Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
------------------------------------------------------------------------------------------------------------------------------------------
The exam is closed book and closed notes.
1. A Newtonian fluid of density ρ and viscosity μ fills the annular gap
between concentric cylinders as shown blow. The inner cylinder of
radius R1 is stationary, and the outer cylinder of radius R2 rotates steadily
at a constant angular speed ω. The flow between the cylinders is laminar
and incompressible. Assume steady, purely circulating motion, 2D flow,
circumferentially symmetric pressure, and neglect gravity. Simplify the
continuity and momentum equations and use appropriate boundary
conditions to determine the velocity distribution in the gap.
The equations of motion of an incompressible Newtonian fluid with constant
density and viscosity in cylindrical coordinates (r, θ, z) with velocity
components (vr, vθ, vz):
Continuity:
1 𝜕
1 𝜕
𝜕
(𝑟𝑣𝑟 ) +
(𝑣𝜃 ) + (𝑣𝑧 ) = 0
𝑟 𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
r-momentum:
𝜕𝑣𝑟
𝜕𝑣𝑟 𝑣𝜃 𝜕𝑣𝑟
𝜕𝑣𝑟 𝑣𝜃2
𝜕𝑝
𝜕 1 𝜕
1 𝜕 2 𝑣𝑟 𝜕 2 𝑣𝑟 2 𝜕𝑣𝜃
(𝑟𝑣𝑟 )) + 2
𝜌(
+ 𝑣𝑟
+
+ 𝑣𝑧
− ) = 𝜌𝑔𝑟 −
+𝜇[ (
+
−
]
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
𝜕𝑟
𝜕𝑟 𝑟 𝜕𝑟
𝑟 𝜕𝜃 2
𝜕𝑧 2 𝑟 2 𝜕𝜃
θ-momentum:
𝜕𝑣𝜃
𝜕𝑣𝜃 𝑣𝜃 𝜕𝑣𝜃
𝜕𝑣𝜃 𝑣𝑟 𝑣𝜃
1 𝜕𝑝
𝜕 1 𝜕
1 𝜕 2 𝑣𝜃 𝜕 2 𝑣𝜃 2 𝜕𝑣𝑟
(𝑟𝑣𝜃 )) + 2
𝜌(
+ 𝑣𝑟
+
+ 𝑣𝑧
+
) = 𝜌𝑔𝜃 −
+𝜇[ (
+
+ 2
]
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
𝑟 𝜕𝜃
𝜕𝑟 𝑟 𝜕𝑟
𝑟 𝜕𝜃 2
𝜕𝑧 2
𝑟 𝜕𝜃
z-momentum:
𝜕𝑣𝑧
𝜕𝑣𝑧 𝑣𝜃 𝜕𝑣𝑧
𝜕𝑣𝑧
𝜕𝑝
1 𝜕
𝜕𝑣𝑧
1 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧
𝜌(
+ 𝑣𝑟
+
+ 𝑣𝑧
) = 𝜌𝑔𝑧 −
+𝜇[
(𝑟
)+ 2
+
]
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝑧
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜃 2
𝜕𝑧 2
2. As wind blows over a chimney, vortices are shedding in the wake as shown in the Figure below.
The dimensional shedding frequency f depends on chimney diameter D, chimney length L, wind
velocity V, and air kinematic viscosity ν. (a) Find dimensionless f which depends on dimensionless
groups. If a 1/10th scale model were to be tested in a wind tunnel and full dynamic similarity was
required: (b) what air velocity would be necessary in the wind tunnel compared to the wind velocity
experienced by the full-scale chimney?; (c) what shedding frequency would be observed in the
wind tunnel compared to the shedding frequency generated by the full-scale chimney?
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
------------------------------------------------------------------------------------------------------------------------------------------
3. A tank of water with depth h is to be drained by a 5-cm-diameter exit pipe. Water density is 998
kg/m3, water viscosity is 0.001 kg/ms. The pipe extends out for 15 m and a turbine and an open
globe valve are located on the pipe. The head provided by the turbine is ht = 10 m. (a) If the exit
flow rate is Q = 0.04 m3/s, calculate h assuming there are no minor losses, the turbine is 100%
efficient, and the pipe is smooth. (b) Calculate Q if h is same as part (a) but there are minor losses
(K = 0.5 for the sharp entrance and K = 6.9 for the open globe valve), the turbine has an efficiency
of 80%, and the pipe is rough with ε = 0.3 mm. Use the value of f from part (a) as initial guess and
stop at the end of the second iteration.
