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RPI / MANE / MAN 4070 / How is the navier-stokes equation applied?

How is the navier-stokes equation applied?

How is the navier-stokes equation applied?

Description

School: Rensselaer Polytechnic Institute
Department: MANE
Course: Aerodynamics
Professor: Professor hirsa
Term: Summer 2015
Tags:
Cost: 75
Name: Test Upload
Description: Final Exam Solutions from 2014
Uploaded: 02/08/2016
12 Pages 15 Views 3 Unlocks
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MANE 4070 – Aerodynamics I:


How is the navier-stokes equation applied?



Final Exam

December 12, 2014

Instructor: Professor Hicken

Time: 180 minutes

Format: Closed-book, no aids, no devices with internet access, no cell phones!

Last Name

First Name

RIN

Question 1

7

Question 2

5

Question 3

5

Question 4

8

Question 5

7

Question 6

7

Question 7

6

Question 8

5

Total

50


What does the navier-stokes equation govern?



Don't forget about the age old question of When did the gunpowder revolution take place?
We also discuss several other topics like What are the elements of consideration?

1

Question 1 (7 marks)

NASA scientists are modeling the Orion spacecraft’s reentry into the earth’s atmosphere. Orion has a radius of 2.5 m.

a. If the scientists want to use a wind-tunnel experiment to predict the forces and moments on the spacecraft, list the requirements on the geometry and the flow that they must meet to achieve dynamic similarity.

b. If the scientists model the reentry flow at high-altitude using the Navier-Stokes partial differential equa tion, what hypothesis do they risk violating and why?

c. A wind-tunnel model is built with a radius of 0.15 m. If the drag force experienced by the model is D = 4.5 N at sea-level conditions and V∞ = 28 m/s, what is the drag force on the actual spacecraft when its velocity is 4000 m/s and the density is ρ∞ = 0.001 kg/m3? Assume the two flows are dynamically similar.


What is computed in kutta-joukowski theorem?



d. Find the Mach number of the spacecraft for the conditions in part c, i.e. V∞ = 4000 m/s and ρ∞ = 0.001 kg/m3, if the atmospheric temperature is T∞ = 282.7 K. You may assume air is an ideal gas at the given conditions. We also discuss several other topics like Who investigated the chemical composition of the nucleus?

e. If the lift on the capsule is negligible, what is the dominant form of inviscid drag on the capsule?

Answer.

a. To achieve dynamic similarity, 1) the similarity parameters, e.g. Re,Ma, must be the same between the actual flow and the experiment and 2) the spacecraft and model must be geometrically similar.

0.5 marks for “similarity parameters must be the same”

0.5 marks for “geometries are similar”

b. The continuum hypothesis will be violated at high altitude, because the gas is rarefied

1 mark for “continuum hypothesis”

1 mark for “gas is rarefied”, or “mean-free path is large”, or similar

c. The coefficient of drag must be the same, since the flows are dynamically similar. For the model, which has an

area of S = πr2 = 0.0707 m2, If you want to learn more check out What are the three groups of population?

CD =D

2ρ∞V2∞S= 0.1326.

Therefore, the drag on Orion for the given conditions would be (Orion has an area of S = 19.64 m2) DOrion = CD12ρ∞V2∞S ≈ 20800 N.

1 mark for CD, (or correctly equating two CD if no value given)

1 mark for drag on spacecraft

d. For the given temperature, a∞ =√γRT∞ = 337 m/s. Therefore, the Mach number is M∞ = 11.9. 1 mark for correct Mach number (0.5 partial mark if a∞ correct) If you want to learn more check out What is vapor-­liquid-­solid method (vls)?

e. Wave drag is the dominant form of inviscid drag on the spacecraft.

1 mark for “wave drag” (drag due to shocks will also be accepted)

2

Question 2 (5 marks)

a. Consider the following partial differential equation:

∂ ρ

∂t+~∇·(ρ~u) = 0. If you want to learn more check out What is the deutsch word of to play the guitar?

Identify the meaning of the two terms in the PDE. Use the table provided below.

b. Write the conservation of mass equation for steady flows.

c. Write the conservation of mass equation for unsteady flows that are incompressible.

d. The Navier-Stokes equations are

∂t+~∇· ρ~u⊗~u+ pI −τ −ρ~f =~0.

