Lesson 15b: Compressible Flow and Choking in Converging Ducts

Key Topics

• Compressible flow in a converging duct
• Calculating critical conditions
• Mass flow rate
• Definition and examples of choking

Compressible Flow in a Converging Duct

• Flow starts at zero speed and accelerates into the converging duct.
• Subsonic flow:
  • Speed increases
  • Mach number increases
  • Pressure decreases
  • Temperature decreases
• Mach number approaches 1 but cannot exceed 1 in a converging duct.

Supersonic Flow

• If Mach number > 1, the behavior is opposite (not valid in a converging duct):
  • Speed decreases
  • Mach number decreases
  • Pressure increases
  • Temperature increases

Choking in Converging Ducts

• Choking: Occurs when Mach number equals 1 at the exit plane.
• Flow cannot go supersonic in a converging duct.
• Conditions at choke point denoted with an asterisk (), e.g., T, P*, ρ*.

Back Pressure (Pb)

• Determines whether flow is subsonic or choked.
• Three cases:
  • Pb > P*: Flow is subsonic, Mach number < 1.
  • Pb = P*: Flow is choked, Mach number = 1.
  • Pb < P*: Flow remains choked, Pe = P*.
• Choked flow continues unchanged despite further decreases in Pb.

Example Problem

• Scenario: Air flows from pressurized tank through a converging nozzle.
• Given: P0, T0, Pb, exit area.
• Need to determine if flow is subsonic, sonic, or supersonic.
• Calculate Mach number and temperature at the exit plane.

Assumptions

• 1D flow, ideal gas, steady, nearly isentropic, adiabatic conditions.
• Large tank with constant conditions.

Findings

• Part A: Flow is subsonic when Pb > P*.
• Part B: Flow becomes sonic at Pb = P* (83.47 kPa).
• Part C: Mach number at exit plane MaE = 0.823, Temperature TE = 458 K.

Mass Flow Rate in Converging Duct

[ \dot{m} = \rho VA ]

• Derived using the ideal gas law and Mach number relations.

• General equation for mass flow rate:

  [ \dot{m} = P_0 A M_A \sqrt{\frac{k}{R}} \left(1 + \frac{k-1}{2}M_A^2\right)^{-\frac{k+1}{2(k-1)}} ]

• For choked flow, simplifies when Ma = 1.

Example Calculation

• Mass flow rate for given conditions: 3.54 kg/s
• Maximum mass flow rate (m_dot_max) for choked flow: 3.64 kg/s

Conclusion

• Understanding the relationship between back pressure and flow conditions is crucial in compressible flow.
• Choked flow is significant as it determines limits of flow rate changes.