Intuition The big picture (WHY these ratios exist)When a gas flows along smoothly (no shocks, no friction, no heat added) and you imagine bringing it to rest reversibly , you get its stagnation (total) properties P 0 , T 0 , ρ 0 P_0,\,T_0,\,\rho_0 P 0 , T 0 , ρ 0 P , T , ρ P,T,\rho P , T , ρ only thing that decides how much is spent is the Mach number M = V / a M=V/a M = V / a M M M
Definition Stagnation (total) propertiesThe stagnation state is the state the fluid would reach if decelerated adiabatically and reversibly (isentropically) to zero velocity . T 0 T_0 T 0 P 0 P_0 P 0 ρ 0 \rho_0 ρ 0 constant along an isentropic flow .
WHAT we want: the three ratios T T 0 , P P 0 , ρ ρ 0 \dfrac{T}{T_0},\ \dfrac{P}{P_0},\ \dfrac{\rho}{\rho_0} T 0 T , P 0 P , ρ 0 ρ M M M γ \gamma γ
The steady adiabatic energy equation (per unit mass) for a perfect gas:
h + 1 2 V 2 = h 0 = const h + \tfrac12 V^2 = h_0 = \text{const} h + 2 1 V 2 = h 0 = const
Why this step? No heat, no work ⇒ stagnation enthalpy h 0 h_0 h 0 V = 0 V=0 V = 0 h 0 h_0 h 0
For a perfect gas h = c p T h=c_p T h = c p T
c p T + 1 2 V 2 = c p T 0 c_p T + \tfrac12 V^2 = c_p T_0 c p T + 2 1 V 2 = c p T 0
Divide by c p T c_p T c p T
T 0 T = 1 + V 2 2 c p T \frac{T_0}{T} = 1 + \frac{V^2}{2 c_p T} T T 0 = 1 + 2 c p T V 2
Now we need V 2 / ( c p T ) V^2/(c_p T) V 2 / ( c p T ) M M M
Speed of sound: a 2 = γ R T ⇒ V 2 = M 2 a 2 = M 2 γ R T a^2=\gamma R T \Rightarrow V^2 = M^2 a^2 = M^2 \gamma R T a 2 = γ R T ⇒ V 2 = M 2 a 2 = M 2 γ R T
c p = γ R γ − 1 c_p = \dfrac{\gamma R}{\gamma-1} c p = γ − 1 γ R c p − c v = R c_p-c_v=R c p − c v = R γ = c p / c v \gamma=c_p/c_v γ = c p / c v
Why this step? These convert velocity into Mach number, the only variable we want.
V 2 2 c p T = M 2 γ R T 2 ⋅ γ R γ − 1 T = γ − 1 2 M 2 \frac{V^2}{2c_p T}=\frac{M^2\gamma R T}{2\cdot\frac{\gamma R}{\gamma-1}\,T}=\frac{\gamma-1}{2}M^2 2 c p T V 2 = 2 ⋅ γ − 1 γ R T M 2 γ R T = 2 γ − 1 M 2
For an isentropic process of a perfect gas:
P 0 P = ( T 0 T ) γ γ − 1 , ρ 0 ρ = ( T 0 T ) 1 γ − 1 \frac{P_0}{P}=\left(\frac{T_0}{T}\right)^{\frac{\gamma}{\gamma-1}},\qquad \frac{\rho_0}{\rho}=\left(\frac{T_0}{T}\right)^{\frac{1}{\gamma-1}} P P 0 = ( T T 0 ) γ − 1 γ , ρ ρ 0 = ( T T 0 ) γ − 1 1
Why this step? From P v γ = Pv^\gamma= P v γ = P v = R T Pv=RT P v = R T v v v P ∝ T γ / ( γ − 1 ) P\propto T^{\gamma/(\gamma-1)} P ∝ T γ / ( γ − 1 ) ρ ∝ T 1 / ( γ − 1 ) \rho\propto T^{1/(\gamma-1)} ρ ∝ T 1/ ( γ − 1 )
Substituting the T 0 / T T_0/T T 0 / T
Consistency check (Feynman test): dividing gives P 0 / P ρ 0 / ρ = ( 1 + γ − 1 2 M 2 ) 1 = T 0 T \dfrac{P_0/P}{\rho_0/\rho}=\left(1+\frac{\gamma-1}{2}M^2\right)^{1}=\dfrac{T_0}{T} ρ 0 / ρ P 0 / P = ( 1 + 2 γ − 1 M 2 ) 1 = T T 0 P / ρ ∝ T P/\rho\propto T P / ρ ∝ T