∂ (ρ~u)

What conservation principle do these equations govern?

e. A 3-dimensional, incompressible, viscous flow can be modeled using the continuity equation and the Navier-Stokes equations only. How many scalar partial differential equations would you need to solve in this case?

(table for answering question 2a)

∂ ρ

∂t

~∇·(ρ~u)

Answer.

a. ∂ ρ/∂t is the rate of change of density at a point

~∇·(ρ~u) is the advective transport of density away from the point.

0.5 marks for meaning of ∂ ρ/∂t

0.5 marks for meaning of ~∇·(ρ~u)

b. For steady flows, the time derivative vanishes from the conservation of mass equation (continuity equation) producing~∇·(ρ~u) = 0.

1 mark for correct equation

c. For incompressible flow, regardless of if they are steady or not, the continuity equation becomes ~∇·~u = 0.

1 mark for divergence of ~u is zero

d. The Navier-Stokes equations govern conservation of (linear) momentum.

1 mark for “conservation of momentum”

e. The Navier-Stokes equations are vector equations, so they consist of 3 PDEs in 3-dimensions. Adding the continuity equation, this gives 4 equations.

1 mark for “4 equations”

3

Question 3 (5 marks)

Consider the inviscid, incompressible, and irrotational 2-dimensional flow described by the stream function ψ(~x) = 12πln(r) m2/s.

Recall that, in polar coordinates, the stream function defines the velocity via

ur =1r∂ψ

∂ θ and uθ = −∂ψ

∂ r.

a. What partial differential equation (PDE) does ψ satisfy? You can give the name of the PDE or write down the equation for the PDE.

b. Sketch the streamlines for the flow described by ψ, and indicate the direction of the flow along the stream lines. Please include the x- y-axes.

c. Find the vorticity of the flow described by ψ for all points~x 6= (0,0).

Hint: If you need it, the vorticity is given by ω =1r∂∂ r(ruθ )−1r∂ur 

∂ θ .

d. For the flow described by ψ, the circulation around any circle centered at the origin is 1 m2/s. Does this contradict what you found in part c? Explain.

e. A uniform free-stream flow of(u, v) = (0,10) m/s is added to the flow described by ψ. If ρ∞ = 1.225 kg/m3, find the force (magnitude and direction) exerted by the flow on the surface of a circle centered at the origin.

Answer.

a. The PDE satisfied by ψ is Laplace’s equation: ∇2ψ =∂2ψ

∂ x2+∂2ψ

1 mark for “Laplace” or its equation

b. 0.5 marks for concentric circles about the origin 0.5 marks for clockwise flow direction indicated

∂ y2= 0.

c. From the given equations for the velocity components, we have ur = 0, and uθ = −12πr. Substituting these values into the vorticity equation we find

∂ θ =1r∂∂ r −12π = 0.

ω =1r∂∂ r(ruθ )−1r∂ur 

Alternatively, we recognize that this is an irrotational flow (except at the origin), so ω = 0 by definition. 1 mark for ω = 0 (0.5 partial credit if velocity components are correct)

d. No, it does not contradict the result from part c, because the velocity field is singular at the origin.

0.5 marks for “No” or “It does not contradict”

0.5 marks for “velocity field is singular” or “velocity blows up” etc

We also gave... 0.5 partial marks for “Yes, because ω = 0 implies Γ = 0

e. By the Kutta-Joukowski theorem, the force magnitude is F = ρ∞V∞Γ = 12.25 N/m. Since the free-stream flow is in the positive y direction and the vortex is clockwise, the force direction is in the negative x direction:

~F = (−12.25,0) N/m.

4

0.5 marks for correct force magnitude

0.5 marks for correct force direction

There was no “carry over” error if you found the vortex was counter-clockwise 5

Question 4 (8 marks)

a. On the airfoil figure below, clearly label: 1) the thickness; 2) the (max) camber; 3) the mean camber line; 4) the angle of attack; 5) the chord length, and; 6) chord line.

b. Place a checkmark beside the statements that are assumptions of thin-airfoil theory:

  the angle of attack must be small and positive

  the thickness-to-chord ratio of the airfoil must be relatively small

  the derivative of the mean camber line, dz/dx, must be relatively small

  the flow is inviscid, irrotational, and incompressible.

c. Consider a symmetric thin airfoil. Estimate its cl at α = 5◦and M∞ = 0.6.

d. Consider a cambered thin airfoil with cm,c/4 = −0.05. Estimate its cm,c/4 at α = 3◦and M = 0.7. e. According to thin airfoil theory, what is significant about the point c/4 for symmetric airfoils?

