Prev
Unit 2 · Lesson 2-1
Stagnation Properties and Isentropic Flow
2h 40m1300 XP
Next

Stagnation Properties and Isentropic Flow

The speed of sound, Mach number, stagnation temperature and pressure, the area–Mach relation and choking

Over the Mojave — 14 October 1947
A B-29 bomber drops a small rocket plane over the Mojave Desert in California. Captain Chuck Yeager lights its engine and climbs, and the Bell X-1 he has named Glamorous Glennis reaches Mach 1.06 — about 1,100 km/h — before gliding down to a landing on the dry lake bed. It is the first crewed supersonic flight (the X-1's history). Pilots called what Yeager crossed the sound barrier. Nothing about the air changes at Mach 1; what changes is how air responds to being pushed. Near the speed of sound it compresses, and the smooth, incompressible fluid of first-year mechanics stops being a good model. Inside a jet engine air crosses that line every second. The fan tips of an airliner engine at take-off move faster than sound relative to the air, the turbine's first row of vanes runs choked at exactly Mach 1, and the nozzle of a fighter accelerates its jet far past it. This unit is the gas dynamics that makes all three possible.
How fast is "near"? Bring a moving gas to rest and its density rises. At Mach 0.3 the change is only 4.4 % (Lab 1 computes it), which is why low-speed aerodynamics treats air as incompressible. Engines live far above that: an airliner cruises at Mach 0.85, its fan face sees about Mach 0.6, and its jets leave near Mach 1.
Compressibility is measured against one speed — the speed at which a small pressure disturbance travels through the gas.
a=γRT,M=Vaa = \sqrt{\gamma R T}, \qquad M = \frac{V}{a}

The speed of sound in a perfect gas and the Mach number. γ is the ratio of specific heats (1.4 for air at ordinary temperatures), R = 287 J/(kg·K) for air.

Derivation
The speed of sound
Ride with a weak pressure pulse moving into still gas. In that frame gas arrives at speed aa with pressure pp and density ρ\rho, and leaves at a−dVa - dV with p+dpp + dp and ρ+dρ\rho + d\rho.
Mass: ρa=(ρ+dρ)(a−dV)\rho a = (\rho + d\rho)(a - dV), so to first order a dρ=ρ dVa\,d\rho = \rho\,dV.
Momentum, using the constant mass flux ρa\rho a:
p+ρa2=(p+dp)+ρa(a−dV)⟹dp=ρa dV.p + \rho a^2 = (p + dp) + \rho a (a - dV) \quad\Longrightarrow\quad dp = \rho a\, dV .
Eliminate dVdV: dp=a2 dρdp = a^2\, d\rho, so a2=dp/dρa^2 = dp/d\rho. The pulse is weak and passes quickly, with no time for heat to flow and nothing to cause friction, so the compression is isentropic: p∝ργp \propto \rho^\gamma and dp/dρ=γp/ρ=γRTdp/d\rho = \gamma p/\rho = \gamma R T. Hence
a=γRT.a = \sqrt{\gamma R T}.
Newton assumed the compression was isothermal and got RT\sqrt{RT}, about 15 % too low; Laplace added the γ\gamma. At 288 K, a=340a = 340 m/s; in the stratosphere at 216.65 K, a=295a = 295 m/s.
Stagnation properties. Slow a gas down without heat or work and its kinetic energy becomes internal energy: it gets hotter. The steady-flow energy equation for an adiabatic flow says the stagnation enthalpy ht=h+V2/2h_t = h + V^2/2 stays the same, and for a perfect gas that fixes a stagnation temperature TtT_t. If the slowing-down is also reversible, the pressure rises to the stagnation pressure ptp_t.
TtT=1+γ−12M2,ptp=(1+γ−12M2)γ/(γ−1)\frac{T_t}{T} = 1 + \frac{\gamma - 1}{2} M^2, \qquad \frac{p_t}{p} = \left(1 + \frac{\gamma - 1}{2} M^2\right)^{\gamma/(\gamma - 1)}

Stagnation temperature and pressure in terms of the static values and the Mach number.

