Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations: Worked examples

3.1.2 · D3 · Physics › Compressible Flow & Aerodynamics › Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

Ye page ek drill ground hai. Parent note Stagnation quantities ne tumhe teen master formulas diye. Yahan hum har tarah ki situation dhundhte hain jinmein ye formulas use ho sakte hain — speed ka har regime, har degenerate input, har trap — aur har ek ko scratch se solve karte hain.

Shuru karne se pehle, tools ko re-anchor karte hain taaki koi bhi symbol bina samjhe use na ho.

Us repeated lump ko $B$ naam dena in problems ke liye sabse useful habit hai: $M$ se $B$ ek baar compute karo, phir jis bhi power ki zaroorat ho use raise karo.

Individual cases mein jaane se pehle, neeche diya master map dekho. Ye teeno ratios ko ek canvas par Mach number ke against plot karta hai — har worked example bas is picture ki ek vertical slice padhna hai. Dhyan do kaise pressure curve (steep power) temperature curve (gentle) se dur rocket karti hai jab $M$ badhta hai: ye ek visual fact examples mein aane wale zyaadatar "surprises" explain karta hai.

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

The scenario matrix

Ye topic jo bhi problem throw kar sakta hai wo in case classes mein se ek (ya blend) hai. Neeche diye examples us cell ke saath tagged hain jo woh hit karta hai; saath milake ye poori grid fill karte hain.

#	Case class	Isme tricky kya hai	Example
A	Given $M$ , find $T_{0}, P_{0}, ρ_{0}$ (forward, supersonic)	Bade exponents errors ko amplify karte hain	Ex 1
B	Given $P_{0} / P$ , find $M$ and $V$ (inverse, subsonic)	Power ko invert karna padta hai	Ex 2
C	Degenerate: $M = 0$ (gas pehle se rest par)	Stagnation = static	Ex 3
D	Limiting: $M \to 0$ chhota par nonzero	Bernoulli vs full formula	Ex 3
E	Sonic point $M = 1$ (throat / reference)	Critical ratios, universal numbers	Ex 4
F	Hypersonic $M ≫ 1$	Enormous heating; asymptotics	Ex 5
G	$T_{0}$ conserved par $P_{0}$ lost (across a shock)	Adiabatic ≠ isentropic	Ex 6
H	Real-world word problem (aircraft nose heating)	Words → $M$ → $T_{0}$ mein translate karna	Ex 7
I	Exam twist: non-air gas $γ \neq = 1.4$ , unit trap	Different $γ$ , kPa vs Pa, $R$	Ex 8

Hum inhe order mein attack karte hain.

Example 1 — Cell A: forward, supersonic

Worked example Air at

$M = 3$ , $T = 220 K$ , $P = 20 kPa$ , $γ = 1.4$ . Find $T_{0}$ , $P_{0}$ , $ρ_{0}$ . Forecast: $T_{0}$ kuch factor se badhega; $P_{0}$ ek bade factor se badhega kyunki $3.5$ power hai. Guess karo: kya $P_{0}$ static pressure se $2 \times$ , $10 \times$ , ya $30 \times$ zyaada hoga? (Padhne se pehle apna guess likho.)

Step 1 — Base factor $B$ compute karo. $B = 1 + \frac{γ - 1}{2} M^{2} = 1 + \frac{0.4}{2} (3)^{2} = 1 + 0.2 \times 9 = 2.8$ Ye step kyun? Har relation $B$ ki ek power hai, isliye hum ise ek baar compute karte hain aur reuse karte hain — yahi poori efficiency trick hai.

Step 2 — Temperature. $T_{0} = B T = 2.8 \times 220 = 616 K$ . Ye step kyun? $T_{0} / T = B^{1}$ ; temperature sabse chhota exponent use karta hai, isliye ye gently climb karti hai.

