AAE 333 Fluid Mechanics Homework #2

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AAE 333 Fluid Mechanics

Fall 2020

Homework #2

Due: Friday 09/11/2020 (3 problems worth 100 points) 1.

[40 pts] A large water tank empties slowly through a small hole under the force of gravity. The flow is steady, and the volumetric flow rate 𝑞𝑞̇ (in m3/s) depends on the exit velocity V, the gravitational acceleration g, the depth of the water h, and the diameter of the hole D. the viscosity µ, and the surface tension σ (force/unit length). (a) Express the nondimensional volumetric flow rate 𝑞𝑞̇ in terms of its dependence on the other nondimensional groups. (b) A test is to be made on a 1/4th scale model. What is the ratio of the model volumetric flow rate to the prototype volumetric flow rate that would be needed to obtain dynamical similarity? (c) (i) If you now decide that, instead of the volumetric flow rate, you are interested in the mass flow rate 𝑚𝑚̇ , and the density 𝜌𝜌 must therefore be included in the dimensional analysis, will the number of nondimensional groups change? (ii) Without going through the full derivation again, use the results of part (a) to express the nondimensional mass flow rate 𝑚𝑚̇ in terms of its dependence on the other nondimensional groups. (d) If you later discover that the mass flow rate 𝑚𝑚̇ depends on the viscosity 𝜇𝜇 and the surface tension 𝜎𝜎 (dimensions of force per unit length), in addition to 𝜌𝜌, 𝑉𝑉, 𝑔𝑔, ℎ, and 𝐷𝐷, find the mass flow rate 𝑚𝑚̇ in terms of its dependence on the other nondimensional groups. Hint: you may avoid the full Buckingham Pi derivation by using the nondimensional groups from part (c) and compiling additional groups with “new” variables and “old” repeating variables simply by inspection.

Solution: (a) 𝑞𝑞̇ = 𝑓𝑓(𝑉𝑉, 𝑔𝑔, ℎ, 𝐷𝐷). N=5 variables.

Denote L=length, T=time, we have L3 L L [𝑞𝑞̇ ] = [ℎ] = L; [𝐷𝐷] = L; [𝑉𝑉] = ; [𝑔𝑔] = 2 .

T T T Of the three fundamental units (mass, length, and time), mass is not present here, so K=2.

Therefore, we should have N-K=3 nondimensional groups.

Select 𝐷𝐷, 𝑉𝑉 as repeating variables. 𝐷𝐷 Show that we can obtain each fundamental unit with these two variables: [𝐷𝐷] = L; �𝑉𝑉 � = T. Form the nondimensional Pi products: Π1 = [𝐷𝐷𝑎𝑎1 𝑉𝑉𝑏𝑏1 ℎ] = 𝐿𝐿𝑎𝑎1 (𝐿𝐿𝑇𝑇 −1 )𝑏𝑏1 𝐿𝐿 = 1 => 𝑎𝑎1 + 𝑏𝑏1 + 1 = 0; 𝑏𝑏1 = 0 => 𝑎𝑎1 = −1 ℎ => Π1 = . 𝐷𝐷 Π2 = [𝐷𝐷 𝑎𝑎2 𝑉𝑉𝑏𝑏2 𝑔𝑔] = 𝐿𝐿𝑎𝑎2 (𝐿𝐿𝑇𝑇 −1 )𝑏𝑏2 𝐿𝐿𝑇𝑇 −2 = 1 => 𝑎𝑎2 + 𝑏𝑏2 + 1 = 0; −𝑏𝑏2 − 2 = 0 𝑔𝑔𝑔𝑔 => 𝑎𝑎2 = 1; 𝑏𝑏2 = −2 => Π2 = 𝐷𝐷𝑉𝑉 −2 𝑔𝑔 = 2 . 𝑉𝑉