Derivation
Stagnation temperature and pressure
For steady adiabatic flow with no shaft work, h+V2/2=hth + V^2/2 = h_t. With h=cpTh = c_p T:
Tt=T+V22cp.T_t = T + \frac{V^2}{2 c_p}.
Divide by TT and use cp=γR/(γ−1)c_p = \gamma R/(\gamma - 1) and V2=M2a2=M2γRTV^2 = M^2 a^2 = M^2 \gamma R T:
TtT=1+V22cpT=1+(γ−1)M2γRT2γRT=1+γ−12M2.\frac{T_t}{T} = 1 + \frac{V^2}{2 c_p T} = 1 + \frac{(\gamma - 1) M^2 \gamma R T}{2\gamma R T} = 1 + \frac{\gamma - 1}{2} M^2 .
For an isentropic process of a perfect gas, p∝Tγ/(γ−1)p \propto T^{\gamma/(\gamma - 1)}, which gives the pressure ratio. Note the difference: TtT_t needs only an adiabatic process, so it survives friction and shocks; ptp_t needs a reversible one, so every loss shows up as a fall in ptp_t.
💡Two bookkeeping quantities
Engine designers follow two numbers along the whole flow path:
  • Stagnation temperature TtT_t tracks energy. It is constant through inlets, nozzles and ducts, rises only where work is done (compressor) or heat is added (combustor), and falls only where work is taken out (turbine).
  • Stagnation pressure ptp_t tracks quality. Compressors raise it; everything else — shocks, friction, mixing, even burning — lowers it. Unit 4 measures each component's losses as a ratio of stagnation pressures.
📐Ram temperature
Air that reaches an inlet at flight speed arrives at its stagnation temperature — the ram temperature. With γ=1.4\gamma = 1.4:
  • Boeing 777 at Mach 0.84 in the stratosphere (T=216.65T = 216.65 K): Tt=247T_t = 247 K, about −26 ∘-26\,^\circC — 30 degrees warmer than the outside air.
  • Concorde at Mach 2.02: Tt=393.5T_t = 393.5 K, about 120 ∘120\,^\circC.
  • SR-71 at Mach 3.2 at 24 km (T=220.65T = 220.65 K): Tt=672.5T_t = 672.5 K, about 400 ∘400\,^\circC.
At Mach 3.2 the air reaches the compressor already at 400 °C, before any compression has been done. The compressor can then add only a little before the turbine's temperature limit is reached. Unit 3 shows that this is what ends the turbojet's speed range, and what makes the ramjet worth having.
Why a nozzle narrows and then widens. In incompressible flow a narrower pipe always means faster flow. Near the speed of sound that stops being true, because the density changes as the flow speeds up. Combining continuity with the momentum equation gives one of the most important results in propulsion:
dAA=(M2−1) dVV.\frac{dA}{A} = (M^2 - 1)\,\frac{dV}{V}.
Below Mach 1 the bracket is negative: to speed the gas up, the duct must converge. Above Mach 1 it is positive: to keep speeding up, the duct must diverge, because the density now falls faster than the velocity rises. Mach 1 can only occur where dA=0dA = 0 — at a minimum area, the throat. That is the shape of every rocket nozzle and every fighter's variable nozzle.
AA∗=1M[2γ+1(1+γ−12M2)]γ+12(γ−1)\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma + 1}\left(1 + \frac{\gamma - 1}{2}M^2\right)\right]^{\frac{\gamma + 1}{2(\gamma - 1)}}

The area–Mach relation: the area the flow needs at Mach M, relative to the sonic throat area A. Every A/A > 1 has one subsonic and one supersonic solution.

