Heat engines — efficiency η = 1 − Q_C - Q_H: Visual walkthrough

1.7.19 · D2 · Physics › Thermodynamics › Heat engines — efficiency η = 1 − Q_C - Q_H

Intuition Yeh page kya karta hai

Parent note (Hinglish) ne derivation bataya tha (us note ke title ka shorthand actually $η = 1 - Q_{C} / Q_{H}$ ko represent karta hai — jisme $Q_{C}$ ko $Q_{H}$ se divide kiya gaya hai; hum woh fraction neeche carefully banayenge). Yahan hum use draw karte hain. Energy ka har arrow, har symbol, aur har "kyun" ka apna ek alag picture hai. Last figure tak aap aankhein band karke engine ko ek shrinking pipe of energy ki tarah dekh paoge — ek wide stream of heat andar aati hai, ek patli stream of work bahar jaati hai — formula memorise nahi karna padega.

Hum maan ke chalte hain ki aapko kuch nahi pata siwaaye is baat ke ki energy joules mein maapi jaati hai aur na kahi se aa sakti hai na kahin gaayab ho sakti hai. Baaki sab hum khud banate hain — jisme "pressure," "volume," aur "state" ka matlab bhi tab define karenge jab pehli baar zaroorat padegi — aur efficiency symbol ka use tab tak nahi karenge jab tak Step 6 mein use earn na kar lein.

Step 1 — Machine ko ek box ki tarah draw karo jisme teen pipes hain

KYA. Equations bhool jao. Ek sealed box imagine karo. Teen cheezein uski boundary cross karti hain: heat andar aati hai kisi hot cheez se, heat bahar jaati hai kisi cold cheez mein, aur ek shaft bahar nikaali hai jo ek wheel ghuma rahi hai (yahi useful work hai).

KYUN. Energy conserve karne se pehle, hum har us channel ka naam lena chahte hain jisse energy andar ya bahar jaati hai. Ek bhi miss hua, toh accounting toot jaayegi. Ek picture guarantee karti hai ki humne sab pakad liya — tum literally box ko cross karne waale arrows gin sakte ho.

PICTURE. Neeche ke figure mein:

Upar ka amber block hai hot reservoir, heat ka ek bada tank temperature $T_{H}$ par.
Neeche ka cyan block hai cold reservoir temperature $T_{C}$ par.
Bada amber arrow jo box mein neeche pour ho raha hai woh hai $Q_{H}$ : woh heat jiske liye hum pay karte hain.
Patla cyan arrow jo neeche se nikal raha hai woh hai $Q_{C}$ : woh heat jise hum dump karne par majboor hain.
White arrow jo side se bahar hai woh hai $W$ : woh work jo hum harvest karte hain.

Hum $Q_{H}$ aur $Q_{C}$ ko magnitudes (hamesha positive) maante hain aur unki direction arrows se track karte hain, signs se nahi. Yeh choice pictures ko clean rakhti hai.

Step 2 — Maango ki machine apni exact starting state par wapas aaye

KYA. Box ke andar ek gas ka chhota sa hissa hai — working substance. Hum demand karte hain ki engine ek cycle mein chale: ek poora loop complete hone ke baad, woh gas usi condition mein wapas aa jaaye jahan se shuru hui thi.

KYUN. Ek car engine ko baar baar fire karna padta hai. Agar gas har loop ke baad zyada hot ya zyada spread out ho jaati, toh eventually "use up" ho jaati. Isliye hum require karte hain ki woh reset ho. Yahi ek requirement woh secret lever hai jo waste heat ko exist karne par majboor karti hai — Step 5 mein dekho yeh kaise kaam karta hai.

Picture se pehle, do words jinhe hum use karenge, isliye ab define karte hain:

PICTURE. Figure gas ki state ko is $P$ – $V$ graph par ek loop ki tarah trace karta hai. Path start dot se nikalta hai, gas ke heated, expand, cool aur compressed hone ke saath bhatakta hai, aur usi dot par wapas aata hai. Woh closed loop "cycle" ka visual meaning hai.