AAE 333 Fluid Mechanics

Fall 2020

Π3 = [𝐷𝐷𝑎𝑎3 𝑉𝑉𝑏𝑏3 𝑞𝑞̇ ] = 𝐿𝐿𝑎𝑎3 (𝐿𝐿𝑇𝑇 −1 )𝑏𝑏3 𝐿𝐿3 𝑇𝑇 −1 = 1 => 𝑎𝑎3 + 𝑏𝑏3 + 3 = 0; −𝑏𝑏3 − 1 = 0 𝑞𝑞̇ => 𝑎𝑎3 = −2; 𝑏𝑏3 = −1 => Π3 = 𝑉𝑉𝐷𝐷2 Therefore, 𝒒𝒒̇ 𝒉𝒉 𝒈𝒈𝒈𝒈 = 𝒇𝒇𝟏𝟏 � , 𝟐𝟐 � 𝟐𝟐 𝑽𝑽𝑫𝑫 𝑫𝑫 𝑽𝑽 Note that once we have nondimensional groups, we can form other nondimensional groups. For 𝐷𝐷 ℎ 𝑔𝑔𝑔𝑔 instance, 1/Π1 = ℎ can be used instead of Π1 = 𝐷𝐷 . Similarly, instead of Π2 = 𝑉𝑉 2 , we can use 𝑉𝑉 2

1/Π2 = 𝑔𝑔𝑔𝑔 or �1/Π2 = 𝑔𝑔𝑔𝑔

𝑉𝑉

�𝑔𝑔𝑔𝑔

ℎ

, or, since Π1 = 𝐷𝐷 is a non-dimensional group, we can use, instead 𝑔𝑔ℎ

1

𝑉𝑉 2

1

of Π2 = 𝑉𝑉 2 , a combination Π1 Π2 = 𝑉𝑉 2 , or Π Π = 𝑔𝑔ℎ or �Π Π = 1 2

1 2

𝑉𝑉

�𝑔𝑔ℎ

.

Note also that a nondimensional number known as the Froude number (Fr) is given by 𝐹𝐹𝐹𝐹 = so one alternative answer is

𝑉𝑉

�𝑔𝑔ℎ

,

𝒒𝒒̇ 𝑫𝑫 𝑽𝑽 𝑫𝑫 = 𝒇𝒇 � , � = 𝒇𝒇 � , 𝑭𝑭𝑭𝑭� 𝟐𝟐 𝟐𝟐 𝑽𝑽𝑫𝑫𝟐𝟐 𝒉𝒉 �𝒈𝒈𝒈𝒈 𝒉𝒉 (b) Dynamic similarity requires that all the nondimensional parameters take the same values in the model and prototype. Therefore 𝑞𝑞̇ 𝑝𝑝 𝑞𝑞̇ 𝑚𝑚 2 = 𝑉𝑉𝑚𝑚 𝐷𝐷𝑚𝑚 𝑉𝑉𝑝𝑝 𝐷𝐷𝑝𝑝 2 Since 𝐷𝐷𝑝𝑝 /𝐷𝐷𝑚𝑚 =4, we have We also require

from which

Therefore, for dynamic similarity

𝑞𝑞̇ 𝑚𝑚 𝑉𝑉𝑚𝑚 = 𝑞𝑞̇ 𝑝𝑝 16𝑉𝑉𝑝𝑝 𝑔𝑔𝐷𝐷𝑚𝑚 𝑉𝑉𝑚𝑚

2 =

𝑔𝑔𝐷𝐷𝑝𝑝 𝑉𝑉𝑝𝑝 2

𝑉𝑉𝑚𝑚 𝐷𝐷𝑚𝑚 1 =� = 𝑉𝑉𝑝𝑝 𝐷𝐷𝑝𝑝 2 𝒒𝒒̇ 𝒎𝒎 𝟏𝟏 = 𝒒𝒒̇ 𝒑𝒑 𝟑𝟑𝟑𝟑

AAE 333 Fluid Mechanics

Fall 2020

(c) 𝑚𝑚̇ = 𝑓𝑓(𝑉𝑉, 𝑔𝑔, ℎ, 𝐷𝐷, 𝜌𝜌) (i) N=6 variables. Since all three fundamental units (mass, length, and time) are present, then K=3. Therefore, we should have N-K=3 nondimensional groups, same as before. (ii) We already have 2 nondimensional groups from part (a) that also work here:

Π1 =

ℎ

𝐷𝐷

and Π2 =

𝑔𝑔𝑔𝑔 𝑉𝑉 2

. We can also easily convert Π3 =

𝑞𝑞̇

𝑉𝑉𝐷𝐷2

by multiplying

and dividing by the density, since 𝑚𝑚̇ = 𝜌𝜌𝑞𝑞̇ , to get a new nondimensional group 𝜌𝜌𝑞𝑞̇

𝜌𝜌𝑉𝑉𝐷𝐷2

=

𝑚𝑚̇

𝜌𝜌𝑉𝑉𝐷𝐷2

Alternatively,

. Therefore, we have

𝒉𝒉 𝒈𝒈𝒈𝒈 𝒎𝒎̇ = 𝒇𝒇 � , � 𝟑𝟑 𝝆𝝆𝑽𝑽𝑫𝑫𝟐𝟐 𝑫𝑫 𝑽𝑽𝟐𝟐 𝑫𝑫 𝑽𝑽 𝑫𝑫 𝒎𝒎̇ = 𝒇𝒇𝟒𝟒 � , � = 𝒇𝒇𝟒𝟒 � , 𝑭𝑭𝑭𝑭� 𝟐𝟐 𝝆𝝆𝑽𝑽𝑫𝑫 𝒉𝒉 �𝒈𝒈𝒈𝒈 𝒉𝒉

(d) 𝑚𝑚̇ = 𝑓𝑓(𝜌𝜌, 𝑉𝑉, 𝑔𝑔, ℎ, 𝐷𝐷, 𝜇𝜇, 𝜎𝜎) N=8 variables. Since all three fundamental units (mass, length, and time) are present, then K=3. Therefore, we should have N-K=5 nondimensional groups. We can select 𝜌𝜌, 𝑉𝑉, 𝐷𝐷 as the set of repeating variables.

We already have 3 nondimensional groups from part (c) that also work here: and

𝑔𝑔𝑔𝑔 𝑉𝑉 2

.

𝑚𝑚̇

𝜌𝜌𝑉𝑉𝐷𝐷2

ℎ

, , 𝐷𝐷

Two additional groups should involve the viscosity and the surface tension. For viscosity, we know the famous nondimensional group: Reynolds number. In this case, we can select the Reynolds number based on D or on h. Both work. Let’s select 𝜌𝜌𝜌𝜌𝜌𝜌 𝑅𝑅𝑅𝑅𝐷𝐷 = 𝜇𝜇 Finally, we need a nondimensional group with the surface tension. With 𝜌𝜌, 𝑉𝑉, 𝐷𝐷 and 𝜎𝜎, we can compile the following nondimensional group (by inspection): 𝜎𝜎 𝜌𝜌𝑉𝑉 2 𝐷𝐷 Therefore, we have

AAE 333 Fluid Mechanics

Fall 2020 𝒎𝒎̇ 𝒉𝒉 𝒈𝒈𝒈𝒈 𝝆𝝆𝝆𝝆𝝆𝝆 𝝈𝝈 = 𝒇𝒇𝟓𝟓 � , 𝟐𝟐 , , � 𝟐𝟐 𝝆𝝆𝑽𝑽𝑫𝑫 𝑫𝑫 𝑽𝑽 𝝁𝝁 𝝆𝝆𝑽𝑽𝟐𝟐 𝑫𝑫

Alternatively,

𝒎𝒎̇ 𝑫𝑫 𝑽𝑽 𝝆𝝆𝝆𝝆𝝆𝝆 𝝈𝝈 𝑫𝑫 𝝈𝝈 = 𝒇𝒇𝟔𝟔 � , , , 𝟐𝟐 � = 𝒇𝒇𝟔𝟔 � , 𝑭𝑭𝑭𝑭, 𝑹𝑹𝑹𝑹, 𝟐𝟐 � 𝟐𝟐 𝝆𝝆𝑽𝑽𝑫𝑫 𝒉𝒉 �𝒈𝒈𝒈𝒈 𝝁𝝁 𝝆𝝆𝑽𝑽 𝑫𝑫 𝒉𝒉 𝝆𝝆𝑽𝑽 𝑫𝑫

2.