Derivation
The area–Mach relation
Steady flow through a duct of slowly varying area: ρVA=\rho V A = const, so
dρρ+dVV+dAA=0.\frac{d\rho}{\rho} + \frac{dV}{V} + \frac{dA}{A} = 0 .
Euler's equation for frictionless flow, dp=−ρV dVdp = -\rho V\, dV, and the isentropic link dp=a2 dρdp = a^2\, d\rho give dρ/ρ=−(V2/a2) dV/V=−M2 dV/Vd\rho/\rho = -(V^2/a^2)\, dV/V = -M^2\, dV/V. Substituting,
dAA=(M2−1)dVV.\frac{dA}{A} = (M^2 - 1)\frac{dV}{V}.
To integrate, compare any station with the sonic one, where the same mass flux passes through A∗A^*: A/A∗=(ρ∗V∗)/(ρV)A/A^* = (\rho^* V^*)/(\rho V). Write ρ∗/ρ=(ρ∗/ρt)(ρt/ρ)\rho^*/\rho = (\rho^*/\rho_t)(\rho_t/\rho) and V∗/V=a∗/(Ma)V^*/V = a^*/(M a) with the isentropic relations ρt/ρ=(Tt/T)1/(γ−1)\rho_t/\rho = (T_t/T)^{1/(\gamma - 1)}, a∗/a=(T∗/T)1/2a^*/a = (T^*/T)^{1/2} and Tt/T∗=(γ+1)/2T_t/T^* = (\gamma + 1)/2. Collecting powers of 1+γ−12M21 + \frac{\gamma - 1}{2}M^2 gives the result. Lab 2 lets SymPy differentiate it and confirm that its only minimum is A/A∗=1A/A^* = 1 at M=1M = 1.
💡Choking and the mass flow parameter
The mass flow through any section can be written with stagnation quantities: m˙TtA pt=γR  M(1+γ−12M2)−γ+12(γ−1).\frac{\dot m \sqrt{T_t}}{A\, p_t} = \sqrt{\frac{\gamma}{R}}\; M \left(1 + \frac{\gamma - 1}{2}M^2\right)^{-\frac{\gamma + 1}{2(\gamma - 1)}} . This mass flow parameter peaks at Mach 1: 0.0404 kg·K½/(N·s) for cold air, 0.0397 for hot gas with γ=1.33\gamma = 1.33. A throat at Mach 1 is choked: lowering the pressure downstream cannot pass more flow; only more ptp_t, less TtT_t or a bigger throat can. The first row of turbine vanes is choked over most of an engine's operating range, so m˙Tt4/pt4\dot m\sqrt{T_{t4}}/p_{t4} is fixed by their throat area. Lab 3 computes the flow through such a vane row, and Unit 8 builds the whole matching of an engine on this fact.
Derivation
The mass flow parameter
m˙=ρVA\dot m = \rho V A with ρ=p/(RT)\rho = p/(RT) and V=MγRTV = M\sqrt{\gamma R T}:
m˙=pRTMγRT A=A p MγRT.\dot m = \frac{p}{RT} M \sqrt{\gamma R T}\, A = A\, p\, M \sqrt{\frac{\gamma}{R T}} .
Replace the static values by stagnation ones, p=pt(T/Tt)γ/(γ−1)p = p_t (T/T_t)^{\gamma/(\gamma - 1)} and T=Tt(T/Tt)T = T_t (T/T_t), with T/Tt=(1+γ−12M2)−1T/T_t = (1 + \frac{\gamma - 1}{2}M^2)^{-1}:
m˙TtApt=γR M(1+γ−12M2)−γγ−1+12,\frac{\dot m \sqrt{T_t}}{A p_t} = \sqrt{\frac{\gamma}{R}}\, M \left(1 + \frac{\gamma - 1}{2}M^2\right)^{-\frac{\gamma}{\gamma - 1} + \frac{1}{2}} ,
and −γγ−1+12=−γ+12(γ−1)-\frac{\gamma}{\gamma - 1} + \frac12 = -\frac{\gamma + 1}{2(\gamma - 1)}. Setting its derivative with respect to MM to zero gives M=1M = 1.
In World War II, in combat in P-51s, during dogfights with 109s and 190s, for the first time we became exposed to the effects of the speed of sound on our airplanes. A Mustang, a P-47, or any of the other fighters that we were using in World War II, the fastest they would go was about 80 percent the speed of sound. They had very thick wings and canopies. That additional distance that the air had to travel to go around that wing that's going at about 80 percent of the speed of sound, brought its relative velocity to the skin of the wing up to the speed of sound.
– Chuck Yeager, interview with the Academy of Achievement (Wikiquote)
💡Why γ is 1.4 in the cold section and 1.33 in the hot
At room temperature an air molecule stores energy in its motion and its rotation, which gives cv=52Rc_v = \tfrac52 R and γ=7/5\gamma = 7/5. At turbine temperatures the molecules also start to vibrate, cpc_p rises and γ\gamma falls; combustion products such as CO₂ and H₂O lower it further. This course uses γ=1.4\gamma = 1.4, cp=1004.5c_p = 1004.5 J/(kg·K) before the combustor and γt=1.33\gamma_t = 1.33, cpt=1156c_{pt} = 1156 J/(kg·K) after it. Lesson 4-3 replaces both with real-gas properties from CoolProp and Cantera.
💡Where you will use this
Lesson 2-2 builds shocks out of these relations. Unit 5 uses them for every inlet and nozzle, Lesson 6-2 for the relative Mach number at a fan's tips, and Unit 8 for the choked vanes and nozzles that fix an engine's operating line. The same equations size the nozzle of every rocket engine.
📚Further reading
↓But there is a problem
Run the area–Mach relation backwards and a supersonic stream should slow down smoothly in a converging duct, pass Mach 1 at the throat and leave subsonic, losing nothing. Every supersonic inlet would like to work that way. Almost none does. In 1886, in Rijeka on the Adriatic, a physicist with an electric spark and a photographic plate caught the reason in mid-air.