Kyunki loop usi dot par wapas aata hai, internal energy exactly wahan khatam hoti hai jahan shuru hui thi: $Δ U_{cycle} = 0$

Step 3 — Box par energy conservation lagao (the First Law)

KYA. Hum ab box ke liye energy conserved hai likh rahe hain. Pehle ek nayi symbol ka naam lo jo hum abhi use karne wale hain: $Q_{net}$ us ==net heat ko represent kare jo cycle ke dauran gas mein flow karta hai== — yaani, jitni bhi heat add ki gayi minus jitni bhi heat remove ki gayi, ek single number ke roop mein jod ke. Andar stored energy phir is net heat se change hoti hai minus jo work bahar gayi.

KYUN. Yeh First Law of Thermodynamics hai — woh ekmaatra physics jo hume actually chahiye. Baaki sab bookkeeping hai. Hum ise isliye use karte hain kyunki yeh "energy create ya destroy nahi ho sakti" ka exact statement hai, ek aisi system ke liye specialised jo heat aur work trade karti hai.

PICTURE. Figure ek balance scale dikhata hai: ek pan mein woh energy jo gas gain karti hai; doosre mein net heat in minus work out. Unhe match karna hi chahiye.

Ab, $Q_{net}$ define ho jaane ke baad, First Law yeh padhta hai:

$change in stored energy of the gas Δ U = net heat that flowed IN Q_{net} - work done BY the gas W$

Ise ek sentence ki tarah padho: andar ki energy utni badhi jitni net heat aayi, minus jo work ek spinning shaft ki tarah bahar gayi.

Step 4 — Do pipes se net heat add karo

KYA. Box mein heat andar aa rahi hai ( $Q_{H}$ ) aur heat bahar ja rahi hai ( $Q_{C}$ ). Net heat $Q_{net}$ (Step 3 mein abhi naam diya) unka difference hai.

KYUN. Step 3 mein $Q_{net}$ term ek single lump tha. Lekin hamare box (Step 1) mein do heat pipes hain jo opposite directions mein point kar rahi hain. Us lump ko fill karne ke liye hume unhe combine karna hoga, direction ka dhyan rakhte hue: andar positive count hoga, bahar negative.

PICTURE. Figure do heat arrows ko ek number line par tip-to-tail align karta hai: ek lamba amber arrow dayi taraf ( $+ Q_{H}$ ), phir ek chhota cyan arrow wapas bayi taraf ( $- Q_{C}$ ). Jo bachta hai woh hai $Q_{net}$ .

$Q_{net} = in from hot (+) Q_{H} - out to cold (-) Q_{C}$

$Q_{C}$ par minus sign real kaam kar raha hai: yeh encode karta hai ki yeh heat ja rahi hai, isliye yeh box jo rakhta hai usme se subtract hoti hai.

Step 5 — Step 2 aur Step 4 ko combine karo aur work nikaalo

KYA. $Δ U = 0$ (Step 2) aur $Q_{net} = Q_{H} - Q_{C}$ (Step 4) ko First Law (Step 3) mein substitute karo, phir $W$ ke liye solve karo.

KYUN. Yeh payoff hai. Cycle condition ( $Δ U = 0$ ) First Law ko stored energy ke baare mein ek statement se badal deti hai ek aisi statement mein jo purely heat aur work ke baare mein hai — exactly woh cheez jo hum pipes par measure kar sakte hain.

PICTURE. Figure Step 3 wala balance scale dikhata hai, lekin ab " $Δ U$ " wala pan zero par pinned hai. Yeh doosre pan ko apne aap balance karne par majboor karta hai: heat-in must equal heat-out plus work-out.

$0 = (Q_{H} - Q_{C}) - W ⟹ W = Q_{H} - Q_{C}$

Shabdon mein: jo work tum harvest karte ho woh exactly woh heat hai jo tumne dump nahi ki. Har joule jo andar aaya ya toh work ki tarah bahar gaya ya waste heat ki tarah — koi teesra exit nahi hai aur kuch peeche nahi raha (kyunki $Δ U = 0$ ).

Step 6 — Efficiency ko "mila ÷ diya" ki tarah define karo

KYA. Efficiency $η$ woh fraction hai jitni heat tumne pay ki woh work ki tarah wapas aayi.

