1. Fluid Mechanics 05

2. Rotating Flow
In the rotating flow ∆l = ∆r and the acceleration will be only in the normal direction.
− 𝑑 𝑑𝑟 𝑃+ɣ𝑧 =ρ∗ − 𝑣 2 𝑟
Where 𝑣=𝑤∗𝑟
𝑑 𝑑𝑟 𝑃+ɣ𝑧 =ρ∗ 𝑤 2 ∗𝑟
𝑃+ɣ𝑧− ρ∗ 𝑤 2 ∗ 𝑟 2 2 =Const

3. 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 A′ A′ shows the surface profile after rotation has been established. Find the elevation difference between the liquid at the center and the wall during rotation.

5. Solution 𝑃1+ɣ1𝑧1− ρ1∗ 𝑤1 2 ∗ 𝑟1 2 2 =𝑃2+ɣ𝑧2− ρ2∗ 𝑤2 2 ∗ 𝑟2 2 2
𝑃1+ɣ1𝑧1− ρ1∗ 𝑤1 2 ∗ 𝑟 =𝑃2+ɣ𝑧2− ρ2∗ 𝑤2 2 ∗ 𝑟2 2 2
Where P1=P2=Patm=Zero gauge, r1=0, r2=0.25 m, w=4 rad/s, ρ=1000Kg/m^3, g=9.81m/s^2
𝑧2−𝑧1= 𝑤 2 ∗ 𝑟2 2 2∗𝑔 = 4 2 ∗ ∗9.81 =0.051𝑚

6. Bernoulli Equation
We apply Bernoulli equation of the tangent component of acceleration.
− 𝑑 𝑑𝑠 𝑃+ɣ𝑧 =ρ∗𝑎𝑡
Where 𝑎𝑡=𝑣∗ 𝑑𝑣 𝑑𝑠 + 𝑑𝑣 𝑑𝑡
Assume the flow is steady then
𝑑𝑣 𝑑𝑡 =0 𝑡ℎ𝑒𝑛 𝑎𝑡=𝑣∗ 𝑑𝑣 𝑑𝑠 = 𝑑( 𝑣 2 2 ) 𝑑𝑠

7. − 𝑑 𝑑𝑠 𝑃+ɣ𝑧 =ρ∗ 𝑑 𝑑𝑠 ( 𝑣 2 2 )
𝑑 𝑑𝑠 𝑃+ɣ𝑧+ρ∗ 𝑣 =0
𝑃+ɣ𝑧+ρ∗ 𝑣 2 2 =Const
𝑃 ɣ +𝑧+ 𝑣 2 2𝑔 =𝑐𝑜𝑛𝑠𝑡

9. 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.

11. Solution ℎ1+ 𝑣1 2 2∗𝑔 =ℎ2+ 𝑣2 2 2∗𝑔 Where v2=2*v1, h1=1m, h2=0.5m
ℎ1+ 𝑣1 2 2∗𝑔 =ℎ2+ 𝑣2 2 2∗𝑔
Where v2=2*v1, h1=1m, h2=0.5m
1+ 𝑣1 2 2∗9.81 =0.5+ (2∗𝑣1) 2 2∗9.81
0.5= 3∗ 𝑣1 2 2∗9.81
v1 = 1.808m/s , v2=3.62m/s

12. 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.

13. Solution 𝑃1 ɣ +𝑧1+ρ∗ 𝑣1 2 2𝑔 = 𝑃2 ɣ +𝑧2+ρ∗ 𝑣2 2 2𝑔
𝑃1 ɣ +𝑧1+ρ∗ 𝑣1 2 2𝑔 = 𝑃2 ɣ +𝑧2+ρ∗ 𝑣2 2 2𝑔
Where P1=P2=Patm, v1=zero, z1-z2=10m
10=1000∗ 𝑣2 2 2∗9.81
Then 𝑣2= 10∗2∗9.81 =14𝑚/𝑠

14. Rate of flow Discharge (Volume flow rate):- General equation 𝑄= 𝑣∗𝑑𝐴
Note that we take only the velocity component normal to the area.
Laminar flow eqn 𝑄=𝑣𝑎𝑣∗𝐴
Where vav is the average velocity

15. Mass flow rate
𝑚 . = ρ∗𝑣∗𝑑𝐴
For constant density across the flow
𝑚 . =ρ 𝑣∗𝑑𝐴
𝑚 . =ρ∗𝑄

16. 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?

