Review of thermodynamics applied to flow — first law for open systems: Visual walkthrough

3.1.1 · D2 · Physics › Compressible Flow & Aerodynamics › Review of thermodynamics applied to flow — first law for ope

Pehle frame se pehle, symbols ke baare mein teen seedhi baatein:

Neeche sab kuch sirf unhi saat ideas aur "energy conserved hoti hai" se bana hai. Koi aur assumption nahi hai.

Step 1 — Seedha sach pakdo: closed-system energy

KYA. Hum abhi ke liye pipes ke baare mein sochne se mana karte hain. Hum gas ka ek akela blob $δ m$ chunte hain aur usi ko follow karte hain. Kyunki yeh hamesha same matter hai, hum woh ek law use kar sakte hain jis par hume poora bharosa hai: $blob ki total energy mein badlaav Δ E = heat in Q - blob dwara kiya gaya sabhi kaam W_{total}$

KYUN. Energy conservation sirf fixed matter ke tukde ke liye — yaani closed system ke liye — simple hoti hai. Ek pipe mein mass tezi se flow karta hai (open system), jo messy hota hai. Isliye hum seedhe aur chhote se shuru karte hain, phir upgrade karte hain.

PICTURE. Figure s01 dekho: blob do baar draw kiya gaya hai — grey jahan se shuru hota hai (inlet 1 ke thoda bahar), magenta jahan khatam hota hai (outlet 2 ke thoda bahar). Machine (violet box) kabhi nahi hilta; sirf hamara blob hilta hai. "Matter ko follow karo" ka yahi poora matlab hai.

Figure — Review of thermodynamics applied to flow — first law for open systems

Step 2 — Blob ke paas har tarah ki energy list karo

KYA. Har kilogram mein, blob teen energies store karta hai: $e = internal (heat/jiggle) u + kinetic (motion) \frac{1}{2} V^{2} + potential (height) g z$ Toh $Δ E = δ m (e_{2} - e_{1})$ — end aur start ka fark.

KYUN. Step 1 ka law blob ki energy mein change chahta hai. Change tabhi calculate kar sakte hain jab pehle sari energies ka naam le lein jo change ho sakti hain. Yeh teen hi poori list hai jo ek chalte hue fluid parcel store kar sakta hai.

PICTURE. Figure s02 teen energies ko coloured bars ke roop mein dikhata hai — state 1 par blob ke liye aur state 2 par blob ke liye. Stack ki total height wahi hai jo Step 1 track karta hai; stack ki shape bars ke beech shift ho sakti hai, jab tak heat aur work total height ke kisi bhi badlaav ko account kare.

Step 3 — Subtle kaam: blob ko darvaazon se guzaarna

KYA. $W_{total}$ shabd mein do alag alag kaam chhupe hain. Pehla kaam: sirf inlet 1 par andar jaane aur outlet 2 par bahar jaane ke liye, blob ko boundary ke paar dhakka dena padta hai. Pressure $p$ par ek face ke paar volume dhakkelna kaam karta hai = force × distance = $p \times (volume) = p (v δ m)$ .

Inlet par, upstream fluid blob ko andar dhakelta hai. Yeh kaam hamare upar hota hai → energy milti hai → hamare liye $+ p_{1} v_{1}$ .
Outlet par, hamare blob ko downstream fluid ko raaste se hatana padta hai. Kaam hamari taraf se hota hai → energy kharch hoti hai → hamare liye $- p_{2} v_{2}$ .

KYUN. Yeh "flow work" $p v$ optional nahi hai aur useful work jaisa nahi hai — yeh boundary cross karne ki unavoidable entry fee hai. Ise miss karo toh energy ka hisaab kabhi balance nahi hoga. Yahi ek insight enthalpy ke exist karne ki wajah hai.

PICTURE. Figure s03: do darvaze. Inlet par upstream gas ka orange piston blob ko length $v_{1}$ ke slab se andar dhakelta hai (work in). Outlet par blob length $v_{2}$ ka magenta slab bahar dhakelta hai (work out). Arrows dikhate hain kaun kise dhakelta hai.

Step 4 — Flow work aur shaft work ko alag karo

KYA. $W_{total}$ ke andar doosra kaam shaft work $w_{s}$ hai: ek turbine blade ya fan jo actually baahri duniya se connected hai. Toh total work per kilogram saaf do hisson mein banta hai: $w_{total} = useful, blade ke zariye w_{s} + flow work: out minus in (p_{2} v_{2} - p_{1} v_{1})$

KYUN. Yeh physically alag taps hain. Flow work ports par entry ki price hai; shaft work real usable energy hai jo spinning axle se bahar nikalti hai. Inhe confuse karna classic double-counting error hai. Inhe alag rakhna hi final formula ko honest banata hai.