KYUN. Hume engines compare karne ke liye ek single fair number chahiye. "Jo mila divided by jo diya" honest measure hai — yeh ek pure ratio hai, unit-free, isliye ek tiny toy engine aur ek giant power plant ko directly compare kiya ja sakta hai.

PICTURE. Figure $Q_{H}$ ko ek full wide pipe ki tarah draw karta hai jo andar aa rahi hai, aur $W$ ko woh narrower stream ki tarah jo output tak survive karti hai. $η$ pipe widths ka ratio hai.

$η \equiv \frac{W}{Q _{H}} = \frac{Q _{H} - Q _{C} work (Step 5)}{heat paid in Q _{H}}$

Ab fraction ko do pieces mein split karo — yeh sirf arithmetic hai, $\frac{a - b}{a} = 1 - \frac{b}{a}$ :

$η = \frac{Q _{H}}{Q _{H}} - \frac{Q _{C}}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$

Term by term: $1$ hai woh sapna (saari heat work ban jaaye); $\frac{Q _{C}}{Q _{H}}$ hai woh waste ki gayi fraction, jo subtract hoti hai. Efficiency hai one minus waste.

Step 7 — Har case check karo, degenerate ones bhi

KYA. Formula ko uske extremes par check karo taaki koi scenario surprise na kare.

KYUN. Jo formula tum par trust karte ho woh woh hai jo tumne apne edges par test kiya ho. Hum $Q_{C}$ ko uske do limits tak push karte hain aur check karte hain ki answer sane rehta hai.

PICTURE. Teen side-by-side "pipe" diagrams: realistic middle case, aur do ends par do impossible extremes.

Case A — $Q_{C} = 0$ (kuch dump nahi kiya): toh $η = 1 - 0 = 1$ , ek perfect engine. Picture mein output stream utni hi wide dikhti hai jitni input. Forbidden by Second Law — yeh woh upper edge hai jise $η$ approach kar sakti hai lekin kabhi reach nahi kar sakti.
Case B — realistic, $0 < Q_{C} < Q_{H}$ : toh $0 < η < 1$ . Output stream genuinely input se narrower hai. Yeh har real engine hai.
Case C — $Q_{C} = Q_{H}$ (sab dump kar do): toh $η = 1 - 1 = 0$ . Saari heat seedha through flow hoti hai, koi work nahi. Picture zero output stream dikhati hai. Useless, lekin impossible nahi.

Toh $η$ hamesha is band mein pinned rehti hai: $0 \leq η < 1$ Yeh kabhi negative nahi jaati (uske liye $Q_{C} > Q_{H}$ chahiye hoga — andar aayi heat se zyada bahar jaaye, heat engine ke liye impossible), aur kabhi $1$ tak nahi pahunchti.

Step 8 — Ceiling: 1 ke kitna paas ja sakte ho?

KYA. Temperatures $T_{H}$ aur $T_{C}$ ke beech heat slide karne wale best-possible (reversible) engine ke liye, waste ratio temperature ratio ke barabar hota hai, $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ , jo efficiency ko seedha $η = 1 - T_{C} / T_{H}$ par le aata hai.

KYUN. Step 7 kehta hai $η < 1$ , lekin kitna kam? Woh gap temperatures par depend karta hai — aur yahaan kyun heat ek temperature "tag" carry karti hai uski intuition hai:

Intuition Best engine ke liye

$Q_{C} / Q_{H} = T_{C} / T_{H}$ kyun hota hai (bina equations ke) Heat ko logon ki ek crowd maano aur temperature ko kitna spread out (disordered) woh crowd hai. Entropy naam ki ek quantity us spread ko measure karti hai; temperature $T$ par exchange ki gayi heat $Q$ ke liye, transfer hone wali entropy ratio $Q / T$ hai — zyada heat, ya zyada thandi handover, matlab zyada disorder move hui.

