3.1.8Compressible Flow & Aerodynamics

Choked flow — condition M = 1 at throat, maximum mass flow

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1. What is choking? (WHAT)

The throat is where AA is minimum. The key fact we'll prove: only at M=1M=1 can the throat carry the maximum possible mass flow.


2. Building blocks (HOW we set it up)

We use isentropic, steady, 1-D flow of a perfect gas, ratio of specific heats γ\gamma, gas constant RR. Stagnation (reservoir) conditions p0,T0,ρ0p_0, T_0, \rho_0 are fixed.

Why the T0/TT_0/T step? Divide h0=h+12V2h_0=h+\frac12V^2 by cpTc_pT: T0T=1+V22cpT\frac{T_0}{T}=1+\frac{V^2}{2c_pT}. Now V22cpT=M2γRT2cpT=M2γR2cp=γ12M2\frac{V^2}{2c_pT}=\frac{M^2\gamma RT}{2c_pT}=\frac{M^2\gamma R}{2c_p}=\frac{\gamma-1}{2}M^2. Done — pure algebra, no memorization.


3. Deriving the mass flow (DERIVATION FROM SCRATCH)

Mass flow through area AA: m˙=ρAV\dot{m}=\rho A V. Express everything in MM and stagnation quantities.

Step 1 — write ρ\rho, VV in stagnation terms.

  • ρ=ρ0(1+γ12M2)1/(γ1)\rho = \rho_0\left(1+\frac{\gamma-1}{2}M^2\right)^{-1/(\gamma-1)}
  • V=Ma=MγRTV = Ma = M\sqrt{\gamma RT}, and T=T0(1+γ12M2)1T = T_0\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}, so V=MγRT0(1+γ12M2)1/2V = M\sqrt{\gamma R T_0}\left(1+\frac{\gamma-1}{2}M^2\right)^{-1/2}.

Why this step? We want m˙\dot m as a function of MM alone (for fixed p0,T0p_0,T_0) so we can find where it maxes out.

Step 2 — combine. Using ρ0=p0/(RT0)\rho_0 = p_0/(RT_0):

  m˙=Ap0T0γR  M(1+γ12M2)γ+12(γ1)  \boxed{\;\dot{m} = \frac{A\,p_0}{\sqrt{T_0}}\sqrt{\frac{\gamma}{R}}\;\frac{M}{\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{\gamma+1}{2(\gamma-1)}}}\;}

Why this step? All the MM-dependence is now in one tidy factor f(M)=M(1+γ12M2)(γ+1)/2(γ1)f(M)=\dfrac{M}{(1+\frac{\gamma-1}{2}M^2)^{(\gamma+1)/2(\gamma-1)}}.

Step 3 — maximize f(M)f(M) over MM at fixed throat area. Set dfdM=0\dfrac{df}{dM}=0. Taking lnf=lnMγ+12(γ1)ln(1+γ12M2)\ln f = \ln M - \frac{\gamma+1}{2(\gamma-1)}\ln(1+\frac{\gamma-1}{2}M^2) and differentiating:

1Mγ+12(γ1)(γ1)M1+γ12M2=0\frac{1}{M} - \frac{\gamma+1}{2(\gamma-1)}\cdot\frac{(\gamma-1)M}{1+\frac{\gamma-1}{2}M^2}=0

1M=(γ+1)M/21+γ12M2    1+γ12M2=γ+12M2    M2=1.\frac{1}{M}=\frac{(\gamma+1)M/2}{1+\frac{\gamma-1}{2}M^2}\;\Rightarrow\; 1+\frac{\gamma-1}{2}M^2=\frac{\gamma+1}{2}M^2\;\Rightarrow\; M^2=1.


