Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

3.1.2 · Physics › Compressible Flow & Aerodynamics

Intuition Badi picture

Ek fast-moving gas ko imagine karo. Usmein do tarah ki "cheezein" hoti hain: thermal energy jo uske warm hone se aati hai, aur uski bulk motion ki kinetic energy. Agar tum bina friction ke (isentropically) aur bina heat loss ke (adiabatically) us gas ko aaram se rest par la sako, to saari kinetic energy thermal energy mein convert ho jaayegi. Jo properties us gas ki us imagined rest state mein hoti, wahi stagnation (total) quantities $T_{0}, P_{0}, ρ_{0}$ hain.

WHY care? Kyunki stagnation temperature $T_{0}$ (equivalently total enthalpy $h_{0}$ ) kisi bhi adiabatic flow mein conserved rehti hai, chahe $T, P$ (yani "static" values) kitna bhi badal jaayein. Stagnation pressure $P_{0}$ sirf reversible (isentropic) adiabatic flow mein conserved hoti hai — jab bhi irreversibility (friction, shocks) hoti hai, yeh drop karti hai. Ye quantities compressible flow ki bookkeeping currency hain.

1. "Static" vs "Total" kya hota hai?

WHAT chahiye humein: formulas jo $T_{0}, P_{0}, ρ_{0}$ ko $T, P, ρ$ aur $M$ se link karein.

2. $T_{0}$ derive karna — energy conservation se

HOW — first principles. Ek adiabatic, steady flow ke liye jisme koi work aur koi elevation change nahi, enthalpy $h$ plus kinetic energy per unit mass conserved rehti hai:

$h + \frac{V ^{2}}{2} = const = h_{0}$

Yeh step kyun? Yeh open system (control volume) ke liye First Law hai: $\dot{Q} - \dot{W} = \overset{m}{˙} [(h_{2} - h_{1}) + \frac{1}{2} (V_{2}^{2} - V_{1}^{2})]$ . $\dot{Q} = \dot{W} = 0$ ke saath, bracket zero ho jaata hai, isliye $h + \frac{1}{2} V^{2}$ constant rehta hai. Stagnation point par $V = 0$ hai, isliye constant $h_{0}$ ke barabar hota hai — yahi stagnation enthalpy hai.

Calorically perfect gas ke liye, $h = c_{p} T$ . Substitute karo:

$c_{p} T + \frac{V ^{2}}{2} = c_{p} T_{0} \Rightarrow T_{0} = T + \frac{V ^{2}}{2 c _{p}}$

Yeh step kyun? $h = c_{p} T$ constant specific heat assume karta hai — moderate temperature ranges par ideal gases ke liye valid hai.

Ab Mach number mein express karo. $a^{2} = γ R T$ (speed of sound) aur relation $c_{p} = \frac{γ R}{γ - 1}$ use karo:

$\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T} = 1 + \frac{V ^{2}}{2 \cdot \frac{γ R}{γ - 1} \cdot T} = 1 + \frac{( γ - 1 )}{2} \frac{V ^{2}}{γ R T}$

Kyunki $γ R T = a^{2}$ aur $M = V / a$ :

3. $P_{0}$ aur $ρ_{0}$ derive karna — isentropy add karke

HOW. Ek ideal gas mein isentropic process ke liye, temperature–pressure aur temperature–density relations hain:

$\frac{P _{0}}{P} = (\frac{T _{0}}{T})^{\frac{γ}{γ - 1}}, \frac{ρ _{0}}{ρ} = (\frac{T _{0}}{T})^{\frac{1}{γ - 1}}$

Yeh step kyun? $P v^{γ} =$ const aur ideal gas $P v = R T$ se: $v$ eliminate karo to $T P^{(1 - γ) / γ} =$ const milta hai, rearrange karo to upar wale exponents milte hain. Usi trick se density relation bhi milta hai.

