Duct Flow with Friction and Heat Addition

Fanno flow, Rayleigh flow, thermal choking and the fundamental pressure loss of burners and afterburners

Concorde's reheat

As Concorde grew heavier during its design, its Olympus 593 engines could no longer give the take-off thrust it needed. The 20 % shortfall was met by adding partial reheat — an afterburner (Olympus 593). Crews lit it only for take-off and to accelerate through the transonic range from Mach 0.95 to 1.7; above that, Concorde cruised supersonically without it (Concorde). An afterburner is the bluntest way to get thrust: spray fuel into the hot gas behind the turbine and burn it. So why not burn more, for longer? Part of the answer is fuel. The rest is this lesson. Heat added to a gas that is already moving costs stagnation pressure, and add enough and the duct chokes: no more flow can pass, however hard the engine pushes.

Real ducts have walls that rub the flow, and some of them have flames inside. To see what each does on its own, take a constant-area duct with steady one-dimensional flow and two idealised cases:

Fanno flow — friction at the walls, no heat transfer.
Rayleigh flow — heat addition, no friction.

Both are irreversible, so both raise the entropy — and in both, the entropy is greatest at Mach 1.

💡Everything heads for Mach 1

Friction and heating both push a flow along its line toward the entropy maximum at Mach 1, from either side:

Subsonic flow speeds up toward Mach 1.
Supersonic flow slows down toward Mach 1.
Both cost stagnation pressure.

Once the flow reaches Mach 1 at the duct exit, the duct is choked: more friction (a longer duct) or more heat cannot be absorbed at the same inlet conditions, and the flow upstream must change — the mass flow falls. Choking by heat is called thermal choking.

\frac{T_t}{T_t^*} = \frac{(\gamma + 1)M^2\left[2 + (\gamma - 1)M^2\right]}{\left(1 + \gamma M^2\right)^2}, \qquad \frac{p_t}{p_t^*} = \frac{\gamma + 1}{1 + \gamma M^2}\left[\frac{2 + (\gamma - 1)M^2}{\gamma + 1}\right]^{\frac{\gamma}{\gamma - 1}}

Rayleigh flow, referred to the sonic state (*): the stagnation temperature and stagnation pressure at Mach M. Heat added per unit mass is q = cp(Tt2 − Tt1).

Derivation

Rayleigh flow

Constant area, no friction: mass gives

\rho V =

const and momentum gives

p + \rho V^2 =

const. With

\rho V^2 = \gamma p M^2

, the momentum equation reads

p(1 + \gamma M^2) =

const, so relative to the sonic state

\frac{p}{p^*} = \frac{\gamma + 1}{1 + \gamma M^2}.

From

\rho V =

const and

V = M\sqrt{\gamma R T}

\rho = p/(RT)

\frac{p M}{\sqrt T} =

const, so

\frac{T}{T^*} = \left(\frac{p}{p^*}\right)^2 M^2 = \frac{(\gamma + 1)^2 M^2}{(1 + \gamma M^2)^2}.

Multiply by

T_t/T = 1 + \frac{\gamma - 1}{2}M^2

and divide by its sonic value

\frac{\gamma + 1}{2}

to get

T_t/T_t^*

; do the same with

p_t/p = (T_t/T)^{\gamma/(\gamma - 1)}

for

p_t/p_t^*

. Heat enters only through the energy equation,

q = c_p(T_{t2} - T_{t1})

: given the inlet Mach number and the heat,

T_{t2}/T_t^*

fixes the exit Mach number.

\frac{4 f L^*}{D} = \frac{1 - M^2}{\gamma M^2} + \frac{\gamma + 1}{2\gamma}\ln\frac{(\gamma + 1)M^2}{2 + (\gamma - 1)M^2}

Fanno flow: the friction length L* that takes a flow at Mach M to Mach 1, for a Fanning friction factor f and a duct of diameter D.

Derivation

Fanno flow

Momentum for a slice

dx

of a duct with wall shear stress

\tau_w = f \cdot \tfrac12\rho V^2

and perimeter

\pi D

A\,dp + \dot m\, dV = -\tau_w \pi D\, dx \quad\Longrightarrow\quad \frac{dp}{p} + \frac{\gamma M^2}{2}\frac{dV^2}{V^2} = -\frac{\gamma M^2}{2}\frac{4 f\, dx}{D}.

