Lecture 5
Contents
• Bernoulli Equation – Terminology
• Bernoulli Equation - Examples
• Dimensional Analysis / Model Testing
(Start)
EXAMPLE
CLASSES
Bring along any
questions you may have …
What Did We Do In Last Lecture?
Derivation of Bernoulli Equation
p
+
ρ
2
v 2 + ρ g z = const.
• Constant along streamline
• All terms have units of pressure
• p is the physically existing pressure
• 2nd term is kinetic energy per unit volume
associated with mean motion of flow
• 3rd term hydrostatic pressure
What Did We Do In Last Lecture?
Keep in mind analogy …
Epot + Ekin = const.
Finished previous lecture
with …
BERNOULLI EQUATION
p
v2
+
+ z = const. = H
ρ g 2g
Alternative Form
p+
ρ
2
v 2 + ρ g z = const.
(7)
• Equations apply along a single stream line.
• Constant for each streamline can be
different.
Continued...
• Eq. (7) from previous lecture / slide …
p+
ρ
2
v 2 + ρ g z = const.
(7)
• If
Eq. (7) applied along streamline between
any two points indicated by suffixes 1 and 2
one gets ...
p1 +
ρ
2
v12 + ρ g z1 = p2 +
ρ
2
v22 + ρ g z2
This is form in which we will usually apply
Bernoulli Equation most of the time.
Continued...
•
Note: For compressible fluid integration of Euler’s
equation ...
1 dp
dv
dz
+v + g =0
ρ ds
ds
ds
can only be partially completed to give ...
dp v 2
∫ ρg + 2 g + z = 0
Now need relation between density and pressure to
continue.
Terminology for Bernoulli Equation
p
(1)
+
ρ
2
v2
(2)
+
ρgz
=
(3)
const.
(4)
(STATIC) PRESSURE
(1) =
The physically existing pressure in fluid. Whenever
reference is made to “pressure” without further
qualification then interpret as this one.
DYNAMIC PRESSURE
(2) =
Not a physically existing pressure. If z=0, such that
Elev. Press. Vanishes, then Dyn. Press. is the difference
between Tot. Press. And Stat. Press.
ELEVATION PRESSURE
(3) =
Can often be made to vanish, i.e. when streamline
horizontal such that we can chose z=0. If streamline not
horizontal, but if change in height is small such that
Elev. Press. is small compared to other terms one
usually neglects it.
TOTAL PRESSURE
(4) =
This has physical significance that it is the pressure at
which fluid comes to rest. Assume z=0, V=0 then one
gets: Stat. Press=Tot. Press.
Terminology for Bernoulli Equation
+
p
•
ρ
2
v2
+
ρgz
=
const.
Note that the STATIC PRESSURE is NOT the
pressure in a static fluid (that is called the
hydrostatic pressure).
•
Neither is the STATIC PRESSURE the pressure
where the fluid is at rest (that is called the stagnation
pressure).
Terminology for Bernoulli Equation
• For ...
p1 +
ρ
2
v12 + ρ g z1 = p2 +
v2 = 0 and
With:
p1 +
•
ρ
2
ρ
2
v22 + ρ g z 2
z1 = z 2
v12 = p2
Hence sum of static pressure at station 1 plus
dynamic pressure at station 1 is equal to static
pressure at station 2 where fluid is at rest
(stagnates). Thus we call ...
p1 +
ρ
2
v12 = Stagnation Pressure
(or Pitot Pressure)
Final Terminology
• When Bernoulli written in form ...
p
ρg
(1)
+
v2
2g
(2)
+
z
=
const.
(3)
(4)
… such that all terms in units of ‘LENGTH’
then one calls ...
(1) =
STATIC HEAD
(2) =
DYNAMIC HEAD
(3) =
ELEVATION HEAD
(4) =
TOTAL HEAD
… and finally ...
(1) + (2) =
STAGNATION HEAD
And recall that ...
Pressure measured relative to atmospheric
pressure is called ...
GAUGE PRESSURE
or
GAGE PRESSURE
… as distinct from
ABSOLUTE PRESSURE
Example 1:
Pitot-Static Tube / Prandtl Tube
Continued …
(1)
(2)
p1
p2
v1 = ?
v2 = 0
Question: What is value of v1?
