Give complete answer, answer of all sub parts a to cQuestion 3- Supersonic Wind Tunnel (25 points): A blow down supersonic wind tunnel has a test section
area of 0.2m” and a design Mach number of 2.5

Question

Give complete answer, answer of all sub parts a to cQuestion 3- Supersonic Wind Tunnel (25 points): A blow down supersonic wind tunnel has a test section
area of 0.2m” and a design Mach number of 2.5. The pressure and temperature in the high pressure air
tankis 50 MPa and 400 K. Ignore all losses between the high pressure tank and the test section.
supersonic ects est section
Setting chor woser
ext those
pressure coy
=| =~ exe
a) Estimate the throat area of the nozzle.
b) Estimate the static pressure and static temperature in the test section.
Estimate the required gas temperature in the high pressure tank to achieve 298 K
temperature in the test section.

VIPSolutions · Accepted Answer

# Supersonic Wind Tunnel Calculations

Given Data:
- Test section area ($A_e$) = .2 m²  
- Mach number in test section ($M_e$) = 2.5  
- High-pressure tank pressure ($P_$) = 50 MPa  
- High-pressure tank temperature ($T_$) = 400 K  
- Assume air as ideal gas ($\gamma = 1.4$, $R = 287 \text{ J/kg·K}$)  
- Ignore losses between tank and test section

---

## a) Estimate the Throat Area of the Nozzle

### **Step 1: Area-Mach Relation**
The area ratio for isentropic flow is:
$$
\frac{A_e}{A^*} = \frac{1}{M_e} \left[ \frac{2}{\gamma+1} (1+ \frac{\gamma-1}{2} M_e^2) \right]^{\frac{\gamma+1}{2(\gamma-1)}}
$$

Where:  
- $A_e$: Test section area (given)  
- $A^*$: Throat area (at Mach 1)  
- $M_e$: Test section Mach number ($=2.5$)

### **Step 2: Substitute Values**
$$
\frac{A_e}{A^*} = \frac{1}{2.5} \left[ \frac{2}{1.4+1} \left( 1 + \frac{.4}{2} \times 2.5^2 \right) \right]^{\frac{2.4}{.8}}
$$
Calculate inside the brackets:
- $\frac{2}{2.4} = .833$
- $\frac{.4}{2} = .2$
- $.2 \times 2.5^2 = .2 \times 6.25 = 1.25$
- $1+1.25=2.25$
- $.833 \times 2.25 = 1.874$
- Exponent: $\frac{2.4}{.8} = 3$

So:
$$
\frac{A_e}{A^*} = \frac{1}{2.5} \left(1.874\right)^3 = \frac{1}{2.5} \times 6.583 = 2.633
$$
$$
A^* = \frac{A_e}{2.633} = \frac{.2}{2.633} = .076 \text{ m}^2
$$

**Final Answer:**  
$$
\boxed{A^* = .076~\text{m}^2}
$$
---

## b) Estimate the Static Pressure and Static Temperature in the Test Section

### **Step 1: Isentropic Relations**

$$
\frac{T_e}{T_} = \frac{1}{1 + \frac{\gamma-1}{2} M_e^2}
$$
$$
\frac{P_e}{P_} = \left(\frac{T_e}{T_}\right)^{\frac{\gamma}{\gamma-1}}
$$

### **Step 2: Substitute Values**
$$
\frac{T_e}{T_} = \frac{1}{1 + .2 \times 6.25} = \frac{1}{1 + 1.25} = \frac{1}{2.25} = .444
$$
$$
T_e = .444 \times 400~\text{K} = 177.8~\text{K}
$$

$$
\frac{P_e}{P_} = (.444)^{\frac{1.4}{.4}} = (.444)^{3.5}
$$
$$
\ln(.444) = -.813
$$
$$
3.5 \times -.813 = -2.845
$$
$$
e^{-2.845} = .0585
$$
$$
P_e = .0585 \times 50 \times 10^6~\text{Pa} = 2.925 \times 10^6~\text{Pa} = 2.93~\text{MPa}
$$

**Final Answers:**  
$$
\boxed{T_e = 177.8~\text{K}},\quad \boxed{P_e = 2.93~\text{MPa}}
$$

---

## c) Required Gas Temperature in High Pressure Tank for $T_e = 298~K$

$$
T_ = \frac{T_e}{T_e/T_}
$$
From (b): $T_e/T_ = .444$

$$
T_ = \frac{298}{.444} = 671~\text{K}
$$

**Final Answer:**  
$$
\boxed{T_ = 671~\text{K}}
$$

---

## **Summary Table**

| Part | Quantity                                   | Value              |
|------|--------------------------------------------|--------------------|
| a    | Throat Area ($A^*$)                      | .076 m²           |
| b    | Static Temperature in Test Section ($T_e$)| 177.8 K            |
|      | Static Pressure in Test Section ($P_e$)  | 2.93 MPa           |
| c    | Required Tank Temperature for $T_e=298K$ | 671 K              |

---

**If you need step-by-step explanations on any part, please let me know!**

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