[30 pts] Tests on a model propeller in a wind tunnel at sea level (air density ρ= 1.2 kg/m3) gave the following results for the thrust at given forward velocities: Velocity (m/s)

0

10

20

30

Thrust (N)

300

278

211

100

The propeller diameter was 0.8 m and it was spun at 2000 rpm. (a) Using dimensional analysis, find the nondimensional parameters which govern this observed

behavior. (b) Using the experimental data given in the table, find the thrust generated by a geometrically similar propeller of diameter 3 m, spinning at 1500 rpm at a forward velocity of 45 m/s, while operating at an altitude where the density is half that at sea level. You may need to interpolate from tabulated values.

Solution: (a) 𝐹𝐹 = 𝑓𝑓(𝑉𝑉, 𝜌𝜌, 𝐷𝐷, 𝜈𝜈) Assumptions: ignore viscosity and sound speed

N=5 variables. Since all three fundamental units (mass, length, and time) are present, then K=3. Therefore, we should have N-K=2 nondimensional groups.

Select 𝜌𝜌, 𝑉𝑉, 𝐷𝐷 as the repeating variables.

Show that we can obtain each fundamental unit with these three variables: 𝐷𝐷 [𝐷𝐷] = L; � � = T; [𝜌𝜌𝐷𝐷3 ] = 𝑀𝑀 𝑉𝑉

Note that units of thrust [𝐹𝐹] = M ∙ L ∙ 𝑇𝑇 −2, and the units of rotational speed [𝜈𝜈] = 𝑇𝑇 −1.

Form the nondimensional Pi products:

AAE 333 Fluid Mechanics

Fall 2020

Π1 = [𝜌𝜌𝑎𝑎1 𝐷𝐷𝑏𝑏1 𝑉𝑉 𝑐𝑐1 𝐹𝐹] = (𝑀𝑀𝐿𝐿−3 )𝑎𝑎1 (𝐿𝐿)𝑏𝑏1 (𝐿𝐿𝑇𝑇 −1 )𝑐𝑐1 (ML𝑇𝑇 −2) = 1 => 𝑎𝑎1 + 1 = 0; −3𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 + 1 = 0; −𝑐𝑐1 − 2 = 0 => 𝑎𝑎1 = −1, 𝑐𝑐1 = −2 , 𝑏𝑏1 = −2 𝐹𝐹 => Π1 = 𝜌𝜌−1 𝐷𝐷−2 𝑉𝑉 −2 𝐹𝐹 = . 𝜌𝜌𝑉𝑉 2 𝐷𝐷2 Π2 = [𝜌𝜌𝑎𝑎2 𝐷𝐷𝑏𝑏2 𝑉𝑉 𝑐𝑐2 𝜈𝜈] = (𝑀𝑀𝐿𝐿−3 )𝑎𝑎2 (𝐿𝐿)𝑏𝑏2 (𝐿𝐿𝑇𝑇 −1 )𝑐𝑐2 (𝑇𝑇 −1 ) = 1 => 𝑎𝑎2 = 0; −3𝑎𝑎2 + 𝑏𝑏2 + 𝑐𝑐2 = 0; −𝑐𝑐2 − 1 = 0 => 𝑎𝑎2 = 0; 𝑏𝑏2 = −1; 𝑐𝑐2 = 1 𝜈𝜈𝜈𝜈 => Π2 = 𝐷𝐷−1 𝑉𝑉𝑉𝑉 = . 𝑉𝑉 Thus, 𝑭𝑭 𝝂𝝂𝝂𝝂 = 𝒇𝒇 � � 𝟏𝟏 𝝆𝝆𝑽𝑽𝟐𝟐 𝑫𝑫𝟐𝟐 𝑽𝑽 (b) We can use the test data in the table to find the function 𝑓𝑓1 and find the thrust of the

prototype by interpolation.

Velocity V (m/s)

0

10

20

30

Thrust F (N)

300

278

211

100

Π1

∞

3.62

0.687

0.145

∞

160

80

53

Π2

For prototype Π2 =

𝜈𝜈𝜈𝜈 1500 × 3 = = 100 𝑉𝑉 45

By interpolation,

Therefore,

Π1 =

𝐹𝐹 = 1.43 𝜌𝜌𝑉𝑉 2 𝐷𝐷2

𝐹𝐹 = 1.43 × 𝜌𝜌𝑉𝑉 2 𝐷𝐷2 = 1.43 ×

1.2 × 452 × 32 N 2

𝑭𝑭 = 𝟏𝟏𝟏𝟏, 𝟔𝟔𝟔𝟔𝟔𝟔 N

AAE 333 Fluid Mechanics

Fall 2020

3.

