Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------The exam is closed book and closed notes.
1. An incompressible, viscous fluid with density, 𝜌, flows past a solid flat plate which has a depth,
𝑏, into the page. The flow initially has a uniform velocity 𝑈 before contacting the plate. The
2𝑦
𝑦 2
velocity profile at location 𝑥 is estimated to have a parabolic shape, 𝑢 = 𝑈 �� � − � � �, for
𝛿
𝛿
𝑦 ≤ 𝛿 and 𝑢 = 𝑈 for 𝑦 ≥ 𝛿 where 𝛿 is the boundary layer thickness. (a) Write the continuity
equation and determine the upstream height from the plate, ℎ, of a streamline which has a height,
𝛿, at the downstream location. Express your answer in terms of 𝛿. (b) Determine the force the
fluid exerts on the plate over the distance 𝑥. Express your answer in terms of 𝜌, 𝑈, 𝑏, and 𝛿. You
may assume that the pressure everywhere is atmospheric pressure.
2. An incompressible fluid flows between two porous, parallel flat plates as shown in the Figure
below. An identical fluid is injected at a constant speed V through the bottom plate and
simultaneously extracted from the upper plate at the same velocity. There is no gravity force in x
and y directions (gx=gy=0). Assume the flow to be steady, fully-developed, 2D, and the pressure
𝜕𝑝
gradient in the x direction to be a constant ( = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡). (a) Write the continuity equation and
𝜕𝑥
show that the y velocity is constant at 𝑣 = 𝑉. (b) Simplify the x-momentum equation and find the
appropriate differential equation for the x velocity component, u. (c) To solve the differential
𝜕𝑝 1
equation, assume that the solution is 𝑢 = 𝐶1 𝑒 𝜆𝑦 − � � 𝑦 + 𝐶2 , where 𝜆 ≠ 0. Replace and
𝜕𝑥 𝜌𝑉
find λ in terms of ρ, V, and μ. (d) Apply boundary conditions and find C1 and C2.
V
y=h
y=0
3. A model scale of a glass sphere is suspended in an upward flow of water moving with a mean
velocity of 1 m/s. The density of the glass is 2360 kg/m3, water density is 1000 kg/m3, and water
viscosity is μ=0.001 kg/m-s. (a) If drag coefficient for sphere is CD ≈ 0.2 for turbulent flow
(𝑅𝑒𝐷 > 5 × 105 ) and CD=0.47 for laminar flow (1 × 104 < 𝑅𝑒𝐷 < 5 × 105 ), calculate the
diameter of the model scale sphere. (b) What would be the water velocity and the drag force for 8
times larger prototype?
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------4. The following data were obtained for flow of 20°C water (ρ=998 kg/m3, μ=0.001 kg/m-s) at 20
m3/hr through a badly corroded 5-cm-diameter pipe with slopes downward at an angle of 8°: p1 =
420 kPa, z1 = 12 m, p2 = 250 kPa, z2 = 3 m. Estimate: (a) the roughness of the pipe ε; and (b) the
percent change in head loss if the pipe were smooth and the flow rate the same.
∆𝑧
(Hint: pipe length 𝐿 =
)
sin 𝜃
5. A small bug rests on the outside of a car side window as shown in the Figure below. The
surrounding air has a density of ρ=1.2 kg/m3 and viscosity of μ=1.8E-5 kg/m-s. Assume that the
flow can be approximated as flat plate flow with no pressure gradient and the start of the
boundary layer begins at the leading edge of the window. (a) Assuming that the flow is turbulent
where the bug is, determine the minimum speed at which the bug will be sheared off of the car
window if the bug can resist a shear stress of up to 1 N/m2. (b) Confirm the turbulent flow
assumption. (c) What is the total skin friction drag acting on the window at this speed?
2𝜏
0.027
0.031
(Turbulent BL: 𝑐𝑓 = 𝜌𝑈𝑤2 ≈ 1/7 ; 𝐶𝐷 ≈ 1/7 )
𝑅𝑒𝑥
𝑅𝑒𝐿
6. Potential flow against a flat plate (Fig. a) can be described with the stream function 𝜓 = 𝐴𝑥𝑦
where A is a constant. This type of flow is commonly called a “stagnation point” flow since it can
be used to describe the flow in the vicinity of the stagnation point at O. By adding a source of
strength m at O (𝜓 = 𝑚𝜃), stagnation point flow against a flat plate with a “bump” is obtained as
illustrated in Fig. b. Determine the bump height, h, as a function of the constant, A, and the source
strength, m.
