Isentropic flow

Most of the flow around the Swift 900 is isentropic: it is adiabatic, because there is no time for heat to flow in or out of a moving parcel of air, and it is reversible, because viscous effects are confined to thin boundary layers and wakes. Outside those layers and away from shock waves, every parcel of air keeps its entropy. That single fact turns the full equations of compressible flow into a handful of algebraic relations between the Mach number and the pressure, temperature and density. Kenji Watanabe asks Arjun to build a small program that reproduces the standard isentropic flow tables, because the group will use those relations every day, from the air data system to the engine inlet.

Stagnation conditions

Lesson 1 introduced the stagnation temperature \(T_0\), the temperature air would reach if brought to rest adiabatically. Dividing the energy equation \(c_pT + \tfrac{1}{2}V^2 = c_pT_0\) by \(c_pT\) and using \(c_p = \gamma R/(\gamma - 1)\) and \(a^2 = \gamma R T\) gives the stagnation temperature ratio as a function of Mach number alone.

If the deceleration is also reversible, it is isentropic, and the isentropic power laws of Lesson 1 give the stagnation pressure and density. The stagnation pressure \(p_0\) is the pressure an ideal pitot tube would read in subsonic flow, and the stagnation density \(\rho_0\) is the density in its mouth.

These ratios are the backbone of compressible aerodynamics. At Mach 0.3 the stagnation pressure is only 6.4 percent above the static pressure. At Mach 0.8 it is 52 percent above it, at Mach 2 nearly eight times, and at Mach 3 almost 37 times. The steep growth explains why supersonic wind tunnels need very high supply pressures, and why the pressures on the nose of a fast aircraft are so large.

The stagnation quantities can be defined at every point in any flow, even one that is never actually brought to rest: they are the values the gas would have if it were. In an isentropic flow they are the same everywhere, so they act as reference values. A wind tunnel with a settling chamber at 300 kPa and 290 K has \(p_0 = 300\ \text{kPa}\) and \(T_0 = 290\ \text{K}\) everywhere in its isentropic nozzle and test section. Across a shock the stagnation temperature stays the same, but the stagnation pressure falls, and that fall is a direct measure of the entropy produced.

Sonic reference conditions

A second set of reference values is the sonic condition, marked with an asterisk: the state the gas would have if it were accelerated or decelerated isentropically to exactly Mach 1. Setting \(M = 1\) in the ratios gives \(T^*/T_0 = 2/(\gamma + 1) = 0.8333\) and \(p^*/p_0 = 0.5283\) for air. The second number is especially important: air flowing from a reservoir reaches the speed of sound when the pressure has fallen to just under 53 percent of the reservoir pressure. Any nozzle whose exit pressure is lower than that can be choked, a condition examined in Lesson 6.

Area and velocity: the reversal at Mach 1

For quasi-one-dimensional flow in a duct whose area changes slowly, continuity says \(\rho V A\) is constant. Differentiating, and using the momentum equation \(dp = -\rho V\,dV\) together with \(dp = a^2\,d\rho\), leads to the area-velocity relation.

This short equation contains one of the most important ideas in gas dynamics. Below Mach 1, the bracket is negative, so a narrowing duct accelerates the flow, as everyday experience suggests. Above Mach 1, the bracket is positive, so a narrowing duct decelerates the flow and a widening duct accelerates it. The reason is density: in supersonic flow the density falls faster than the velocity rises, so the stream tube must widen to carry the same mass flow. At exactly Mach 1 the area change must be zero, which means sonic flow can occur only at a throat, a minimum area. To accelerate a gas from rest to supersonic speed, the duct must converge to a throat and then diverge. This is the converging-diverging nozzle invented for steam turbines by Gustaf de Laval in the 1880s, and it is used in every rocket engine and supersonic wind tunnel.

The area-Mach relation

Integrating the area-velocity relation for isentropic flow of a perfect gas gives the ratio of the local area to the sonic throat area \(A^*\), the area at which the flow would just reach Mach 1. The result depends only on Mach number.

The relation is easy to evaluate forward: given the Mach number, find the area ratio. Design usually needs the inverse: given the area of a nozzle station, find the Mach number. There is no closed form for the inverse, and every area ratio above 1 has two solutions, one subsonic and one supersonic. Which one occurs depends on the pressures at the two ends of the duct, as Lesson 6 shows. Arjun's program solves the inverse by Newton's method, starting on the requested branch and stepping toward the root using a numerical derivative. For an area ratio of 2.5 it finds Mach 0.240 and Mach 2.443 in a few iterations.

