Refrigerators and heat pumps — COP: Visual walkthrough

1.7.20 · D2 · Physics › Thermodynamics › Refrigerators and heat pumps — COP

Step 1 — Do reservoirs aur natural downhill flow

KYA. Do bade boxes banao. Upar wala box hot hai — iska temperature $T_{H}$ rakho (chhota $H$ bas "hot" ka matlab hai). Neeche wala box cold hai — temperature $T_{C}$ ( $C$ matlab "cold"). "Temperature" yahan bas kitni cheez garam hai, kelvin scale pe measure kiya gaya hai (Step 7 mein dekhenge kyun kelvin aur Celsius nahi).

KYU. Pehle hum heat ko upar pump karein, hume agree karna hoga ki "upar" ka matlab kya hai. Heat woh energy hai jo temperature difference ki wajah se move karti hai, aur apne aap sirf hot box se cold box ki taraf jaati hai — kabhi doosri taraf nahi. Yahi ek-taraf ka rule puri wajah hai ki fridge ko plug in karna padta hai.

PICTURE. Lal arrow hai free, natural flow: hot → cold.

Step 2 — Arrow ko reverse karo: yahi machine hai

KYA. Ab heat ko galat direction mein force karo: ek chunk $Q_{C}$ cold box se bahar kheecho aur upar le jao. $Q_{C}$ mein subscript $C$ ka matlab hai "yeh heat cold reservoir se li gayi hai."

KYU. Yahi exactly ek refrigerator karta hai: andar ko thanda rakhta hai heat hatakar. Lekin Step 1 ne kaha ki heat akele nahi chadh sakti — toh kuch push karna hoga. Woh push hai work.

PICTURE. Upar jaata lal arrow $Q_{C}$ ab gravity se lad raha hai, agar yun kahein. Yeh akele nahi chadh sakta.

Step 3 — Sab kuch kahan jaata hai? Energy balance

KYA. Energy ko follow karo. Machine mein har cycle mein: $Q_{C}$ (cold box se) plus $W$ (wall socket se). Machine se bahar: ek chunk hot box mein dump hota hai, ise $Q_{H}$ kahte hain ("heat to hot reservoir"). Cycle ka matlab hai machine ka andar ka hissa exactly apni shuru ki state mein wapas aata hai, toh woh kuch extra store nahi karta.

KYU. Energy na appear ho sakti hai na vanish (First Law). Agar machine kuch nahi rakhti, toh jo andar gaya woh bahar aana chahiye: $andar cold se Q_{C} + andar socket se W = bahar hot mein Q_{H}$

PICTURE. Do patle arrows andar jaate hain (neeche + side se), ek mota arrow upar se bahar aata hai. Mota = patla + patla.

Yeh First Law of Thermodynamics se exactly match karta hai: $Δ U_{cy c l e} = 0$ .

Step 4 — "Kitna accha hai?" ek benefit ÷ cost ratio hai

KYA. Hum ek number chahte hain jo accha ya bura bataye. Ek accha fridge bahut saari heat move karta hai thodi work pay karke. Toh ratio banao $COP_{r e f} = \frac{jo tum chahte ho}{jo tum pay karte ho} = \frac{Q _{C}}{W} .$ "COP" ka full form hai Coefficient of Performance — bas is ratio ka naam.

KYU ratio, aur kyun yahi wala? Kyunki "accha" comparative hai: 800 J move karna impressive hai sirf us work ke relative jo lagi. Benefit ko cost se divide karna ek leverage number deta hai jo fridge ke size se independent ho. Cooling job ke liye benefit hai $Q_{C}$ (khane se kheenchi gayi heat).

PICTURE. Ek balance beam: mota $Q_{C}$ arrow "want" pan mein, patla $W$ arrow "pay" pan mein. Bada COP "want" ki taraf heavily tilt karta hai.

Step 5 — Heat-pump twist (same machine, alag wish)

KYA. Ab fridge ke andar ki jagah room mein khado. Tumhe ab $Q_{C}$ ki parwah nahi; tumhe $Q_{H}$ ki parwah hai, woh heat jo room mein deliver hoti hai. Same machine, naya benefit: $COP_{h p} = \frac{Q _{H}}{W} = \frac{Q _{H}}{Q _{H} - Q _{C}} .$

KYU. Hardware mein kuch nahi badla — bas kaunsa arrow tumhe valuable lagta hai yeh badla. "Heat pump" ek fridge hai jiske mote upar wale arrow ke neeche tum khade ho warm rehne ke liye.

PICTURE. Same box jaise Step 3 mein, lekin hot room mein ek insaan draw hai jo mote $Q_{H}$ arrow (red) ko pakad raha hai; cold side fade hai.

Step 6 — Best possible machine: koi entropy waste nahi

KYA. Ab tak $Q_{C}$ aur $Q_{H}$ kuch bhi ho sakte the jo Step 3 maane. Ab demand karo best machine — ek reversible (Carnot) jo kuch waste nahi karta. "Waste" measure karne ke liye hum entropy use karte hain: jab heat $Q$ temperature $T$ pe cross karta hai, toh woh entropy $Q / T$ carry karta hai. Entropy ek bookkeeping quantity hai jo universe ke liye sirf badhti hai (Second Law); ek perfect machine ise flat rakhti hai.

