AE4451
Winter 2002
Review
Test
1 sheet of your handwriting
allowed.
(Answers in italics)
1. (28 points)
Plot the following quantities
along the axis of a
converging-diverging nozzle, starting from the reservoir, where the flow is not
moving at any appreciable speed, going
through the throat, and downstream, through a normal shock and out to
the nozzle exit. On the plot, also show the level of the outside pressure for
reference.
Stagnation temperature: constant
– unless the Mach number is so high that you have to consider changes in
specific heats across the shock.
Stagnation enthalpy: constant –
adiabatic process, no work extracted.
Stagnation pressure: constant
until the shock (isentropic expansion), drops across the shock and remains
constant at that level (isentropic deceleration) thereafter.
Static temperature: decreases
(from the stagnation value in the reservoir) through the nozzle until the
shock, shoots up across the shock; increases towards the stagnation value as
the flow slows down after the shock.
Static pressure: decreases
(from the stagnation value in the reservoir) through the nozzle until the
shock, shoots up across the shock; increases towards the new stagnation value
as the flow slows down after the shock. Much more than the temperature. The
final value of static temperature at the exit must be equal to the outside
(ambient) pressure because the region downstream of the shock is subsonic –
information can propagate upstream in the nozzle up to the shock, causing the
pressure at the nozzle exit to equalize with the outside pressure.
Mach number: increases from 0
at the reservoir to 1.0 at the throat, increases further in the supersonic
region in the divergent part of the nozzle (otherwise there can’t be a shock).
Drops across the shock to a subsonic value, then decreases further further
downstream in the nozzle.
Flow velocity: increases from 0
at the reservoir , and until the shock; drops across the shock, and further
decreases downstream of the shock as the area increases in subsonic flow in the
divergent part of the nozzle.
2. (30
points)
·
Given that the composition of air is 79% diatomic nitrogen,
20%diatomic oxygen and 1% rare gases with a molecular weight of 44, and the
ratio of specific heats for this air is 1.3, find the specific heats of air at
constant pressure and at constant volume.
Answer:
Molecular weight comes out to be 28.97 – gas constant becomes 287. Specific
heat at constant pressure is
, which comes out to
be 1244 m2 s-1 K-1
·
A stagnation probe is placed in an air flow where the
velocity is 200 m/s, static temperature is 500K, and static pressure is 1
atmosphere. What is the static enthalpy of the flow? What is the stagnation
enthalpy?
Answer:
;
which works out to
be
50,250 +
200*200/2 = 522,250 m2 s-2
·
Calculate the Mach angle at Mach 2.5
Answer: 23.58 degrees.
·
Calculate the speed of sound at the surface of the planet
Xylon (pressure: 0.1 million Newtons per square meter) where the atmosphere is
100% Xenon and the temperature is 400K.
Answer:
which is 206.25 m/s
3. (12
points)
Plot the section lift coefficient as a function of angle of
attack for a 2-D, low-speed, symmetric airfoil. Also plot the lift coefficient
versus angle of attack for a 3-D rectangular wing with a symmetric section
(incompressible flow). What is the slope of this line? Why, physically, are the
two slopes similar / different? What happens when the angle of attack gets
large?
Both are straight lines passing through (0,0) since the
airfoil section is symmetric; the 2-D lift curve slope is less than or equal to
per radian; the 3-D lift curve slope is lower because
tip losses and downwash increase as angle of attack increases. At high angle of
attack, the airfoil and the 3-D wing both stall. As angle of attack keeps on increasing, the lift coefficient
does increase again, reaching a maximum somewhere around 45 degrees (still much
less than the peak reached at best L/D). The drag coefficient is also very high at high angle of attack.
4. (30
points)
A shock “reflects”
as a shock from a solid wall (or plane of symmetry) and as an expansion
from a free surface.
Kutta condition – stagnation point at the trailing edge.
Same as the conditions at the edge of a supersonic jet exiting a nozzle – slip line, velocity vectors parallel on either side of the slip line, pressure same across the slip line.
In supersonic case, there is no information passing between the upper and lower surfaces, hence no mechanism for making each adjust to the other.