4. Critical (sonic, "starred") conditions

Plug M=1M=1 into the isentropic relations to get the critical ratios — the conditions at the throat when choked:


Figure — Choked flow — condition M = 1 at throat, maximum mass flow

5. Worked examples


6. Common mistakes (Steel-man + fix)


7. Active-recall flashcards

#flashcards/physics

What Mach number occurs at the throat of a choked nozzle?
M=1M=1 (sonic).
Why does mass flow stop increasing once choked?
Throat is sonic; pressure signals can't travel upstream past M=1M=1, so the throat can't "feel" lower back pressure.
Critical pressure ratio p/p0p^*/p_0 for air (γ=1.4\gamma=1.4)?
0.528\approx 0.528.
Critical temperature ratio T/T0T^*/T_0 for air?
0.833\approx 0.833 (=2/(γ+1)=2/(\gamma+1)).
General formula T/T0T^*/T_0?
2γ+1\dfrac{2}{\gamma+1}.
General formula p/p0p^*/p_0?
(2γ+1)γ/(γ1)\left(\dfrac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)}.
At what MM is m˙\dot m per unit throat area maximized?
At M=1M=1 (where dm˙/dM=0d\dot m/dM=0).
How does m˙max\dot m_{\max} scale with p0p_0, T0T_0, AA^*?
p0\propto p_0, A\propto A^*, 1/T0\propto 1/\sqrt{T_0}.
Choked-flow max mass-flow formula?
m˙max=Ap0γ/(RT0)(2/(γ+1))(γ+1)/2(γ1)\dot m_{\max}=A^*p_0\sqrt{\gamma/(RT_0)}\,(2/(\gamma+1))^{(\gamma+1)/2(\gamma-1)}.
To get supersonic exit, what nozzle shape is required?
Converging–diverging (de Laval).
When choked, where does remaining expansion to ambient occur?
Outside the nozzle (under-expanded jet).

Recall Feynman: explain to a 12-year-old

Imagine squeezing toothpaste through the narrowest part of a tube. As you suck harder on the far end, more comes out — for a while. But once the toothpaste at the narrow spot is moving as fast as a "message" can travel through it (the speed of sound), the narrow spot can't hear your sucking anymore. So no matter how hard you pull, the same amount keeps coming out. That "as fast as a sound message" speed is M=1M=1, and the steady stuck amount is the maximum mass flow.

Connections

Concept Map

increases

expressed via

written in Mach terms

gives

substitute rho, V

maximize df/dM = 0

defines

yields

because sonic blocks

so

location of

Lower back pressure

Mass flow m-dot

m-dot = rho A V

f of M factor

Isentropic 1-D perfect gas

Stagnation relations T0/T, p0/p

M = 1 at throat

Choked flow

Maximum m-dot

Downstream cannot signal upstream

Further lowering Pb does nothing

Throat = minimum area A

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek converging nozzle hai jisme reservoir (stagnation) se gas bahar nikal rahi hai. Jaise-jaise tum back pressure pbp_b kam karte ho, pehle to mass flow m˙\dot m badhta jaata hai — zyada pressure drop, zyada flow. Lekin ek limit aati hai: jab nozzle ke sabse patle hisse (throat) pe flow ki speed sound ke barabar ho jaati hai, yaani M=1M=1. Iske baad chahe tum back pressure kitna bhi giraao, mass flow nahi badhega. Isi ko choked flow kehte hain.

WHY aisa hota hai? Pressure ki "information" sirf sound ki speed se travel karti hai. Jab throat pe flow already sonic hai, to downstream ka low pressure ka signal upstream throat tak pahunch hi nahi sakta — throat ko "sunai" nahi deta. Isliye throat respond nahi kar paata, aur m˙\dot m apne maximum pe lock ho jaata hai. Maths bhi yahi kehti hai: m˙\dot m ko MM ke against maximize karo, to derivative zero M=1M=1 pe hi aata hai. Pure coincidence nahi — exact maximum.

Air ke liye yaad rakhne wali numbers: critical pressure ratio p/p00.528p^*/p_0 \approx 0.528 aur temperature ratio T/T00.833T^*/T_0 \approx 0.833. Matlab jaise hi pb/p0p_b/p_0 giirke 0.528 tak aaya, nozzle choke ho gaya, aur throat ka pressure 0.528p00.528\,p_0 pe atak gaya. Ek important point: m˙max1/T0\dot m_{max} \propto 1/\sqrt{T_0} — yaani garam reservoir se kam mass flow milta hai (kyunki density gir jaati hai). Aur dhyan rakho: sirf converging nozzle se supersonic flow nahi milta — uske liye converging–diverging (de Laval) nozzle chahiye.

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