Question

In a low-speed airplane, a venturimeter with a 1.3:1 area ratio is used for airspeed measurement. The airplane’s maximum speed at sea level is 90 m/s. If the density of air at sea level is 1.225 kg/m³, the maximum pressure difference between the inlet and the throat of the venturimeter is \_\_\_\_ kPa (rounded off to two decimal places).

Hint:

To calculate the pressure difference using a venturimeter, use the Bernoulli’s equation along with the continuity equation to relate the velocities at the inlet and the throat. The pressure difference is proportional to the square of the velocity difference.

Answer 1

The pressure difference in a venturimeter is related to the velocity of the air through Bernoulli’s equation. The equation for the pressure difference \( \Delta P \) is: \[ \Delta P = \frac{1}{2} \rho (V_1^2 - V_2^2) \] where:
\( \rho \) is the air density (1.225 kg/m³),
\( V_1 \) is the velocity at the inlet,
\( V_2 \) is the velocity at the throat.
Since the area ratio is given as 1.3:1, the velocity at the throat \( V_2 \) can be calculated using the continuity equation: \[ A_1 V_1 = A_2 V_2 \] where \( A_1 \) and \( A_2 \) are the areas at the inlet and throat, respectively. The area ratio \( A_1/A_2 = 1.3 \), so: \[ V_2 = \frac{A_1}{A_2} \cdot V_1 = 1.3 \cdot V_1 \] Now, substitute \( V_2 = 1.3 \cdot V_1 \) into the pressure difference equation: \[ \Delta P = \frac{1}{2} \rho \left( V_1^2 - (1.3 V_1)^2 \right) \] \[ \Delta P = \frac{1}{2} \rho \left( V_1^2 - 1.69 V_1^2 \right) \] \[ \Delta P = \frac{1}{2} \rho \left( -0.69 V_1^2 \right) \] \[ \Delta P = -0.345 \rho V_1^2 \] Substituting the given values \( \rho = 1.225 \, {kg/m}^3 \) and \( V_1 = 90 \, {m/s} \): \[ \Delta P = -0.345 \times 1.225 \times (90)^2 \] \[ \Delta P = -0.345 \times 1.225 \times 8100 \] \[ \Delta P \approx -3400.4 \, {Pa} \] Converting to kPa: \[ \Delta P \approx 3.40 \, {kPa} \] Thus, the maximum pressure difference between the inlet and the throat of the venturimeter is approximately 3.40 kPa.