PICTURE. Figure s04 violet machine par do taps label karta hai: orange doorway arrows (flow work, ports par) versus ek violet spinning shaft upar (shaft work). Same machine, do bilkul alag energy channels.

Step 5 — Raw balance banao

KYA. Steps 1–4 ko milao. Closed-system law ko $δ m$ se divide karo taaki sab kuch per kilogram ho jaye ( $q = Q / δ m$ ): $Δ E / δ m (e_{2} - e_{1}) = q - [w_{s} + (p_{2} v_{2} - p_{1} v_{1})]$ $e = u + \frac{1}{2} V^{2} + g z$ expand karo: $(u_{2} + \frac{1}{2} V_{2}^{2} + g z_{2}) - (u_{1} + \frac{1}{2} V_{1}^{2} + g z_{1}) = q - w_{s} - (p_{2} v_{2} - p_{1} v_{1})$

KYUN. Kuch naya nahi — bas Step 2 aur Step 4 ko Step 1 mein substitute kiya. Yeh "cleanup se pehle" ki equation hai: sab sach yahan hai, lekin $p v$ terms galat side par bikhar gaye hain, view clutter kar rahe hain.

PICTURE. Figure s05 balance ko see-saw ki tarah dikhata hai: state 1 ki energies (plus heat in) left pan mein, state 2 ki energies (plus work out) right pan mein, aur do loose $p v$ flow-work blocks awkwardly side mein pade hain — saaf hone ki guzaarish kar rahe hain.

Step 6 — Magic move: $p v$ ko $u$ se chipkao aur enthalpy banao

KYA. Flow-work terms ko equals sign ke paar slide karo taaki $p_{1} v_{1}$ $u_{1}$ ke saath jude, aur $p_{2} v_{2}$ $u_{2}$ ke saath jude: $\equiv h_{1} (u_{1} + p_{1} v_{1}) + \frac{1}{2} V_{1}^{2} + g z_{1} + q = \equiv h_{2} (u_{2} + p_{2} v_{2}) + \frac{1}{2} V_{2}^{2} + g z_{2} + w_{s}$ $u + p v$ ka pair itna reliably aata hai ki hum ise ek naam dete hain:

KYUN. Har baar jab fluid boundary cross karta hai, $p v$ $u$ ke saath welded hokar aata hai. Inhe hamesha alag alag likhne ki bajaye, hum unka sum ek baar define karte hain. Equation turant clean lagti hai — flow work ko dobara mention nahi karna padta; woh safely $h$ ke andar chhupa hai.

PICTURE. Figure s06: Step 5 ke loose $p v$ blocks $u$ bars par snap ho jaate hain, $h_{1}$ aur $h_{2}$ ke taller "enthalpy" columns banaate hain. Same energy, sundar accounting.

Step 7 — Edge case: aerodynamic collapse (koi shaft nahi, koi heat nahi, gas)

KYA. Nozzle, diffuser, ya free stream ke liye hum teen terms band karte hain:

$w_{s} = 0$ — koi blade nahi, bas ek duct.
$q = 0$ — flow itni tez hai ki heat leak hone ka time hi nahi (adiabatic).
$g z$ negligible — halka gas height par dhyan nahi deta.

SFEE simat jaati hai: $h + \frac{1}{2} V^{2} = h_{0} = const, h_{0} = ==stagnation enthalpy==$ aur calorically perfect gas ke liye $h = c_{p} T$ ke saath: $c_{p} T + \frac{1}{2} V^{2} = c_{p} T_{0} \Rightarrow \frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T} .$

KYUN. Yeh approximations hawa se nahi aayi — har dropped term ek physical feature correspond karta hai jo device mein nahi hai. Jo bachta hai woh master trade hai: enthalpy $\leftrightarrow$ kinetic energy. Gas ko speed up karne se woh thanda ho jaata hai. (Use slow karne se — jaise probe ki nose par — woh garam hota hai: yahi Mach-2 stream ka $T_{0}$ hai.)

PICTURE. Figure s07: jaise jaise duct narrow hoti hai aur $V$ left-to-right badhta hai, orange kinetic bar badata hai jabki violet enthalpy bar exactly utna hi ghatta hai — unka sum (dashed line $h_{0}$ ) flat rehta hai. Temperature, $h$ par sawaar, girta hai.

Step 8 — Degenerate checks: kya formula extreme inputs par survive karta hai?

KYA. SFEE ko uske corners par push karo aur confirm karo ki woh sensible rehti hai.