𝐴
(Hint: 𝜓 = 𝐴𝑥𝑦 in Cylindrical Coordinates is 𝜓 = 𝐴(𝑟 cos 𝜃)(𝑟 sin 𝜃) = 𝑟 2 sin 2𝜃)
2
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
------------------------------------------------------------------------------------------------------------------------------------------
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
------------------------------------------------------------------------------------------------------------------------------------------
Solution:
1.
(a) Continuity Equation:
ℎ
𝛿
(3)
� 𝑈 𝑏 𝑑𝑦 = � 𝑢(𝑦) 𝑏 𝑑𝑦
0
0
𝛿
𝑦 2
2𝑦
𝑏ℎ𝑈 = � 𝑏𝑈 �� � − � � � 𝑑𝑦
𝛿
𝛿
0
= 𝑏𝑈 �
𝛿
2𝛿
𝑦2 𝑦3
− 2 � = 𝑏𝑈 � �
𝛿 3𝛿 0
3
(1.5)
(0.5)
2
2
𝑏ℎ𝑈 = 𝑏𝑈𝛿 ⇒ ℎ = 𝛿 (0.5)
3
3
(b) Linear Momentum Equation in x Direction:
� 𝐹𝑥 = − � 𝑢(𝑦)𝜌[𝑢(𝑦)] 𝑏 𝑑𝑦 + � 𝑢(𝑦)𝜌[𝑢(𝑦)] 𝑏 𝑑𝑦 (2)
1
ℎ
𝛿
2
−𝐹 = − � 𝜌𝑈 2 𝑏 𝑑𝑦 + � 𝜌𝑢2 (𝑦) 𝑏 𝑑𝑦
0
0
𝛿
−𝐹 = −𝜌𝑈 2 𝑏ℎ + 𝜌𝑏 � 𝑢2 (𝑦)𝑑𝑦
𝛿
𝛿
2
= −𝜌𝑈2 𝑏ℎ + 𝜌𝑏𝑈 2 � �
0
0
(1)
(0.5)
2
2𝑦 𝑦 2
(0.5)
− 2 � 𝑑𝑦
𝛿
𝛿
𝛿
𝛿
4𝑦 2 𝑦 4 4𝑦 3
4𝑦 3 𝑦 5 𝑦 4
4
1
8
2𝑦 𝑦 2
𝛿
� � − 2 � 𝑑𝑦 = � � 2 + 4 − 3 � 𝑑𝑦 = � 2 + 4 − 3 � = 𝛿 + 𝛿 − 𝛿 =
𝛿
𝛿
𝛿
𝛿
3𝛿
5𝛿
𝛿 0 3
5
15
𝛿
0
0
−𝐹 = −𝜌𝑈 2 𝑏ℎ + 𝜌𝑏𝑈 2 �
8
𝛿�
15
2
8
= −𝜌𝑈 2 𝑏 � 𝛿� + 𝜌𝑏𝑈 2 � 𝛿�
3
15
∴ 𝐹=
2
𝜌𝑈 2 𝑏𝛿
15
(0.5)
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------2.