Answer.

a. See (annotated) Lecture 8, slide 4. 1/3 mark for each correctly labeled term

b. All are assumptions of thin-airfoil theory except for the first: the angle of attack does not need to be positive. 0.5 marks for each correct

c. For a symmetric airfoil, αL=0 = 0. Therefore, according to thin-airfoil theory, we have cl,0 = 2π(5◦π/180◦) = 0.5483. Using this value with the Prandtl-Glauret correction we find

cl = pcl,0 

1− M2∞= 0.6854.

1 mark for incompressible cl,0 

1 mark for corrected value, cl 

d. According to thin-airfoil theory, c/4 is the aerodynamic center, so cm,c/4 = −0.05, regardless of α. At M∞ = 0.7 we find, using the Prandtl-Glauret correction,

cm,c/4 =cm,c/4,0 

p1− M2∞= −0.07.

1 mark for corrected value, cm,c/4 

e. According to thin-airfoil theory, c/4 is the center of pressure and aerodynamic center for symmetric airfoils.

0.5 for center of pressure

0.5 for aerodynamic center

6

Question 5 (7 marks)

Consider two wings with the following circulation distributions: N

wing 1 Γ(θ) = 2bV∞ wing 2 Γ(θ) = 2bV∞ 

An sin(nθ) = 2bV∞ [0.01 sin(θ)−0.005 sin(2θ) +0.0003 sin(3θ)] n=1

N

An sin(nθ) = 2bV∞ [0.01 sin(θ)]

n=1

The two wings have rectangular planforms, with spans of b = 21 m and chord lengths of c = 3 m. a. Find the coefficient of lift for both wings.

b. Find the coefficient of induced drag for both wings

c. What is unusual about wing 1? Hint: A2 is nonzero.

d. What is special about wing 2?

e. List a disadvantage of wing 2?

f. Suppose an aircraft is designed to operate at standard sea-level conditions. How would you have to change its span to get the same induced drag if it operated at an altitude where the density is half the value of the sea-level value? You can assume the velocity, lift, and span efficiency remain unchanged.

Answer.

a. The coefficient of lift is defined by CL = A1πAR. The two wings have the same aspect ratio of AR = b2/S = b/c = 7, and the same A1 = 0.01 coefficient; therefore, the coefficient of lift for both wings is

CL = 0.22

1 mark for correct CL (0.5 partial credit if AR is correct)

No “carry over” if AR was incorrect

b. To find the coefficient of induced drag for wing 1, we first need to compute its span efficiency factor. Using the

given formula we find

e1 =

"

1+

3∑ n=2

n

 An A1 

 2#−1 

≈ 0.6655

For the second wing, we can either compute its span efficiency factor explicitly, or recognize that it is an elliptical distribution (since An = 0,∀n > 1). Either way, we find e2 = 1. Adapting the given formula for C(elliptic) 

D,i, the induced-drag coefficients are

CD,i,1 =C2L 

πARe1≈ 0.00330 and CD,i,2 =C2L 

πARe2≈ 0.00220

1 mark for correct CD,i,1 (0.5 partial credit if e1 correct)

1 mark for correct CD,i,2 

No “carry over” if CL or AR were incorrect

c. Wing 1 is unusual because A2 6= 0, which means that its lift distribution is nonsymmetric about the root. 1 mark for “nonsymmetric lift distribution” or similar

7

d. Wing 2 has an elliptical lift distribution and/or has a minimum induced drag for all (planar) wings with the same parameters.

1 mark for “elliptical lift” or “minimum induced drag

e. A disadvantage of wing 2 is that it increases the root bending moment, which will likely increase the necessary structural weight. Wing 2 may also be more difficult to manufacture.

