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

1.7.6 · D4 · Physics › Thermodynamics › Heat transfer — conduction (Fourier's law k), convection, ra

Definition Heat flow ke liye ek symbol, ek sign convention

Is poore page mein hum har rate of heat transfer — conduction, convection, ya radiation — ko same symbol $H$ se likhenge (yeh heat current hai, units watts mein). Parent note radiated rate ko $P$ kehta hai; yahan hum use $H$ rename karte hain taaki jab hum channels add karein (conduction + convection + radiation) toh hum hamesha like-for-like watts add kar rahein.

Sign convention: $H > 0$ ka matlab hai ki heat hot body se door cooler surroundings ki taraf flow ho rahi hai (hot → cold, jaisa Second law of thermodynamics demand karta hai). Negative $H$ ka matlab hoga ki body net heat gain kar rahi hai.
Area convention: $A$ ka matlab hamesha woh area hai jiske through woh particular channel actually heat transfer karta hai. Jab ek body ek se zyada faces se radiate ya convect kare, toh hum subscript se spell out karte hain, e.g. $A_{rad}$ = total radiating area (sab faces milake), $A_{conv}$ = total convecting area.

Level 1 — Recognition

L1.1 — Mode pehchano

Har situation mein dominant heat-transfer mode batao: (a) Sardi ke din dhoop mein tumhara chehra garam hona. (b) Chai ki mug mein rakha metal ka chammach garam hona. (c) Ubalne se pehle kettle mein garam paani ka upar uthna.

Recall Solution

HUM KYA CHECK KAR RAHE HAIN: har case mein energy kya lekar chal rahi hai.

(a) Radiation — energy space ke vacuum mein electromagnetic waves ki tarah travel karti hai; Suraj aur Dharti ke beech koi medium hi nahi hai.
(b) Conduction — chammach ek solid hai, isliye atoms/electrons khud flow kiye bina energy aage paas karte hain.
(c) Convection — garam paani ki density kam ho jaati hai, woh upar uthta hai, aur thanda paani neechhe aata hai, yeh ek physically moving fluid loop hai.

L1.2 — Fourier's law padho

Thermal conductivity $k$ ki SI units batao, aur yeh bhi batao ki wall ki thickness $Δ x$ badhane se heat current $H$ badhega ya ghategaa.

Recall Solution

$H = k A \frac{Δ T}{Δ x}$ se $k$ solve karo: $k = \frac{H Δ x}{A Δ T}$ . Units: $\frac{W \cdot m}{m ^{2} \cdot K} = W m^{- 1} K^{- 1}$ . Kyunki $Δ x$ denominator mein hai, zyada moti wall se $H$ chhota hoga — moti wall better insulate karti hai. Yeh intuition se match karta hai "lamba rasta = dheema flow."

Level 2 — Application

L2.1 — Brick wall se heat

Ek brick wall ka $k = 0.6 W m^{- 1} K^{- 1}$ , area $A = 15 m^{2}$ , thickness $Δ x = 0.20 m$ hai. Andar $2 2^{\circ}$ C hai, bahar $2^{\circ}$ C hai. Heat current $H$ nikalo.

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

Upar wali figure dikhati hai ki "heat current" kaisa dikhta hai: parallel arrows (ek heat-flow field) warm inside face se cold outside face ki taraf streaming kar rahe hain, aur colour gradient magenta-hot se navy-cold ki taraf fade ho rahi hai. Sabhi arrows same length ke hain kyunki steady state mein same current $H$ wall ke har plane ko cross karta hai — kuch pile up nahi hota.

Recall Solution

KYA HAI: Fourier's law ka direct plug-in ek single slab ke liye. Kelvin conversion kyun nahi: humein sirf temperature ka difference chahiye, aur $^{\circ}$ C mein difference K mein difference ke equal hota hai ( $Δ T = 22 - 2 = 20 K$ ). $H = k A \frac{Δ T}{Δ x} = 0.6 \times 15 \times \frac{20}{0.20} = 900 W$ Sign positive hai, isliye hamare convention se heat bahar flow ho rahi hai (inside → outside) — bilkul expected, jab andar bahar se garam ho.

L2.2 — Radiating sphere

Ek blackbody sphere ( $e = 1$ ) jiska radius $r = 0.05 m$ hai, $T = 400 K$ par hai aur surroundings $T_{0} = 300 K$ par hain. Net radiated heat current nikalo.

Recall Solution

KYA HAI: sphere ki surface area $A = 4 π r^{2}$ hai, phir net Stefan–Boltzmann apply karo (sab ek closed surface se radiate ho raha hai). $A = 4 π (0.05)^{2} = 4 π \times 0.0025 = 0.031416 m^{2}$ Net kyun, sirf $T^{4}$ kyun nahi: sphere emit bhi karta hai ( $e σ A T^{4}$ ) aur room se absorb bhi karta hai ( $e σ A T_{0}^{4}$ ); observable heat loss in dono ka difference hai. $H = 1 \times 5.67 \times 1 0^{- 8} \times 0.031416 \times (40 0^{4} - 30 0^{4})$ $40 0^{4} = 2.56 \times 1 0^{10}$ , $30 0^{4} = 8.1 \times 1 0^{9}$ , difference $= 1.75 \times 1 0^{10}$ . $H = 5.67 \times 1 0^{- 8} \times 0.031416 \times 1.75 \times 1 0^{10} \approx 31.2 W$ $H > 0$ : zyada garam sphere thande room ko heat lose karta hai, jo haari sign rule se consistent hai.