[30 pts] The lift force (F) produced by a fly depends on the wing beat frequency (f), its forward velocity (V), the density (ρ) and viscosity (µ) of the fluid, the length or “span” (S), and width or “chord” (c) of the wings, and the Young’s modulus of the wing material (E, dimensions of stress). (a) Express the non-dimensional lift force as a function of the other non-dimensional groups. (b) A fly is observed to travel at 1 m/s when it beats its wings at 120 Hz. If a dynamically similar robotic fly was built 100 times larger than a real fly, what would the forward velocity and wing beat frequency be if it was tested in silicon oil with a density equal to that of water, but with a viscosity 50 times that of water? (Assume µwater = 1×10-3 Pa·s) Solution: (a) Eight parameters are named, with the following units: F:

𝑀𝑀𝑀𝑀

µ:

𝑀𝑀

𝑇𝑇 2

𝐿𝐿𝐿𝐿

f:

1

𝐿𝐿

ρ:

c: L

E:

V:

𝑇𝑇

S: L

𝑇𝑇

𝑀𝑀

𝐿𝐿3 𝑀𝑀

𝐿𝐿𝐿𝐿 2

Three fundamental quantities are used (M, L, and T) so K = 3. There are 8 dimensional variables, so N = 8.

N – K = 8 – 3 = 5, so we are looking for 5 Π groups. Select ρ (M/L3), V (L/T), and S (L) as repeating variables. Indeed, together they contain Mass, Length, and Time, and no nondimensional variable can be formed from ρ, V and S, so these are suitable choices. We look for 5 Π groups in the system: 𝛱𝛱1 = ρ𝑎𝑎11 𝑉𝑉 𝑎𝑎12 𝑆𝑆 𝑎𝑎13 𝐹𝐹 𝛱𝛱2 = ρ𝑎𝑎21 𝑉𝑉 𝑎𝑎22 𝑆𝑆 𝑎𝑎23 𝑓𝑓 𝛱𝛱3 = ρ𝑎𝑎31 𝑉𝑉 𝑎𝑎32 𝑆𝑆 𝑎𝑎33 µ 𝛱𝛱4 = ρ𝑎𝑎41 𝑉𝑉 𝑎𝑎42 𝑆𝑆 𝑎𝑎43 𝑐𝑐 𝛱𝛱5 = ρ𝑎𝑎51 𝑉𝑉 𝑎𝑎52 𝑆𝑆 𝑎𝑎53 𝐸𝐸

Remember, the goal will be to find the appropriate three exponents (an1, an2, an3) with each variable to make the Π group non-dimensional. 𝛱𝛱1 (𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑡𝑡ℎ𝑒𝑒 "𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓”): 𝑀𝑀 𝑎𝑎11 𝐿𝐿 𝑎𝑎12 𝑎𝑎 𝑀𝑀𝑀𝑀 0 0 0 𝑀𝑀 𝐿𝐿 𝑇𝑇 = � 3 � � � [𝐿𝐿] 13 2 𝐿𝐿 𝑇𝑇 𝑇𝑇 M: 1+𝑎𝑎11 = 0 ⇒ 𝒂𝒂𝟏𝟏𝟏𝟏 = −𝟏𝟏 L: 1+𝑎𝑎12 + 𝑎𝑎13 − 3𝑎𝑎11 = 0 ⇒ 𝑎𝑎12 + 𝑎𝑎13 = −4 T: −2 − 𝑎𝑎12 = 0 ⇒ 𝒂𝒂𝟏𝟏𝟏𝟏 = −𝟐𝟐 and 𝒂𝒂𝟏𝟏𝟏𝟏 = −𝟐𝟐 𝛱𝛱1 =

𝐹𝐹 ρ𝑉𝑉 2 𝑆𝑆 2

AAE 333 Fluid Mechanics

𝛱𝛱2 :

