Question

Let us start with the example wind tunnel of the lecture, but with a small change: the test section should still have sea-level conditions, but with M = 4.

What reservoir pressure is required? Enter your answer in bar
unanswered

What reservoir temperature is required? Enter your answer in K
unanswered

What is the flow speed in the test section, in m/s?
unanswered

What is the flow speed in the throat, in m/s?
unanswered

What is the expansion ratio, /

Answer 1

To determine the required reservoir pressure, temperature, flow speeds, and expansion ratio for a wind tunnel with a test section at sea-level conditions and a Mach number of 4 (M=4), we can use the relationships and equations from compressible flow.

### Given:
- **Mach Number (M)**: 4
- **Sea-level conditions** (typical values):
- Pressure \( P_0 \): 101.325 kPa (or 1.01325 bar)
- Temperature \( T_0 \): 288.15 K
- Specific heat ratio \( \gamma \): 1.4 (for air)

### 1. Required Reservoir Pressure

The relationship between reservoir and test section pressures for supersonic flow can be derived from the isentropic flow relations. The following can be used:

\[
P_t = P_0 \left(1 + \frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}
\]

Where:
- \( P_t \) is the total (reservoir) pressure
- \( M \) is the Mach number (M=4)
- Pressures must be in the same units.

Substituting the values:

\[
P_t = 101.325 \left(1 + \frac{1.4 - 1}{2} \cdot 4^2\right)^{\frac{1.4}{1.4 - 1}}
\]

Calculating inside the parentheses:

\[
= 101.325 \left(1 + 0.2 \cdot 16\right)^{3.5}
= 101.325 \left(1 + 3.2\right)^{3.5}
= 101.325 \left(4.2\right)^{3.5}
\]

Now calculate \(4.2^{3.5}\):
\[
4.2^{3.5} \approx 152.114
\]
Calculating \( P_t \):
\[
P_t \approx 101.325 \cdot 152.114 \approx 15473.31 \, \text{Pa} \approx 15.47 \, \text{bar}
\]

### 2. Required Reservoir Temperature

Using the relationship between reservoir temperature and Mach number:
\[
T_t = T_0 \left(1 + \frac{\gamma - 1}{2} M^2\right)
\]

Substituting values:
\[
T_t = 288.15 \left(1 + \frac{1.4 - 1}{2} \cdot 4^2\right)
= 288.15 \left(1 + 0.2 \cdot 16\right)
= 288.15 \left(1 + 3.2\right)
= 288.15 \cdot 4.2
\]
Calculating:
\[
T_t \approx 288.15 \cdot 4.2 = 1216.23 \, \text{K}
\]

### 3. Flow Speed in the Test Section

The flow speed in the test section can be calculated using:
\[
V = M \cdot a
\]

Where \( a \) is the speed of sound at sea-level conditions:

\[
a = \sqrt{\gamma R T} \quad \text{(where R is the specific gas constant)}
\]
For air, \( R \approx 287 \, \text{J/(kg·K)} \):

Calculating \( a \):
\[
a = \sqrt{1.4 \cdot 287 \cdot 288.15} \approx \sqrt{116388.6} \approx 341 \, \text{m/s}
\]

Now calculate \( V \):
\[
V = 4 \cdot 341 \approx 1364 \, \text{m/s}
\]

### 4. Flow Speed in the Throat

The flow speed in the throat (where Mach number is 1) can be calculated as:
\[
V_t = a_t = 341 \, \text{m/s} \quad \text{(at M = 1)}
\]

### 5. Expansion Ratio

The expansion ratio \( \epsilon \) can be determined as follows:
\[
\epsilon = \frac{A_t}{A_0} = \frac{1}{M} \cdot \left(\frac{P_0}{P_t}\right)^{\frac{(1 - 1/\gamma)}{2}}
\]

For M=4, we first calculate the area ratio from the isentropic relations, which typically follow the trend of high Mach flow ratios.

But since this is a theoretical evaluation, the specific method to determine this is not straightforward without simulation or empirical data but can be approximated as \( A_t \) approaching the limit.

### Summary

- **Reservoir Pressure (P_t)**: ≈ 15.47 bar
- **Reservoir Temperature (T_t)**: ≈ 1216.23 K
- **Flow Speed in the Test Section**: ≈ 1364 m/s
- **Flow Speed in the Throat**: ≈ 341 m/s
- **Expansion Ratio**: Theoretical calculations (would require more details for exact numbers).

Please verify all calculations and assumptions based on your specific context and settings.