Question
Consider an aircraft flying at 250 m/s and -20 C. At one point on the wing, the temperature is 240 K. Hint: is 1.008
What is the velocity at this point?
unanswered
Suppose now the free-stream pressure is 38794 Pa.
Calculate the pressure in the point on the wing, in Pa
unanswered
If it is also given that the free-stream density is 0.53396 kg/m, we can calculate the density.
Calculate the density at this point on the wing
What is the velocity at this point?
unanswered
Suppose now the free-stream pressure is 38794 Pa.
Calculate the pressure in the point on the wing, in Pa
unanswered
If it is also given that the free-stream density is 0.53396 kg/m, we can calculate the density.
Calculate the density at this point on the wing
Answers
GPT-4o mini
To solve the problem, we will use the following relationships from fluid dynamics and thermodynamics.
### 1. Velocity at the point on the wing
Using the temperature ratio, we can find the relationship between the velocities. The relationship between the temperatures and velocities can be described by the following equation:
\[
\frac{T_1}{T_2} = \left(\frac{V_1}{V_2}\right)^k
\]
where:
- \(T_1\) is the temperature in the free stream,
- \(T_2\) is the temperature at the point on the wing,
- \(V_1\) is the free stream velocity,
- \(V_2\) is the velocity at the point on the wing,
- \(k\) is the specific heat ratio.
### Given:
- \(T_1 = -20 \, \text{C} = 253.15 \, \text{K}\)
- \(T_2 = 240 \, \text{K}\)
- \(V_1 = 250 \, \text{m/s}\)
- \(k = 1.008\)
### Plugging in the values, we find \(V_2\):
\[
\frac{253.15}{240} = \left(\frac{250}{V_2}\right)^{1.008}
\]
Now, we can solve for \(V_2\):
Rearranging gives:
\[
\left(\frac{250}{V_2}\right) = \left(\frac{253.15}{240}\right)^{\frac{1}{1.008}}
\]
Calculating \(\frac{253.15}{240}\):
\[
\frac{253.15}{240} \approx 1.0547
\]
Now taking to the power of \(\frac{1}{1.008}\):
\[
\left(1.0547\right)^{\frac{1}{1.008}} \approx 1.045
\]
Thus, we have:
\[
\frac{250}{V_2} \approx 1.045
\]
Now solving for \(V_2\):
\[
V_2 \approx \frac{250}{1.045} \approx 239.04 \, \text{m/s}
\]
### 2. Calculate the pressure at the wing point
Using Bernoulli’s equation, we can estimate the dynamic pressures. The pressure \(P_2\) at the point on the wing can be calculated as:
\[
P_2 = P_1 + \frac{1}{2} \rho (V_1^2 - V_2^2)
\]
Where:
- \(P_1\) is the free-stream pressure (38794 Pa),
- \(\rho\) is the density,
- \(V_1\) and \(V_2\) are the free stream and local velocities respectively.
### First, we need to find the local density \( \rho \).
We can use the ideal gas law to calculate the density based on the free-stream conditions and the Mach number relationship.
### Given:
The free-stream density is 0.53396 kg/m³.
Thus, we will assume:
\[
\rho_1 = 0.53396 \, \text{kg/m}^3
\]
Assuming \(P_1\) and \(T_1\) are from the ideal gas law, we can find \(P_2\).
Now substituting numbers in pressure equation:
\[
P_2 = P_1 - \frac{1}{2} \rho_1 (V_1^2 - V_2^2)
\]
Calculating:
\[
dP = \frac{1}{2} \cdot 0.53396 \, \text{kg/m}^3 \cdot (250^2 - (239.04)^2)
\]
Calculating the velocities:
- \(250^2 = 62500\),
- \((239.04)^2 \approx 57100.81\)
Now,
\[
dP = \frac{1}{2} \cdot 0.53396 \cdot (62500 - 57100.81) \approx \frac{1}{2} \cdot 0.53396 \cdot 5399.19 \approx 1434.73 \, \text{Pa}
\]
Thus:
\[
P_2 = 38794 - 1434.73 \approx 37359.27 \, \text{Pa}
\]
### 3. Calculate the density
To calculate the local density at the point on the wing, we can use the ideal gas law:
\[
\rho = \frac{P}{RT}
\]
Where \(R\) for air can be approximated to be \(287.05 \, \text{J/(kg K)}\) and at temperature \(T_2 = 240 K\):
\[
\rho_2 = \frac{P_2}{RT_2} = \frac{37359.27}{287.05 \cdot 240}
\]
Calculating \(R T_2\):
\[
287.05 \cdot 240 \approx 68826 \, \text{J/m}^3
\]
Now:
\[
\rho_2 = \frac{37359.27}{68826} \approx 0.542 \, \text{kg/m}^3
\]
### Summary:
1. Velocity at the point on the wing \(V_2 \approx 239.04 \, \text{m/s}\)
2. Pressure in the point on the wing \(P_2 \approx 37359.27 \, \text{Pa}\)
3. Density at the point on the wing \(\rho_2 \approx 0.542 \, \text{kg/m}^3\)
### 1. Velocity at the point on the wing
Using the temperature ratio, we can find the relationship between the velocities. The relationship between the temperatures and velocities can be described by the following equation:
\[
\frac{T_1}{T_2} = \left(\frac{V_1}{V_2}\right)^k
\]
where:
- \(T_1\) is the temperature in the free stream,
- \(T_2\) is the temperature at the point on the wing,
- \(V_1\) is the free stream velocity,
- \(V_2\) is the velocity at the point on the wing,
- \(k\) is the specific heat ratio.
