Transcript Document
Fluid Mechanics 05
Rotating Flow
In the rotating flow βl = βr and the acceleration
will be only in the normal direction.
π
βπ£ 2
β
π + Ι£π§ = Ο β
ππ
π
Where π£ = π€ β π
π
π + Ι£π§ = Ο β π€ 2 β π
ππ
Ο β π€2 β π2
π + Ι£π§ β
= Const
2
Example
A cylindrical tank of liquid shown in the figure is
rotating as a solid body at a rate of 4 rad/s. The
tank diameter is 0.5 m. The line AA depicts the
liquid surface before rotation, and the line Aβ² Aβ²
shows the surface profile after rotation has
been established. Find the elevation difference
between the liquid at the center and the wall
during rotation.
Solution
Ο1 β π€12 β π12
π1 + Ι£1π§1 β
2
Ο2 β π€22 β π22
= π2 + Ι£π§2 β
2
Where P1=P2=Patm=Zero gauge, r1=0,
r2=0.25 m, w=4 rad/s, Ο=1000Kg/m^3,
g=9.81m/s^2
π€ 2 β π22 42 β 0.252
π§2 β π§1 =
=
= 0.051π
2βπ
2 β 9.81
Bernoulli Equation
We apply Bernoulli equation of the tangent
component of acceleration.
π
β
π + Ι£π§ = Ο β ππ‘
ππ
ππ£
ππ£
Where ππ‘ = π£ β +
ππ
ππ‘
Assume the flow is steady then
π£2
ππ£
ππ£ π( 2 )
= 0 π‘βππ ππ‘ = π£ β
=
ππ‘
ππ
ππ
π
π π£2
β
π + Ι£π§ = Ο β ( )
ππ
ππ 2
π
π£2
π + Ι£π§ + Ο β
=0
ππ
2
π£2
π + Ι£π§ + Ο β
= Const
2
π
π£2
+π§+
= ππππ π‘
Ι£
2π
Example
Piezometric tubes are tapped into a venturi
section as shown in the figure. The liquid is
incompressible. The upstream piezometric
head is 1 m, and the piezometric head at the
throat is 0.5 m. The velocity in the throat
section is twice large as in the approach
section. Find the velocity in the throat section.
Solution
π£12
π£22
β1 +
= β2 +
2βπ
2βπ
Where v2=2*v1, h1=1m, h2=0.5m
π£12
(2 β π£1)2
1+
= 0.5 +
2 β 9.81
2 β 9.81
3 β π£12
0.5 =
2 β 9.81
v1 = 1.808m/s , v2=3.62m/s
Example
A open tank filled with water
and drains through a port at
the bottom of the tank. The
elevation of the water in the
tank is 10 m above the drain.
The drain port is at
atmospheric pressure. Find
the velocity of the liquid in
the drain port.
Solution
π1
π£12 π2
π£22
+ π§1 + Ο β
=
+ π§2 + Ο β
Ι£
2π
Ι£
2π
Where P1=P2=Patm, v1=zero, z1-z2=10m
10= 1000 β
π£22
2β9.81
Then π£2 = 10 β 2 β 9.81 = 14π/π
Rate of flow
Discharge (Volume flow rate):General equation π = π£ β ππ΄
Note that we take only the velocity
component normal to the area.
Laminar flow eqn π = π£ππ£ β π΄
Where vav is the average velocity
Mass flow rate
π. =
Ο β π£ β ππ΄
For constant density across the flow
π. = Ο
π£ β ππ΄
π. = Ο β π
Example
The water velocity in the channel shown in
the accompanying figure has a distribution
across the vertical section equal to u/umax =
(y/d)1/2. What is the discharge in the
channel if the water is 2 m deep (d = 2 m),
the channel is 5 m wide, and the maximum
velocity is 3 m/s?
Solution
π¦ 1/2
π’ = π’πππ₯ β ( )
π
π=
π=
π’πππ₯ β
π=
π β ππ΄
1 1/2
( ) *π¦1/2
π
1
1 2
3β
2
Q = 10.606 β
β 5 β ππ¦
β5β
1
π¦2
β ππ¦
π¦1/2 β ππ¦
2
Q = 10.606 β β π¦ 3/2
3
3
2
π3
Q = 10.606 β β 2 2 = 20
3
π
Example
A jet of water discharges into an open tank, and
water leaves the tank through an orifice in the
bottom at a rate of 0.003 m3/s. If the crosssectional area of the jet is 0.0025 m2 where the
velocity of water is 7 m/s, at what rate is water
accumulating in (or evacuating from) the tank?
Solution
π3
πππ = π΄1 β π£1 = 0.0025 β 7 = 0.0175
π
π3
πππ’π‘ = 0.003
π
π3
πππ β πππ’π‘ = 0.0175 β 0.003 = 0.0145
π
πΎπ
.
π = Ο β πππ β πππ’π‘ = 14.5
π
Example
A 10 cm jet of water issues from a 1 m
diameter tank. Assume that the velocity in
the jet is 2πβ m/s where h is the
elevation of the water surface above the
outlet jet. How long will it take for the
water surface in the tank to drop from h0 =
2 m to hf = 0.50 m?
Solution
πππ’π‘ =
ππ
π£ β ππ΄ =
ππ‘
ππ
π΄π‘ β πβ
ππ‘ =
=
πππ’π‘
2 β π β β β π΄π
Ξ
π΄π‘ = β π·π‘ 2 = 0.785 π2
4
Ξ
π΄π = β π·π2 = 0.00785 π2
4
π‘
β
π΄π‘
πβ
ππ‘ =
β
2 β π β π΄π
0
β0 β
π‘=
π‘=
2 β π΄π‘
2 β π β π΄π
2 β 0.785
β
2 β 9.81 β 0.00785
t= 31.9 s
β0 β β
β
2 β 0.5
Example
Water with a density of 1000 kg/m3 flows
through a vertical venturimeter as shown. A
pressure gage is connected across two taps in
the pipe (station 1) and the throat (station 2).
The area ratio Athroat/Apipe is 0.5. The
velocity in the pipe is 10 m/s. Find the
pressure difference recorded by the pressure
gage.
Solution
π1 = π2 = π΄1 β π£1 = π΄2 β π£2
π΄1 π£2
π£2
=
=2=
π΄2 π£1
10
20π
π£2 =
π
π£12
π£22
β1 +
= β2 +
2βπ
2βπ
π£22 β π£12 202 β 102
β1 β β2 =
=
= 15.3π
2βπ
2 β 9.8
π1 β π2 = Ο β π β β1 β β2 = 150πππ
Rotation and Irrotation flow
Assume 2-D flow (x,y)
Ξ²
ΞΈ = + ΞΈπ΄
2
Ξ
Ξ² = + ΞΈπ΅ β ΞΈπ΄
2
Ξ (ΞΈπ΄ + ΞΈπ΅)
ΞΈ= +
4
2
The rotational rate
(ΞΈΚΉπ΄ + ΞΈΚΉπ΅)
ΞΈΚΉ =
2
If ΞΈΚΉ=0 the flow is irrotational.
The rotational rate in terms of velocity.
1
ππ€
ππ£
Ξ©π₯ = β ( β )
2
ππ¦
ππ§
1
ππ’
ππ€
Ξ©π¦ = β ( β )
2
ππ§
ππ₯
1
ππ£
ππ’
Ξ©π§ = β ( β )
2
ππ₯
ππ¦
Vorticity (w) w=2*Ξ©