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Fanno Flow
- 1. FANNO CURVE &FANNO EQUATION Dhaval Chauhan Mechanical Engineer
- 2. Fanno Curve Considera flow of a fluid in a perfectly insulated (Q=0) constant area duct. Continuity equation: ṁ = ρ ∙ AC ∴ ṁ A = ρ ∙ C = Constant, G …(1) Where G = Mass flow density. Energy equation: h + C2 2 = h0 …(2) Where h0= Stagnation enthalpy
- 3. But from Equation(1), C= 𝐺 𝜌 ∴ h + G2 2ρ2 = h0 …(3) If the properties of ℎ0 and G are known on upstream side, the values of h and ρ can be obtained at any section of the duct. Fig. 1 shows the relation defined by Equation (1) for a single value of ℎ0 and various values of G on the plot of h Vs V or 1 ρ. These curves are called Fanno lines. According to second law of thermodynamics, the entropy during the adiabatic process with friction always increases.
- 4. Therefore, thepath of states for portion of curves must be from left to right. It follows that during subsonic flow, the effect of friction will be to increase the velocity and Mach number with reduction in enthalpy and pressure. While, during supersonic flow, the effect of friction will be to decrease the velocity and Mach number with increase in enthalpy and pressure.
- 5. Fig. (1) Fannolines on (h-s) diagram
- 6. Fanno flow equation: Continuity equation: dρ ρ + d∁ ∁ = 0 Perfect gas equation: dp p = dρ ρ + dT T By definition of Mach number: 1 M2 ∙ dM2 = 1 C2 ∙ dC2 − dT T Energy equation: dT T + γ_1 2 M2 ∙ dC2 C2 = 0
- 7. Momentum equation γM2 2 × dC2 C2 +4f ∙ dx D × γM2 2 + dp p = 0 Stagnation pressure – Mach number relationship dp0 p0 = dp p + γM2 2 1 1 + γ − 1 2 M2 ∙ dM2 M2 Impulse function dF F = dp p + γM2 1 + γM2 ∙ d M2 M2
- 8. Solution of FannoLines Equations and Effect of Wall Friction on Fluid Properties We have discussed above simultaneous Equations which relate to following eight different variables: 𝑑ρ ρ , 𝑑𝐶2 𝐶2 , 𝑑𝑀2 𝑀2 , 𝑑𝑇 𝑇 , 𝑑𝑝 𝑝 , 𝑑𝑝0 𝑝0 , 𝑑𝐹 𝐹 and 4𝑓 ∗ 𝑑𝑥 𝐷 Out of the above, variable 4𝑓 ∗ 𝑑𝑥 𝐷 is independent variable which is responsible for changes in flow properties during the flow in a duct. Now we can solve the simultaneous equation discussed in terms of removing seven variables as follows: ∴ 𝑑𝑝 𝑝 = 𝑑𝜌 𝜌 + 𝑑𝑇 𝑇 …(1)
- 9. and, 𝑑𝑇 𝑇 + 𝛾−1 2 𝑀2 ∗ 𝑑𝐶2 𝐶2 =0 …(2) On substituting the value of 𝑑𝑇 𝑇 from Eq.(2) in Eq.(1), 𝑑𝑝 𝑝 = 𝑑𝜌 𝜌 - 𝛾−1 2 𝑀2 ∗ 𝑑𝐶2 𝐶2 …(3) But, 𝑑𝜌 𝜌 = − 1 2 ∗ 𝑑𝐶2 𝐶2 from Continuity equation, hence, 𝑑𝑝 𝑝 = − 1+ 𝛾−1 𝑀2 2 ∗ 𝑑𝐶2 𝐶2 …(4) On substituting the value of Eq.(4) in Momentum equation we get, 𝑑𝑃 𝑝 = − 1+ 𝛾−1 𝑀2 2 ∗ 4𝑓 ∗𝑑𝑥 𝐷 ∗ 𝛾𝑀2 1−𝑀2 …(5)
- 10. On substitutingthe value of Eq.(5) in (4) we get, 1 2 𝑑𝐶2 𝐶2 = 𝑑𝐶 𝐶 = 4𝑓 ∗𝑑𝑥 𝐷 ∗ 𝛾𝑀2 2 1−𝑀2 …(6) Since, from Continuity equation, therefore substituting the value of 𝑑𝐶 𝐶 from Eq.(6) in Continuity equation we get, 𝑑𝜌 𝜌 = − 4𝑓 ∗𝑑𝑥 𝐷 ∗ 𝛾𝑀2 2 1−𝑀2 …(7) From Perfect gas equation we have, 𝑑𝜌 𝜌 = 𝑑𝜌 𝜌 + 𝑑𝑇 𝑇 …(8) On substituting the value of Eq.(5) & (7) we get, 𝑑𝑇 𝑇 = − 4𝑓 ∗𝑑𝑥 𝐷 ∗ 𝛾𝑀4(𝛾−1) 2 1−𝑀2 …(9)
- 11. From Equationby definition of Mach number, 1 M2 ∙ dM2 = 1 C2 ∙ dC2 − dT T On substituting the values from Eq.(6) & (9) we get, 𝑑𝑀2 𝑀2 = 4𝑓 ∗𝑑𝑥 𝐷 ∗ 𝛾𝑀2 1−𝑀2 ∗ 1 + 𝛾−1 2 𝑀2 …(10) From Momentum equation, 𝑑𝑝0 𝑝0 = 𝑑𝑝 𝑝 + 𝛾𝑀2 2 ∗ 1 1+ 𝛾−1 2 𝑀2 ∗ 𝑑𝑀2 𝑀2 …(11) On substituting the values from Eq.(5) & (10) we get, 𝑑𝑝0 𝑝0 = −4𝑓 ∗ 𝑑𝑥 𝐷 ∗ 𝛾𝑀2 2 …(12)
- 12. From equationof Impulse Function, dF F = dp p + γM2 1+γM2 ∙ d M2 M2 …(13) On substituting the values from Eq.(5) & (10) we get, dF F = − 4𝑓∗𝑑𝑥 𝐷 ∗ 𝛾𝑀2 2 1+𝛾𝑀2 …(14)
- 13. Integration of Fannoequations Variation of Mach number with duct length f = 1 Lmax 0 Lmax f ∙ dx Temperature ratio T T∗ = γRT 2 γRT∗ 2 = a2 a∗2 Density ratio ρ ρ∗ = C2 C = 1 M 2 1 + γ − 1 2 M2 γ − 1
- 14. Velocity ratio C C∗ =M ∙ γ + 1 2 1 + γ − 1 2 M2 1 2 Pressure ratio p p∗ = 1 M ∙ γ + 1 2 1 + γ − 1 2 M2 1 2 Stagnation pressure ratio p0 p0 ∗ = 1 M ∙ 2 1 + γ − 1 2 M2 γ + 1 γ−1 2 γ−1
- 15. Impulse functionratio F F∗ = 1 + γM2 M 2 1 + γ 1 + γ − 1 2 M2 1 2 Change in entropy s − s∗ = −R ∙ ln 2 1 + γ − 1 2 M2 γ + 1 γ−1 2 γ−1 × 1 M