### Given:
- \(T_1 = -20 \, \text{C} = 253.15 \, \text{K}\)
- \(T_2 = 240 \, \text{K}\)
- \(V_1 = 250 \, \text{m/s}\)
- \(k = 1.008\)
### Plugging in the values, we find \(V_2\):
\[
\frac{253.15}{240} = \left(\frac{250}{V_2}\right)^{1.008}
\]
Now, we can solve for \(V_2\):
Rearranging gives:
\[
\left(\frac{250}{V_2}\right) = \left(\frac{253.15}{240}\right)^{\frac{1}{1.008}}
\]
Calculating \(\frac{253.15}{240}\):
\[
\frac{253.15}{240} \approx 1.0547
\]
Now taking to the power of \(\frac{1}{1.008}\):
\[
\left(1.0547\right)^{\frac{1}{1.008}} \approx 1.045
\]
Thus, we have:
\[
\frac{250}{V_2} \approx 1.045
\]
Now solving for \(V_2\):
\[
V_2 \approx \frac{250}{1.045} \approx 239.04 \, \text{m/s}
\]
### 2. Calculate the pressure at the wing point
Using Bernoulli’s equation, we can estimate the dynamic pressures. The pressure \(P_2\) at the point on the wing can be calculated as:
\[
P_2 = P_1 + \frac{1}{2} \rho (V_1^2 - V_2^2)
\]
Where:
- \(P_1\) is the free-stream pressure (38794 Pa),
- \(\rho\) is the density,
- \(V_1\) and \(V_2\) are the free stream and local velocities respectively.
### First, we need to find the local density \( \rho \).
We can use the ideal gas law to calculate the density based on the free-stream conditions and the Mach number relationship.
### Given:
The free-stream density is 0.53396 kg/m³.
Thus, we will assume:
\[
\rho_1 = 0.53396 \, \text{kg/m}^3
\]
Assuming \(P_1\) and \(T_1\) are from the ideal gas law, we can find \(P_2\).
Now substituting numbers in pressure equation:
\[
P_2 = P_1 - \frac{1}{2} \rho_1 (V_1^2 - V_2^2)
\]
Calculating:
\[
dP = \frac{1}{2} \cdot 0.53396 \, \text{kg/m}^3 \cdot (250^2 - (239.04)^2)
\]
Calculating the velocities:
- \(250^2 = 62500\),
- \((239.04)^2 \approx 57100.81\)
Now,
\[
dP = \frac{1}{2} \cdot 0.53396 \cdot (62500 - 57100.81) \approx \frac{1}{2} \cdot 0.53396 \cdot 5399.19 \approx 1434.73 \, \text{Pa}
\]
Thus:
\[
P_2 = 38794 - 1434.73 \approx 37359.27 \, \text{Pa}
\]
### 3. Calculate the density
To calculate the local density at the point on the wing, we can use the ideal gas law:
\[
\rho = \frac{P}{RT}
\]
Where \(R\) for air can be approximated to be \(287.05 \, \text{J/(kg K)}\) and at temperature \(T_2 = 240 K\):
\[
\rho_2 = \frac{P_2}{RT_2} = \frac{37359.27}{287.05 \cdot 240}
\]
Calculating \(R T_2\):
\[
287.05 \cdot 240 \approx 68826 \, \text{J/m}^3
\]
Now:
\[
\rho_2 = \frac{37359.27}{68826} \approx 0.542 \, \text{kg/m}^3
\]
### Summary:
1. Velocity at the point on the wing \(V_2 \approx 239.04 \, \text{m/s}\)
2. Pressure in the point on the wing \(P_2 \approx 37359.27 \, \text{Pa}\)
3. Density at the point on the wing \(\rho_2 \approx 0.542 \, \text{kg/m}^3\)