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AAE 333 Fall 2014, Sample Final Exam Today’s simple ... · PDF fileAAE 333 Fall 2014,...
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AAE 333 Fall 2014, Sample Final Exam
Today’s simple instructions:
120 minutes for this exam
●DO YOUR OWN WORK or score zero on the exam. ● Fill in name and student number on the answer sheet.
● Three doublesided, handwritten original formula sheets are permitted. No electronics.
● All questions in this exam concern the incompressible steady flow of air or water unless specified in the problem.
● You may write on these pages as you wish, they are yours to keep.
● At the end of the testing period the class will be instructed to put down your pencil and stop writing. Failure to heed this very simple instruction will result in a zero on this exam.
● At the end, hand in your answer sheet, keep these exam pages and your formula sheets.
● Unless otherwise specified in the problem statement, the velocity field is denoted by:
or(x, ) (x, )i (x, )jV→
y = u y ˆ + v y ˆ .(r, ) (r, )e (r, )eV
→θ = ur θ ˆr + uθ θ ˆθ
● Unless noted, a boundary layer question assumes a flat plate along other xaxis
and with uniform inflow, .(x, ) iV→
y = U∞ˆ
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1. In a laminar boundary layer, the rate of change of tangential velocity with respect to the
distance normal to the surface, usually written as , at the separation point is∂y∂u
A. Zero B. 1 C. Infinite D. Approximately / δ ue E. None of the above
2. Very few exact solutions of the NavierStokes equations exist because:
A. NavierStokes equations are not sufficiently accurate B. Not enough boundary conditions are known C. The equations are nonlinear in ways that make them very difficult to solve D. Air is too squishy E. None of the above
3. While a turbulent boundary layer generally has greater skin friction drag than a laminar
boundary layer on the same body, a turbulent boundary layer can actually reduce drag on blunt bodies by:
A. Moving the forward stagnation point randomly B. Moving the separation point towards the aft end of the body C. Pressurizing the wake D. Heating the air in front of the body E. None of the above
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4. For Poiseuille flow, the velocity profile across the duct is A. Linear B. Parabolic C. Cubic D. Gaussian E. None of the above
5. Order of magnitude analysis for a 2D steady laminar boundary flow shows that, for ( ) O
indicating the order of magnitude of the item in the parentheses,
A. O (∂x∂u)≪ O(∂y∂v) B. O (∂x∂u)≫ O(∂y∂v) C. O (∂x∂u) ≃ δ D. O (∂x∂u) = O(∂y∂v) E. None of the above
(1), (1)u = u*
U0 * = O ∂x
∂u = ∂(x /L )* 0*
∂(u /U )* 0*= O
(1) ∂x∂u + ∂y
∂v = 0 ⇒ ∂y∂v = − ∂x
∂u = O 6. In a laminar boundary layer, the distance from the wall to the point where the velocity is
99% of the edge velocity is called the A. Displacement thickness B. Boundary layer thickness C. Momentum thickness D. Energy thickness E. None of the above
7. The Blasius boundary layer solution method
A. Assumes a laminar flow B. Assumes a turbulent flow C. Assumes a separated flow D. Assumes reversed flow E. None of the above
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8. The inviscid solution for the velocity profile in Couette flow is A. Velocity is zero B. Velocity varies linearly across the channel C. Velocity varies quadratically across the channel D. No solution exists in the inviscid limit E. None of the above
9. The energy equation is only required to solve incompressible flows if there is heat addition.
A. True B. False
∇ ∙ V→ = 0
V ) V ∇P f∂t∂v→
+ (→∙ ∇ →
= − ρ1 + ρ
1→visc + . . .
depends on velocity so there are 4 unknowns and 4 equations:f→
visc τ )( xy = μ ∂y∂u
nknowns u, , , u : v w P quations continuity, x omentum, y omentum, z omentum e : −m −m −m
10. To apply Thwaites’s method to the flow over a Rankine oval in a uniform flow of , iV→ =U∞ˆ you need to know the edge velocity for the boundary layer at every point around the body. For this you would use:
A. ue = U∞
B. from the surface velocity of the circular cylinder potential flow solution ue
C. / U ue ∞ = 1 D. from the surface velocity of the potential flow solution for the Rankine ue
oval E. None of the above
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11. We solved the planar Couette flow where the wall at y=0 was stationary and the wall at y=h was moving to the right at velocity . If now the lower wall were to move at velocity ue ,− ue the boundary conditions for the flow are:
A. and (0) − u = ue (h) u = ue B. and (0) v = 0 (h) v = 0
C. (0) (h) ω→ = ω→ = 0 D. A and B E. None of the above
12. Curves that are drawn such that the tangents at every point along the curve are in the same
direction as the velocity vectors at those points are called A. Streamlines B. Streaklines C. Pathlines D. Timelines E. None of the above
13. The dynamic pressure at the stagnation point on an airfoil subject to freestream conditions,
, is01.3 kPa, V 0 m/s, ρ .2754 kg/m P∞ = 1 ∞ = 6 ∞ = 1 3 A. Zero B. 2,296 Pa C. 4,591 Pa D. 101.3 kPa E. None of the above
At a stagnation point the velocity is zero and therefore the dynamic pressure, , is zero.ρVq = 2
1 2 14. Choose the correct statement about pressure in a fluid.
