Fanno flow: adiabatic flow with friction

Why it matters

Gas pipelines, engine bleed ducts, wind-tunnel supply lines and the ducts of supersonic test rigs are long enough for wall friction to change the Mach number substantially. Fanno flow tells you how far a gas can flow before it chokes, how much pressure is lost, and why a long duct can limit the mass flow that a compressor or nozzle can deliver.

Key ideas

Model. Steady, one-dimensional flow of a perfect gas in a constant-area duct, adiabatic (no heat transfer), no work, with wall shear stress τw described by an average friction factor. Because the flow is adiabatic, T₀ is constant; friction is irreversible, so entropy rises and p₀ falls.

Fanno line. Combining continuity (ρV = constant) and energy (h₀ = constant) gives a curve on the h–s (or T–s) diagram. It has a point of maximum entropy, and at that point M = 1. Since entropy can only increase along the duct:

• Subsonic inlet: friction accelerates the flow towards M = 1. p and T fall, V rises, ρ falls.
• Supersonic inlet: friction decelerates the flow towards M = 1. p, T, ρ rise, V falls.
• In both cases p₀ falls. Friction drives the flow towards M = 1 from either side; it can never carry it through M = 1.

Choking length L*. For a given inlet Mach number, L* is the duct length that brings the flow exactly to M = 1 at the exit. The function f·L*/D depends only on M. For any length L between two Mach numbers, f·L/D = (f·L*/D)₁ − (f·L*/D)₂.

Ducts longer than L.*

• Subsonic: the flow adjusts upstream — the inlet Mach number (and mass flow, for a given supply) decreases until the exit is just sonic. This is frictional choking.
• Supersonic: a normal shock forms inside the duct; the subsonic flow behind it then accelerates back to M = 1 at the exit. If the duct is much longer, the shock moves into the supply nozzle and the duct flow is subsonic throughout.

Limits. For supersonic flow f·L*/D cannot exceed 0.822 (γ = 1.4, M → ∞), so supersonic friction ducts are short — typically tens of diameters at most.

Friction factor convention. This lesson uses the Darcy factor f (≈ 0.02 for typical turbulent flow). Many textbooks use the Fanning factor f_F = f/4 and write 4·f_F·L/D. Check which one your table uses. Friction factors come from the Moody chart or a correlation — take them from your data book.

Formulas

f·L*/D = (1 − M²)/(γ·M²) + ((γ+1)/(2γ))·ln[ (γ+1)·M² / (2 + (γ−1)·M²) ]

T/T* = (γ+1) / (2 + (γ−1)·M²)

p/p* = (1/M)·√[ (γ+1) / (2 + (γ−1)·M²) ]

ρ/ρ* = V*/V = (1/M)·√[ (2 + (γ−1)·M²) / (γ+1) ]

p₀/p₀* = (1/M)·[ (2 + (γ−1)·M²) / (γ+1) ]^((γ+1)/(2(γ−1))) (same expression as A/A* in isentropic flow).

s* − s = R·ln(p₀/p₀*) — entropy rise to the choked state (J/(kg·K)).

f·L/D = (f·L*/D)₁ − (f·L*/D)₂ — length between two sections.

Symbols: f Darcy friction factor (dimensionless); L duct length (m); D diameter or hydraulic diameter 4A/P (m); M Mach number; superscript * = values at the (possibly hypothetical) section where M = 1 for the same mass flow and T₀; T, p, ρ, V, p₀ in K, Pa, kg/m³, m/s, Pa; γ ratio of specific heats; R gas constant.

Worked examples

Example 1 (standard). Air enters a 50 mm diameter pipe at M₁ = 0.3, p₁ = 200 kPa, T₁ = 300 K. Take f = 0.02. Find the length to choking and the exit pressure and temperature at that length. γ = 1.4.

1. f·L*/D at M = 0.3: (1 − 0.09)/(1.4 × 0.09) + (2.4/2.8)·ln(2.4 × 0.09/(2 + 0.4 × 0.09)) = 7.222 + 0.8571 × ln(0.10609) = 7.222 − 1.923 = 5.299.
2. L* = 5.299 × 0.05/0.02 = 13.25 m.
3. p₁/p* = (1/0.3)·√(2.4/2.036) = 3.619, so p* = 200/3.619 = 55.3 kPa.
4. T₁/T* = 2.4/2.036 = 1.1788, so T* = 300/1.1788 = 254.5 K.

Answer: L ≈ 13.2 m; exit p ≈ 55.3 kPa, T ≈ 254.5 K* (the flow has accelerated and cooled).

Example 2 (GATE level). Air enters a 30 mm diameter duct at M₁ = 2.5 with f = 0.02. Find L* and the Mach number 0.30 m from the inlet.

1. f·L*/D at M = 2.5: (1 − 6.25)/(1.4 × 6.25) + 0.8571·ln(2.4 × 6.25/(2 + 0.4 × 6.25)) = −0.6000 + 0.8571 × ln(3.3333) = −0.6000 + 1.0320 = 0.4320.
2. L* = 0.4320 × 0.03/0.02 = 0.648 m.
3. For L = 0.30 m: f·L/D = 0.02 × 0.30/0.03 = 0.200. Remaining f·L*/D = 0.4320 − 0.200 = 0.2320.
4. Inverting on the supersonic branch: M₂ ≈ 1.77.

Answer: L ≈ 0.65 m; M ≈ 1.77 at 0.30 m*. Note how short a supersonic friction duct is.

Common mistakes

• Thinking friction always slows the flow; subsonic Fanno flow accelerates.
• Mixing Darcy and Fanning friction factors (a factor of 4 error).
• Subtracting the wrong way: f·L/D = (f·L*/D)inlet − (f·L*/D)exit.
• Applying isentropic p₀ relations along the duct — p₀ falls with friction.
• Expecting a long subsonic duct to give supersonic exit flow; it chokes at M = 1 and the inlet adjusts.
• Using the supersonic branch for a subsonic problem when inverting f·L*/D.

For GATE AE

Expect: L* for a given inlet Mach number, exit Mach number for a given length, exit pressure and temperature via the starred ratios, the effect of increasing length beyond L*, and true/false statements on trends of p, T, V and s along the duct in subsonic and supersonic flow. Fanno tables are usually provided; practise using them and the subtraction rule.

Quick check

1. In subsonic Fanno flow, does static temperature rise or fall along the duct?
2. Which property is constant in Fanno flow: T₀, p₀ or s?
3. What happens in a subsonic duct longer than L*?
4. What is f·L*/D at M = 1?
5. What is the maximum f·L*/D for supersonic flow of air?

Answers: 1. falls; 2. T₀; 3. the inlet Mach number and mass flow reduce so the exit is just sonic; 4. zero; 5. about 0.822.