Step 3 — Pressure. $P_{0} = P B^{γ / (γ - 1)} = 20 \times (2.8)^{3.5}$ . $(2.8)^{3.5} \approx 36.73$ , isliye $P_{0} \approx 734.6 kPa$ . Ye step kyun? Exponent $γ / (γ - 1) = 1.4/0.4 = 3.5$ — steep wala — isliye pressure explode karta hai. (" $30 \times$ " guess karna sahi tha, actually $\approx 37 \times$ .)

Step 4 — Density. $ρ_{0} = ρ B^{1/ (γ - 1)}$ . Pehle ideal gas law $ρ = P / (R T)$ se static density nikalo, $R = 287 J/kg\cdotK$ ke saath: $ρ = \frac{20000}{287 \times 220} = 0.3168 kg/m^{3}$ Phir $ρ_{0} = 0.3168 \times (2.8)^{2.5}$ , aur $(2.8)^{2.5} \approx 13.12$ , isliye $ρ_{0} \approx 4.157 kg/m^{3}$ . Ye step kyun? Exponent $1/ (γ - 1) = 2.5$ . Note karo $P$ Pa mein tha (20000) isliye density SI mein aati hai.

Verify: Ideal-gas consistency check hold karna chahiye: $\frac{P _{0}}{P} = \frac{ρ _{0}}{ρ} \cdot \frac{T _{0}}{T}$ . $36.73 = ? 13.12 \times 2.8 = 36.74$ ✓ (rounding). Teeno mutually consistent hain — ek powerful self-check.

Teeno answers exactly wahi teen heights hain jo tum master figure par $M = 3$ par read karte: temperature $2.8$ par, density $13.1$ par, pressure $36.7$ par — pressure curve baaki sab ke upar towering karti hui.

Example 2 — Cell B: inverse, subsonic

Worked example Ek subsonic

Pitot tube deta hai $P_{0} = 150 kPa$ , static $P = 120 kPa$ , static $T = 260 K$ , air. $M$ aur $V$ find karo. Forecast: $P_{0} / P = 1.25$ — modest, isliye $M$ kahin $0.5$ – $0.7$ range mein hona chahiye ( $25%$ pressure rise fair chunk hai par $M = 1$ par milne wale $\sim 89%$ rise se kaafi kam). Abhi ek number guess karo.

Step 1 — Power ko invert karke base factor nikalo. $B = (\frac{P _{0}}{P})^{(γ - 1) / γ} = (1.25)^{0.4/1.4} = (1.25)^{0.2857} \approx 1.0658$ Ye step kyun? Pressure humein $B^{3.5}$ deta hai; exponent ko hatane ke liye hum reciprocal $1/3.5 = (γ - 1) / γ$ se raise karte hain. Ye "arctan undoing tan" move jaisa hai — hum us operation ko invert karte hain jisne formula banaya tha.

Step 2 — $B$ se $M$ solve karo. $1 + 0.2 M^{2} = 1.0658 \Rightarrow M^{2} = \frac{0.0658}{0.2} = 0.329 \Rightarrow M = 0.574$ Ye step kyun? $B = 1 + \frac{γ - 1}{2} M^{2}$ jahan $\frac{γ - 1}{2} = 0.2$ ; algebra se $M$ isolate hota hai.

Step 3 — Mach ko speed mein convert karo. Local sound speed $a = γ R T = 1.4 \times 287 \times 260$ : $a = 104, 468 \approx 323.2 m/s, V = M a = 0.574 \times 323.2 \approx 185.5 m/s$ Ye step kyun? Mach dimensionless hai; real speed paane ke liye hum local $a$ se multiply karte hain — dekho Speed of sound a = √(γRT).

Verify: $M = 0.574$ se $P_{0} / P$ rebuild karo: $B = 1 + 0.2 (0.329) = 1.0658$ , $B^{3.5} = 1.065 8^{3.5} \approx 1.250$ ✓. Also $M = 0.574 < 1$ , isliye "subsonic" assumption hold karta hai aur isentropic Pitot formula legitimate hai.