Fall 2020

𝑀𝑀 𝑎𝑎21 𝐿𝐿 𝑎𝑎22 𝑎𝑎 1 𝑀𝑀 𝐿𝐿 𝑇𝑇 = � 3 � � � [𝐿𝐿] 23 𝐿𝐿 𝑇𝑇 𝑇𝑇 0 0 0

M: 𝒂𝒂𝟐𝟐𝟐𝟐 = 𝟎𝟎 L: −3𝑎𝑎21 + 𝑎𝑎21 + 𝑎𝑎22 = 0 ⇒ 𝑎𝑎21 = −𝑎𝑎22 T: −1 − 𝑎𝑎22 = 0 ⇒ 𝒂𝒂𝟐𝟐𝟐𝟐 = −𝟏𝟏 and 𝒂𝒂𝟐𝟐𝟐𝟐 = 𝟏𝟏

𝛱𝛱2 =

𝛱𝛱3 :

𝑓𝑓𝑓𝑓 𝑉𝑉

𝑀𝑀 𝑎𝑎31 𝐿𝐿 𝑎𝑎32 𝑎𝑎 𝑀𝑀 𝑀𝑀 𝐿𝐿 𝑇𝑇 = � 3 � � � [𝐿𝐿] 33 𝐿𝐿 𝑇𝑇 𝐿𝐿𝐿𝐿 M: 1 + 𝑎𝑎31 = 0 ⇒ 𝒂𝒂𝟑𝟑𝟑𝟑 = −𝟏𝟏 L: −1 + 𝑎𝑎32 + 𝑎𝑎33 − 3𝑎𝑎31 = 0 ⇒ 𝑎𝑎32 + 𝑎𝑎33 = −2 T: −1 − 𝑎𝑎32 = 0 ⇒ 𝒂𝒂𝟑𝟑𝟑𝟑 = −𝟏𝟏 and 𝒂𝒂𝟑𝟑𝟑𝟑 = −𝟏𝟏 0 0 0

𝛱𝛱3 =

𝛱𝛱4 :

𝑀𝑀 𝑎𝑎41 𝐿𝐿 𝑎𝑎42 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0 = � 3 � � � [𝐿𝐿]𝑎𝑎43 𝐿𝐿 𝐿𝐿 𝑇𝑇

M: 𝒂𝒂𝟒𝟒𝟒𝟒 = 𝟎𝟎 T: 𝒂𝒂𝟒𝟒𝟒𝟒 = 𝟎𝟎 L: 1 + 𝑎𝑎42 + 𝑎𝑎43 − 3𝑎𝑎41 = 0 ⇒

𝛱𝛱5 :

µ ρ𝑉𝑉𝑉𝑉

𝒂𝒂𝟒𝟒𝟒𝟒 = −𝟏𝟏

𝛱𝛱4 =

𝑐𝑐 𝑆𝑆

𝑀𝑀 𝑎𝑎51 𝐿𝐿 𝑎𝑎52 𝑀𝑀 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0 = � 3 � � � [𝐿𝐿]𝑎𝑎53 2 𝐿𝐿 𝑇𝑇 𝐿𝐿𝐿𝐿

M: 1 + 𝑎𝑎51 = 0 ⇒ 𝒂𝒂𝟓𝟓𝟓𝟓 = −𝟏𝟏 T: −2 − 𝑎𝑎52 = 0 ⇒ 𝒂𝒂𝟓𝟓𝟓𝟓 = −𝟐𝟐 L: −1 + 𝑎𝑎52 + 𝑎𝑎53 − 3𝑎𝑎51 = 0 ⇒ −1 − 2 + 𝑎𝑎53 + 3 = 0 ⇒ 𝒂𝒂𝟓𝟓𝟓𝟓 = 𝟎𝟎 𝛱𝛱5 =

𝐸𝐸 ρ𝑉𝑉 2

AAE 333 Fluid Mechanics

Fall 2020

Now construct a function from the 𝛱𝛱 groups (we use ϕ as we’ve already used f for frequency): 𝛱𝛱1 = 𝜙𝜙(𝛱𝛱2 , 𝛱𝛱3 , 𝛱𝛱4 , 𝛱𝛱5 )