Solution:
•
Bernoulli for points (1) and (2)
p1 +
•
ρ
2
v12 + ρ g z1 = p2 +
ρ
2
v22 + ρ g z2
Neglect hydrostatic pressure; horizontal streamline …
p1 +
ρ
2
v12 + ρ g z1 = p2 +
p1 +
ρ
2
v12 = p2 +
ρ
2
ρ
2
v22 + ρ g z2
v22
Continued …
•
(1)
(2)
p1
p2
v1 = ?
v2 = 0
Point (2) is stagnation point where velocity is zero …
p1 +
ρ
2
v12 = p2 +
p1 +
•
ρ
2
ρ
2
v22
v12 = p2
Solve for velocity …
v1 =
2
ρ
( p2 − p1 )
Continued …
(1)
(2)
p1
p2
v1 = ?
v2 = 0
Stagnation
Pressure
•
Free-Stream
Static Pressure.
Solve for velocity …
v1 =
2
ρ
( p2 − p1 )
[Pressure here
Equal to p1 at
point (1) since
it is assumed
that probe does
not affect flow.]
Continued …
Pitot-Static Tube / Prandtl Tube
• This type of probe/methodology is used in my Vortex /
•
Wind tunnel Lab L5 to measure free stream velocity via
pressures!
In lab I will assume that you are familiar with this!
• In lab will work with arrangement similar to that
shown below. The difference being that aeroplane
is replaced by a circular cylinder, and probe not in
wake of cylinder but in region where flow is not
affected by presence of cylinder.
Pitot-Static /
Prandtl Tube
Example 2:
•
Water flowing in open channel at depth of 2m and
velocity 3m/s. It then flows downs contracting chute
into another channel where depth is 1m and velocity
10m/s. Assuming frictionless flow, determine difference
in elevation of the channels.
(1)
2m 3m/s
y=?
( 2)
1m
10m/s
Solution:
•
Bernoulli for points (1) and (2)
p1 +
•
ρ
2
v12 + ρ g z1 = p2 +
ρ
2
v22 + ρ g z2
(A)
Since (1) and (2) on free surface:
p1 = p2 = patm.
Eq. (A) becomes...
ρ
2
v12 + ρ g z1 =
ρ
2
v22 + ρ g z2
(B)
Continued ...
(1)
2m 3m/s
y=?
1m
•
2
v12 + ρ g z1 =
ρ
2
v22 + ρ g z2
•
From sketch we see... z1 = y + 2 m
•
Introduce this into (B) ...
ρ
2
v12 + ρ g ( y + 2 ) =
ρ
2
(B)
v22 + ρ g z2 (C)
Solve (C) for y ...
y=
•
10m/s
Last line from previous page repeated...
ρ
•
( 2)
(
)
1 2 2
v2 − v1 + ( z2 − 2 m )
2g
(D)
With given values ...
v1 = 3 m/s , v2 = 10 m/s ,
z2 = 1 m
Continued...
•
Eq. (D) yields...
y=
2
2
(
)
[
]
[
]
10
m/s
−
3
m/s
+ (1 m − 2 m )
2
2 ⋅ 9.81 m/s
1
= 3.64 m
Example 3: VENTURI TUBE
Example 3: VENTURI TUBE
•
Venturi tube is device where flow rate in a pipe line is
measured by narrowing a part of tube. In narrowed
part flow velocity increases. By measuring resultant
decreasing pressure flow rate in pipe line can be
determined.
ρ m (with ρ < ρ )
m
w
(3)
D2 with Area, A2
( 2)
ρw
Water
z2
z1
(1)
D1 with Area, A1
z=0
Solution:
•
Bernoulli for points (1) and (2)
ρ
ρ
p1 + w V12 + ρ w g z1 = p2 + w V22 + ρ w g z2 (1)
2
2
•
Mass conservation yields...
A1 V1 = A2 V2
A
V2 = 1 V1
A2
(2)
Continued ...
•
Substituting Eq. (2) into Eq. (1) gives...
