Problem Statement
A gas is flowing through a horizontal pipe of cross-sectional area of 30 cm². At a point the pressure is 30 N per cm² (gauge) and temperature 20°C. At another section the area of cross-section is 15 cm² and pressure is 25 N/cm² gauge. If the mass rate of flow of gas is 0.15 kg/s, find the velocities of the gas at these two sections, assuming an isothermal change. Take R = 287 J/kg K, and atmospheric pressure 10 N/cm².
Given Data & Constants
- Mass flow rate, \(\dot{m} = 0.15 \, \text{kg/s}\)
- Gas constant, \(R = 287 \, \text{J/kg K}\)
- Temperature, \(T = 20^\circ\text{C} = 293.15 \, \text{K}\)
- Atmospheric pressure, \(P_{atm} = 10 \, \text{N/cm}^2 = 100,000 \, \text{N/m}^2\)
- Section 1:
- Area, \(A_1 = 30 \, \text{cm}^2 = 0.003 \, \text{m}^2\)
- Gauge Pressure, \(P_{g1} = 30 \, \text{N/cm}^2 = 300,000 \, \text{N/m}^2\)
- Section 2:
- Area, \(A_2 = 15 \, \text{cm}^2 = 0.0015 \, \text{m}^2\)
- Gauge Pressure, \(P_{g2} = 25 \, \text{N/cm}^2 = 250,000 \, \text{N/m}^2\)
Solution
1. Calculate Absolute Pressures
The ideal gas law requires absolute pressure (\(P_{abs} = P_{gauge} + P_{atm}\)).
2. Calculate Gas Densities (\(\rho\))
Using the ideal gas law \(P = \rho R T\), we find the density at each section. The process is isothermal, so temperature is constant.
3. Calculate Velocities (\(V\))
Using the mass continuity equation, \(\dot{m} = \rho A V\), we can solve for the velocity at each section.
The velocity at the first section is approximately 10.52 m/s.
The velocity at the second section is approximately 24.04 m/s.
Explanation of the Process
This problem demonstrates the principles of compressible fluid flow. Unlike liquids (which are nearly incompressible), the density of a gas changes significantly with pressure. The key steps are:
- Find Absolute Pressure: All gas law calculations must use absolute pressure, which is the gauge pressure plus the surrounding atmospheric pressure.
- Find Density: The ideal gas law allows us to calculate the density of the gas at each section. Because the pressure is lower at the second section, the gas expands and its density decreases.
- Apply Mass Continuity: The mass flow rate (\(\dot{m}\)) must be constant throughout the pipe. The equation \(\dot{m} = \rho A V\) shows that if both the density (\(\rho\)) and the area (\(A\)) decrease, the velocity (\(V\)) must increase significantly to maintain the same mass flow. This is why the velocity is much higher at the second section.
