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ASEN 3111

Aerodynamics

Exam 1, Sept 18 Fall 2001

Name:

This is a open book, closed note exam.

Units of the answers must be consistent with those of the problem statement.

Give vectorial solutions as vectors.

1. At entry to circular orbit, the space shuttle is at an altitude of 200km and traveling at7.8km/s. The standard atmosphere predicts a molecular mean free path of 240m and adensity of 1.5 1010kg/m3 at this altitude. The shuttle cross-sectional area, normal tothe direction of flight is about 131 m2. What is the first approximation to the drag underthese conditions? Justify the model you use and the origin of your first approximation.There are two parts to the answer:

1. Which flow regime is experienced by the shuttle?

Given that we are not given the length or span of the space shuttle, we can obtain acharacteristic length by taking: L =

131 = 11.4m Thus, Kn = /L = 240/11.4 =

21 is much bigger than one and therefore we must use a molecular approach.2. What is the drag?

We are not given any detailed shape of the space shuttle, but we know that inthe case of a perfect absorption of the incoming molecules, the drag coefficientwould be 2. This is the first approximation to the drag coefficient and it tends tounderestimate the drag by 10 to 20 percent. Using Cd = 2, we obtain:

D =1

2U

2

SCd = 0.5 1.5 1010 (7.8 103)2 131 2 = 1.2N

2. Show whether the following incompressible flow satisfy the mass conservation equation:

V = r cos2 er r sin2 eThe equation of continuity in an incompressible flow (in steady state or unsteady flow)

is given by:.V = 0

Substituting the value for and V:

.V = (er .r

+e

r

.

).(r cos2 er r sin2 e)

= er. r

(r cos2 er r sin2 e) + er.

(r cos2 er r sin2 e)

= er.(cos 2 er sin2 e) +e

r.(2r sin2 er + r cos2 e 2r cos2 e r sin2 er)

= cos2 + cos 2 2cos2 = 0

The equation of continuity is therefore satisfied.

Show all work to receive credit Page 1 of 3 Use back of page if necessary

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ASEN 3111

Aerodynamics


3. A nozzle is directing water at a fixed wall. The velocity of water at the exit of the nozzleis 35m/s. The nozzle cross-section is circular with a diameter of 2.4in. The density ofwater is = 1000kg/m3 and pressure of the air surrounding the nozzle is 1atm.

Nozzle

PP

ControlVolume

(a) What is the total pressure force exerted by the flow on the wall? We can write theintegral form of the momentum equation (equation 2.13) and we note the following:

As the flow is assumed steady , the derivative with respect to time is zero. The surface force Fsurface is equal to the pressure force exerted by the fluid on

the plate, but it is opposite in direction

The plate being vertical, any pressure force on it must be horizontal and weneed consider only the horizontal component of the equation

The body force which is due to gravity is vertical and therefore can be ignored,following the previous remark

Therefore, we have:

Fplate = Fsurface,x = AVx(V .n)dA

=

A

ui(V .n)dA

We then note that the integral on the surface has two contributions where the flowcrosses the boundary. The first contribution is at the nozzle and the second is faraway from the center of the flow. This second contribution has a zero velocity alongthe x axis and therefore:

Fplate = uuAnozzlei = 35

1000

35

(1.2

2.54 102)2i = 3575iN


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ASEN 3111

Aerodynamics


(b) What is the total pressure at point P ?

At point A, the pressure of the flow must be the atmospheric pressure (PA = 1atm =1.01325 105Pa, and the velocity of the flow must be VA = 35m/s. At point P, bysymmetry, the flow cannot go up nor down, therefore its velocity must be VP = 0.Writing Bernoullis Equation for these two points:

PA + gZA +1

2V2A = PP + gZP +

1

2V2P

and by rearranging and noting that ZA = ZP:

PP = PA + g(ZA ZP) + 12(V2A V2P)

= 1.01325 105 + 1000 9.81 0 + 0.5 1000 (352 0)= 7.13825 105 Pa


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