A. Pressure is a body force B. Pressure acts normal to a surface C. Pressure is a frictional force D. Pressure acts parallel to a surface E. None of the above
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15. A 0.5 meter long spinning rod with a diameter of 10 cm is in an air flow of 50 meters per second. the rod spins at a rate of 200 rpm. Air density is 1.2754 kilograms per cubic meter. If the section coefficient of drag is known to be 0.3, then the total drag on the rod is
A. Zero B. 24 N C. 96 N D. 480 N E. None of the above
The total drag on the rod is related to the coefficient of drag by . The freestream CD = D
q S∞
dynamic pressure is computed as ρ V (1.2754 kg/m )(50 m/s) 594 Paq∞ = 2
1∞ ∞
2 = 21 3 2 = 1
The reference area, , isS
0.5 m)(0.1 m) 0.05 m S = ( = 2
Therefore, the total drag over the rod is the product of the coefficient of drag, freestream dynamic pressure, and reference area.
q S 0.3)(1594 N/m )(0.05 m ) 4 N D = CD ∞ = ( 2 2 = 2
16. A pool of water has a depth of 10 meters. How much greater is the pressure at a depth of 4
meters than at the surface? A. 39 kg m1 s2 B. 39 kg m2 s2 C. 39x103 kg m1 s2 D. 39x103 kg m2 s2 E. None of the above
The pressure varies linearly with depth according to the hydrostatic pressure equation,
g ∂y∂P = − ρ
Therefore, regardless of the depth of the pool the pressure difference caused by a depth variation of 4 meters is
P g(ΔH) 1, 00 kg/m )(9.81 m/s )(4 m) 39 0 kg m s Δ = ρ = ( 0 3 2 = × 1 3 −1 −2 17. A point source of strength m is located at the origin with a uniform flow given by i.V
→= U∞ ˆ
A stagnation point: A. Can not exist B. Exists directly upstream of the source C. Exists at the source D. Exists at ±y→ ∞ E. None of the above
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18. A point vortex of strength is located at the origin with a uniform flow given by Γ i.V→= U∞ ˆ
A stagnation point: A. Can not exist B. Exists at the vortex C. Exists at ± y → ∞ D. Exists directly above or below the vortex E. None of the above
19. The substantial derivative of temperature, , is equal to what value for steady flow?DtDT
A. Zero B. T∂t
∂T + V→∙ ∇
C. T V→∙ ∇
D. T ∇ E. None of the above
The substantial derivative of temperature is equal to
TDtDT = ∂t
∂T + V→∙ ∇
and if the flow is steady, then the temporal derivative is zero such that
TDtDT = V
→∙ ∇
20. An aircraft wing is placed in a wind tunnel with a freestream velocity of 40 m/s, air density
of 1.2754 kg/m3, and static pressure of 101.3 kPa. If a pressure tap located on the surface of the wing records a pressure of 75 kPa, then the velocity at the edge of the boundary layer is
A. 40 m/s B. 207 m/s C. 234 m/s D. Cannot be determined from the information given E. None of the above
At higher Reynolds numbers where the boundary layers are thin, the pressure gradient across the boundary layer is negligible such that . Therefore, the static pressure measured on the P /∂y ∂ ≈ 0 surface of the wing is approximately equal to the pressure at the edge of the boundary layer
. Outside the boundary layer, the flow is inviscid so Bernoulli’sP 5 kPa) ( e = P surface = 7 equation may be used to obtain the edge velocity.
01.3 0 Pa (1.2754 kg/m )(40 m/s) 02.32 kPaP∞ + q∞ = P 0 = 1 × 1 3 + 21 3 2 = 1
07 m/s V e =√ ρ∞2(P −P )0 e =√ 1.2754 kg/m3
2(102.32 kPa−75 kPa) = 2
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21. Three containers connected at the base are filled with a fluid. The top of each container is open to the atmosphere and surface tension is negligible. The container shapes are all different. Choose the letter for the figure that shows the correct fluid levels in the containers at equilibrium conditions.
