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

True — ek shock adiabatic hoti hai (thin front ke across koi heat nahi jaati), aur $T_0$ total enthalpy ko track karti hai, jise energy conservation kisi bhi adiabatic process mein constant rakhti hai, chahe reversible ho ya nahi.

False — ek shock irreversible hoti hai, entropy upar jaati hai, aur $P_0$ hamesha uske across drop karti hai, chahe $T_0$ na kare.

True — isentropic ka matlab hai reversible aur adiabatic, toh energy conservation $T_0$ fix karti hai aur entropy-constant-rehna $P_0$ aur ${\rho }_0$ ko bhi fix karta hai.

False — saath-saath move karne wala thermometer static values padhta hai; stagnation values woh hain jo gas reach karti agar use isentropically rest pe laya jata, ek imagined state hai.

True — $V=0$ ke saath gas pehle se rest mein hai, toh convert karne ke liye koi kinetic energy nahi hai aur $T_0=T$, $P_0=P$, ${\rho }_0=\rho$.

True — dono ek hi base $1+\frac{\gamma -1}{2}{M}^2$ ke saath badhte hain, lekin $P_0/P$ mein exponent $\gamma /(\gamma -1)=3.5$ (air ke liye) hai versus $T_0/T$ ke liye $1$, isliye pressure bahut zyada tezi se chadhta hai.

False — yeh compressible relation ka $M\to 0$ limit hai; low speeds ke liye bilkul sahi hai aur sirf tab noticeably inaccurate hota hai jab $M\gtrsim 0.3$ ho jaha $M^4$ compressibility term matter karta hai.

True — wall ke across koi heat nahi jaati toh total enthalpy (hence $T_0$) fixed hai, lekin friction irreversible hai, entropy badhata hai, aur $P_0$ ko bleed off karta hai.

True — isentropically rokne se gas sabse zyada compress hoti hai, aur koi bhi real (irreversible) stop kam recovery deta hai, isliye ${\rho }_0$ ceiling hai.

Bernoulli sirf incompressible limit hai; $M=2$ pe yeh $P_0$ ko bahut underestimate karta hai (roughly true value ka aadha), isliye tumhe poori isentropic $P_0/P=(1+\frac{\gamma -1}{2}{M}^2{)}^{\gamma /(\gamma -1)}$ use karni chahiye.

Yeh $c_p$ hona chahiye, $c_v$ nahi, kyunki steady-flow energy equation enthalpy $h=c_{p}T$ use karti hai, internal energy $u=c_{v}T$ nahi.

Adiabatic hona akela sirf $T_0$ (energy) guarantee karta hai; $P_0$ ke liye process ko reversible bhi hona chahiye, aur ek shock strongly irreversible hoti hai, isliye $P_0$ drop karta hai.

Local Mach number us point pe static temperature use karta hai, kyunki fluid jo sound speed actually experience karta hai woh uski static $T$ pe depend karti hai, imagined stagnation $T_0$ pe nahi.

Ise sirf flow adiabatic hone ki zaroorat hai; yeh friction ke saath ya shocks ke through bhi hold karta hai, unlike pressure aur density relations jo isentropy require karti hain.

Ye ideal gas law ke through temperature ratio se differ karte hain: $P_0/P=({\rho }_0/\rho )(T_0/T)$, isliye base pe exponents respectively $\gamma /(\gamma -1)$ aur $1/(\gamma -1)$ hain, equal nahi.

Sound-speed substitution missing hai: $\frac{1}{2}\gamma P{M}^2=\frac{1}{2}\gamma P{V}^2/(\gamma RT)=\frac{1}{2}(P/RT)V^2=\frac{1}{2}\rho V^2$, toh $\gamma$ aur $P/RT$ ka ek factor $\rho$ mein combine hota hai, $P$ mein nahi.

Temperature total energy reflect karta hai, aur energy kisi bhi adiabatic stop mein conserved hoti hai; pressure us energy ki ordering reflect karta hai, jo depend karta hai kaise gas ko roka jaata hai, isliye ise reversibility (isentropy) assumption chahiye.

Ek isentropic stop kinetic energy ko entropy mein waste nahi karta, toh yeh maximum possible pressure recover karta hai; koi bhi real, irreversible deceleration kam recover karta hai.

Kyunki $P_0$ sirf same reh sakta hai ya gir sakta hai, aur yeh tab girata hai jab irreversibility (friction, shocks, mixing) hoti hai, isliye $P_0$ mein drop directly lost useful energy ko measure karta hai.

Heat loss ke bina energy gas mein rahti hai ($T_0$ fix hota hai); gentle/frictionless hona entropy constant rakhta hai ($P_0$ aur ${\rho }_0$ fix hote hain) — dono mil ke ek single, well-defined reference state define karte hain.

Bernoulli nearly-incompressible fluid ke liye mechanical-energy statement hai, aur $M\to 0$ pe density changes vanish ho jaate hain, isliye compressible relation ka leading term exactly $P_0-P=\frac{1}{2}\rho V^2$ pe collapse karta hai.

Gas ko speed up karne se thermal/pressure energy bulk motion mein convert hoti hai, static pressure aur temperature drop karte hain; phir ideal gas law density ko force karke neeche laata hai, isliye $\rho /{\rho }_0$ $M=1$ pe lagbhag $0.63$ tak gir jaata hai.

Har factor $1+\frac{\gamma -1}{2}{M}^2\to 1$, toh $T_0/T$, $P_0/P$, aur ${\rho }_0/\rho$ sab $1$ approach karte hain — static aur stagnation states merge ho jaate hain.

Haan, jab tak flow adiabatic hai; yeh relation energy conservation ka pure algebraic consequence hai aur kisi bhi Mach number ke liye hold karta hai, subsonic ho ya supersonic.

Nahi, ratio $\gamma$ pe depend karta hai $\frac{\gamma -1}{2}$ factor ke through, toh different $\gamma$ same Mach number pe bhi different temperature ratio dega.

Nahi — shock ne pehle hi kuch $P_0$ irreversibly destroy kar diya, aur baad ka isentropic stage use recover nahi kar sakta, isliye final $P_0$ upstream value se neeche hai jabki $T_0$ unchanged hai.

Haan — agar density truly change nahi ho sakti toh fluid ko rokne se $\rho$ unchanged rahta hai, aur compressible ${\rho }_0/\rho$ relation smoothly $M\to 0$ limit mein $1$ return karta hai jo iske consistent hai.

Lagbhag $M\approx 0.3$ pe, jahan $M^4$ correction term pehli baar kuch percent exceed karta hai; usse neeche Bernoulli ek achha approximation hai, usse upar poori relation use karo.