A table is just these three formulas evaluated for γ = 1.4 \gamma=1.4 γ = 1.4 M M M
M M M T / T 0 T/T_0 T / T 0 P / P 0 P/P_0 P / P 0 ρ / ρ 0 \rho/\rho_0 ρ / ρ 0
0
1.000
1.000
1.000
0.5
0.9524
0.8430
0.8852
1.0
0.8333
0.5283
0.6339
2.0
0.5556
0.1278
0.2301
P P 0 < ρ ρ 0 < T T 0 \dfrac{P}{P_0}<\dfrac{\rho}{\rho_0}<\dfrac{T}{T_0} P 0 P < ρ 0 ρ < T 0 T M > 0 M>0 M > 0 γ γ − 1 = 3.5 \frac{\gamma}{\gamma-1}=3.5 γ − 1 γ = 3.5
Worked example Example 1 — Air atM = 2 M=2 M = 2 T 0 = 300 T_0=300 T 0 = 300 P 0 = 500 P_0=500 P 0 = 500 γ = 1.4 \gamma=1.4 γ = 1.4 γ − 1 2 M 2 = 0.2 × 4 = 0.8 \dfrac{\gamma-1}{2}M^2=0.2\times4=0.8 2 γ − 1 M 2 = 0.2 × 4 = 0.8 = 1.8 =1.8 = 1.8
T = 300 / 1.8 = 166.7 T = 300/1.8 = 166.7 T = 300/1.8 = 166.7 Why? T 0 / T = 1.8 T_0/T=1.8 T 0 / T = 1.8 P = 500 / 1.8 3.5 = 500 / 7.824 = 63.9 P = 500/1.8^{3.5}=500/7.824=63.9 P = 500/1. 8 3.5 = 500/7.824 = 63.9 Why? exponent γ / ( γ − 1 ) = 3.5 \gamma/(\gamma-1)=3.5 γ / ( γ − 1 ) = 3.5 ρ / ρ 0 = 1.8 − 2.5 = 0.2301 \rho/\rho_0 = 1.8^{-2.5}=0.2301 ρ / ρ 0 = 1. 8 − 2.5 = 0.2301 Why? exponent 1 / ( γ − 1 ) = 2.5 1/(\gamma-1)=2.5 1/ ( γ − 1 ) = 2.5
Worked example Example 3 — Inverse problem: givenP / P 0 = 0.3 P/P_0=0.3 P / P 0 = 0.3 M M M ( 1 + 0.2 M 2 ) 3.5 = 1 / 0.3 = 3.333 \left(1+0.2M^2\right)^{3.5}=1/0.3=3.333 ( 1 + 0.2 M 2 ) 3.5 = 1/0.3 = 3.333 1 + 0.2 M 2 = 3.333 1 / 3.5 = 1.4036 1+0.2M^2 = 3.333^{1/3.5}=1.4036 1 + 0.2 M 2 = 3.33 3 1/3.5 = 1.4036 M 2 = 0.4036 / 0.2 = 2.018 ⇒ M = 1.42 M^2 = 0.4036/0.2=2.018\Rightarrow M=1.42 M 2 = 0.4036/0.2 = 2.018 ⇒ M = 1.42 Why? We invert the same formula — tables are just this lookup precomputed.
T / T 0 = ( 1 + γ − 1 2 M 2 ) T/T_0=(1+\frac{\gamma-1}{2}M^2) T / T 0 = ( 1 + 2 γ − 1 M 2 ) reciprocal )
Why it feels right: the formula 1 + γ − 1 2 M 2 1+\frac{\gamma-1}{2}M^2 1 + 2 γ − 1 M 2 T / T 0 T/T_0 T / T 0 The fix: the clean factor equals T 0 / T T_0/T T 0 / T T T T smaller than stagnation T T T T / T 0 ≤ 1 T/T_0\le1 T / T 0 ≤ 1
Common mistake Using the same exponent forP P P ρ \rho ρ T T T Why it feels right: they all come from the same base factor. The fix: exponents differ — T T T ρ \rho ρ 1 γ − 1 = 2.5 \frac{1}{\gamma-1}=2.5 γ − 1 1 = 2.5 P P P γ γ − 1 = 3.5 \frac{\gamma}{\gamma-1}=3.5 γ − 1 γ = 3.5
Common mistake Applying these across ashock .
Why it feels right: they're "the gas tables". The fix: they require isentropic flow. Across a shock entropy jumps, so P 0 P_0 P 0 P 0 P_0 P 0
Recall Feynman: explain to a 12-year-oldImagine running with a balloon. If you suddenly stop the air in front of you, it piles up — gets a little hotter, squishier, and pushes harder. The faster you run (bigger Mach number), the more it piles up when stopped. The "stopped" values are T 0 , P 0 , ρ 0 T_0,P_0,\rho_0 T 0 , P 0 , ρ 0 how much smaller depends only on your speed compared to the speed of sound.