$T_{0} / T$ ka result daalo:

Ye teenon isentropic flow relations hain. Note karo ki ye ideal gas law ke saath consistent hain: $\frac{P _{0}}{P} = \frac{ρ _{0}}{ρ} \cdot \frac{T _{0}}{T}$ . ✓

Figure — Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

4. Low-speed limit — sanity check (Forecast-then-Verify)

Small $M^{2}$ ke liye binomial series use karke $(1 + \frac{γ - 1}{2} M^{2})^{γ / (γ - 1)}$ expand karo:

$\frac{P _{0}}{P} \approx 1 + \frac{γ}{2} M^{2} + \frac{γ}{8} M^{4} + \dots$

Toh $P_{0} - P \approx \frac{γ}{2} P M^{2} = \frac{γ}{2} P \frac{V ^{2}}{γ R T} = \frac{1}{2} \frac{P}{R T} V^{2} = \frac{1}{2} ρ V^{2}$ .

Verified: incompressible Bernoulli, $M \to 0$ limit hai. $M^{4}$ term compressibility correction hai.

5. Worked Examples

6. Common Mistakes (Steel-man)

7. Active Recall

Recall Quick self-test (answers hide karo)

Shock ke across kaun sa stagnation quantity conserved hai? → $T_{0}$ (adiabatic). $P_{0}$ nahi — irreversibility se yeh girta hai.
$P_{0} / P$ par exponent kya hai? → $γ / (γ - 1)$ .
$P_{0}$ kaunsa process define karta hai? → Rest par Isentropic deceleration.
$M \to 0$ ke saath $P_{0} - P$ ka limit? → $\frac{1}{2} ρ V^{2}$ (Bernoulli).

Recall Feynman: ek 12-saal ke bacche ko explain karo

Ek tez hawa mein uski rush mein chupi energy hoti hai. Agar tum apna haath bahar nikalo aur hawa ko poori tarah rok do, to woh rush garmi aur squeeze mein badal jaati hai — tumhara haath door ki shaant hawa se zyada garam aur zyada push feel karta hai. Yeh "ruki hui" temperature aur pressure total values hain. Hawa jitni tez, utna bada jump. Garmi ka jump sirf iski zaroorat hai ki koi heat bahar na jaaye; pressure ke jump ke liye use smoothly bina rubbing ke rokna padta hai — agar ise roughly roko, to kuch squeeze hamesha ke liye kho jaata hai.

8. Flashcards

Stagnation temperature ratio formula

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

Stagnation pressure ratio formula

P_{0} / P = (1 + \frac{γ - 1}{2} M^{2})^{γ / (γ - 1)}

Stagnation density ratio formula

ρ_{0} / ρ = (1 + \frac{γ - 1}{2} M^{2})^{1/ (γ - 1)}

T_{0}

kis conservation law se milta hai?

Steady-flow energy (adiabatic flow ke liye enthalpy + KE conserved) se.

P_{0}

aur

ρ_{0}

define karne ke liye kaun sa process chahiye?

Isentropic (reversible adiabatic) deceleration to rest.

Kya

T_{0}

shock ke across conserved hai?

Haan (shock adiabatic hai).

P_{0}

nahi — irreversibility use giraa deti hai.

Kis condition mein

P_{0}

conserved hota hai?

Sirf reversible (isentropic) adiabatic flow mein; friction/shocks ise reduce karte hain.

P_{0} - P

ka low-Mach limit?

\frac{1}{2} ρ V^{2}

— incompressible Bernoulli.

T_{0} = T + V^{2} /2 c_{p}

mein

c_{v}

nahi

c_{p}

kyun?

Kyunki energy equation enthalpy

h = c_{p} T

use karti hai.

Teenon stagnation ratios ko link karne waala relation?

P_{0} / P = (ρ_{0} / ρ) (T_{0} / T)

(ideal gas law).

T_{0} / T

M = 2

γ = 1.4

1.8

Connections

Speed of sound a = √(γRT)
Mach number and flow regimes
Isentropic flow relations
Normal shock waves — total pressure loss
Bernoulli's equation (incompressible limit)
Steady-flow energy equation (First Law for control volumes)
Pitot tube measurement