Example 3 — Cells C & D: rest aur near-rest (degenerate + limit)

Neeche di figure $M = 0$ ke paas master map ka "zoom-in" hai: ye exact pressure curve ko straight Bernoulli line ke saath overlay karta hai taaki tum dekh sako kaise woh origin par kiss karte hain aur sirf $M \approx 0.3$ ke baad alag hote hain. Ye visual hi is example ka poora content hai.

Worked example (a) Gas ek tank mein

$T = 300 K$ , $P = 101 kPa$ par still baitha hai. $T_{0}, P_{0}$ kya hain? (b) Ek slow duct flow mein $V = 30 m/s$ , $T = 300 K$ , $P = 101 kPa$ , $ρ = 1.173 kg/m^{3}$ hai. Full $P_{0}$ vs Bernoulli compare karo. Forecast: (a) Agar kuch move nahi kar raha, stagnation static ke barabar honi chahiye — difference speed mein stored hoti hai, aur yahan koi speed hai hi nahi. (b) $30 m/s$ par ( $M \approx 0.086$ ) Bernoulli almost perfect hona chahiye. Error guess karo: $1%$ se kam?

Part (a) — $M = 0$ degenerate case (Cell C). Step 1. $M = 0 \Rightarrow B = 1 + 0.2 (0) = 1$ . Step 2. $T_{0} = 1 \times 300 = 300 K$ , $P_{0} = 101 \times 1^{3.5} = 101 kPa$ . Ye step kyun? Koi bulk motion na hone se dump karne ke liye koi ordered kinetic energy nahi hai — stagnation hi static hai. Ye poori theory ka sanity boundary hai, aur figure ka left edge hai jahan saari curves $1$ par milti hain.

Part (b) — small- $M$ limit (Cell D). Step 1 — $M$ nikalo. $a = 1.4 \times 287 \times 300 = 347.2 m/s$ , $M = 30/347.2 = 0.0864$ . Step 2 — full compressible $P_{0}$ . $B = 1 + 0.2 (0.0864)^{2} = 1.001494$ , $P_{0} = 101 \times B^{3.5} = 101 \times 1.005238 = 101.529 kPa$ . Toh $P_{0} - P = 0.529 kPa$ . Ye step kyun? Exact relation, koi approximation nahi — ye figure ke solid curve par ek point hai. Step 3 — Bernoulli Bernoulli's equation (incompressible limit): $P_{0} - P = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.173) (30)^{2} = 527.9 Pa = 0.5279 kPa$ . Ye step kyun? Ye parent note mein derive kiya gaya $M \to 0$ limit hai — dashed straight line — aur ise is tiny $M$ par almost match karna chahiye.

Verify: Full deta hai $529 Pa$ , Bernoulli deta hai $528 Pa$ — difference $\approx 0.2%$ , exactly jaisa itne low $M$ ke liye forecast kiya tha. Bernoulli yahan safe hai; ye tabhi fail karta hai jab $M ≳ 0.3$ .

Example 4 — Cell E: sonic reference $M = 1$

Worked example Air (

$γ = 1.4$ ) ke liye, $M = 1$ par critical ratios $T_{0} / T^{*}$ , $P_{0} / P^{*}$ , $ρ_{0} / ρ^{*}$ compute karo. Ye universal constants hain jinhe tumhe recognize karna chahiye. Forecast: $M = 1$ par base factor $1.2$ hai. Pressure ratio famous " $1.893$ " hoga jo tables mein dikhai deta hai. Kya tumhe iska reciprocal yaad hai (static/total ≈ 0.528)?

Step 1 — base factor. $B = 1 + 0.2 (1)^{2} = 1.2$ . Ye step kyun? $M = 1$ reference point (star $^{*}$ ) hai jo Isentropic flow relations aur nozzle theory mein har jagah use hota hai.

Step 2 — teen ratios. $\frac{T _{0}}{T ^{*}} = 1.2, \frac{P _{0}}{P ^{*}} = 1. 2^{3.5} \approx 1.8929, \frac{ρ _{0}}{ρ ^{*}} = 1. 2^{2.5} \approx 1.5774$ Ye step kyun? Same $B$ , teen exponents — lump ko naam dene ka payoff.