Assumptions:
𝜕
1) Steady flow ( = 0)
𝜕𝑡
2) Incompressible flow (ρ=constant)
𝜕𝑢
3) Fully developed ( = 0)
4) 2D flow (𝑤 = 0,
𝜕𝑥
𝜕
𝜕𝑧
= 0)
𝜕𝑝
5) Constant pressure gradient ( = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝜕𝑥
6) gx=gy=0
(a)
Continuity:
𝜕𝑢 𝜕𝑣 𝜕𝑤
+
+
=0
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝜕𝑣
𝜕𝑦
(b)
0(3) +
𝜕𝑣
𝜕𝑦
(1.5)
+ 0(4) = 0
= 0 ⇒ 𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(0.5)
(0.5)
𝑣 = 𝑉 𝑎𝑡 𝑦 = 0 𝑎𝑛𝑑 𝑦 = ℎ ⇒ 𝑣 = 𝑉 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 (0.5)
x-momentum:
𝜕𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑝
𝜕2𝑢 𝜕2𝑢 𝜕2𝑢
𝜕𝑢
+𝑣
+ 𝑤 � = 𝜌𝑔𝑥 −
+ 𝜇 � 2 + 2 + 2�
𝜌� +𝑢
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑡
𝜌 �0(1) + 0(3) + 𝑉
(c)
𝜕𝑢
𝜕𝑦
+ 0(4)� = 0(6) −
𝜌𝑉
𝜕𝑝
𝜕𝑥
+ 𝜇 �0(3) +
𝜕𝑝
𝜕2𝑢
𝜕𝑢
=−
+𝜇 2
𝜕𝑥
𝜕𝑦
𝜕𝑦
(0.5)
𝜕𝑝 1
𝑢 = 𝐶1 𝑒 𝜆𝑦 − � �
𝑦 + 𝐶2
𝜕𝑥 𝜌𝑉
𝜕𝑢
𝜕𝑦
𝜕𝑝 1
𝜕𝑥 𝜌𝑉
= 𝐶1 𝜆𝑒 𝜆𝑦 − � �
𝜕2𝑢
= 𝐶1 𝜆2 𝑒 𝜆𝑦
𝜕𝑦 2
(0.5)
𝜕2 𝑢
𝜕𝑦 2
+ 0(4)�
(2)
(0.5)
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------Replace in the differential equation:
𝜕𝑝
𝜕𝑝 1
𝜌𝑉 �𝐶1 𝜆𝑒 𝜆𝑦 − � � � = −
+ 𝜇�𝐶1 𝜆2 𝑒 𝜆𝑦 �
𝜕𝑥
𝜕𝑥 𝜌𝑉
⇒ 𝜆=
Therefore:
(d)
𝜌𝑉
𝜇
(0.5)
𝜌𝑉
𝜕𝑝 1
𝑦
𝑢 = 𝐶1 𝑒 𝜇 − � �
𝑦 + 𝐶2
𝜕𝑥 𝜌𝑉
Boundary conditions:
𝑢 = 0 𝑎𝑡 𝑦 = 0 𝑎𝑛𝑑 𝑦 = ℎ (1.5)
𝑦 = 0: 0 = 𝐶1 + 𝐶2 ⇒ 𝐶2 = −𝐶1
𝑦 = ℎ: 0 = 𝐶1 𝑒
𝜌𝑉ℎ
𝜇
𝜕𝑝 ℎ
𝜕𝑥 𝜌𝑉
−� �
𝐶1 �1 − 𝑒
⇒ 𝐶1 = −
𝜌𝑉ℎ
𝜇 �
+ 𝐶2 = 𝐶1 𝑒
Replace find C1 and C2 to find the final solution:
𝜕𝑝 ℎ
𝜕𝑥 𝜌𝑉
−� �
− 𝐶1
𝜕𝑝 ℎ
= −� �
𝜕𝑥 𝜌𝑉
𝜕𝑝 ℎ
� � 𝜌𝑉
𝜕𝑥
1−
𝜌𝑉ℎ
𝜇
(0.5)
𝜌𝑉ℎ
𝑒 𝜇
, 𝐶2 =
𝜌𝑉𝑦
�
𝜕𝑝 ℎ
�
𝜕𝑥 𝜌𝑉
1−
𝜌𝑉ℎ
𝑒 𝜇
ℎ 𝜕𝑝 1 − 𝑒 𝜇
𝑦
𝑢=
� ��
− �
𝜌𝑉ℎ
𝜌𝑉 𝜕𝑥
ℎ
1−𝑒 𝜇
(0.5)
(0.5)
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------3.
