Post on 19-Jan-2021
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
The exam is closed book and closed notes.
1. A helicopter rotor rotates at 𝜔 =20.94 rad/s in air (ρ=1.2 kg/m3 and μ=1.8E-5 kg/m-s). Each
blade has a chord length of 53 cm and extends a distance of 7.3 m from the center of the rotor hub.
Assume that the blades can be modeled as very thin flat plates at a zero angle of attack. (a) At what
radial distance from the hub center is the flow at the blade trailing edge turbulent (Recrit = 5E5).
(b) Find the boundary layer thickness at the blade tip trailing edge (c) At what rotor angular
velocity does the wall shear stress at the blade tip trailing edge become 80 N/m2?
Hint : 𝑅𝑒 =𝜌𝑈𝐶
𝜇 where 𝑈 = 𝑟𝜔 , 𝐶: 𝐶ℎ𝑜𝑟𝑑 𝐿𝑒𝑛𝑔𝑡ℎ
Turbulent BL : 𝛿
𝑥≈
0.16
𝑅𝑒1/7 , 𝑐𝑓 =2𝜏𝑤
𝜌𝑈2 ≈0.027
𝑅𝑒𝑥1/7
2. A parachute of a new design is tested in standard air (ρ=1.2 kg/m3 and μ=1.8E-5 kg/m-s) with a
total weight of the load and parachute of 200 N. The diameter for the tested prototype is 5 m. The
results showed that the parachute reaches a constant velocity of 3 m/s. (a) Use the prototype data
and find the drag coefficient for the parachute. (b) If you are to repeat the experiment for a 2.5
times smaller model using Reynolds similarity, and the weight of the model parachute comes out
as 40 N, how much load do you need to add?
(Hint : 𝐶𝐷 =𝐷𝑟𝑎𝑔
1
2𝜌𝑉2𝐴
, 𝑅𝑒 = 𝜌𝑉𝐷
𝜇 )
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
3. 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.
Hint : 𝜓𝑎 = 𝐴𝑥𝑦 corresponds to 𝜓 = 𝐴(𝑟 cos 𝜃)(𝑟 sin 𝜃) =𝐴
2𝑟2 sin 2𝜃 in Cylindrical Coordinates
𝜓𝑏 =𝐴
2𝑟2 sin 2𝜃 + 𝑚𝜃
4. A viscous, incompressible fluid flows between two infinite, vertical, parallel plates distance h
apart, as shown in the Figure. Use the given coordinate system and assume that the flow is
laminar, steady, fully developed (𝜕𝑢
𝜕𝑥= 0), and planar (𝑤 = 0;
𝜕
𝜕𝑧= 0). Pressure and gravity are
not negligible. (a) Simplify the continuity equation and show that 𝑣 = 0. (b) Using the y-
momentum equation show that pressure is only a function of x. (c) Find the velocity distribution
𝑢(𝑦) in terms of 𝜇, 𝜌, 𝑔,𝑑𝑝
𝑑𝑥, and ℎ.
Incompressible Continuity Equation: 𝜕𝑢
𝜕𝑥+
𝜕𝑣
𝜕𝑦+
𝜕𝑤
𝜕𝑧= 0
Incompressible Navier-Stokes Equations in Cartesian Coordinates:
𝜌 (𝜕𝑢
𝜕𝑡+ 𝑢
𝜕𝑢
𝜕𝑥+ 𝑣
𝜕𝑢
𝜕𝑦+ 𝑤
𝜕𝑢
𝜕𝑧) = 𝜌𝑔𝑥 −
𝜕𝑝
𝜕𝑥+ 𝜇 (
𝜕2𝑢
𝜕𝑥2 +𝜕2𝑢
𝜕𝑦2 +𝜕2𝑢
𝜕𝑧2)
𝜌 (𝜕𝑣
𝜕𝑡+ 𝑢
𝜕𝑣
𝜕𝑥+ 𝑣
𝜕𝑣
𝜕𝑦+ 𝑤
𝜕𝑣
𝜕𝑧) = 𝜌𝑔𝑦 −
𝜕𝑝
𝜕𝑦+ 𝜇 (
𝜕2𝑣
𝜕𝑥2 +𝜕2𝑣
𝜕𝑦2 +𝜕2𝑣
𝜕𝑧2)
𝜌 (𝜕𝑤
𝜕𝑡+ 𝑢
𝜕𝑤
𝜕𝑥+ 𝑣
𝜕𝑤
𝜕𝑦+ 𝑤
𝜕𝑤
𝜕𝑧) = 𝜌𝑔𝑧 −
𝜕𝑝
𝜕𝑧+ 𝜇 (
𝜕2𝑤
𝜕𝑥2 +𝜕2𝑤
𝜕𝑦2 +𝜕2𝑤
𝜕𝑧2 )
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
5. Consider an experiment in which the drag on a two-dimensional body immersed in a steady
incompressible flow can be determined from measurement of velocity distribution far upstream
and downstream of the body as shown in Figure below. Velocity far upstream is the uniform flow
𝑈∞, and that in the wake of the body is measured to be 𝑢(𝑦) =𝑈∞
2(
𝑦2
𝑏2 + 1), which is less than 𝑈∞
due to the drag of the body. Assume that there is a stream tube with inlet height of 2H and outlet
height of 2b as shown in Figure below. (a) Determine the relationship between H and b using the
continuity equation. (b) Find the drag per unit length of the body as a function of 𝑈∞, b and 𝜌.