Step 3 — stagnation ke fractions ke roop mein padhna. Sonic par static/total: $T^{*} / T_{0} = 0.8333$ , $P^{*} / P_{0} = 0.5283$ , $ρ^{*} / ρ_{0} = 0.6339$ . Ye step kyun? Interpretation: nozzle ke throat tak pahunchne ke liye, gas ko apne reservoir pressure ka $\approx 53%$ tak drop karna padta hai. Ye " $0.528$ " choking criterion hai.

Verify: Consistency: $P_{0} / P^{*} = ? (ρ_{0} / ρ^{*}) (T_{0} / T^{*}) = 1.5774 \times 1.2 = 1.8929$ ✓.

Example 5 — Cell F: hypersonic $M ≫ 1$

Worked example Ek re-entry probe

$M = 8$ par air mein $T = 250 K$ , $γ = 1.4$ ke through fly kar raha hai. $T_{0}$ estimate karo. Ye dangerous kyun hai? Forecast: $M = 8$ par $M^{2} = 64$ term dominate karta hai. $T_{0}$ enormous hoga — hazaron kelvin. Order of magnitude guess karo.

Step 1 — base factor. $B = 1 + 0.2 (8)^{2} = 1 + 12.8 = 13.8$ . Ye step kyun? " $1$ " ab $12.8$ ke next negligible hai — hypersonic asymptotics: $B \to \frac{γ - 1}{2} M^{2}$ .

Step 2 — stagnation temperature. $T_{0} = B T = 13.8 \times 250 = 3450 K$ . Ye step kyun? Sirf adiabatic relation use karta hai — $T_{0}$ ke liye smooth deceleration ka assumption nahi chahiye (Cell G humein remind karata hai kyun).

Step 3 — physical meaning. $3450 K$ zyaadatar metals ke melting point ( $\sim 1800 K$ ) se zyaada hai. Isliye re-entry vehicles ko heat shields chahiye hoti hain: nose par rest mein laya gaya air colossal kinetic energy heat mein dump kar deta hai. Ye step kyun? Number ko engineering reality se connect karta hai.

Verify: Asymptotic check — $\frac{γ - 1}{2} M^{2} = 0.2 \times 64 = 12.8$ , aur $B / (0.2 M^{2}) = 13.8/12.8 = 1.078$ , matlab " $+ 1$ " sirf $\approx 7%$ contribute karta hai $M = 8$ par, confirming karta hai ki high- $M$ approximation $T_{0} \approx \frac{γ - 1}{2} M^{2} T$ reasonable hai. Also $3450 K > 1800 K$ , isliye "dangerous" justified hai.

Example 6 — Cell G: $T_{0}$ survive karta hai, $P_{0}$ lost ho jaata hai (across a shock)

Neeche di figure is example ka heart hai. Use left-to-right padhte hain: orange arrow = flow jo violet dashed shock line cross kar rahi hai. Magenta bar shock ke across bilkul flat rehti hai — yahi $T_{0} = 540 K$ conserved hai. Navy bar shock par neeche step down karta hai — yahi $P_{0}$ ka $391$ se $283 kPa$ tak drop karna hai. Vertical dotted drop aur violet " $27.7%$ loss" label exactly woh jagah mark karte hain jahan second law bite karta hai. Numbers follow karte waqt un dono bars par nazar rakhna.

Worked example Air ek

normal shock mein $M_{1} = 2$ , static $T_{1} = 300 K$ , $P_{1} = 50 kPa$ par enter karta hai. Downstream (shock tables se) $M_{2} = 0.5774$ , $P_{2} = 225 kPa$ , $T_{2} = 506.25 K$ hai. Dikhao ki $T_{0}$ unchanged hai par $P_{0}$ drop karta hai. Forecast: Shock adiabatic hai (koi heat cross nahi karta) par violently irreversible hai. Isliye $T_{0, 1} = T_{0, 2}$ exactly, jabki $P_{0, 2} < P_{0, 1}$ . Pressure loss guess karo: $10%$ ? $30%$ ?