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
-----------------------------------------------------------------------------------------------------------------------------------------Solution 1:
(1)
ASSUMPTIONS:
𝜕
1. Steady flow (𝜕𝑡 = 0)
2. Purely circumferential flow (vr=vz=0)
𝜕
3. 2D flow (𝜕𝑧 = 0)
𝜕𝑝
4. circumferentially symmetric pressure (𝜕𝜃 = 0)
5. Negligible gravity
ANALYSIS:
Continuity:
1 𝜕
1 𝜕
𝜕
(𝑟𝑣𝑟 ) +
(𝑣𝜃 ) + (𝑣𝑧 ) = 0
𝑟 𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
0(2) +
𝜕𝑣𝜃
𝜕𝜃
θ-momentum:
𝜌(
1 𝜕
(𝑣𝜃 )
𝑟 𝜕𝜃
+ 0(2) = 0
(2)
=0
(6)
𝜕𝑣𝜃
𝜕𝑣𝜃 𝑣𝜃 𝜕𝑣𝜃
𝜕𝑣𝜃 𝑣𝑟 𝑣𝜃
+ 𝑣𝑟
+
+ 𝑣𝑧
+
)
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
1 𝜕𝑝
𝜕 1 𝜕
1 𝜕 2 𝑣𝜃 𝜕 2 𝑣𝜃 2 𝜕𝑣𝑟
(𝑟𝑣𝜃 )) + 2
= 𝜌𝑔𝜃 −
+𝜇[ (
+
+ 2
]
𝑟 𝜕𝜃
𝜕𝑟 𝑟 𝜕𝑟
𝑟 𝜕𝜃 2
𝜕𝑧 2
𝑟 𝜕𝜃
𝜌(0(1) + 0(2) + 0(6) + 0(2,3) + 0(2)) = 0(5) − 0(4) + 𝜇 [
𝜕 1 𝜕
(𝑟𝑣𝜃 )) + 0(6) + 0(3) + 0(2)]
(
𝜕𝑟 𝑟 𝜕𝑟
𝜕 1 𝜕
(𝑟𝑣𝜃 )) = 0
(
𝜕𝑟 𝑟 𝜕𝑟
(2)
Integrate:
1 𝜕
(𝑟𝑣𝜃 ) = 𝐶1
𝑟 𝜕𝑟
𝜕
(𝑟𝑣𝜃 ) = 𝐶1 𝑟
𝜕𝑟
(1)
Integrate again:
𝑟2
+ 𝐶2
2
𝑟 𝐶2
𝑣𝜃 = 𝐶1 +
2 𝑟
𝑟𝑣𝜃 = 𝐶1
B.C.:
1) At r = R1, vθ = 0
2) At r = R2, vθ = R2ω
(1)
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
-----------------------------------------------------------------------------------------------------------------------------------------Replace:
𝑅1 𝐶2
+
2 𝑅1
𝑅2 𝐶2
𝑅2 𝜔 = 𝐶1
+
2 𝑅2
0 = 𝐶1
And find:
𝐶1 =
𝐶2 =
2𝜔
(1)
𝑅 2
1 − ( 1)
𝑅2
−𝜔𝑅1 2
(1)
𝑅 2
1 − (𝑅1 )
2
Substitute into the expression for 𝑣𝜃 :
𝑣𝜃 =
𝜔𝑅1
1−
𝑟
(
𝑅1 2 𝑅1
(𝑅 )
2
−
𝑅1
)
𝑟
(1)
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
------------------------------------------------------------------------------------------------------------------------------------------
Solution 2:
ASSUMPTIONS: the problem is only a function of the above dimensional variables
ANALYSIS:
(a)
𝑓 = 𝑓𝑐𝑛(𝐷, 𝐿, 𝑉, 𝜈 ) ;
𝑓 = {𝑇 −1 }
𝐷 = {𝐿}
n= = 5
𝐿 = {𝐿} 𝑉 = {𝐿𝑇 −1 }
∴𝑘 =𝑛−𝑗 =3
𝜈 = {𝐿2 𝑇 −1 } ;
(0.5)
𝑟𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = 𝐷, 𝑉
𝜋1 = 𝑓𝐷 𝑎1 𝑉𝑏1 = {(𝑇 −1 )(𝐿)𝑎1 (𝐿𝑇 −1 )𝑏1 } = {𝐿0 𝑇 0 }
𝑎1 = 1; 𝑏1 = −1
𝜋1 =
𝑓𝐷
𝑉
(1.5)
𝜋2 = 𝐿𝐷 𝑎2 𝑉𝑏2 = {(𝐿)(𝐿)𝑎2 (𝐿𝑇 −1 )𝑏2 } = {𝐿0 𝑇 0 }
𝑎2 = −1; 𝑏2 = 0
𝜋2 =
𝐿
𝐷
(1.5)
𝜋3 = 𝜈𝐷 𝑎3 𝑉𝑏3 = {(𝐿2 𝑇 −1 )(𝐿)𝑎3 (𝐿𝑇 −1 )𝑏3 } = {𝐿0 𝑇 0 }
𝑎3 = −1; 𝑏3 = −1
𝜋3 =