Add mass (

d\rho/\rho + dV/V = 0

), the gas law and constant

T_t

(adiabatic walls), and eliminate everything but

M

\frac{dM^2}{M^2} = \frac{\gamma M^2\left(1 + \frac{\gamma - 1}{2}M^2\right)}{1 - M^2}\;\frac{4 f\, dx}{D}.

The right-hand side is positive below Mach 1 and negative above it: friction drives

M

toward 1 either way. Separating variables and integrating from

M

to 1 gives

4fL^*/D

Both at once. Shapiro's generalised one-dimensional flow puts the two effects side by side:

\frac{dM^2}{M^2} = \frac{1 + \frac{\gamma - 1}{2}M^2}{1 - M^2}\left[\left(1 + \gamma M^2\right)\frac{dT_t}{T_t} + \gamma M^2\,\frac{4 f\, dx}{D}\right].

The bracket is positive for heating and for friction; the factor

1/(1 - M^2)

flips the sign of the response at Mach 1 and blows up there. Lab 3 integrates this equation through a burner with solve_ivp, checks it against the Rayleigh result, and finds where the flow chokes.

📐The fundamental pressure loss of a burner

Double the stagnation temperature, 850 K to 1700 K, in a constant-area burner (

\gamma = 1.33

, Lab 1). The stagnation pressure lost to the heating alone:

inlet Mach 0.05: 0.17 %
inlet Mach 0.10: 0.67 %
inlet Mach 0.20: 2.8 %
inlet Mach 0.30: 6.7 %

An afterburner raising

T_t

by 1.9 times loses 2.5 % at Mach 0.2, 12.7 % at Mach 0.4 — and at Mach 0.45 it chokes. This "hot" loss comes on top of the "cold" loss from friction and mixing in the flame tube (Lesson 7-3). Compressors deliver air far too fast to burn, so every combustor and afterburner starts with a diffuser that slows it down.

📐How long can a duct be?

A 5 cm pipe with a Fanning friction factor of 0.005 (Lab 2): entering at Mach 0.3 the flow chokes after 13.25 m; entering at Mach 2 it is down to Mach 1 after only 0.76 m. Supersonic ducts must be short. A 3 m pipe of the same size chokes any inlet flow faster than Mach 0.485 — and no supersonic flow can get through it without a shock, because

4fL^*/D

never exceeds 0.82 for

\gamma = 1.4

I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science, whatever the matter may be.

– William Thomson, Lord Kelvin, lecture on "Electrical Units of Measurement", 3 May 1883 (Wikiquote)

⚠Supersonic combustion runs the other way

Heat drives supersonic flow down toward Mach 1. A scramjet, which burns its fuel in a supersonic stream, must therefore widen its combustor as the heat goes in, or the flow would slow to Mach 1, choke and push a shock out of the engine. In a supersonic duct, a diverging area accelerates the flow and offsets the heating.

💡Where you will use this

Lesson 3-3 puts the burner pressure loss into ramjets and afterburners. Lesson 4-1 turns it into the burner and afterburner pressure ratios

\pi_b

and

\pi_{AB}

of a real cycle. Lesson 7-3 designs combustors around the hot and cold losses — and around the diffuser that keeps the inlet Mach number low.

📚Further reading

Ascher H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, Vol. 1 (Ronald Press, 1953), the chapters on friction, heat exchange and generalised one-dimensional flow.
John D. Anderson, Modern Compressible Flow: With Historical Perspective (3rd ed., McGraw-Hill, 2003), chapter 3.
Fanno flow and Rayleigh flow on Wikipedia, with the T–s diagrams.

↓But there is a problem

We can now follow a parcel of air through every part of an engine: slow it in the inlet, squeeze it, burn fuel in it at low Mach number, expand it through the turbine and the nozzle — and account for every bit of stagnation pressure lost on the way. In January 1930 a 22-year-old RAF officer named Frank Whittle patented an engine built from exactly those steps (Frank Whittle). It took eleven years to fly it. The next unit asks the question he had to answer first: if every part were perfect, how good could such an engine be?