# Isentropic flow: stagnation ratios and the area-Mach relation solved by Newton's method
import math
G = 1.4
def T0_T(M): return 1 + (G - 1) / 2 * M * M
def p0_p(M): return T0_T(M) ** (G / (G - 1))
def A_Astar(M): return (1 / M) * ((2 / (G + 1)) * T0_T(M)) ** ((G + 1) / (2 * (G - 1)))
def mach_from_area(ratio, supersonic):
    M = 2.2 if supersonic else 0.3          # start on the wanted branch
    for _ in range(50):
        f = A_Astar(M) - ratio
        d = (A_Astar(M + 1e-6) - A_Astar(M - 1e-6)) / 2e-6
        M_new = M - f / d
        M = max(M_new, 1.0001) if supersonic else min(max(M_new, 1e-4), 0.9999)
        if abs(f) < 1e-12: break
    return M
print(" M     T0/T    p0/p    rho0/rho  A/A*")
for M in (0.3, 0.8, 1.0, 1.5, 2.0, 3.0):
    print(f"{M:4.1f} {T0_T(M):7.4f} {p0_p(M):8.4f} {T0_T(M) ** (1 / (G - 1)):8.4f} {A_Astar(M):7.4f}")
# the two answers for one area ratio
for ratio in (1.5, 2.5):
    print(f"\nA/A* = {ratio}: subsonic M = {mach_from_area(ratio, False):.4f}, supersonic M = {mach_from_area(ratio, True):.4f}")
# ram temperature and total pressure at the cruise point: M 0.80 at 12 500 m
T = 216.65; p = 22632 * math.exp(-1500 / (287.05 * T / 9.80665))   # ISA at 12 500 m
print(f"\ncruise: T0 = {T * T0_T(0.8):.1f} K (rise {T * (T0_T(0.8) - 1):.1f} K), p0 = {p * p0_p(0.8):.0f} Pa (p = {p:.0f} Pa)")
# engine face: the inlet slows the air from M 0.80 to M 0.50 at the fan face
A_fan = 0.385                                   # fan face area, m2 (study value)
A_capture = A_fan * A_Astar(0.80) / A_Astar(0.50)
print(f"capture area for M 0.50 at the fan: {A_capture:.3f} m2 (fan face {A_fan} m2)")

 M     T0/T    p0/p    rho0/rho  A/A*
 0.3  1.0180   1.0644   1.0456  2.0351
 0.8  1.1280   1.5243   1.3514  1.0382
 1.0  1.2000   1.8929   1.5774  1.0000
 1.5  1.4500   3.6710   2.5317  1.1762
 2.0  1.8000   7.8244   4.3469  1.6875
 3.0  2.8000  36.7327  13.1188  4.2346

A/A* = 1.5: subsonic M = 0.4303, supersonic M = 1.8541

A/A* = 2.5: subsonic M = 0.2395, supersonic M = 2.4428

cruise: T0 = 244.4 K (rise 27.7 K), p0 = 27232 Pa (p = 17865 Pa)
capture area for M 0.50 at the fan: 0.298 m2 (fan face 0.385 m2)

Air data: measuring Mach number

The air data computer of the Swift 900 measures the total pressure from a pitot probe and the static pressure from flush ports on the fuselage. In subsonic flight the probe brings the air to rest almost isentropically, so the stagnation pressure ratio can be inverted to give the Mach number directly.

The true airspeed then needs the temperature. A probe on the fuselage measures the total air temperature, which is close to \(T_0\) because the air is brought nearly to rest in the probe. The computer divides by the stagnation temperature ratio to obtain the static temperature, then the speed of sound, then the true airspeed. Real probes do not quite recover the full stagnation temperature: a recovery factor of about 0.95 to 1.0 is applied, determined by calibration. The method fails above Mach 1, because a pitot probe in supersonic flow has a shock in front of it. Lesson 3 gives the correct formula for that case.

The program shows that at the cruise point, Mach 0.80 at 12,500 m, the ambient air is at 216.65 K and 17.9 kPa, while the stagnation temperature is 244.4 K and the stagnation pressure 27.2 kPa. The 28 K ram temperature rise is not small. It warms the leading edges, it reduces the ice threat slightly, and it must be accounted for in the cooling of avionics that use ram air. The ram pressure rise of over 9 kPa is what the engine inlet and the cabin ventilation system can use.

The engine inlet as an isentropic diffuser

A turbofan's fan works best when the air reaches it at about Mach 0.5 to 0.6, much slower than the Mach 0.8 flight speed. The inlet must slow the air with as little loss of stagnation pressure as possible, because every percent of pressure lost costs roughly a percent of thrust. In subsonic flight much of this deceleration happens ahead of the inlet: the stream tube that enters the engine at cruise is smaller than the inlet opening, and it widens as it approaches, slowing the air without any wall friction. The area-Mach relation gives the required capture area. If the fan face has an area of 0.385 m² and the air must arrive at Mach 0.50, the free-stream tube that feeds it at Mach 0.80 has an area of only 0.298 m². The inlet lip, designed for low speed and take-off, is larger still, so at cruise the air spills around the lip, and the shape of the outer cowl must be designed so that the spilled air does not form a shock.

This is a good example of a compressible-flow idea that has no incompressible counterpart. In incompressible flow the capture area would be set simply by the ratio of speeds, 0.50 to 0.80, giving 0.241 m². The compressible answer is 24 percent larger, because the air is denser at the fan face than in the free stream. An inlet sized with incompressible thinking would be too small for the engine.

Where isentropic flow breaks down

The isentropic relations fail in three places. Inside boundary layers, friction generates entropy, and the stagnation pressure falls toward the wall. Across shock waves the change is sudden and irreversible, and the next lesson gives the relations that replace the isentropic ones. Where heat is added, as in a combustor, the entropy rises and Lesson 8 applies. Outside those regions the isentropic relations are extremely accurate, and they are used even in computational fluid dynamics as checks on the far field and on boundary conditions.