Best engine reversible hota hai: yeh koi extra disorder create nahi karta. Isliye jo disorder woh cold side par chhodta hai woh exactly us disorder ke barabar hona chahiye jo usne hot side se uthaya tha.
Equal disorder in and out ka matlab hai $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ (li gayi entropy = di gayi entropy).
Us single balance ko rearrange karo aur tumhe milega $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ . Real engine extra disorder banata hai, isliye woh isse zyada dump karta hai — uska $Q_{C}$ bada hota hai aur efficiency kam.

Toh is best-case ratio ko $η = 1 - Q_{C} / Q_{H}$ mein substitute karte hue:

PICTURE. Figure Carnot ceiling $η_{Carnot} = 1 - T_{C} / T_{H}$ ko fixed $T_{C}$ ke liye $T_{H}$ ke against plot karta hai: yeh $1$ ki taraf badhta hai lekin sirf tab $1$ reach karta hai jab $T_{H} \to \infty$ ya $T_{C} \to 0$ .

$η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} (T in kelvin)$

Jitna bada temperature gap, utna hi perfect $1$ ke paas creep karte ho — lekin figure mein flat asymptote woh wall hai jo tum kabhi cross nahi kar sakte. (Poori machine Carnot Cycle hai.)

Ek-picture summary

Pehle ke figures ne har ek step ko freeze kiya. Yeh final figure kuch naya karta hai: yeh poori logic chain ko ek single energy river par overlay karta hai, colour-coding karta hai ki kaun sa step kis piece ka owner hai — taaki tum derivation ko ek nazar mein upar se neeche padh sako, nau alag cheezein dekhne ki jagah. Heat $Q_{H}$ andar aati hai (Step 1), cycle condition use split karti hai (Steps 2–5), work $W$ branch off hoti hai, waste $Q_{C}$ drain ho jaati hai, aur work branch ki inflow ke saath width ratio hi efficiency hai (Step 6), Step 8 ki temperature wall upar mark ki gayi hai.

Recall Feynman retelling — poora walkthrough plain words mein

Ek box imagine karo jisme upar se ek hot pipe andar aa rahi hai, neeche se ek cold pipe bahar ja rahi hai, aur side se ek shaft nikal raha hai jo ek wheel ghuma raha hai. Heat upar se andar rush karti hai; kuch heat shaft pakad ke wheel ghuma deti hai (woh tumhari work hai); baaki neeche cold sink mein tumble ho jaati hai.

Ab trick: hum demand karte hain ki box ke andar ka kaam har loop ke baad exactly waisa ho jaaye jaisa pehle tha, taaki koi energy andar chhupi na rahe. Iska matlab hai ki jo bhi joule upar se aayi woh bahar jaani hi chahiye — ya toh spin ki tarah, ya dumped heat ki tarah. Isliye work = heat in minus heat dumped. Simple counting.

Efficiency sirf yeh poochhna hai: jitni heat maine kharid, uska kitna fraction spin ki tarah bahar aaya? Woh hai work over heat-in, jo same hai one minus (dumped over bought). Dumped part kabhi zero nahi ho sakta — nature tumhe saari heat ko spin mein nahi badalne degi — isliye efficiency hamesha perfect $1$ se kam rehti hai. Aur jo absolute best tum kabhi kar sakte ho woh do temperatures se set hota hai: best engine utna hi disorder spread karta hai jitna soakhta hai, jo uski waste ratio ko $T_{C} / T_{H}$ par pin karta hai, giving $η = 1 - T_{C} / T_{H}$ . Zyada hot hot aur zyada cold cold tumhe zyada spin deta hai, lekin $1$ par woh wall kabhi nahi hilti.

Connections

First Law of Thermodynamics — Steps 3–5 ka energy-balance scale.
Second Law of Thermodynamics — kyun $Q_{C} = 0$ case (Step 7A) forbidden hai.
Carnot Cycle — Step 8 ki ceiling ke peeche reversible engine.
Entropy — disorder ratio $Q / T$ jo $Q_{C} / Q_{H} = T_{C} / T_{H}$ pin karta hai.
PV diagrams and Work — Step 2 ka closed loop, aur $W$ uske enclosed area ki tarah.
Refrigerators and Heat Pumps — summary figure mein har arrow ulta kar do.