Question

Suppose it is your job to make a subsonic, incompressible wind tunnel. The test section should be square, 2x2m. The conditions in the test section are given to be standard sea level conditions and a flow speed of 50 m/s.


Image courtesy of NASA , CC - BY

What should be the mass flow through the tunnel?

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The flow passes through a circular cross-section with a radius of 2 meters. What is the flow speed in this cross-section?

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Answer 1

To calculate the mass flow through the wind tunnel and the flow speed in the circular cross-section, we will follow these steps:

1. **Calculate the mass flow rate through the test section.**
2. **Determine the flow speed in the circular cross-section.**

### Step 1: Calculate the Mass Flow Rate

The mass flow rate (\( \dot{m} \)) can be calculated using the equation:
\[
\dot{m} = \rho \cdot A \cdot V
\]
where:
- \( \rho \) is the air density at standard sea level conditions (approximately \( 1.225 \, \text{kg/m}^3 \)),
- \( A \) is the cross-sectional area of the test section,
- \( V \) is the flow velocity.

For a square test section with dimensions \( 2 \times 2 \, \text{m} \):
\[
A = 2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{m}^2
\]

Substituting the values into the equation:
\[
\dot{m} = 1.225 \, \text{kg/m}^3 \cdot 4 \, \text{m}^2 \cdot 50 \, \text{m/s}
\]
Calculating this gives:
\[
\dot{m} = 1.225 \cdot 4 \cdot 50 = 245 \, \text{kg/s}
\]

### Step 2: Determine the Flow Speed in the Circular Cross-Section

Next, we need to find the flow speed in the circular cross-section. The area of the circular cross-section is given by:
\[
A_c = \pi r^2
\]
Given that the radius (\( r \)) is 2 m:
\[
A_c = \pi (2 \, \text{m})^2 = 4\pi \, \text{m}^2 \approx 12.566 \, \text{m}^2
\]

Now, using the principle of conservation of mass, the mass flow rate must remain constant. Therefore, we can say:
\[
\dot{m} = \rho \cdot A_c \cdot V_c
\]
where \( V_c \) is the flow speed in the circular cross-section.

Rearranging gives:
\[
V_c = \frac{\dot{m}}{\rho \cdot A_c}
\]
Substituting for \( \dot{m} \), \( \rho \), and \( A_c \):
\[
V_c = \frac{245 \, \text{kg/s}}{1.225 \, \text{kg/m}^3 \cdot 12.566 \, \text{m}^2}
\]
Calculating this gives:
\[
V_c \approx \frac{245}{15.372} \approx 15.95 \, \text{m/s}
\]

### Summary of Results
- **Mass Flow Rate through the Tunnel:** \( \dot{m} \approx 245 \, \text{kg/s} \)
- **Flow Speed in the Circular Cross-Section:** \( V_c \approx 15.95 \, \text{m/s} \)