Bilkul flow nahi ( $V_{1} = V_{2} = 0$ , koi heat nahi, ek port sealed): SFEE deta hai $h_{1} = h_{2} + w_{s}$ . Yeh bas ek stationary closed device balance hai — kuch cross nahi hua, toh yeh enthalpy form mein closed-system law par reduce ho jaata hai. ✓
Incompressible, slow, no work, no heat ( $v =$ const, chhota $V$ ): $h = u + p v$ aur near-constant $u$ ke saath, balance ban jaata hai $p_{1} v + \frac{1}{2} V_{1}^{2} + g z_{1} = p_{2} v + \frac{1}{2} V_{2}^{2} + g z_{2}$ — Bernoulli equation. ✓
Turbine vs compressor (signs): turbine kaam deta hai → $w_{s} > 0$ ; compressor kaam leta hai → $w_{s} < 0$ . Parent ke Example 2 mein, $w_{s} = c_{p} (T_{1} - T_{2}) = 1005 (300 - 500) = - 201 kJ/kg$ , correctly negative. ✓

KYUN. Jis law par aap bharosa karte ho, usse apni boundaries par pehle se jaane-maane simpler laws mein gracefully degrade karna chahiye. SFEE ka teeno checks pass karna tumhara evidence hai ki Step 6 mein bundling legitimate thi, koi trick nahi thi.

PICTURE. Figure s08: teen mini-panels — ek sealed box (→ closed law), ek smooth incompressible tube (→ Bernoulli), aur ek shaft with $\pm$ arrows (→ turbine/compressor sign) — ek hi equation ko teen jaane-pehchaane bheshon mein dikhata hai.

Ek-picture summary

Poori film ek frame mein: ek blob 1 par andar aata hai, flow work $p_{1} v_{1}$ se andar dhakela jaata hai, ek machine se guzarta hai jo heat $q$ add kar sakti hai aur shaft work $w_{s}$ nikal sakti hai, phir 2 par flow work $p_{2} v_{2}$ deke bahar nikalata hai. Har $u + p v$ ko $h$ mein bundle karo aur balance clean padha jaata hai: jo andar jaata hai woh bahar zaroor aana chahiye.

$h_{1} + \frac{1}{2} V_{1}^{2} + g z_{1} + q = h_{2} + \frac{1}{2} V_{2}^{2} + g z_{2} + w_{s}$

Recall Feynman retelling — seedhi baaton mein poora walkthrough

Hum poori windy pipe ke baare mein sochna nahi chahte the, toh humne cheating ki: humne apni aankhein hawa ke ek chhote se puff par chipka diin aur usi ke saath saath chale. Ek fixed puff ke saath chalte hue, energy simple hai — heat in, minus jitna kaam woh karta hai, equals kitni uski energy badli (Step 1). Puff teen tarah ki energy carry karta hai: andar jiggle-heat, tez-daudne ki energy, aur kitna-upar-hai energy (Step 2). Phir humne sneaky part notice kiya: sirf agले darvaze se andar ghusne ke liye, peeche ki bheed puff ko dhakkelti hai (woh kaam uske upar karte hain); aur peeche wale darvaze se bahar nikalne ke liye, puff aage ki bheed ko dhakkelti hai (woh kaam karta hai). Yeh darvaze-se-dhakkelne ki cost "flow work," $p v$ hai, aur yeh kisi bhi useful kaam se bilkul alag hai jo koi fan ya turbine blade add ya le sakta hai (Steps 3–4). Humne sab kuch likh diya (Step 5) aur dekha ki door-cost $p v$ hamesha andar waali energy $u$ ke saath chipka rehta tha — toh humne unhe hamesha ke liye chipka diya aur bundle ko enthalpy, $h = u + p v$ naam diya (Step 6). Ab hisaab saaf hai: bundle-in + motion + height + heat = bundle-out + motion + height + shaft work. Bina blades aur bina heat lose karne ke time ke saade duct ke liye, jo bachta hai woh sirf ek trade hai: puff sirf apna bundle kharch karke speed up hota hai, toh woh daudne mein thanda ho jaata hai (Step 7). Aur jab hum equation ko ajeeb corners mein dhakelnte hain — kuch bhi flow nahi, ya slow paani, ya blade ek taraf ya doosri taraf — woh vinrmy se purani closed-system law, ya Bernoulli, ya sahi turbine/compressor sign ban jaati hai (Step 8). Ek imandaar law, kai bhesh.

Connections

Parent: First Law for Open Systems
Stagnation properties & isentropic relations
Speed of sound and Mach number
Conservation of mass — continuity equation
Closed-system first law of thermodynamics
Bernoulli equation as low-speed limit of SFEE
Nozzles and diffusers