KYU yeh tool — entropy $Q / T$ — aur sirf energy nahi? Energy balance (Step 3) ke infinitely many solutions hain; yeh kabhi best nahi chunti. Hume ek doosri equation chahiye. Second Law of Thermodynamics ise supply karta hai: cold se kheenchi gayi entropy hot mein dump ki gayi entropy ke barabar honi chahiye, warna universe ki entropy giregi (forbidden) ya badhegi (wasteful).

PICTURE. Do entropy "packets" $Q_{C} / T_{C}$ (cold se nikalte hue) aur $Q_{H} / T_{H}$ (hot mein enter karte hue), size mein equal draw kiye gaye hain (lal balance line).

Punchline: heat ratio ab sirf temperatures se fix hai, gas ki messy details se nahi.

Step 7 — Substitute karo: temperatures heats ki jagah lete hain

KYA. $COP_{r e f} = \frac{Q _{C}}{Q _{H} - Q _{C}}$ lo aur upar aur neeche dono ko $Q_{C}$ se divide karo: $COP_{r e f} = \frac{Q _{C} / Q _{C}}{( Q _{H} - Q _{C} ) / Q _{C}} = \frac{1}{\frac{Q _{H}}{Q _{C}} - 1} = \frac{1}{\frac{T _{H}}{T _{C}} - 1} = \frac{T _{C}}{T _{H} - T _{C}} .$

KYU $Q_{C}$ se divide karo? Kyunki Step 6 ne sirf ek ratio $Q_{H} / Q_{C}$ diya tha, heats alag alag kabhi nahi diye. $Q_{C}$ se divide karne se har heat us known ratio ban jaati hai, jise hum phir $T_{H} / T_{C}$ se swap karte hain.

PICTURE. Temperature ki ek number line: $T_{C}$ aur $T_{H}$ mark karo; in dono ke beech ka gap "climb" $T_{H} - T_{C}$ hai. COP hai low temperature "climb" ke upar.

Step 8 — Edge cases: formula extremes pe kya kehta hai

KYA & KYU. Ek formula jis par tum trust karte ho usse apni limits pe sahi behave karna chahiye.

Chhota gap $T_{H} \to T_{C}$ : denominator $\to 0$ , toh $COP \to \infty$ . Heat ko koi hill nahi chahiye toh almost kuch nahi lagta. Sense banta hai.
Bada gap $T_{C} \to 0$ (ya $T_{H}$ bahut upar $T_{C}$ se): numerator chhota, denominator bada, $COP \to 0$ . Absolute zero ke paas kisi cheez se heat pump karna almost impossible hai — Second Law yahan sabse zyada kaata hai.
$W = 0$ with $Q_{C} > 0$ : $COP = \infty$ chahiye ek finite gap par — forbidden. Heat free mein nahi chadh sakti; tumhe pay karna hi hoga.

PICTURE. COP plot kiya gap $T_{H} - T_{C}$ ke against: ek steep curve jo infinity ki taraf rocket karti hai jab gap band hota hai, zero ki taraf sink karti hai jab gap bada hota hai (lal curve).

Ek-picture summary

Upar saari baatein, ek frame mein: heat cold se upar force ki gayi, work andar pay ki gayi, surplus hot mein dump kiya gaya; do entropy packets balanced hain (reversible); aur final fraction $T_{C} / (T_{H} - T_{C})$ temperature gap pe print kiya gaya.

Recall Poore walkthrough ki Feynman retelling

Heat aalsi hai — sirf neeche roll karti hai, hot se cold (Step 1). Ek fridge cold food se heat ka ek scoop pakadta hai aur use upar warm room mein dhakel deta hai, jo woh sirf electricity ki push pay karke kar sakta hai (Steps 2–3). "Kitna accha hai?" bas heat upar le jaana ÷ push kharch hai — bada number accha hai (Step 4). Fridge ki jagah room mein khado aur tum mote top arrow ko value karte ho, jisme tumhari push bhi hamesha hoti hai, toh ek heat pump exactly ek point zyada score karta hai (Step 5). Best possible machine koi entropy waste nahi karta: jo entropy woh cold se uthata hai woh entropy hot mein drop karta hai, aur woh ek balance heat ratio ko $T_{H} / T_{C}$ se fix kar deta hai (Step 6). Use swap karo aur messy heats gayab ho jaate hain, ek clean fraction bachta hai: cold temperature over the climb (Step 7). Ise edges pe test karo — koi hill nahi matlab free lifting (COP → ∞), giant hill matlab near-impossible lifting (COP → 0) — aur sab sense banta hai (Step 8). Bas yaad rakho: kelvin, hamesha kelvin.

Connections

First Law of Thermodynamics — Step 3 ka energy balance.
Second Law of Thermodynamics — kyun work pay karni padti hai, aur Step 6 ka entropy balance.
Entropy — woh $Q / T$ bookkeeping jo reversible heat ratio deta hai.
Carnot Cycle — woh reversible machine jo ceiling tak pahunchti hai.
Heat Engines and Efficiency — same $Q_{H}, Q_{C}, W$ aage ki taraf chalaaye.