𝑓𝐷
𝑉
𝜈
𝑉𝐷
(1.5)
𝐿 𝜈
)
𝐷 𝑉𝐷
= 𝑓𝑐𝑛 ( ,
(b)
𝜈𝑝
𝜈𝑚
𝑉𝑚 𝜈𝑚 𝐷𝑝
=
,
=
𝑉𝑚 𝐷𝑚 𝑉𝑝 𝐷𝑝
𝑉𝑝
𝜈𝑝 𝐷𝑚
𝑉𝑚
= (1)(10) = 10
𝑉𝑝
(2)
(c)
𝑓𝑚 𝐷𝑚 𝑓𝑝 𝐷𝑝
=
,
𝑉𝑚
𝑉𝑝
𝑓𝑚 𝐷𝑝 𝑉𝑚
=
𝑓𝑝 𝐷𝑚 𝑉𝑝
𝑓𝑚
= (10)(10) = 100
𝑓𝑝
Solution 3:
(2)
j=2
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
-----------------------------------------------------------------------------------------------------------------------------------------ANALYSIS:
Energy equation
between free-surface (1) and exit (2):
𝑝
𝑉2
𝑝
𝑉2
+ 𝑧) = ( +
+ 𝑧) + ℎ𝑓 − ℎ𝑝 + ℎ𝑡
( +
𝜌𝑔 2𝑔
𝜌𝑔 2𝑔
1
2
𝑝1 = 𝑝2 = 𝑝𝑎𝑡𝑚
𝑧1 − 𝑧2 = ℎ
𝑉1 = 0; ℎ𝑝 = 0
ℎ𝑓 =
𝑉22
𝐿
(𝑓 + ∑ 𝐾)
2𝑔 𝐷
(1.5)
Replace and find h:
𝑉22
𝐿
ℎ=
(1 + 𝑓 + ∑ 𝐾) + ℎ𝑡
2𝑔
𝐷
(1.5)
Find velocity using the flow rate and then Re:
𝑄
(0.04 m3 /s)
𝑉2 = 𝜋
=𝜋
= 20.4 m/s
𝐷2
(0.05 m)2
4
4
(1)
𝜌𝑉2 𝐷 (998 kg/m3 )(20.4 m/s)(0.05 m)
𝑅𝑒 =
=
= 1.02E6 (𝑡𝑢𝑟𝑏. )
𝜇
(0.001 kg/ms)
(0.5)
(a)
Find the friction factor from the moody diagram using Re:
𝑓𝑠𝑚𝑜𝑜𝑡ℎ ~0.011
ℎ=
(0.5)
𝑉22
𝐿
(1 + 𝑓𝑠𝑚𝑜𝑜𝑡ℎ ) + ℎ𝑡
2𝑔
𝐷
(1)
m 2
(20.4 s )
(15 m)
(0.011)
ℎ=
[1
+
] + (10 m) = 101 m
m
(0.05 m)
(2) (9.81 2 )
s
(1)
Name: ------------------
Exam 2
Time: 60 minutes
ME:5160
Fall 2019
-----------------------------------------------------------------------------------------------------------------------------------------(b)
Consider minor losses, turbine efficiency, and roughness of the pipe with h from part (a).
ℎ=
𝑉22
𝐿
ℎ𝑡
(1 + 𝑓𝑟𝑜𝑢𝑔ℎ + 𝐾𝑒𝑛𝑡 + 𝐾𝑣𝑎𝑙𝑣𝑒 ) +
2𝑔
𝐷
𝜂
(0.5)
Find an expression of 𝑉2 as a function of 𝑓𝑟𝑜𝑢𝑔ℎ :
ℎ
10
2𝑔 (ℎ − 𝜂𝑡 )
(2)(9.81) (101 −
)
0.8
𝑉2 = √
=√
𝐿
(15)
1 + 𝑓𝑟𝑜𝑢𝑔ℎ 𝐷 + 𝐾𝑒𝑛𝑡 + 𝐾𝑣𝑎𝑙𝑣𝑒
1 + 𝑓𝑟𝑜𝑢𝑔ℎ
+ 0.5 + 6.9
(0.05)
1736.37
𝑉2 = √
300𝑓𝑟𝑜𝑢𝑔ℎ + 8.4
(0.5)
Use 𝑓𝑠𝑚𝑜𝑜𝑡ℎ as initial guess to compute new velocity and Re:
1736.37
m
𝑉2 = √
= 12.18
300(0.011) + 8.4
s
→
𝑅𝑒 = 6.08E5
Find the friction factor from the Moody diagram using Re and relative roughness and iterate twice.
𝜀 (0.0003 m)
=
= 0.006
𝐷
(0.05 m)
(0.5)
Iteration 1:
𝑓𝑟𝑜𝑢𝑔ℎ ~0.032
1736.37
m
𝑉2 = √
= 9.82
300(0.032) + 8.4
s
→
𝑅𝑒 = 4.90E5
(0.5)
Iteration 2:
𝑓𝑟𝑜𝑢𝑔ℎ ~0.032 (converged)
(0.5)
Compute flow rate:
𝜋
m 𝜋
𝑄 = 𝑉2 ( 𝐷 2 ) = (9.82 ) (0.05 m)2 = 0.019 m3 /s
4
s 4
Compared to part (a), the flow rate is reduced by 52.5% due to additional losses.
(0.5)
0