𝑭𝑭 𝒇𝒇𝒇𝒇 𝝁𝝁 𝒄𝒄 𝑬𝑬 = 𝝓𝝓 � , , , � ρ𝑽𝑽𝟐𝟐 𝑺𝑺𝟐𝟐 𝑽𝑽 ρ𝑽𝑽𝑽𝑽 𝑺𝑺 ρ𝑽𝑽𝟐𝟐

Note that since the ratio c/S is one of the nondimensional groups and a combination of two nondimensional groups is also a proper nondimensional group, alternative answers can contain, for 𝑭𝑭 𝑭𝑭 𝑭𝑭 𝒇𝒇𝒇𝒇 𝒇𝒇𝒇𝒇 𝝁𝝁 𝝁𝝁 example, ρ𝑽𝑽𝟐𝟐 𝒄𝒄𝟐𝟐 or ρ𝑽𝑽𝟐𝟐 𝒄𝒄𝒄𝒄 instead of ρ𝑽𝑽𝟐𝟐 𝑺𝑺𝟐𝟐 , 𝑽𝑽 instead of 𝑽𝑽 , and ρ𝑽𝑽𝒄𝒄 instead of ρ𝑽𝑽𝑽𝑽. ρ𝑽𝑽𝑽𝑽

ρ𝑽𝑽𝑽𝑽

Note also that 𝝁𝝁 = 𝑹𝑹𝑹𝑹𝑺𝑺 – the Reynolds number based on S, and 𝝁𝝁 = 𝑹𝑹𝑹𝑹𝒄𝒄 – the Reynolds 𝝁𝝁

number based on c. Therefore, either of these Reynolds numbers can be used instead of ρ𝑽𝑽𝑽𝑽.

(b) Other than the repeating variables, the problem involves varying viscosity (µ) and frequency (f).

Therefore to maintain dynamic similarity, we must match 𝛱𝛱2 and 𝛱𝛱3 between the two cases.

Since frequency is what we seek but we are given the ratio of viscosity, start with 𝛱𝛱3 to look for velocity. We will notate the original case as “1” and larger fly in silicon oil as “2”. µ1 µ2 = ρ1 𝑉𝑉1 𝑆𝑆1 ρ2 𝑉𝑉2 𝑆𝑆2

First, assume the “original” fly moves in air. Now note that the density is equal to water (1000 kg/m3; for air 1.225 kg/m3), but the fly is 100 times larger and the fluid 50 times more viscous than water (water: 1×10-3 Pa·s; for air 1.8×10-4 Pa·s) in case “2”, plug in 𝑉𝑉1and solve for 𝑉𝑉2: 0.05 0.00018 = 1.225 ∙ 1 ∙ 1 1000 ∙ 𝑉𝑉2 ∙ 100

And similarly for 𝛱𝛱2 :

𝑽𝑽𝟐𝟐 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 m/s 𝑓𝑓1 𝑆𝑆1 𝑓𝑓2 𝑆𝑆2 = 𝑉𝑉1 𝑉𝑉2

𝑉𝑉2 was solved above so we plug in the values and write:

(A very unusual fly!)

120 ∙ 1 𝑓𝑓2 ∙ 100 = 1 0.0034 𝒇𝒇𝟐𝟐 =0.004 Hz

AAE 333 Fluid Mechanics

Fall 2020

Assume now that the “original” fly moves in water.

Then 50 1 = 1 ∙ 𝑉𝑉2 ∙ 100 1∙1∙1 And similarly for 𝛱𝛱2 :

𝑽𝑽𝟐𝟐 = 𝟎𝟎. 𝟓𝟓 m/s 𝑓𝑓1 𝑆𝑆1 𝑓𝑓2 𝑆𝑆2 = 𝑉𝑉1 𝑉𝑉2

𝑉𝑉2 was solved above so we plug in the values and write:

120 ∙ 1 𝑓𝑓2 ∙ 100 = 1 0.5

(Again, a very unusual fly!)

𝒇𝒇𝟐𝟐 =0.6 Hz

The solution to this part obtained with either assumptions is viewed as the correct solution.