A. Figure A B. Figure B C. Figure C D. Figure D E. Figure E
22. The divergence of the velocity field given by isx z) i x ) j z k V
→= ( + y ˆ + ( 2 + z2 ˆ + 2 ˆ
A. Zero B. z i j 2x ) k − 2 ˆ + y ˆ + ( − z ˆ C. x z2 + y − 3 D. 3 E. None of the above
The divergence of a vector results in a scalar and is defined as
∇ ∙ V→ = ∂x∂u + ∂y
∂v + ∂z∂w = ∂x
∂(x+yz) + ∂y∂(x +z )2 2
+ ∂z∂(2z) = 1 + 0 + 2 = 3
23. The KuttaJoukowski theorem (or law, rule, equation, etc.) states that
A. Velocities above and below a trailing edge are equal at the trailing edge. B. Lift per span is the product of air density, velocity, and circulation C. A circular cylinder potential flow solution can be transformed into an airfoil
solution in a complexnumber plane. D. Total pressure is the sum of static and dynamic pressures. E. None of the above
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24. Lift and drag in aerodynamics are in general caused by: A. Normal forces that air exerts on the body B. Nonzero viscosity C. Normal and tangential forces that air exerts on the body D. Pressure gradients E. None of the above
25. The Kutta condition states that
A. Velocities above and below a trailing edge are equal at the trailing edge. B. Lift is the product of air density, velocity, and circulation C. A circular cylinder potential flow solution can be transformed into an airfoil
solution in a complexnumber plane. D. Total pressure is the sum of static and dynamic pressures. E. None of the above
26. Consider the flow over two circular cylinders, one having four times the diameter of the
other. The flow over the smaller cylinder has a freestream density and temperature given by and , respectively. The flow over the larger cylinder has a freestream density,ρ1 T 1
velocity, and temperature given by , and . Assume that / 4, V 00 m/s ρ2 = ρ1 2 = 1 TT 2 = 4 1 both and are proportional to . For the two flows to be dynamically similar, the μ a T 1/2 velocity over the smaller cylinder, , must be equal toV 1
A. 25 m/s B. 50 m/s C. 100 m/s D. 200 m/s E. No dynamic similarity can be obtained from the given conditions
For the two flows to be dynamically similar, the Reynolds and Mach numbers should be the same. Comparing Reynolds numbers first,
T11/2
ρ V D1 1 1 = (4T )11/2
(ρ /4)V (4D )1 2 1 Rearranging to obtain the velocity over the smaller cylinder and cancelling similar terms on either side,
0 m/sV 1 = 2V 2 = 5
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27. For the square 2D differential control volume sketched on to the right, the velocity entering the left side is labeled u. Then the velocity leaving the right side is,
A. xu + d B. dx∂x
∂u C. dxu + ∂x
∂u
D. dyu + ∂y∂u
E. None of the above 28. According to Buckingham Pi Theorem, the number of fundamental dimensions required to
describe the physical variables in a process is 3. The number of physical variables in the relation is 5. The number of dimensionless products needed to reexpress the physical relation is
A. 1 B. 2 C. 3 D. 5 E. None of the above
The number of Pi groups is N M, where N is the number of physical variables and M is the number of fundamental dimensions. Therefore N M is equal to 53 = 2. 29. A water sprinkler in steady state operation discharges a total of 0.002 cubic meters of water
each second. The inlet diameter of the sprinkler is 2.5 cm. The force required to anchor the sprinkler in place, , isF y
A. Zero B. 8 N C. 8 N D. 16 N E. None of the above
In steady state operation, the volumetric flow rate into the sprinkler is equal to the flow rate out of the sprinkler. We are only interested in the resultant force in the ydirection, and therefore use the ymomentum equation assuming steady, incompressible flow. Also, neglect viscous losses through the sprinkler.
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(V ) dSρ∫
∫
v
→∙ n̂ = F y
The water is discharged in the xdirections only, so the only contributions to the ymomentum are the resultant force and the convection into the sprinkler. The volumetric flow rate is already given as
.002 m /s A Q̇in = 0 3 = vin in ⇒ vin = AinQ̇in
The outward normal is always taken for the normal vector, and the inlet velocity is in the opposite direction of the normal vector. Therefore, the resultant force required to keep the sprinkler anchored is a negative value pushing the sprinkler down.
