Question
                In a wind tunnel a flow is sped up. In the reservoir, the velocity is approximately zero; in the test section, the air is at standard conditions with a flow speed of 30 m/s.
What is the total pressure in the test section?
unanswered
 
What is the static pressure in the reservoir?
            
        What is the total pressure in the test section?
unanswered
What is the static pressure in the reservoir?
Answers
                                                    There are no human answers yet.
                                            
                
                    Answered by
            GPT-4o mini
            
    To determine the total pressure in the test section of the wind tunnel, we can use the Bernoulli equation. The total pressure \( P_t \) is defined as the sum of the static pressure \( P \) and the dynamic pressure \( q \):
\[
P_t = P + q
\]
where dynamic pressure \( q \) is given by:
\[
q = \frac{1}{2} \rho v^2
\]
In standard conditions (approximately sea level), the density of air \( \rho \) is about \( 1.225 \, \text{kg/m}^3 \) and the flow velocity \( v \) in the test section is given as \( 30 \, \text{m/s} \).
### Step 1: Calculate the dynamic pressure
Using the formula for dynamic pressure:
\[
q = \frac{1}{2} \rho v^2 = \frac{1}{2} (1.225) (30)^2
\]
Calculating this step-by-step:
1. Calculate \( v^2 \):
\[
(30)^2 = 900 \, \text{m}^2/\text{s}^2
\]
2. Now substitute:
\[
q = \frac{1}{2} (1.225) (900) = 0.6125 \times 900 = 551.25 \, \text{Pa}
\]
### Step 2: Determine static pressure in the test section
At standard atmospheric conditions, the static pressure \( P \) is approximately \( 101325 \, \text{Pa} \) or \( 101.325 \, \text{kPa} \).
### Step 3: Calculate total pressure in the test section
Now we can substitute our values into the formula for total pressure:
\[
P_t = P + q = 101325 \, \text{Pa} + 551.25 \, \text{Pa} = 101876.25 \, \text{Pa}
\]
### Final Answer for Total Pressure in Test Section
The total pressure in the test section is approximately:
\[
P_t \approx 101876.25 \, \text{Pa} \quad (\text{or } \approx 101.88 \, \text{kPa})
\]
### Regarding the Static Pressure in Reservoir
In the wind tunnel's reservoir, if we assume the flow velocity is effectively zero (as stated), and that it is at the same temperature and density, we can consider the static pressure in the reservoir to be the same as the standard atmospheric pressure:
\[
P \approx 101325 \, \text{Pa}
\]
### Summary of Answers:
- Total Pressure in the Test Section: \( 101876.25 \, \text{Pa} \) (or \( \approx 101.88 \, \text{kPa} \))
- Static Pressure in the Reservoir: \( 101325 \, \text{Pa} \) (or \( 101.325 \, \text{kPa} \))
    
\[
P_t = P + q
\]
where dynamic pressure \( q \) is given by:
\[
q = \frac{1}{2} \rho v^2
\]
In standard conditions (approximately sea level), the density of air \( \rho \) is about \( 1.225 \, \text{kg/m}^3 \) and the flow velocity \( v \) in the test section is given as \( 30 \, \text{m/s} \).
### Step 1: Calculate the dynamic pressure
Using the formula for dynamic pressure:
\[
q = \frac{1}{2} \rho v^2 = \frac{1}{2} (1.225) (30)^2
\]
Calculating this step-by-step:
1. Calculate \( v^2 \):
\[
(30)^2 = 900 \, \text{m}^2/\text{s}^2
\]
2. Now substitute:
\[
q = \frac{1}{2} (1.225) (900) = 0.6125 \times 900 = 551.25 \, \text{Pa}
\]
### Step 2: Determine static pressure in the test section
At standard atmospheric conditions, the static pressure \( P \) is approximately \( 101325 \, \text{Pa} \) or \( 101.325 \, \text{kPa} \).
### Step 3: Calculate total pressure in the test section
Now we can substitute our values into the formula for total pressure:
\[
P_t = P + q = 101325 \, \text{Pa} + 551.25 \, \text{Pa} = 101876.25 \, \text{Pa}
\]
### Final Answer for Total Pressure in Test Section
The total pressure in the test section is approximately:
\[
P_t \approx 101876.25 \, \text{Pa} \quad (\text{or } \approx 101.88 \, \text{kPa})
\]
### Regarding the Static Pressure in Reservoir
In the wind tunnel's reservoir, if we assume the flow velocity is effectively zero (as stated), and that it is at the same temperature and density, we can consider the static pressure in the reservoir to be the same as the standard atmospheric pressure:
\[
P \approx 101325 \, \text{Pa}
\]
### Summary of Answers:
- Total Pressure in the Test Section: \( 101876.25 \, \text{Pa} \) (or \( \approx 101.88 \, \text{kPa} \))
- Static Pressure in the Reservoir: \( 101325 \, \text{Pa} \) (or \( 101.325 \, \text{kPa} \))
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