Question
A Boeing 747 is flying at M = 0.7 at an altitude where the temperature is 223.15 K
and the density is 0.41268 kg/m3. The pressure coefficient on a point of the wing is -2.
Calculate the pressure (in Pa) at this point on the wing:
and the density is 0.41268 kg/m3. The pressure coefficient on a point of the wing is -2.
Calculate the pressure (in Pa) at this point on the wing:
Answers
GPT-4o mini
To calculate the pressure at a point on the wing of the Boeing 747, we can use the pressure coefficient, which is defined as:
\[
C_p = \frac{P - P_{\infty}}{\frac{1}{2} \rho_{\infty} V_{\infty}^2}
\]
Where:
- \( C_p \) = pressure coefficient
- \( P \) = pressure at the point on the wing
- \( P_{\infty} \) = free-stream pressure
- \( \rho_{\infty} \) = free-stream density
- \( V_{\infty} \) = free-stream velocity
Given:
- \( C_p = -2 \)
- \( \rho_{\infty} = 0.41268 \, \text{kg/m}^3 \)
- The altitude temperature of the air is \( T = 223.15 \, \text{K} \)
### Step 1: Calculate the free-stream velocity \( V_{\infty} \)
The speed of sound \( a \) at the given temperature can be calculated using the formula:
\[
a = \sqrt{\gamma R T}
\]
Where:
- \( \gamma \) for air = 1.4 (approx)
- \( R \) for air = 287.05 J/(kg·K)
Now, calculate \( a \):
\[
a = \sqrt{1.4 \times 287.05 \times 223.15}
\]
Calculating:
\[
a \approx \sqrt{1.4 \times 287.05 \times 223.15} \approx \sqrt{113157.774} \approx 336.67 \, \text{m/s}
\]
Now, calculate the free-stream velocity \( V_{\infty} \) using the Mach number \( M = 0.7 \):
\[
V_{\infty} = M \cdot a = 0.7 \cdot 336.67 \approx 235.67 \, \text{m/s}
\]
### Step 2: Calculate the free-stream pressure \( P_{\infty} \)
We can use the ideal gas law to find \( P_{\infty} \):
\[
P = \rho R T
\]
Calculating \( P_{\infty} \):
\[
P_{\infty} = 0.41268 \, \text{kg/m}^3 \times 287.05 \, \text{J/(kg \cdot K)} \times 223.15 \, \text{K}
\]
Calculating:
\[
P_{\infty} \approx 0.41268 \times 287.05 \times 223.15 \approx 25735.76 \, \text{Pa}
\]
### Step 3: Calculate the pressure at the point on the wing \( P \)
Now, we can rearrange the equation for \( C_p \) to solve for \( P \):
\[
P = P_{\infty} + C_p \cdot \frac{1}{2} \rho_{\infty} V_{\infty}^2
\]
Calculating \( \frac{1}{2} \rho_{\infty} V_{\infty}^2 \):
\[
\frac{1}{2} \rho_{\infty} V_{\infty}^2 = \frac{1}{2} \times 0.41268 \times (235.67)^2
\]
Calculating:
\[
= 0.20634 \times 55646.71 \approx 11455.77 \, \text{Pa}
\]
Now substituting back into the equation for \( P \):
\[
P = 25735.76 + (-2) \times 11455.77
\]
\[
P = 25735.76 - 22911.54 \approx 2824.22 \, \text{Pa}
\]
### Final Result:
The pressure at the point on the wing is approximately **2824.22 Pa**.
\[
C_p = \frac{P - P_{\infty}}{\frac{1}{2} \rho_{\infty} V_{\infty}^2}
\]
Where:
- \( C_p \) = pressure coefficient
- \( P \) = pressure at the point on the wing
- \( P_{\infty} \) = free-stream pressure
- \( \rho_{\infty} \) = free-stream density
- \( V_{\infty} \) = free-stream velocity
Given:
- \( C_p = -2 \)
- \( \rho_{\infty} = 0.41268 \, \text{kg/m}^3 \)
- The altitude temperature of the air is \( T = 223.15 \, \text{K} \)
### Step 1: Calculate the free-stream velocity \( V_{\infty} \)
The speed of sound \( a \) at the given temperature can be calculated using the formula:
\[
a = \sqrt{\gamma R T}
\]
Where:
- \( \gamma \) for air = 1.4 (approx)
- \( R \) for air = 287.05 J/(kg·K)
Now, calculate \( a \):
\[
a = \sqrt{1.4 \times 287.05 \times 223.15}
\]
Calculating:
\[
a \approx \sqrt{1.4 \times 287.05 \times 223.15} \approx \sqrt{113157.774} \approx 336.67 \, \text{m/s}
\]
Now, calculate the free-stream velocity \( V_{\infty} \) using the Mach number \( M = 0.7 \):
\[
V_{\infty} = M \cdot a = 0.7 \cdot 336.67 \approx 235.67 \, \text{m/s}
\]
### Step 2: Calculate the free-stream pressure \( P_{\infty} \)
We can use the ideal gas law to find \( P_{\infty} \):
\[
P = \rho R T
\]
Calculating \( P_{\infty} \):
\[
P_{\infty} = 0.41268 \, \text{kg/m}^3 \times 287.05 \, \text{J/(kg \cdot K)} \times 223.15 \, \text{K}
\]
Calculating:
\[
P_{\infty} \approx 0.41268 \times 287.05 \times 223.15 \approx 25735.76 \, \text{Pa}
\]
### Step 3: Calculate the pressure at the point on the wing \( P \)
Now, we can rearrange the equation for \( C_p \) to solve for \( P \):
\[
P = P_{\infty} + C_p \cdot \frac{1}{2} \rho_{\infty} V_{\infty}^2
\]
Calculating \( \frac{1}{2} \rho_{\infty} V_{\infty}^2 \):
\[
\frac{1}{2} \rho_{\infty} V_{\infty}^2 = \frac{1}{2} \times 0.41268 \times (235.67)^2
\]
Calculating:
\[
= 0.20634 \times 55646.71 \approx 11455.77 \, \text{Pa}
\]
Now substituting back into the equation for \( P \):
\[
P = 25735.76 + (-2) \times 11455.77
\]
\[
P = 25735.76 - 22911.54 \approx 2824.22 \, \text{Pa}
\]
### Final Result:
The pressure at the point on the wing is approximately **2824.22 Pa**.