Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴): Visual walkthrough

1.7.6 · D2 · Physics › Thermodynamics › Heat transfer — conduction (Fourier's law k), convection, ra

Hum ek solid slab se heat flow ki rate banate hain — kuch nahi aur. Chalte hain, har letter kama ke lete hain.

Step 1 — Hum measure kya kar rahe hain? (heat current $H$ )

KYA. Ek solid wall ki picture karo. Baayein taraf garam hai, daayein taraf thanda. Energy steadily garam se thande ki taraf risti hai. Hum chahte hain ek single number jo bataaye ki energy kitni tezi se wall paar kar rahi hai.

KYUN. "Kitna garam hai" kaafi nahi — hum flow se matlab rakhte hain, joules jo har second cross kar rahe hain. Physicists kisi cheez ke "per second" flow ko current kehte hain. Isliye hum heat current banate hain:

$H = \frac{d Q}{d t}$

$Q$ ::: heat energy ki maatra (joules, J mein maapi jaati hai) jo wall paar kar chuki hai.
$t$ ::: time seconds mein.
$\frac{d Q}{d t}$ ::: " $Q$ har second mein kitna badalta hai" — ise joules per second padho, jo ki ek watt (W) hota hai. Chote $d$ 's ka matlab hai "mein ek tiny change".

Toh $H$ watts mein measure hota hai, bilkul ek lightbulb ki power ki tarah. Yahi poora target hai: $H$ ke liye ek formula dhundho.

PICTURE. Neeche, yellow arrows woh heat hain jo wall se daayein taraf risi ja rahi hai. Woh number jo hum dhundh rahe hain woh hai: har second us dashed line se kitne joules cross karte hain?

Step 2 — Zyada wall, zyada flow ( $H \propto A$ )

KYA. Sab kuch waisa hi rakho lekin wall ko zyada chauda aur zyada uncha banao — use zyada face area $A$ do (square metres, m² mein maapa jaata hai).

KYUN. Heat wall ke face se hoke cross karti hai. Face double karo aur tumne simply do identical walls ko side by side chipka diya; har ek utna hi ristaati hai, toh saath milke do guna zyada ristaati hain. Yahan kuch bhi subtle nahi — yeh pure "do cheezein ek cheez se do guna hoti hain" hai.

$H \propto A$

$\propto$ ::: "proportional hai" — saath-saath badhta hai. Agar $A$ teen guna ho, $H$ teen guna ho jaata hai.
$A$ ::: cross-sectional area jisse hoke heat flow karti hai, thickness nahi.

PICTURE. Do identical wall-tiles ko stack karo toh double area ki wall banti hai aur arrow count bhi double ho jaata hai.

Step 3 — Zyada steep temperature drop, zyada flow ( $H \propto Δ T$ )

KYA. Ab area fix karo aur dono face temperatures badlo. Garam face ko $T_{1}$ aur thanda face ko $T_{2}$ kaho, jahan $T_{1} > T_{2}$ hai. Unka gap hai

$Δ T = T_{1} - T_{2} .$

$Δ$ (Greek "delta") ::: "mein antar" — hamesha baad wala minus pehle wala, ya yahan garam minus thanda.
$Δ T$ ::: left face kitne degrees zyada garam hai right face se.

KYUN. Heat isliye flow karti hai kyunki temperature difference hota hai (second law: energy garam se thande ki taraf khisakti hai). Koi difference nahi → koi flow nahi. Zyada difference ek zyada steep "pahaadi" hai energy ke liye neeche girane ke liye, isliye yeh tezi se flow karti hai. Experiment confirm karta hai ki yeh directly proportional hai — gap double karo, flow double ho jaata hai.

$H \propto Δ T$

PICTURE. Halki temperature slope ek trickle drive karti hai; steep slope ek torrent drive karti hai. Blue line wall ke across plot ki gayi temperature hai — iska steepness hi matter karta hai.

Step 4 — Moti wall, kam flow ( $H \propto 1/Δ x$ )

KYA. Area aur temperatures fix rakho, lekin wall ko mota banao. Thickness ko $Δ x$ (metres) kaho — woh distance jo heat ko garam face se thande face tak cross karni hoti hai.

KYUN. Energy atom se atom tak hops karti hai, jaise ek bucket chain. Ek moti wall ek lambi bucket chain hai same temperature drop ke liye, isliye har point par "slope" halka hota hai aur flow dheema hota hai. Thickness double karo aur (same faces ke liye) flow aadha ho jaata hai:

$H \propto \frac{1}{Δ x}$

$\frac{1}{Δ x}$ ::: "inversely proportional" — jaise $Δ x$ badhta hai, $H$ ghatta hai. Yeh Steps 2 aur 3 ka ulta hai.

PICTURE. Same garam aur thande faces, lekin ek moti wall same $Δ T$ ko lambe path par spread kar deti hai — slope flat ho jaata hai, arrows thin ho jaate hain.

Step 5 — Teeno facts ko jodo ( $k$ ka janam)

KYA. Ab hamare paas teen alag-alag proportionalities hain. Inhe ek statement mein multiply karo:

$H \propto A \frac{Δ T}{Δ x} .$

" $\propto$ " ko " $=$ " mein badalne ke liye hum ek single constant $k$ daalaate hain jo capture karta hai ki wall kis material ki bani hai:

$H = k A \frac{Δ T}{Δ x}$

EK CONSTANT KYUN CHAHIYE. Proportionality humein law ki shape batati hai lekin scale nahi. Copper aur wood mein law ki same shape hoti hai lekin wildly different flow — woh difference $k$ mein baki hai, jo thermal conductivity hai.

