Analogy of ...
• Low(Minimum) Velocity
• High (Max.) Potential Energy
• Velocity Increases
• Potential Energy Decreases
• Maximum Velocity
• Minimum Potential Energy
• Velocity Decreases
• Potential Energy Increases
• Low (Min.) Velocity
• High (Max.) Potential Energy
… Bernoulli Equation expresses …
CONSERVATION OF ENERGY FOR
FLOW!
Continued...
•
Bernoulli Equation can be derived from
what is known as the Steady-Flow Energy
Equation (SFEE).
•
While derivation via SFEE may make it
more obvious why Bernoulli equation
represents conservation of energy we will
however, follow the derivation of Douglas et
al. (2001, Fluid Mech. 4th Ed., Prentice Hall,
p. 141-141) that is reproduced in your Tech.
Sci. 2 book.
Continued...
•
•
•
Consider streamtube surrounding a streamline shown
below.
Cross-sectional area at entrance and exit small enough
that flow velocities constant over cross-sections.
Assume that fluid is inviscid.
D
δs
A
C
θ
A
v
p
pside
B
A + δA
v + δv
p + δp
mg
z + δz
z
•
Mass per unit time flowing =
•
Momentum per unit time =
ρ Av
(ρ A v) v
Rate of increase of momentum from AB to CD
= ( ρ A v ) (v + δv ) − ( ρ A v ) v
(1)
= ( ρ A v ) δv
Continued...
Forces acting to produce this increase of
momentum in direction of motion are:
cosθ =
δz
δs
•
Force due to p in direction of motion =
•
Force due to ( p + δp )opposing motion =
•
Force due to pside producing a component
motion in the direction of motion =
pside δA
•
Force due to m g producing a
component opposing motion =
m g cosθ
pA
( p + δp ) ( A + δ A )
(2)
Resultant Force in direction of motion
= p A − ( p + δp ) ( A + δA) + pside δA − m g cosθ
Continued...
Forces acting to produce this increase of
momentum in direction of motion are:
cosθ =
•
Force due to pside producing a component
motion in the direction of motion =
δz
δs
pside δA
(2)
Resultant Force in direction of motion
= p A − ( p + δp ) ( A + δA) + pside δA − m g cosθ
Continued...
•
Value of pside will vary from p at AB to ( p + δp ) at CD
and can be taken as:
pside = p + k δp
where k is a fraction.
•
Weight of element:
1 ⎞
⎛
mg = ρ g × Volume = ρ g ⎜ A + δA ⎟ δs
2 ⎠
⎝
•
Component of weight in direction of element:
δz
1 ⎞
⎛
mg cosθ = ρ g ⎜ A + δA ⎟ δs ⋅
δs
2 ⎠
⎝
•
(3)
Multiplying out Eq. (2) from previous slide and
substitute in Eq. (3):
Resultant Force in direction of motion
= − p δA − Aδp − δpδA + pδA + kδp ⋅ δA
1 ⎞
δz
⎛
− ρ g ⎜ A + δ A ⎟ δs ⋅
2 ⎠
δs
⎝
Continued...
• Two terms cancel each other out…
Resultant Force in direction of motion
= − p δA − Aδp − δpδA + pδA + kδp ⋅ δA
1 ⎞
δz
⎛
− ρ g ⎜ A + δ A ⎟ δs ⋅
2 ⎠
δs
⎝
Continued...
• and
now we further neglect small terms of
order
δ2
Resultant Force in direction of motion
= − p δA − Aδp − δpδA + pδA + kδp ⋅ δA
1 ⎞
δz
⎛
− ρ g ⎜ A + δ A ⎟ δs ⋅
2 ⎠
δs
⎝
Continued...
• So, what is left is….
Resultant Force in direction of motion
= − Aδp − ρ g A δz (4)
Continued...
• Applying
Newton’s second law, i.e. equating
Eq. (4) with Eq. (1)
Resultant Force in direction of motion
= − Aδ p − ρ g A δ z
(4)
Rate of increase of momentum
from AB to CD
= ( ρ A v ) δv
(1)
( ρ A v ) δ v = − Aδ p − ρ g A δ z
• Divide by ρ A δs
…
δv
δz
1 δp
+v + g =0
ρ δs δs
δs
• Let δs → 0
…
Continued...
• Limit yields ...
EULER’s EQUATION
1 dp
dz
dv
+v + g =0
ρ ds
ds
ds
• The
•
(5)
equation states, in differential form,
relationship between pressure, density and
elevation along a streamline for steady flow.
It cannot be integrated until relation
between density and pressure is known!
Continued...
HOWEVER...
• For
incompressible flow, i.e. for flow with
constant density...
1
ρ
dp + v dv + g dz = 0
∫ ρ dp + ∫ v dv + ∫ g dz = const.
1
dp + v dv + g dz = const.
∫
∫ ∫
ρ
1
1
1 2
p + v + g z = const.
ρ
2
• Terms
mass.
(6)
in Eq. (6) represent energy per unit
Continued...
• Dividing Eq. (5) through by g gives:
BERNOULLI EQUATION
p
v2
+
+ z = const. = H
ρ g 2g
• Here
•
the terms represent energy per unit
weight.
All terms are all in units of (Length).
• Alternative form is...
Alternative Form
p+
• Here
ρ
2
v 2 + ρ g z = const.
(7)
terms represent energy per unit
volume.
Continued...
This term is
evidently
Energy / Volume.
Note:
Hence, other
terms must have
same meaning
m 2
v
p+ 2
+ ρ g z = const.
Volume
Alternative Form
p+
• Here
ρ
2
v 2 + ρ g z = const.
(7)
terms represent energy per unit
volume. We can see this from above note!