1 mark for “increased root moment” or “increased structural weight” or “difficult to manufacture”

f. Recall that induced drag behaves as Di ∝ L2/(ρ∞V2∞b2e); this proportionality can also be derived from the given CD,i expression if necessary. If the two designs have the same induced drag, Di,1 = Di,2, then

ρ∞V2∞b21e=L2 

L2 

(ρ∞/2)V2∞b22e

⇒ b22 = 2b21.

Therefore, the span would need to be increased by a factor of √2.

1 mark for “increase by √2”

8

Question 6 (7 marks)

A Pitot tube is mounted on an aircraft that is flying at an altitude where

p∞ = 5530 Pa, ρ∞ = 0.0888 kg/m3, and T∞ = 217 K.

a. List 2 of the 3 assumptions necessary in order to apply the Bernoulli equation along a streamline.

b. Assume the conditions in part a hold. Use the Bernoulli equation to find the velocity of the aircraft if the Pitot tube measures p0 = 5814 Pa.

c. What is the aircraft’s Mach number for the velocity found in part b?

d. The aircraft increases velocity until p0 = 7055 Pa. The assumptions of the Bernoulli equation no longer hold. Find the aircraft’s velocity assuming the flow is isentropic.

e. The aircraft increases its velocity further to V∞ = 370 m/s. For this velocity, what total pressure does the Pitot tube measure?

Answer.

a. Flow must be 1) steady, 2) inviscid, and 3) incompressible along the streamline.

1 mark (partial credit of 0.5 for each correct assumption)

b. Rearranging the Bernoulli equation, we find

V∞ =

1 mark for correct velocity

s

2(p0 − p∞)

ρ∞≈ 80 m/s.

c. The free-stream speed of sound is a∞ =√γRT∞ ≈ 295 m/s; therefore, M∞ = V∞/a∞ ≈ 0.271. 1 mark for correct Mach number (0.5 partial credit for a∞)

d. Since the flow is isentropic, we can use the total-to-static pressure ratio to find the Mach number:

M∞ =

vuut2 γ −1

" p0 p

 γ−1γ−1#≈ 0.6

The free-stream speed of sound is still a∞ ≈ 295 m/s; therefore, V∞ = a∞M∞ ≈ 177 m/s. 1 mark for correct free-stream Mach number

1 mark for correct velocity

We also accepted correct answers that used total enthalpy.

e. Here the Mach number is V∞/a∞ = 1.25, so there is a normal shock ahead of the Pitot tube. The Mach number and pressure downstream of the shock are given by the normal-shock relations:

M2 =

s

1+[(γ −1)/2]M2∞ 

γM2∞ −(γ −1)/2≈ 0.81, p2 = p∞ 

 

1+2γ γ +1

M2∞ −1  ≈ 9210 Pa.  

Finally, using the total-to-static pressure ratio between the shock and the Pitot tube gives

 

p0 = p2 

1+γ −1

2M22

  γ

γ−1≈ 14200 Pa.

0.5 marks (each) for correct M2 and p2 after shock 1 mark for correct p0 for Pitot tube

9

Question 7 (6 marks)

Consider the M∞ = 1.9 flow around a wedge with half-angle 9◦and angle of attack 11◦; see the figure below. The free-stream pressure is p∞ = 1 atm.

9o 11o 

When answering the questions below, give your answers to 2 significant digits (unless indicated otherwise). a. Find the shock angle β for the shock that forms on the lower surface.

b. Find the Mach number and pressure along the lower surface of the wedge.

c. Find the Mach number and pressure along the upper surface of the wedge.

d. For M∞ = 1.9, at what angle of attack will the oblique shock detach? Give your answer to the nearest whole number.

Answer.

a. The flow deflection angle is θ = 11◦ +9◦ = 20◦. From the given θ-β-M curves, we find that β = 58◦. 1 mark for correct shock angle

b. The normal component of the incoming Mach number is Mn,∞ = M∞ sin(β) = 1.61. Therefore, using the normal

shock relations Mn,2 =

s

 

1+[(γ −1)/2]M2n,∞

γM2n,∞ −(γ −1)/2≈ 0.665, p2 = p∞ 

1+2γ γ +1

M2n,∞ −1  ≈ 2.9 atm.  

The Mach number after the oblique shock is M2 = Mn,2/sin(β −θ) ≈ 1.1.

