Note that the accuracy required for the questions is high, on most questions a margin of 2% has

been set (just as in the homework exercises). This could mean that an answer of 0.3 is marked
wrong when the actual answer is 0.31. Therefore, always enter a su�cient number of signi�cant
�gures (usually 3 or 4 su�ce).
Consider a Cessna C208B Grand Caravan, �ying at 3000m at a relatively low speed of 245 km/hr
to allow a load of skydivers to exit the plane. Here you see the aircraft just after take-o�:
Figure 1: A Cessna Grand Caravan. Image CC - BY 2.0 by Gerard van der Schaaf
At 3000m, the following properties of air are known:
Pressure: ph = 70121 N/m2 Density: �h = 0:90926 kg/m3 Temperature: Th = 268:67 K
A) Calculate the Mach number to show that the compressibility of the �ow can be neglected.
B) At a point B on top of the wing, just outside the boundary layer, the air�ow velocity relative
to the wing is VB= 85 m/s. Calculate the pressure in point B, in Pa.
C) At a point C on the wing, the pressure is 68000 Pa. Calulate the pressure coe�cient in point
C.
D) Suppose now that under a certain angle of attack, the minimum pressure coe�cient at a very
low Mach number is -2. Calulate the critical Mach number for this angle of attack.

1 answer

To solve the questions step by step, we will use the relevant formulas and values provided.

A) Calculate the Mach Number

The Mach number (M) is defined as the ratio of the speed of the object (in this case, the aircraft) to the speed of sound in the surrounding medium (air):

\[ M = \frac{V}{a} \]

Where:

  • \( V \) is the speed of the aircraft in m/s.
  • \( a \) is the speed of sound in air at the given conditions.
  1. Convert speed from km/hr to m/s:

\[ V = 245 \frac{km}{hr} \times \frac{1000 , m}{1 , km} \times \frac{1 , hr}{3600 , s} = 68.06 , m/s \]

  1. Calculate the speed of sound \( a \):

The speed of sound in air can be calculated using the following formula:

\[ a = \sqrt{\gamma \cdot \frac{R}{M} \cdot T} \]

Where:

  • \( \gamma \) (ratio of specific heats) for air is approximately 1.4.
  • \( R \) (specific gas constant for dry air) = 287 J/(kg·K).
  • \( T \) is the temperature in Kelvin (given as \( T_h = 268.67 , K \)).

Now substituting the values:

\[ a = \sqrt{1.4 \cdot 287 \cdot 268.67} \approx \sqrt{108496.48} \approx 329.77 , m/s \]

  1. Calculate the Mach number:

\[ M = \frac{V}{a} = \frac{68.06}{329.77} \approx 0.206 \]

Since \( M < 0.3\), we can say that the compressibility of the flow can be neglected.

B) Calculate the pressure at point B

The pressure at point B (just outside the boundary layer) can be calculated using Bernoulli's equation, which states that in a streamline flow, the total mechanical energy is conserved:

\[ p_B = p_h + \frac{1}{2} \rho_h V_h^2 - \frac{1}{2} \rho_h V_B^2 \]

Where:

  • \( p_h = 70121 , N/m^2 \) (pressure at height)
  • \( \rho_h = 0.90926 , kg/m^3 \) (density)
  • \( V_h \) is the flow velocity around the aircraft (which can be approximated as the speed of the aircraft since it's at a relatively low angle of attack), so \( V_h = 68.06 , m/s\).
  • \( V_B = 85 , m/s \) (velocity at point B)

Substituting the values into Bernoulli's equation:

\[ p_B = 70121 + \frac{1}{2} \cdot 0.90926 \cdot (68.06^2) - \frac{1}{2} \cdot 0.90926 \cdot (85^2) \]

Calculating \( \frac{1}{2} \cdot 0.90926 \):

  • \( \frac{1}{2} * 0.90926 = 0.45463 \)

Calculating \( 68.06^2 \) and \( 85^2 \):

  • \( 68.06^2 \approx 4636.52 \)
  • \( 85^2 \approx 7225 \)

Substituting back in:

\[ p_B = 70121 + 0.45463 \cdot 4636.52 - 0.45463 \cdot 7225 \]

Calculating:

  • \( 0.45463 \cdot 4636.52 \approx 2106.55 \)
  • \( 0.45463 \cdot 7225 \approx 3288.49 \)

Now:

\[ p_B = 70121 + 2106.55 - 3288.49 \approx 70121 - 1181.94 \approx 68939.06 , Pa \]

C) Calculate the pressure coefficient at point C

The pressure coefficient \( C_p \) is defined as:

\[ C_p = \frac{p - p_h}{\frac{1}{2} \rho_h V_h^2} \]

Where \( p_C = 68000 , Pa \).

Substituting the values we have:

\[ C_p = \frac{68000 - 70121}{\frac{1}{2} \cdot 0.90926 \cdot (68.06^2)} \]

Calculating \( \frac{1}{2} \cdot \rho_h \cdot V_h^2 \):

\[ \frac{1}{2} \cdot 0.90926 \cdot 4636.52 \approx 2106.55 , Pa \]

So now:

\[ C_p = \frac{68000 - 70121}{2106.55} = \frac{-121}{2106.55} \approx -0.0574 \]

D) Calculate the critical Mach number

The critical Mach number \( M_{crit} \) is defined by the relationship with the minimum pressure coefficient:

\[ C_{p, min} = \frac{p_{critical} - p_h}{\frac{1}{2} \rho_h V_{crit}^2} \]

Given \( C_{p,min} = -2 \).

From this we can derive the critical Mach number \( M_{crit} \):

\[ C_{p,min} = \frac{68000 - 70121}{\frac{1}{2} \cdot 0.90926 \cdot V_{crit}^2} \]

Rearranging gives:

\[ -2 = \frac{-121}{\frac{1}{2} \cdot 0.90926 \cdot V_{crit}^2} \]

Thus,

\[ 2 \cdot \frac{1}{2} \cdot 0.90926 \cdot V_{crit}^2 = 121 \]

\[ 0.90926 \cdot V_{crit}^2 = 121 \]

\[ V_{crit}^2 = \frac{121}{0.90926} \approx 133.4 \]

Taking the square root gives:

\[ V_{crit} = \sqrt{133.4} \approx 11.55 , m/s \]

Now we can find the critical Mach number using the previously found speed of sound \( a \):

\[ M_{crit} = \frac{V_{crit}}{a} = \frac{11.55}{329.77} \approx 0.035 \]

This critical Mach number indicates that the aircraft can operate at speeds below this range without encountering compressibility effects.

Summary of Answers:

A) Mach number: \( M \approx 0.206 \) (compressibility can be neglected).

B) Pressure at point B: \( p_B \approx 68939.06 , Pa\).

C) Pressure coefficient at point C: \( C_p \approx -0.0574 \).

D) Critical Mach number: \( M_{crit} \approx 0.035 \).

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