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)