ρw
ρ w ⎛ A1
2
⎞
⎜⎜ V1 ⎟⎟ + ρ w g z2
p1 +
v12 + ρ w g z1 = p2 +
2
2 ⎝ A2 ⎠
•
Rearrange ...
⎛ ⎛ A ⎞2 ⎞
V12 ⎜ ⎜⎜ 1 ⎟⎟ − 1⎟ = ( p1 − p2 ) + ρ w g ( z1 − z2 )
⎟
⎜ ⎝ A2 ⎠
2
⎝
⎠
ρw
V1 =
( p1 − p2 ) + ρ w g ( z1 − z2 )
ρ w ⎛⎜ ⎛ A1 ⎞
⎞
⎜⎜ ⎟⎟ − 1⎟
⎟
2 ⎜ ⎝ A2 ⎠
⎝
⎠
2
=
This solution is maximum velocity V1 which would
exist for frictionless flow. In reality there is
viscosity present and one introduces a discharge
coefficient cd such that ...
(V1 )actual = cd (V1 )ideal
•
To apply above equation still need to evaluate pressure
difference ...
Continued ...
•
How to get pressure difference ….
∆h
ρm
h
l
Water, ρ w
z2 − z1
( 2)
(1)
•
Use pressure-depth ideas developed earlier. Summing
up all pressure contributions on path from (1) to (2) ...
p1 − ρ w g ( z2 − z1 ) − ρ w gl − ρ w gh − ρ m g∆h
+ ρ m g∆h + ρ m gh + ρ w gl = p2
p1 − p2 = ρ w g ( z2 − z1 ) + ρ w gh − ρ m gh
p1 − p2 = ρ w g ( z 2 − z1 ) + ( ρ w − ρ m )gh
Example 4:
Beavers with
engineering degrees
discovered
use of Bernoulli
principle to ventilate
burrows !
Example 4:
•
Beavers with engineering degrees discovered use of
Bernoulli principle to ventilate burrows. If burrow as
below than at one entrance wind velocity increases,
creating a pressure differential that induces air flow
through burrow. Calculate the pressure difference
between entrance and exit for the friendly beaver...
V = 22 km/h
Streamline
( 2)
V = 18 km/h
(1)
p 2 < p1
ρ air = 1.225 kg/m3
Solution:
•
Bernoulli for points (1) and (2)
p1 +
•
ρ
2
v12 + ρ g z1 = p2 +
ρ
2
v22 + ρ g z2
(A)
Assume negligible elevation change, hence: z1 = z2
p1 +
•
ρ
2
Eq. (A) becomes...
v12 = p2 +
ρ
2
v22
Solve (B) for pressure difference ...
(B)
Continued...
•
Last equation from previous slide repeated...
p1 +
ρ 2
ρ
v1 = p2 + v22
2
2
∆p = p1 − p2 =
•
(C)
With given values...
v1 = 18 km/h = 5 m/s
•
(
v22 − v12 )
2
ρ
v2 = 22 km/h = 6.11 m/s
Eq. (C) yields ...
(
1.225 kg/m3
[6.11 m/s]2 − [5 m/s]2
∆p = p1 − p2 =
2
)
kg m
2
N
s
= 7.55
= 7.55
= 7.55
= 7.55 Pa
2
2
2
ms
m
m
kg
•
p1 = p2 + 7.55 Pa pressure at exit (1) is
So, since
higher than pressure at exit (2). This pressure
difference drives a venting flow from (1) towards (2).
Example 5: Homework/ Background Reading
•
A hurricane is a tropical storm formed over ocean by
low atmospheric pressures. As hurricane approaches
land, inordinate ocean swells (very high tides)
accompany hurricane. A Class-5 hurricane features
winds in excess of 240 km/h, although wind velocity at
center “eye” is very low.
In sketch below atmospheric pressure 310 km from eye
is 101,591 Pa (558.8 mm Hg at point 1, generally
normal for ocean) and winds are calm. Hurricane
atmospheric pressure at eye at point (3) is 74,500 Pa
(762 mm Hg).
(I) Estimate ocean swell at eye of the hurricane at (3)
(II)Estimate ocean swell at (2) where wind velocity is
310 km/h.