(Hint : Momentum Equation ∑𝐹𝑥 = ∫ 𝑢𝜌(𝑉 ⋅ 𝑛)𝑑𝐴 )
6. The parallel galvanized-iron pipe system (𝜖 = 0.15 𝑚𝑚) delivers water at 200C (𝜌 =
998 𝑘𝑔/𝑚3 and 𝜇 = 0.001 𝑘𝑔/𝑚 ⋅ 𝑠) with a total flow rate of 0.036 m3/s. (a) Find out the relation
between 𝑉1 and 𝑉2. If the pump is wide open and not running, with a loss coefficient of K=1.5,
(b) determine the velocity in each pipe(𝑉1 and 𝑉2). Use 𝑓1 = 𝑓2 = 0.02 for your initial guess.
(Hint : ℎ𝑓 = 𝑓𝐿
𝑑
𝑉2
2𝑔 )
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 1:
a)
𝑈 = 𝑟𝜔
𝑅𝑒 =𝜌𝑈𝑐
𝜇=
𝜌𝑟𝜔𝑐
𝜇
𝑅𝑒𝑐𝑟𝑖𝑡 =𝜌𝑟𝑐𝑟𝑖𝑡𝜔𝑐
𝜇
𝑟𝑐𝑟𝑖𝑡 =𝑅𝑒𝑐𝑟𝑖𝑡𝜇
𝜌𝜔𝑐=
(5𝐸5)(1.8𝐸 − 5)
(1.2)(20.94)(0.53)= 0.68 𝑚
b) At the tip trailing edge:
𝑅𝑒 =𝜌𝑟𝑡𝑖𝑝𝜔𝑐
𝜇=
(1.2)(7.3)(20.94)(0.53)
(1.8𝐸 − 5)= 5,401,124
𝛿
𝑥≈
0.16
𝑅𝑒1/7
𝛿 ≈0.16(0.53)
(5401124)17
= 0.00926 𝑚 = 9.26 𝑚𝑚
c)
𝜏𝑤 = 0.0135𝜌𝑈2
𝑅𝑒𝑥1/7
=0.0135𝜇1/7𝜌6/7𝑈13/7
𝑥1/7
Re-arrange to find U:
𝑈 = 𝑟𝑡𝑖𝑝𝜔 = (𝜏𝑤 𝑥1/7
0.0135𝜇1/7𝜌6/7)
7/13
𝜔 =1
(7.3)(
(80.0)(0.53)1/7
0.0135 (1.8𝐸 − 5)1/7(1.2)6/7)
7/13
= 29.88 𝑟𝑎𝑑/𝑠
(1)
(1)
(1)
(2)
(1)
(2)
(1)
(1)
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 2:
(a)
𝐶𝐷 =𝐷
12
𝜌𝑉2𝐴
𝐴 =𝜋𝐷2
4
If in equilibrium at constant velocity, then:
𝐷 = 𝑊
𝐶𝐷 =𝑊
12 𝜌𝑉2𝐴
=(200)
12
(1.2)(3)2 𝜋4
(5)2= 1.89
(b)
To repeat the experiment, it has to be designed to reach same Reynolds number as prototype:
𝑅𝑒𝑚 = 𝑅𝑒𝑝
𝑉𝑚𝐷𝑚
𝜈𝑚=
𝑉𝑝𝐷𝑝
𝜈𝑝 → 𝑉𝑚 = 𝑉𝑝
𝐷𝑝
𝐷𝑚= 𝑉𝑝𝜆 = (3)(2.5) = 7.5 𝑚/𝑠
At this speed the drag coefficient should be the same as prototype since 𝐶𝐷 = 𝑓(𝑅𝑒). Therefore:
𝐶𝐷𝑚 = 𝐶𝐷𝑝 = 1.89
𝑊𝑡𝑜𝑡𝑎𝑙 = 𝐶𝐷
1
2𝜌𝑉2𝐴 = (1.89)(0.5)(1.2)(7.5)2
𝜋
4(5/2.5)2 = 200 𝑁
𝑊𝑙𝑜𝑎𝑑 = 𝑊𝑡𝑜𝑡𝑎𝑙 − 𝑊𝑃𝑎𝑟𝑎𝑐ℎ𝑢𝑡𝑒 = 200 − 40 = 160
(1)
(1)
(1)
(1)
(2)
(2)
(1)
(1)
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 3:
𝜓 =𝐴
2𝑟2 sin 2𝜃 + 𝑚𝜃
𝑣𝜃 = −𝜕𝜓
𝜕𝑟= −𝐴𝑟 sin 2𝜃
𝑣𝑟 =1
𝑟
𝜕𝜓
𝜕𝜃= 𝐴𝑟 cos 2𝜃 +
𝑚
𝑟
For the bump, the stagnation point occurs at:
𝑟 = ℎ, 𝜃 =𝜋
2
(𝑣𝜃)𝑠𝑡𝑎𝑔 = −𝐴ℎ sin 𝜋 = −𝐴ℎ (0) = 0
(𝑣𝑟)𝑠𝑡𝑎𝑔 = 𝐴ℎ cos 𝜋 +𝑚
ℎ= 𝐴ℎ (−1) +
𝑚
ℎ= 0
𝐴ℎ =𝑚
ℎ ⇒ ℎ = √
𝑚
𝐴
(2)
(1)
(1)
(1)
(2)
(2)
(1)
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 4:
(a)
Continuity: 𝜕𝑢
𝜕𝑥+
𝜕𝑣
𝜕𝑦+
𝜕𝑤
𝜕𝑧= 0
0(3) +𝜕𝑣
𝜕𝑦+ 0(4) = 0
𝜕𝑣
𝜕𝑦= 0 ⇒ 𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑣 = 0 𝑎𝑡 𝑦 =ℎ
2 𝑎𝑛𝑑 𝑦 = −
ℎ
2 ⇒ 𝑣 = 0 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 (5)
(b) y - Momentum:
𝜌 (𝜕𝑣
𝜕𝑡+ 𝑢
𝜕𝑣
𝜕𝑥+ 𝑣
𝜕𝑣
𝜕𝑦+ 𝑤
𝜕𝑣
𝜕𝑧) = 𝜌𝑔𝑦 −
𝜕𝑝
𝜕𝑦+ 𝜇 (
𝜕2𝑣
𝜕𝑥2+
𝜕2𝑣
𝜕𝑦2+
𝜕2𝑣
𝜕𝑧2)
𝜌(0(1) + 0(5) + 0(5) + 0(4)) = 0 −𝜕𝑝
𝜕𝑦+ 𝜇(0(5) + 0(5) + 0(5))
𝜕𝑝
𝜕𝑦= 0 ⇒ 𝑝 = 𝑝(𝑥)
(c) x- Momentum:
𝜌 (𝜕𝑢
𝜕𝑡+ 𝑢
𝜕𝑢
𝜕𝑥+ 𝑣
𝜕𝑢
𝜕𝑦+ 𝑤
𝜕𝑢
𝜕𝑧) = 𝜌𝑔𝑥 −
𝜕𝑝
𝜕𝑥+ 𝜇 (
𝜕2𝑢
𝜕𝑥2+
𝜕2𝑢
𝜕𝑦2+
𝜕2𝑢
𝜕𝑧2)
𝜌(0(1) + 0(3) + 0(5) + 0(4)) = −𝜌𝑔 −𝑑𝑝
𝑑𝑥+ 𝜇 (0(3) +
𝜕2𝑢
𝜕𝑦2+ 0(4))
𝜕2𝑢
𝜕𝑦2=
1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
Integrate: 𝜕𝑢
𝜕𝑦=
1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥) 𝑦 + 𝐶1
Integrate again:
𝑢 =1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
𝑦2
2+ 𝐶1𝑦 + 𝐶2
Boundary conditions:
𝑢 = 0 𝑎𝑡 𝑦 =ℎ
2 𝑎𝑛𝑑 𝑦 = −
ℎ
2
0 =1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
ℎ2
8+ 𝐶1
ℎ
2+ 𝐶2
0 =1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
ℎ2
8− 𝐶1
ℎ
2+ 𝐶2
𝐶1 = 0
𝐶2 = −1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
ℎ2
8
Replace and find final solution:
𝑢 =1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
𝑦2
2−
1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥)
ℎ2
8 ⇒ 𝑢 =
1
𝜇(𝜌𝑔 +
𝑑𝑝
𝑑𝑥) (
𝑦2
2−
ℎ2
8)
(1.5)
(1)
(1)
(1)
(1)
(1)
(0.5)
(1)
(1)
(1)