Step 1 — upstream stagnation. $B_{1} = 1 + 0.2 (4) = 1.8$ . $T_{0, 1} = 1.8 \times 300 = 540 K$ ; $P_{0, 1} = 50 \times 1. 8^{3.5} = 50 \times 7.824 = 391.2 kPa$ . Ye step kyun? Upstream state par standard forward computation — figure mein dono bars ka left end.

Step 2 — downstream stagnation temperature. $B_{2} = 1 + 0.2 (0.5774)^{2} = 1.0667$ . $T_{0, 2} = 1.0667 \times 506.25 = 540.0 K$ . Ye step kyun? Hum downstream static values aur Mach use karke $T_{0}$ conservation test karte hain — ye flat magenta bar ka right end hai.

Step 3 — downstream stagnation pressure. $P_{0, 2} = P_{2} B_{2}^{3.5} = 225 \times 1.066 7^{3.5} = 225 \times 1.2569 = 282.8 kPa$ . Ye step kyun? Same isentropic relation shock ke downstream smooth flow ke andar apply hoti hai, par post-shock state reference karke — navy bar ka lower step.

Step 4 — the loss. $\frac{P _{0, 2}}{P _{0, 1}} = \frac{282.8}{391.2} = 0.723$ — total pressure ka $27.7%$ loss. Ye step kyun? Ye ratio shock ki entropy penalty hai; energy ( $T_{0}$ ) conserve hoti hai, order ( $P_{0}$ ) nahi. "Temp is Tame, Press is Power" — aur Power woh cheez hai jo tum lose kar sakte ho.

Verify: $T_{0, 1} = T_{0, 2} = 540 K$ ✓ (digit tak conserved). $P_{0, 2} / P_{0, 1} = 0.723 < 1$ ✓ (ek genuine loss, jaisa second law kisi bhi irreversible adiabatic process ke liye demand karta hai).

Example 7 — Cell H: real-world word problem

Worked example Ek jet

$V = 250 m/s$ par air mein $T = 223 K$ (lagbhag $11 km$ altitude) ke through cruise karta hai. Nose par stagnation point par mounted sensor kya temperature read karta hai? (Wahan air fully rest par laya jaata hai.) Forecast: Nose sensor $T_{0}$ read karta hai, $- 5 0^{\circ} C$ ambient se zyaada garam. Kitna zyaada — kuch degrees, ya tens of degrees?

Step 1 — words ko $M$ mein translate karo. Nose stagnation point woh hai jahan $V \to 0$ , isliye humein free-stream Mach chahiye. $a = 1.4 \times 287 \times 223 = 89, 577 = 299.3 m/s$ ; $M = 250/299.3 = 0.8353$ . Ye step kyun? "Stagnation point par sensor" ka matlab hai rest par decelerate karna ka physical realisation — sensor $T_{0}$ read karta hai.

Step 2 — base factor. $B = 1 + 0.2 (0.8353)^{2} = 1 + 0.1396 = 1.1396$ . Ye step kyun? $B$ woh single quantity hai jo Mach number ko har stagnation ratio mein turn karta hai; hum temperature relation use karne se pehle ise ek baar build karte hain.

Step 3 — stagnation temperature. $T_{0} = 1.1396 \times 223 = 254.1 K$ . Ye step kyun? Sirf adiabatic $T_{0}$ relation chahiye — nose par friction $T_{0}$ change nahi karta.

Verify: Direct energy form $T_{0} = T + \frac{V ^{2}}{2 c _{p}}$ se cross-check karo, $c_{p} = \frac{γ R}{γ - 1} = \frac{1.4 \times 287}{0.4} = 1004.5 J/kg\cdotK$ ke saath: $T_{0} = 223 + \frac{25 0 ^{2}}{2 \times 1004.5} = 223 + 31.1 = 254.1 K$ ✓. Nose ambient se $\approx 31 K$ zyaada garam run karta hai — "ram heating", real aur measurable.