Eye
Hurricane
Calm
Ocean
(1) Level
( 2)
h2
(3)
h1
Ocean
Continued ...
Assumptions
•
Steady, incompressible, irrotational (so that
Bernoulli applicable)
(Note: This is certainly a very questionable
assumption for a highly turbulent flow)
•
The effect of water drifted into air is negligible
Properties
Density air
ρ A = 1.225 kg/m
Density sea water
3
ρ SW = 1023 kg/m 3
Density Mercury
ρ M = 13,560 kg/m 3
Continued, Cengel p. 199 ...
Solution, Part (I):
•
Reduced atmospheric pressure over the water causes to
rise. Thus decreased pressure at (3) relative to (1)
causes ocean water to rise at (3). Hence...
∆p = ( ρgh )Hg = ( ρgh )SW
ρ Hg ghHg = ρ SW ghSW
hSW =
hSW =
13,560 kg/m 3
1023 kg/m
3
ρ Hg
hHg
ρ SW
(A)
(762 mm Hg − 558.8 mm Hg )
= 2.693 m
•
This result is equivalent to the storm surge at the eye of
the hurricane since the velocity there is negligible and
there are no dynamic effects.
Continued...
Solution Part (II):
•
Here we have to take the high wind speed into account
and need Bernoulli. We write down Bernoulli equation
for points A and B...
p A v A2
p B vB2
+
+ zA =
+
+ zB
ρg 2 g
ρg 2 g
•
Since zA=zB and vB=0 we have...
p A v A2
p
+
= B
ρg 2 g
ρg
2
⎛ 240,000 m/s ⎞
⎜
⎟
pB − p A v A2 ⎝ 3600
⎠ = 226.5 m
=
=
2g
ρg
2 ⋅ 9.81 ms-2
•
Density value to be used is air density in
hurricane...
ρ Hur =
PAir
Patm, Air
ρ atm, Air =
74,500 Pa
⋅1.225 kgm 3
101,591 Pa
= 0.898 kgm 3
Continued...
•
Using relation (A) from Part (I) we get ...
hdynamic =
hSW =
ρ Air
hAir
ρ SW
0.898 kg/m 3
1023 kg/m
3
⋅ 226.5 m
= 0 .2 m
•
Therefore pressure at (2) is 0.2m seawater column
lower than pressure at point (3) due to high wind
velocities, causing ocean to rise an additional 0.2 m.
Then total storm surge at point (2) becomes...
h2 = h1 + h dynamic = 2.693 m + 0.2 m = 2.893 m
DISCUSSION
Problem involves highly turbulent flow and intense
breakdown of streamlines. Thus, applicability of
Bernoulli equation in Part (II) is questionable.
Furthermore, flow in eye of storm is not irrotational,
and Bernoulli constant changes across streamlines.
Bernoulli analysis can be thought of as limiting, ideal
case, and shows that rise of seawater due to highvelocity winds cannot be more than 2.893m.
Dimensional Analaysis and Model Testing
•Introduction to Dimensional Analysis
Consider drag D of sphere ….
On what quantities does it depend?
Diameter, d
Flow Speed, V
Fluid Density, ρ
Fluid Viscosity, µ
Write
D = F (d ,V , ρ , µ )
(1)
• Note: Eq.(1) reads … Drag, D, is a Function of ...
What does the above mean in terms of the
measurements we have to carry out to collect
data for all possible spheres in all types of fluids?
Continued ...
WE NEED ...
1 page for Drag as
function of 2 variables
(e.g. velocity and
diameter)
d increases
from curve
to curve
1 page for
each value
of ρ
1 book for Drag as
function of 3 variables
(e.g. velocity, diameter,
density)
Shelf of books for Drag as
a function of 4 variables
(velocity, diameter,
density, viscosity)
If we want 10 data points per curve, at £10 each
experiment, this will cost...
10 × 10 × 10 × 10 × £10 = £100,000
THERE MUST BE
A BETTER WAY !?!?
Continued...
DRAG
GOAL IS TO COMPRESS SHELF
OF BOOKS INTO ONE SINGLE
GRAPH...
4 Independent
Experimental Parameters
How Could We Possibly
Achieve This?