T-R-P = 1, 2.5, 3.5 "
The exponents on the base factor ( 1 + γ − 1 2 M 2 ) (1+\frac{\gamma-1}{2}M^2) ( 1 + 2 γ − 1 M 2 ) T emperature, R ho (density), P ressure are 1, 2.5, 3.5 for air. Alphabetical-ish T<R<P with increasing exponent ⇒ pressure drops fastest.
What conditions define a stagnation state? Decelerate the flow to zero velocity isentropically (adiabatic + reversible).
Formula for T 0 / T T_0/T T 0 / T M M M T 0 / T = 1 + γ − 1 2 M 2 T_0/T = 1+\frac{\gamma-1}{2}M^2 T 0 / T = 1 + 2 γ − 1 M 2 .
Formula for P 0 / P P_0/P P 0 / P ( 1 + γ − 1 2 M 2 ) γ / ( γ − 1 ) \left(1+\frac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)} ( 1 + 2 γ − 1 M 2 ) γ / ( γ − 1 ) .
Formula for ρ 0 / ρ \rho_0/\rho ρ 0 / ρ ( 1 + γ − 1 2 M 2 ) 1 / ( γ − 1 ) \left(1+\frac{\gamma-1}{2}M^2\right)^{1/(\gamma-1)} ( 1 + 2 γ − 1 M 2 ) 1/ ( γ − 1 ) .
Exponents for T, ρ, P (air, γ=1.4)? 1, 2.5, 3.5.
Why is T 0 T_0 T 0 Stagnation enthalpy is conserved (energy eq.); but
P 0 P_0 P 0 drops if irreversible.
Critical pressure ratio P ∗ / P 0 P^*/P_0 P ∗ / P 0 1.2 − 3.5 ≈ 0.528 1.2^{-3.5}\approx0.528 1. 2 − 3.5 ≈ 0.528 .
Which ratio falls fastest with M and why? Pressure, because its exponent
γ / ( γ − 1 ) = 3.5 \gamma/(\gamma-1)=3.5 γ / ( γ − 1 ) = 3.5 is largest.
Can you use these tables across a normal shock? No — entropy rises,
P 0 P_0 P 0 is not conserved; treat each side separately.
Given P / P 0 = 0.3 P/P_0=0.3 P / P 0 = 0.3 Solve
( 1 + 0.2 M 2 ) 3.5 = 1 / 0.3 ⇒ M ≈ 1.42 (1+0.2M^2)^{3.5}=1/0.3\Rightarrow M\approx1.42 ( 1 + 0.2 M 2 ) 3.5 = 1/0.3 ⇒ M ≈ 1.42 .
Adiabatic energy eq h + half V2 = h0
Stagnation state V=0 isentropic
Isentropic relations P v gamma const
P0/P ratio power gamma/ gamma-1
rho0/rho ratio power 1/ gamma-1
Consistency P ratio / rho ratio = T0/T
Intuition Hinglish mein samjhoDekho, idea simple hai: jab gas smoothly (bina shock, bina friction) flow karti hai, toh uska ek "total energy account" hota hai jise hum stagnation properties T 0 , P 0 , ρ 0 T_0, P_0, \rho_0 T 0 , P 0 , ρ 0 T , P , ρ T, P, \rho T , P , ρ Mach number M M M
Derivation ka core: energy equation c p T + 1 2 V 2 = c p T 0 c_pT+\frac12 V^2=c_pT_0 c p T + 2 1 V 2 = c p T 0 T 0 / T = 1 + γ − 1 2 M 2 T_0/T=1+\frac{\gamma-1}{2}M^2 T 0 / T = 1 + 2 γ − 1 M 2 P ∝ T γ / ( γ − 1 ) P\propto T^{\gamma/(\gamma-1)} P ∝ T γ / ( γ − 1 ) ρ ∝ T 1 / ( γ − 1 ) \rho\propto T^{1/(\gamma-1)} ρ ∝ T 1/ ( γ − 1 ) γ = 1.4 \gamma=1.4 γ = 1.4
Yeh "tables" koi magic nahi — bas γ = 1.4 \gamma=1.4 γ = 1.4 P / P 0 = 0.3 P/P_0=0.3 P / P 0 = 0.3 nozzle choking — jab pressure 0.528 P 0 0.528\,P_0 0.528 P 0 M = 1 M=1 M = 1
Ek galti se bachna: yeh formulas sirf isentropic flow ke liye valid hain. Shock ke aar-paar mat lagana , kyunki entropy badhti hai aur P 0 P_0 P 0 T / T 0 T/T_0 T / T 0 T 0 / T T_0/T T 0 / T T / T 0 T/T_0 T / T 0