(a)
B FD
𝑊 = 𝜌𝑔𝑙𝑎𝑠𝑠 𝑔𝑉
(0.5)
𝐵 = 𝜌𝑤𝑎𝑡𝑒𝑟 𝑔𝑉
1
𝐹𝐷 = 𝐶𝐷 𝜌𝑤𝑎𝑡𝑒𝑟 𝐴𝑈 2 (0.5)
2
W
𝑉=
𝑊 = 𝐵 + 𝐹𝐷 ⇒ 𝜌𝑔𝑙𝑎𝑠𝑠 𝑔
𝜋𝐷 3
,
6
𝜋𝐷3
6
�𝜌𝑔𝑙𝑎𝑠𝑠 − 𝜌𝑤𝑎𝑡𝑒𝑟 �𝑔
𝐷=
Assume turbulent:
𝐷=
𝑅𝑒𝐷 =
(0.5)
𝐴=
𝜋𝐷 2
4
= 𝜌𝑤𝑎𝑡𝑒𝑟 𝑔
𝜋𝐷3
6
1
2
+ 𝐶𝐷 𝜌𝑤𝑎𝑡𝑒𝑟
𝜋𝐷 3
1
𝜋𝐷 2 2
= 𝐶𝐷 𝜌𝑤𝑎𝑡𝑒𝑟
𝑈
6
4
2
𝜋𝐷2 2
𝑈
4
(2)
3𝐶𝐷 𝜌𝑤𝑎𝑡𝑒𝑟 𝑈 2
4�𝜌𝑔𝑙𝑎𝑠𝑠 − 𝜌𝑤𝑎𝑡𝑒𝑟 �𝑔
3(0.2)(1000)(1)2
= 0.0112 𝑚
4(2360 − 1000)(9.81)
𝑈𝐷 (1)(0.0112)
=
= 11,200 = 1.12 × 104 < 5 × 105 ⇒ 𝑇𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑣𝑎𝑙𝑖𝑑 (0.5)
(1𝐸 − 6)
𝜈
Assume laminar:
(0.5)
𝐷=
(0.5)
𝑅𝑒𝐷 =
(b)
(1)
3(0.47)(1000)(1)2
= 0.0264 𝑚
4(2360 − 1000)(9.81)
𝑈𝐷 (1)(0.0264)
=
= 264,000 = 2.64 × 105
(1𝐸 − 6)
𝜈
𝐷𝑝
𝑅𝑒𝑝 = 𝑅𝑒𝑚 ⇒
𝐷𝑚
𝑶𝑲, 𝑳𝒂𝒎𝒊𝒏𝒂𝒓 (0.5)
𝑈𝑝 𝐷𝑝 𝑈𝑚 𝐷𝑚
1
𝐷𝑚
=
⇒ 𝑈𝑝 = 𝑈𝑚
= (1.0) � � = 0.125 𝑚/𝑠 (0.5)
𝜈
𝜈
𝐷𝑝
8.0
(1) 𝐶 = 𝐶
𝐷𝑝
𝐷𝑚 ⇒
𝐹𝐷 𝑝
= 8.0
(0.5)
𝐹𝐷 𝑝
1
𝜌𝐴 𝑈 2
2 𝑝 𝑝
=
𝐹𝐷 𝑚
1
𝜌𝐴 𝑈 2
2 𝑚 𝑚
𝐷𝑝 2 𝑈𝑝 2
𝐴𝑝 𝑈𝑝 2
1 2
2
= 𝐹𝐷 𝑚
� � = 𝐹𝐷 𝑚 � � � � = 𝐹𝐷 𝑚 (8.0) � � = 𝐹𝐷 𝑚
𝐴𝑚 𝑈𝑚
𝐷𝑚
𝑈𝑚
8.0
(1)
𝜋(0.0264)2
1
𝜋𝐷𝑚 2
(1.0)2 = 0.13 𝑁 (0.5)
= 𝐶𝐷 𝜌𝑤𝑎𝑡𝑒𝑟
𝑈𝑚 2 = (0.47)(0.5)(1000)
4
4
2
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------4.