1 mark for pressure p2 ∈ [2.8,2.9] (0.5 partial credit for equation)

1 mark for Mach number M2 ∈ [1.08,1.1] (0.5 partial credit for equation)

We gave 0.5 partial credit if Mn,∞ was correct

No “carry over” for incorrect β and/or θ.

c. The flow is deflected by θ2 = 2◦ on the upper surface. Using the Prandtl-Meyer table, we find ν(M∞) = 23.59◦. Denoting the region after the expansion with 3, we find

ν(M3) = ν(M∞) +θ2 = 25.59◦.

Interpolating between the data points in the table, we find M3 ≈ 1.97. Using this value and the total-to-static pressure ratio (the expansion is isentropic), we find

p3 =p3/p0,3 

p∞/p0,∞p∞ =

 1+ (γ −1)M2∞/2 1+ (γ −1)M23/2

 γ/(γ−1) 

× p∞ = 0.897 atm.

1 mark for Mach number M3 ∈ [1.97,2.0]

1 mark for pressure p3 ∈ [0.88,0.9] (0.5 for correct equation)

No “carry over” if M3 was incorrect

d. For M∞ = 1.9, the oblique shock becomes detached if the deflection angle is greater than θmax ≈ 21◦(found from the θ-β-M curve). From the geometry, this corresponds to an angle of attack of α ≈ 12◦.

1 mark for 12◦(0.5 partial credit for θmax ≈ 21◦)

10

Question 8 (5 marks)

Consider the supersonic wind tunnel depicted below. The exit area of the nozzle, which is the same as the cross sectional area of the test section, is Ae = 16 m2. In the following, you can assume there are no shocks in the converging-diverging nozzle or the test section.

test section

a. To achieve M = 2 in the test section, what area does the throat need to have? Use the closest entry in the appropriate table.

b. If the reservoir (to the left of the nozzle above) is at p0 = 5 atm, what is the static pressure in the test section when the test section’s Mach number is M = 2?

c. If the reservoir temperature is T0 = 288 K, what is the flow velocity in the test section when M = 2?

d. When the Mach number is M = 2 in the test section, the mass flow rate is 0.109 kg/s. What is the mass flow rate when the Mach number in the test section is increased to M = 3?

Answer.

a. For supersonic flow, the throat must be sonic, so At = A∗. At M = 2, the area ratio is A/A∗ ≈ 1.75. Therefore, if Ae = 16 m2, then At = A∗ = 9.14.

1 mark for correct At value

b. The flow is isentropic, so we can use the total-to-static pressure ratio:

 −γ

γ−1≈ 0.639 atm.

1 mark for correct pressure

 

p = p0 

1+γ −1

2M2 

c. Using the total-to-static temperature ratio, we find

 

T = T0 

1+γ −1

2M2 

 −1 

≈ 160 K.

Therefore, the speed of sound in the test-section is a =√γRT ≈ 254 m/s, and the velocity is V = aM = 507 m/s.

1 mark for correct temperature in test section

1 mark for correct velocity

d. Two possible solutions:

Solution 1: Ae changes, but At = A∗stays the same.

The flow is choked (the throat is sonic), so the mass flow rate is the same: ˙m = 0.109 kg/s. 11

Solution 2: At = A∗changes, but Ae stays the same.

First we need to know the stagnation density in the reservoir. Using the mass-flow rate for the M = 2 case we have

ρeVeAe = 0.109 kg/s ⇒ ρe = 1.34×10−5kg/m3.

Using the total-to-static ratio for the density (the flow is isentropic), we find the stagnation/total density in the

reservoir to be

ρ0 = ρ

 

1+γ −1

2M2 

  1γ−1= 5.84×10−5kg/m3.

Next, we use the total-to-static ratios to find the density and temperature in the test section at M = 3:

 

ρe = ρ0 

1+γ −1

2M2 

 −1γ−1= 4.45×10−6kg/m3,

 

Te = T0 

1+γ −1

2M2 

 −1 

= 103 K.

Finally, we can use the temperature and Mach number to find the velocity, and then, with the density and area, the mass flow rate:

m˙ = ρeVeAe =4.45×10−6  3pγRTe 16 kg/s = 0.0434 kg/s

1 mark for correct value of m˙

12

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