17. Solution 𝑢=𝑢𝑚𝑎𝑥∗ ( 𝑦 𝑑 ) 1/2 𝑄= 𝑉∗𝑑𝐴
𝑄= 𝑉∗𝑑𝐴
𝑄= 𝑢𝑚𝑎𝑥∗ ( 1 𝑑 ) 1/2 * 𝑦 1/2 ∗ 5∗𝑑𝑦
𝑄= 3∗ ∗5∗ 𝑦 ∗𝑑𝑦
Q=10.606∗ 𝑦 1/2 ∗𝑑𝑦

18. Example Q=10.606∗ 2 3 ∗𝑦 3/2 Q=10.606∗ 2 3 ∗ 2 3 2 =20 𝑚 3 𝑠
A jet of water discharges into an open tank, and water leaves the tank through an orifice in the bottom at a rate of m3/s. If the cross-sectional area of the jet is m2 where the velocity of water is 7 m/s, at what rate is water accumulating in (or evacuating from) the tank?

20. Solution 𝑄𝑖𝑛=𝐴1∗𝑣1=0.0025∗7=0.0175 𝑚 3 𝑠 𝑄𝑜𝑢𝑡=0.003 𝑚 3 𝑠
𝑄𝑖𝑛=𝐴1∗𝑣1=0.0025∗7= 𝑚 3 𝑠
𝑄𝑜𝑢𝑡= 𝑚 3 𝑠
𝑄𝑖𝑛−𝑄𝑜𝑢𝑡=0.0175−0.003= 𝑚 3 𝑠
𝑚 . = ρ∗ 𝑄𝑖𝑛−𝑄𝑜𝑢𝑡 =14.5 𝐾𝑔 𝑠

21. Example
A 10 cm jet of water issues from a 1 m diameter tank. Assume that the velocity in the jet is 2𝑔ℎ 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?

23. Solution 𝑄𝑜𝑢𝑡= 𝑣∗𝑑𝐴 = 𝑑𝑉 𝑑𝑡 𝑑𝑡= 𝑑𝑉 𝑄𝑜𝑢𝑡 = 𝐴𝑡∗𝑑ℎ 2∗𝑔∗ℎ ∗𝐴𝑛
𝑄𝑜𝑢𝑡= 𝑣∗𝑑𝐴 = 𝑑𝑉 𝑑𝑡
𝑑𝑡= 𝑑𝑉 𝑄𝑜𝑢𝑡 = 𝐴𝑡∗𝑑ℎ 2∗𝑔∗ℎ ∗𝐴𝑛
𝐴𝑡= Π 4 ∗ 𝐷𝑡 2 = 𝑚 2
𝐴𝑛= Π 4 ∗ 𝐷𝑛 2 = 𝑚 2
0 𝑡 𝑑𝑡= 𝐴𝑡 2∗𝑔 ∗𝐴𝑛 ∗ ℎ0 ℎ 𝑑ℎ ℎ

24. 𝑡= 2∗𝐴𝑡 2∗𝑔 ∗𝐴𝑛 ∗ ℎ0 − ℎ
𝑡= 2∗ ∗9.81 ∗ ∗ 2 − 0.5
t= 31.9 s

25. 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.

27. Solution 𝑄1=𝑄2=𝐴1∗𝑣1=𝐴2∗𝑣2 𝐴1 𝐴2 = 𝑣2 𝑣1 =2= 𝑣2 10 𝑣2= 20𝑚 𝑠
ℎ1+ 𝑣1 2 2∗𝑔 =ℎ2+ 𝑣2 2 2∗𝑔
ℎ1−ℎ2= 𝑣2 2 − 𝑣1 2 2∗𝑔 = − ∗9.8 =15.3𝑚
𝑃1−𝑃2=ρ∗𝑔∗ ℎ1−ℎ2 =150𝑘𝑝𝑎

28. Rotation and Irrotation flow
Assume 2-D flow (x,y)
θ= β 2 +θ𝐴
β= Π 2 +θ𝐵−θ𝐴
θ= Π 4 + (θ𝐴+θ𝐵) 2

29. The rotational rate
θʹ= (θʹ𝐴+θʹ𝐵) 2
If θʹ=0 the flow is irrotational.
The rotational rate in terms of velocity.
Ω𝑥= 1 2 ∗( 𝑑𝑤 𝑑𝑦 − 𝑑𝑣 𝑑𝑧 )
Ω𝑦= 1 2 ∗( 𝑑𝑢 𝑑𝑧 − 𝑑𝑤 𝑑𝑥 )
Ω𝑧= 1 2 ∗( 𝑑𝑣 𝑑𝑥 − 𝑑𝑢 𝑑𝑦 )
Vorticity (w) w=2*Ω