(a)
The flow rate:
(0.5)
20
𝑄 = 20 𝑚3 /ℎ =
𝑚3 /𝑠 = 0.00556 𝑚3 /𝑠
3600
The average velocity:
(0.0056)
𝑄
𝑄
(0.5)
=𝜋
= 2.83 𝑚/𝑠
𝑉= =𝜋
𝐴
(0.05)2
𝑑2
4
4
The pipe length:
∆𝑧
12 − 3
(0.5)
𝐿=
=
= 64.7 𝑚
sin 𝜃
sin 8°
The steady flow energy equation:
�
𝑝
𝑉2
𝑝
𝑉2
+
+ 𝑧� = � +
+ 𝑧� + ℎ𝑓 (2)
𝜌𝑔 2𝑔
𝜌𝑔 2𝑔
1
2
𝑝
𝑝
(0.5)
𝑉1 = 𝑉2 ⇒ � + 𝑧� = � + 𝑧� + ℎ𝑓
𝜌𝑔
𝜌𝑔
1
2
(420,000)
(250,000)
+ 12 =
+ 3 + ℎ𝑓 (0.5)
(998)(9.81)
(998)(9.81)
(1)
ℎ𝑓 = 𝑓
ℎ𝑓 = 26.36 𝑚
(64.7) (2.83)2
𝐿 𝑉2
= 26.36 = 𝑓
⇒ 𝑓 = 0.05 (0.5)
(0.05) 2(9.81)
𝑑 2𝑔
(0.5)
𝜌𝑉𝑑 (998)(2.83)(0.05)
=
= 141,217 ≈ 1.4 × 105
𝑅𝑒𝑑 =
(0.001)
𝜇
From Moody chart, read: (1)
𝜀
≈ 0.021
𝑑
𝜀 = (0.021)(0.05) = 0.00105 𝑚 = 1.05 𝑚𝑚 (0.5)
(b)
From Moody chart at the same 𝑅𝑒𝑑 = 1.4 × 105 the friction factor smooth pipe reads:
𝑓𝑠𝑚𝑜𝑜𝑡ℎ = 0.017
Percent change is:
(1)
(0.5)
(64.7) (2.83)2
𝐿 𝑉2
= (0.017)
= 8.98 𝑚
�ℎ𝑓 �𝑠𝑚𝑜𝑜𝑡ℎ = 𝑓
(0.05) 2(9.81)
𝑑 2𝑔
(26.36) − (8.98)
× 100 ≈ 66% (0.5)
(26.36)
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------5.
(a)
𝑈2
(0.5)
𝑐
𝜏𝑤 = 𝜌
2 𝑓
𝑐𝑓 ≈
𝜏𝑤 = 𝜌
=𝜌
=
Re-arrange to find U:
(1)
(c)
𝑅𝑒𝑥 =
(0.5)
1/7
𝑅𝑒𝑥
𝑈 2 0.027 (1)
2 𝑅𝑒 1/7
𝑥
(1)
𝑈 2 0.027
1/7
2 𝜌𝑈𝑥
�
�
𝜇
0.0135𝜇 1/7 𝜌6/7 𝑈13/7
𝑥 1/7
𝜏𝑤 𝑥 1/7
𝑈=�
�
0.0135𝜇 1/7 𝜌6/7
(1.0)(0.4)1/7
=�
�
0.0135 (1.8𝐸 − 5)1/7 (1.2)6/7
(b)
(1)
0.027
7/13
7/13
= 20.16 𝑚/𝑠
(0.5)
𝜌𝑈𝑥 (1.2)(20.16)(0.4)
=
= 537,600 > 5 × 105 𝑂𝐾, 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
(1.8𝐸 − 5)
𝜇
𝐶𝐷 ≈
0.031
1/7
𝑅𝑒𝐿
(1)
(1)
𝜌𝑈𝐿 (1.2)(20.16)(1.0)
𝑅𝑒𝐿 =
=
= 1,344,000
(1.8𝐸 − 5)
𝜇
𝐶𝐷 =
0.031
= 0.00413
(1344000)1/7
1
𝐷 = 𝐶𝐷 𝜌𝑈 2 𝑏𝐿 (1)
2
= (0.00413)(0.5)(1.2)(20.16)2 (0.7)(1.0) = 0.7 𝑁 (0.5)
(1)
Fluids-ID:-----------------
Final Exam
Time: 120 minutes
Name: -----------------Course: 58:160, Fall 2013
-----------------------------------------------------------------------------------------------------------------------------------------6.
𝐴
𝜓 = 𝑟 2 sin 2𝜃 + 𝑚𝜃 (2.5)
2
𝑣𝜃 = −
𝑣𝑟 =
𝜕𝜓
= −𝐴𝑟 sin 2𝜃 (1)
𝜕𝑟
1 𝜕𝜓
𝑚
= 𝐴𝑟 cos 2𝜃 + (1)
𝑟 𝜕𝜃
𝑟
For the bump, the stagnation point occurs at:
𝑟 = ℎ,
𝜃=
𝜋
2
(1)
(1) (𝑣 )
𝜃 𝑠𝑡𝑎𝑔 = −𝐴ℎ sin 𝜋 = −𝐴ℎ (0) = 0 (1)
(1) (𝑣𝑟 )𝑠𝑡𝑎𝑔 = 𝐴ℎ cos 𝜋 +
𝐴ℎ =
𝑚
𝑚
= 𝐴ℎ (−1) + = 0 (1)
ℎ
ℎ
𝑚
𝑚
⇒ ℎ=�
ℎ
𝐴
(0.5)