Question

4. Imagine that you have designed a low-speed airplane with a maximum velocity at sea level of 90 m/s. For your airspeed instrument, you plan to use a venturi tube with a 1.3:1 area ratio. Inside the cockpit is an airspeed indicatora dial that is connected to a pressure gauge sensing the venturi tube pressure difference (p 1 ​ −p 2 ​ ) and properly calibrated in terms of velocity. What is the maximum pressure difference you would expect the gauge to experience?

Answer 1

The maximum pressure difference the gauge would experience is approximately 64.62 Pa.

Step-by-step explanation:

In the design of the low-speed airplane, a venturi tube is used as the airspeed instrument. The venturi tube has a 1.3:1 area ratio, meaning the cross-sectional area of the throat (A2) is 1.3 times smaller than the cross-sectional area of the inlet (A1). The relationship between velocity and cross-sectional area in a venturi tube is given by the equation
`\(v_1 A_1 = v_2 A_2\), where \(v_1\) and \(v_2\)`are the velocities at the inlet and throat, respectively.

Given that the maximum velocity at sea level is 90 m/s, we can use the area ratio to find the velocity at the throat
`(\(v_2\)).`

`\[v_2 = (v_1)/(1.3) \]`

Substitute the given values:

`\[v_2 = (90)/(1.3) \approx 69.23 \, \text{m/s} \]`

Now, to find the pressure difference
`(\(p_1 - p_2\))`, we can use Bernoulli's equation:

`\[p_1 + (1)/(2) \rho v_1^2 = p_2 + (1)/(2) \rho v_2^2\]`

Where
`\(\rho\)` is the air density. Since we are at sea level,
`\(\rho\)` is approximately
`\(1.225 \, \text{kg/m}^3\)`. Rearranging the equation to solve for
`\(p_1 - p_2\)`:

`\[p_1 - p_2 = (1)/(2) \rho (v_2^2 - v_1^2)\]`

Substitute the values:

`\[p_1 - p_2 = (1)/(2) * 1.225 * (69.23^2 - 90^2) \approx 64.62 \, \text{Pa}\]`

Therefore, the maximum pressure difference the gauge would experience is approximately 64.62 Pa.