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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

View more...
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

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