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 5: a) Continuity:
2𝜌𝐻𝑈∞ = 𝜌 ∫ 𝑢(𝑦)𝑑𝑦𝑏
−𝑏
= 𝜌 ∫𝑈∞
2(
𝑦2
𝑏2+ 1) 𝑑𝑦
𝑏
−𝑏
2𝜌𝐻𝑈∞ = 𝜌𝑈∞
2∫ (
𝑦2
𝑏2+ 1) 𝑑𝑦
𝑏
−𝑏
= 𝜌𝑈∞
2(
𝑦3
3𝑏2+ 𝑦)]
−𝑏
𝑏
2𝐻 =1
2(
𝑏3
3𝑏2+ 𝑏 +
𝑏3
3𝑏2+ 𝑏) =
1
2(
8
3𝑏) =
4
3𝑏
𝐻 =2𝑏
3
b) x-momentum:
∑𝐹𝑥 = ∫ 𝑢𝜌(𝑉 ⋅ 𝑛)𝑑𝐴
Drag per unit length:
−𝐹𝐷 = −𝜌𝑈∞2 (2𝐻) + 𝜌 ∫ 𝑢2(𝑦)𝑑𝑦
𝑏
−𝑏
𝐹𝐷 = 𝜌𝑈∞2 (2𝐻) − 𝜌 ∫ [
𝑈∞
2(
𝑦2
𝑏2+ 1)]
2
𝑑𝑦𝑏
−𝑏
= 𝜌𝑈∞2 (2𝐻) − 𝜌
𝑈∞2
4∫ (
𝑦2
𝑏2+ 1)
2
𝑑𝑦𝑏
−𝑏
Calculating integral:
∫ (𝑦2
𝑏2+ 1)
2
𝑑𝑦𝑏
−𝑏
= ∫ (𝑦4
𝑏4+
2𝑦2
𝑏2+ 1) 𝑑𝑦
𝑏
−𝑏
=𝑦5
5𝑏4+
2𝑦3
3𝑏2+ 𝑦]
−𝑏
𝑏
= 2 (𝑏
5+
2𝑏
3+ 𝑏) =
56
15𝑏
Entering into the momentum equation:
𝐹𝐷 = 2𝜌𝐻𝑈∞2 −
1
4𝜌𝑈∞
2 (56
15𝑏) = 𝜌𝑈∞
2 (4
3𝑏 −
14
15𝑏) = 𝜌𝑈∞
22𝑏
5
(2)
(1)
(1)
(1)
(2)
(1)
(1)
(1)
Name: ------------------ Final Exam Time: 120 minutes
ME:5160 Fall 2019 ------------------------------------------------------------------------------------------------------------------------------------------
Solution 6:
a) Continuity:
𝑄1 + 𝑄2 =𝜋
4𝑑1
2𝑉1 +𝜋
4𝑑2
2𝑉2 = 𝑄𝑡𝑜𝑡𝑎𝑙; 𝑉2 =4
𝜋𝑑22 𝑄𝑡𝑜𝑡𝑎𝑙 −
𝑑12
𝑑22 𝑉1
𝑉2 =4
𝜋0.0420.036 −
0.052
0.042 𝑉1
𝑉2 = 28.65 − 1.56 𝑉1
(b)
Same head loss for parallel pipes:
ℎ𝑓1 = ℎ𝑓2 + ℎ𝑚2
𝑓1
𝐿1
𝑑1
𝑉12
2𝑔−
𝑉22
2𝑔(𝑓2
𝐿2
𝑑2+ 𝐾) = 0
𝑓1
60
0.05
𝑉12
2 × 9.81−
𝑉22
2 × 9.81(𝑓2
55
0.04+ 1.5) = 0
61.16𝑓1𝑉12 − (28.65 − 1.56 𝑉1)2(70.08𝑓2 + 0.076) = 0
Reynolds Number:
𝑅𝑒1 =𝜌𝑉1𝐷1
𝜇=
998 × 0.05
0.001𝑉1 = 49900 𝑉1
𝑅𝑒2 =𝜌𝑉1𝐷1
𝜇=
998 × 0.04
0.001𝑉2 = 39920𝑉2
Relative roughness:
𝜖
𝐷1=
0.15
50= 0.003
𝜖
𝐷2=
0.15
40= 0.00375
Guessing 𝑓1 = 𝑓2 = 0.02
𝑓1 = 0.02, 𝑓2 = 0.02 → 𝑉1 = 11.59 → 𝑉2 = 10.54 → 𝑅𝑒1 = 57800, 𝑅𝑒2 = 421000
𝑓1 = 0.0264, 𝑓2 = 0.0282 → 𝑉1 = 11.69 → 𝑉2 = 10.37
∴ 𝑉1 = 11.69 𝑚/𝑠
(1)
(1)
(2)
(1)
(0.5)
(1)
(0.5)
(0.5)
(0.5)
(1)
(1)