$k$ ::: yeh material kitni eagerly conduct karta hai. Bada $k$ (metal) → zyada flow; chota $k$ (foam, air) → achha insulator.
$k$ ki Units. Rearrange karo: $k = \frac{H Δ x}{A Δ T}$ → $\frac{W \cdot m}{m ^{2} \cdot K} = W m^{- 1} K^{- 1}$ . $k$ exist karta hai precisely isliye taaki dono sides ke units match hon.

PICTURE. Teeno influences (blue area, pink drop, yellow inverse-thickness) ek mixing funnel mein jaate hain; $k$ woh aakhri ingredient hai jo final number fix karta hai.

Step 6 — Slab se ek point tak: derivative aur minus sign

KYA. Abhi tak $Δ T /Δ x$ poori wall ka average slope tha. Lekin temperature seedhi line mein nahi bhi gir sakti. Wall ke ek single point par flow describe karne ke liye, slab ko zero thickness tak shrink karo. Jab tum $Δ x \to 0$ hone do, toh ratio $Δ T /Δ x$ derivative $\frac{d T}{d x}$ ban jaata hai.

SIRF RATIO NAHI, DERIVATIVE KYUN. Ek derivative exactly yeh sawaal ka jawaab deta hai ki " $T$ yahan kitni steeply change ho rahi hai?" — blue temperature curve ka local slope. Ise use karna law ko tab bhi kaam karne deta hai jab wall layered ho ya slope bend kare. Yeh tool "instantaneous steepness" ke liye bana hai, isliye yeh sahi wala hai.

MINUS SIGN AB KYUN AATA HAI. $x$ ko us direction mein badhta define karo jis direction mein heat flow karti hai. Kyunki yeh thande side ki taraf flow karti hai, temperature neeche jaati hai jaise $x$ upar jaata hai, isliye $\frac{d T}{d x}$ negative hai. Lekin aage ki taraf pointing karne wali heat current ek positive number honi chahiye. Negative slope ko positive current mein flip karne ke liye hum aage minus lagaate hain:

PICTURE. Blue curve downhill slope karti hai (negative $d T / d x$ ); minus use flip karta hai taaki yellow current arrow aage pointing aur positive aaye.

Step 7 — Har case, check kiya gaya (signs, zero, degenerate walls)

Hume kabhi koi scenario unchecked nahi chhodna chahiye. Corners dekho:

Heat doosri taraf flow kar rahi hai ( $T_{2} > T_{1}$ ): ab $d T / d x$ positive hai, isliye $- k A d T / d x$ negative hai → $H < 0$ . Negative $H$ ka matlab simply hai "flow backward hai", yaani right se left. Sign auto-correct ho jaata hai; tumhe direction guess karne ki zaroorat nahi.
Koi temperature difference nahi ( $T_{1} = T_{2}$ ): $d T / d x = 0 \Rightarrow H = 0$ . Flat temperature, zero flow — intuition se match karta hai (kuch bhi drive nahi kar raha).
Perfect insulator ( $k \to 0$ , e.g. vacuum-flask foam): $H \to 0$ chahe gap kitna bhi bada ho. Isliye insulation kaam karti hai — $k$ khatam karo aur current khatam ho jaati hai.
Zero thickness (slab form mein $Δ x \to 0$ ): $H \to \infty$ . Bina thickness ki wall koi resistance nahi offer karti — infinite flow, isliye derivative form (ek finite local slope) ek real wall ki honest description hai.
Zero area ( $A = 0$ ): $H = 0$ . Koi face nahi, koi flow nahi. Trivial hai lekin real — ek wire ka end-face apne tiny $A$ se conduct karta hai, isliye thin wires thodi total heat conduct karti hain.

PICTURE. Paanch corner cases ka ek chota "map": reversed flow, flat $T$ , dead insulator, zero thickness, zero area — har ek ka apna verdict.

Ek-picture summary

Upar sab kuch ek single frame mein: wall, downhill temperature line, teeron influences apne arrows ke saath, constant $k$ , aur final boxed law jisme har symbol tagged hai ki woh kya karta hai.

Recall Feynman retelling (zor se bolo, koi symbols nahi)

Ek wall imagine karo, ek taraf garam, doosri taraf thanda. Heat garam se thande ki taraf chhupna chahti hai. Kitni tezi se yeh chhupti hai — yahi hamaari heat current hai, joules per second, watts. Teeno plain facts pace set karte hain: chaudi wall zyada jaane deti hai (zyada doors), bada temperature gap zyada dhakka deta hai (steeper hill), aur moti wall isko slow karti hai (lambi crawl). Teeno ko multiply karo aur ek material number $k$ chidkao — metals ke liye bada, foam ke liye chota — aur tumhare paas flow hai. Aakhir mein, agar tum ek exact spot par flow chahte ho, tum temperature ka slope (derivative) use karte ho, aur aage minus sign lagaate ho sirf isliye taaki number positive aaye jab heat natural taraf flow kare. Gap khatam karo, ya $k$ khatam karo, ya area shrink karo, aur flow mar jaata hai — bilkul waisa jaise common sense demand karta hai. Wahi poori kahaani hai Fourier's law.

Recall Quick self-test

Same face temperatures ke liye moti wall kam heat kyun lete jaati hai? ::: Same $Δ T$ bade $Δ x$ par spread → gentler slope $\frac{d T}{d x}$ → chota $H$ (kyunki $H \propto 1/Δ x$ ). $H$ ki negative value tumhe kya batati hai? ::: Heat $- x$ direction mein (backward) flow kar rahi hai; sign direction encode karta hai, "negative amount" of heat nahi. Constant $k$ kahan se aaya aur yeh kyun zaruri hai? ::: Yeh proportionality $H \propto A Δ T /Δ x$ ko equation mein convert karta hai aur material encode karta hai; yeh units bhi fix karta hai.