Problem Statement
Find the mass rate of flow of air through a venturimeter having inlet diameter as 400 mm and throat diameter 200 mm. The pressure at the inlet of the venturimeter is 27.468 N/cm² (abs.) and temperature of air at inlet is 20°C. The pressure at the throat is given as 25.506 N/cm² absolute. Take k = 1.4 and R = 287 J/kg K.
Given Data & Constants
- Inlet (Section 1):
- Diameter, \(D_1 = 400 \, \text{mm} = 0.4 \, \text{m}\)
- Pressure, \(P_1 = 27.468 \, \text{N/cm}^2 = 274680 \, \text{N/m}^2\)
- Temperature, \(T_1 = 20^\circ\text{C} = 293.15 \, \text{K}\)
- Throat (Section 2):
- Diameter, \(D_2 = 200 \, \text{mm} = 0.2 \, \text{m}\)
- Pressure, \(P_2 = 25.506 \, \text{N/cm}^2 = 255060 \, \text{N/m}^2\)
- Gas Properties:
- Adiabatic index, \(k = 1.4\)
- Gas constant, \(R = 287 \, \text{J/kg K}\)
Solution
1. Calculate Areas and Densities
First, we calculate the areas and the density of the air at the inlet and throat.
$$ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.4)^2 \approx 0.12566 \, \text{m}^2 $$
$$ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.2)^2 \approx 0.031416 \, \text{m}^2 $$
$$ \rho_1 = \frac{P_1}{R T_1} = \frac{274680}{287 \times 293.15} \approx 3.265 \, \text{kg/m}^3 $$
For an adiabatic (isentropic) process, we find the temperature and density at the throat.
$$ T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = 293.15 \left(\frac{255060}{274680}\right)^{\frac{0.4}{1.4}} \approx 286.96 \, \text{K} $$
$$ \rho_2 = \frac{P_2}{R T_2} = \frac{255060}{287 \times 286.96} \approx 3.097 \, \text{kg/m}^3 $$
2. Calculate Velocity at the Throat (\(V_2\))
We use the steady flow energy equation for a compressible fluid, relating the velocities at the two sections.
$$ V_2 = \sqrt{\frac{2 \frac{k}{k-1} \frac{P_1}{\rho_1} \left[1 - \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}\right]}{1 - \left(\frac{A_2}{A_1}\right)^2 \left(\frac{P_2}{P_1}\right)^{\frac{2}{k}}}} $$
$$ V_2 = \sqrt{\frac{2 \times 3.5 \times \frac{274680}{3.265} \left[1 - (0.92857)^{0.2857}\right]}{1 - (0.25)^2 \times (0.92857)^{2/1.4}}} $$
$$ V_2 = \sqrt{\frac{589226 \times [1 - 0.9789]}{1 - 0.0625 \times 0.899}} = \sqrt{\frac{12432.6}{0.9438}} \approx \sqrt{13172.9} $$
$$ V_2 \approx 114.77 \, \text{m/s} $$
3. Calculate the Mass Flow Rate (\(\dot{m}\))
The mass flow rate is the product of the density, area, and velocity at the throat.
$$ \dot{m} = \rho_2 A_2 V_2 $$
$$ \dot{m} = 3.097 \, \text{kg/m}^3 \times 0.031416 \, \text{m}^2 \times 114.77 \, \text{m/s} \approx 11.16 \, \text{kg/s} $$
Final Result:
The mass rate of flow of air is approximately 11.16 kg/s.