Example 8 — Cell I: exam twist (non-air gas + unit trap)

Worked example Helium (

$γ = 5/3$ , $R = 2077 J/kg\cdotK$ ) $M = 1.5$ , static $T = 400 K$ , static $P = 80, 000 Pa$ par flow kar raha hai. $T_{0}$ aur $P_{0}$ find karo. Trap: $γ = 1.4$ mat use karo aur pressure units dhyan se dekho. Forecast: Helium ka $γ = 5/3 \approx 1.667$ larger hai, isliye coefficient $\frac{γ - 1}{2} = 0.333$ bada hai — $T_{0}$ air ke mukable $M$ ke saath faster badhta hai. Guess karo ki kya $T_{0} / T$ yahan same $M$ par air ki value ( $1.45$ ) se zyaada hai.

Step 1 — coefficient. $\frac{γ - 1}{2} = \frac{5/3 - 1}{2} = \frac{2/3}{2} = \frac{1}{3} = 0.3333$ . Ye step kyun? Yahan $0.2$ (air) use karna classic exam mistake hogi. Coefficient $γ$ -dependent hai, isliye har naye gas ke liye ise rebuild karna padta hai.

Step 2 — base factor. $B = 1 + 0.3333 (1.5)^{2} = 1 + 0.3333 \times 2.25 = 1.75$ . Ye step kyun? Same " $B$ once" habit — par ab helium ka coefficient baked in hai, air ka nahi.

Step 3 — temperature. $T_{0} = 1.75 \times 400 = 700 K$ . Ye step kyun? $T_{0} / T = 1.75 > 1.45$ — confirm hua: heavier $γ$ faster heat karta hai, jaisa forecast kiya tha.

Step 4 — pressure exponent. $\frac{γ}{γ - 1} = \frac{5/3}{2/3} = 2.5$ . Toh $P_{0} = P B^{2.5} = 80, 000 \times 1.7 5^{2.5}$ . $1.7 5^{2.5} \approx 4.051$ , giving $P_{0} \approx 324, 100 Pa \approx 324.1 kPa$ . Ye step kyun? Exponent $γ / (γ - 1) = 2.5$ air ke $3.5$ se different hai; $3.5$ use karna galat hota. Pressure Pa mein nikla kyunki humne $P$ ko Pa mein rakha.

Verify: Consistency $P_{0} / P = (ρ_{0} / ρ) (T_{0} / T)$ . Density exponent $\frac{1}{γ - 1} = 1.5$ , toh $ρ_{0} / ρ = 1.7 5^{1.5} = 2.315$ . Phir $2.315 \times 1.75 = 4.051 = 1.7 5^{2.5}$ ✓. Teeno relations kisi bhi $γ$ ke liye mutually consistent rehte hain.

Recall

Recall Is matrix ke one-line reflexes

Base factor definition ::: $B = 1 + \frac{γ - 1}{2} M^{2}$ ; phir $T_{0} / T = B$ , $P_{0} / P = B^{γ / (γ - 1)}$ , $ρ_{0} / ρ = B^{1/ (γ - 1)}$ . $M = 0$ par $T_{0}, P_{0}$ kya hain? ::: Static $T, P$ ke barabar — koi motion nahi, kuch stored nahi. Air ke liye critical pressure ratio $P^{*} / P_{0}$ ? ::: $\approx 0.528$ (choking value). Normal shock ke across, kaun si total quantity conserve hoti hai? ::: $T_{0}$ (adiabatic); $P_{0}$ drop karta hai (irreversible). Same $M$ par: helium ( $γ = 5/3$ ) ya air ( $γ = 1.4$ ) ka $T_{0} / T$ bada hota hai? ::: Helium ka, kyunki $\frac{γ - 1}{2}$ larger hai. $P_{0}$ ke liye Bernoulli kab use kar sakte ho? ::: Sirf $M ≲ 0.3$ par (error chhota rehta hai); uske upar full power law use karo.