AE1110x - Introduction to Aeronautical Engineering
Question 1 A) The temperature in the channel decreases. Using the energy equation: if the speed goes up, the temperature must drop. Note that even for flows with friction the effect of the area increase is big enough to counter the effect of friction.
Question 2 C) 5%, as has been treated in the lectures.
Question 3 C) Both 3 and 4 are true; see lecture 4 on flow separation.
Question 4 D) The pressure decreases and the temperature decreases. This can be seen from the graphs in lecture 2, or calculated using the second form of the isentropic relations.
Question 5 D) Is difficult to establish without additional information. Due to the Mach number the lift curve slope goes up. Due to the wing the lift curve slope goes down. As long as you do not have any information of the aspect ratio of the wing and the span effectiveness factor you do not know how much it goes down. (it may even very well remain constant).
Test Module B - Answers
1
AE1110x - Introduction to Aeronautical Engineering
Cessna Grand Caravan A) The Mach number can be calculated by calculating the speed in m/s and the speed of sound, and dividing the two. The speed of sound is given by : a =
The speed of the aircraft is:
γRT =
√
1.4 · 287 · 268.67 = 328.56 m/s 245 = 68.1 m/s 3.6
V =
Then: M =
68.1 V = = 0.207 328.56 a
So M < 0.3 and the flow is incompressible. B) For this question, we can use Bernoulli’s equation: p + ∞
1 2 1 ρV = p B + ρV B2 2 2 ∞
Rearranging for p B gives: 1 ρ V 2 − V B2 2 1 = 70121 + · 0.90926 · 68.12 − 852 2 = 68942 Pa
p B = p + ∞
∞
C) The pressure coefficient is defined as: C p =
p − p 0 1 2 0 2 ρV
Substituting values: C p =
68000 − 70121 1 2 2 · 0.90926 · 68.1
= -1.007
D) The critical Mach number can be found by taking the intersection of the airfoils-specific curve (the Prandtl-Glauert correction) and the airfoil-independent curve (for the critical pressure coefficient, given the free-stream Mach number). We can solve this for the free-stream Mach number using, for instance, a graphical calculator: C p ,crit =
2 γM 2
∞
2
2 + ( γ − 1) M γ + 1
∞
γ γ −1
For C p,0 = −2, we find a critical Mach number of 0.4863.
2
−1
=
C p, 0
− 1
M 2
∞
Test Module B - Answers
AE1110x - Introduction to Aeronautical Engineering
Boeing 727 √
√
A) For the Mach number we first need the speed of sound, a = γRT = 1.4 · 287 · 255.7 = 320.5 m/s. We also need to convert the speed of the aircraft: V = 810 3.6 = 225 m/s. Then the Mach number is M =
V a
=
225 320.5
=0.702. ∆C l ∆α
B) The lift gradient of the profile is simply a0 =
=
0.6−0 4−(−2)
=0.10 /degree.
C) The airfoil is mounted on a aircraft, and then we start flying at a higher Mach number. It is important to note this order: if you apply the corrections the other way around, you will get incorrect results. First, let us calculate the aspect ratio: 34.52 b 2 = = 7.988 A = 149 S Then, we apply a correction for a finite wing: awing =
=
a0
1+
57.3a0 πAe
0.11 1+
57.3 0.11 π 7.988 0.82
·
·
= 0.0842
·
Then, we apply the Prandtl-Glauert correction for the Mach number: awing,M =0.6 =
=
a
√ wing 2 1 − M 0 0842 √ . 1 − 0.62
= 0.10526
D) For the lift-to-drag ratio, we need the lift and the drag, or alternatively, the lift coefficient and the drag coefficient. The lift coefficient at 3 degrees can be found, using the zero-lift angle of -2 degrees: C L = a · (α − α0 ) = 0.1179 · (3 − (−2)) = 0.5895 The drag coefficient is the sum of the profile drag and the induced drag: C D = C d +
C L2 πAe
0.5895 2 = 0.0062 + π · 7.988 · 0.82 = 0.023088 The lift-drag ratio is then: 0.5895 L C L = = 0.023088 D C D = 25.53
Test Module B - Answers
3
AE1110x - Introduction to Aeronautical Engineering
E) In the stagnation point, the velocity is zero. Since the flow is compressible we can use the energy equation: 1 1 C p T 0 + V 02 = C p T 1 + V 12 2 2 If we take the free-stream conditions for the point 1, we get: T 0 =
C p T 1 +
1 2
V 12 − V 02
C p
1008 · 255.7 + 12 2252 − 02 = 1008 = 280.8 K
F) An equation to relate temperatures and densities for compressible flows is one of the isentropic relations: ρ2 = ρ1
1
· T 2 T 1
ρ2 = ρ 1
γ −1
T 2 T 1
= 0.736
1
γ −1
250 255.7
1 1.4−1
= 0.6957
4
Test Module B - Answers