Q v NF y = − ρ ˙ in in = − AinρQ̇in
2
= − π(0.025 m)24(1,000 kg/m )(0.002 m /s)3 3 2
= − 8 30. Two objects, object 1 and object 2, are flying with the same velocity through air, but object
2 is flying at higher altitudes where the fluid temperature is much lower. The Mach number of object 2 will be
A. The same as for object 1 B. Less than that of object 1 C. More than that of object 1 D. Cannot be determined from the information given E. None of the above
The Mach number depends on flow velocities as well as the speed of sound. The flow velocities are the same, but the speed of sound depends on the gas (air in this case) and the temperature as
a = √γRT where is the ratio of specific heats and is the gas constant. Therefore, a flow with / C γ = Cp v R lower temperatures and keeping everything else the same will have lower speed of sound. This means the Mach number will be higher. 31. The cross product, , is nonzero for an irrotational flow.∇ )V
→× ( × V
→
A. True B. False
For irrotational flow the quantity, , is zero. Therefore the cross product of velocity with ∇ × V→ vorticity (curl of velocity) is also zero.
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32. Viscosity brings two processes into the fluid model. These two are:
A. Heat transfer and momentum transfer B. Momentum transfer and momentum thickness C. Momentum deficits and wakes D. Enthalpy and entropy E. None of the above
33. The boundary conditions of flow tangency and noslip are
A. Tangential velocity and normal velocity are zero at the wall, respectively
B. and ,(y ) 0 v = 0 = V(→∙ n)ˆ = 0
respectively C. Both A and B are correct D. Normal velocity and
tangential velocity are zero at the wall, respectively E. None of the above
34. Which of the following are true about stream functions?
A. The volumetric flow rate is defined as Ψ2 −Ψ1 B. along a streamlineΨd = 0 C. Streamlines are everywhere orthogonal to equipotential lines D. Both A and B E. A, B, and C are true
35. A stream of water with a crosssectional area of 2 cm2 moves in the direction shown below
with a velocity of = 30 m/s. The bucket is moving in the same direction as the stream V s of water, but with a fixed velocity of = 20 m/s. The time rate of change in water V b volume inside the bucket is
A. 0.002 m3 / s B. 0.003 m3 / s C. 0.003 m3 / s D. 0.070 m3 / s E. None of the above
This is an unsteady mass conservation problem with constant
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density. The conservation equation is written as
dV (V ) dS∂∂t ∫
∫
∫
ρ = − ∫
∫
ρ
→ rel∙ n̂
Density is constant and may therefore be factored and cancelled on either side. The relative velocity is the difference between the flow velocity in the inertial frame relative to the reference frame,
,V→ rel
= V→ I− V
→ ref
with the reference frame being the moving bucket. Therefore, the relative velocity is a constant
. The rate of change in volume in the bucket is then0 m/s (− 0 m/s) 0 m/sV→ rel
= − 3 − 2 = − 1 − V )A 10 m/s)(2 0 m ) .002 m /s∂t
∂V = (→ rel
∙ n̂ s = ( × 1 −4 2 = 0 3 36. Consider a velocity field where the x and y components of velocity are given by and xu = c
, where is a constant. The equation of the streamline isy v = − c c A. xy = c B. x y = − c C. y = c2 lnx D. x y = c2 −2 E. None of the above
The streamline equation is
udx = v
dy Substituting the velocities and integrating,
cxdx = dy
−cy ln x = − ln y +C1
Rearrange to solve for y as a function of x.
y x lny = C1 − lnx ⇒ = C2−1
37. Flow tangency is satisfied in the panel methods through the use of
A. Vortices B. Sources C. Sinks D. Both B and C E. None of the above
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38. A function, , is called a potential if it is related to a field, , byϕ V
→
A. ϕV→= ∇
B. ϕ = ∇ ∙ V→ C. ϕV
→= ∇ × ∇
D. ϕ onstantV→= ∇ + c
E. None of the above 39. Viscous effects can be neglected in aerodynamics when
A. Reynolds number is small B. Reynolds number is large C. Mach number is small D. The flow is incompressible E. None of the above
40. Air of density 1 kilogram per cubic meter flows at a speed of 20 meters per second. What
is the dynamic pressure? A. 20 Pascals B. 40 Pascals C. 200 Pascals D. 400 Pascals E. None of the above
The dynamic pressure is
ρV (1 kg/m )(20 m/s) 00 Pa q = 21 2 = 2
1 3 2 = 2
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