Level 3 — Analysis

L3.1 — Composite (series) wall

Ek wall mein equal area $A = 10 m^{2}$ ke do slabs back-to-back hain: slab 1 (brick): $k_{1} = 0.6$ , $Δ x_{1} = 0.10 m$ ; slab 2 (foam): $k_{2} = 0.04$ , $Δ x_{2} = 0.05 m$ . Andar $2 0^{\circ}$ C, bahar $0^{\circ}$ C. $H$ aur brick–foam interface par temperature nikalo.

Figure ko ek story ki tarah padho: magenta line actual temperature hai jab tum hot inside face (left) se cold outside face (right) ki taraf chalte ho. Yeh brick ke across dheere se girta hai (low resistance = chhoti drop) aur foam ke across tezi se girta hai (high resistance = badi drop). Neeche same-length flow arrows yaad dilate hain ki current $H$ dono slabs mein identical hai — slope badalta hai, flow nahi.

Recall Solution

Resistances kyun add hote hain: steady state mein slabs ke beech koi energy pile up nahi hoti, isliye same heat current $H$ dono mein se flow karta hai — bilkul series resistors ki tarah (dekho Electrical resistance Ohm's law analogy). Series ⇒ resistances add karo. $R_{1} = \frac{Δ x _{1}}{k _{1} A} = \frac{0.10}{0.6 \times 10} = 0.016667 K/W$ $R_{2} = \frac{Δ x _{2}}{k _{2} A} = \frac{0.05}{0.04 \times 10} = 0.125 K/W$ $R_{tot} = 0.141667 K/W, H = \frac{Δ T}{R _{tot}} = \frac{20}{0.141667} \approx 141.2 W$ Interface temperature: same $H$ slab 1 cross karta hai, isliye brick ke across temperature drop $Δ T_{1} = H R_{1} = 141.2 \times 0.016667 = 2.35 K$ hai. Hot face ( $2 0^{\circ}$ C) se shuru karke: $T_{interface} = 20 - 2.35 \approx 17. 6^{\circ} C$ Dhyan do ki foam (bahut bada $R_{2}$ ) almost pura drop lekar chal jaata hai — yahi ek insulator karta hai: temperature difference ko khud mein sameta leta hai. Yeh wahi plunge hai jo figure mein dikhti hai.

L3.2 — Drop kaun jeet ta hai?

L3.1 mein poore $20 K$ drop ka kitna fraction foam ke across hota hai? Interpret karo.

Recall Solution

$Δ T_{2} = H R_{2} = 141.2 \times 0.125 = 17.65 K$ . Fraction $= 17.65/20 = 0.88 = 88%$ . Interpretation: jis slab ka resistance zyada hoga, woh zyada temperature drop lega — bilkul series circuit mein bade resistor par zyada voltage drop ki tarah. Insulation is tarah kaam karti hai ki $Δ T$ khud apne andar concentrate kar le.

Level 4 — Synthesis

L4.1 — Conduction jo radiation ko feed kare (steady state)

Ek thin blackbody plate ka geometric face area $A_{face} = 0.02 m^{2}$ hai aur yeh dono faces se radiate karti hai, isliye iska total radiating area $A_{rad} = 2 A_{face} = 0.04 m^{2}$ hai. Isko ek conductive rod se back mein heat diya ja raha hai. Steady state mein plate $T = 500 K$ par hai, surroundings $T_{0} = 300 K$ par hain, $e = 1$ . Rod ko kitna conductive heat current $H$ deliver karna hoga?

Recall Solution

KYU: steady state mein energy in = energy out. Rod ka conduction in dono faces se net radiation out ke equal hona chahiye, warna plate ka temperature drift karega. $H = e σ A_{rad} (T^{4} - T_{0}^{4}), A_{rad} = 2 A_{face} = 0.04 m^{2}$ $50 0^{4} = 6.25 \times 1 0^{10}$ , $30 0^{4} = 8.1 \times 1 0^{9}$ , difference $= 5.44 \times 1 0^{10}$ . $H = 1 \times 5.67 \times 1 0^{- 8} \times 0.04 \times 5.44 \times 1 0^{10} \approx 123.4 W$ Sanity check: agar rod zyada deliver kare, toh plate garam hoti jaayegi jab tak uska $T^{4}$ radiation match na kare; agar kam kare, toh thandi hoti jaayegi. Steady state woh balance point hai.

L4.2 — Combined convection + radiation loss

Ek freestanding vertical panel $T_{s} = 350 K$ par hai, iska geometric face area har face ke liye $A_{face} = 1 m^{2}$ hai aur emissivity $e = 0.9$ hai. Air aur surroundings dono $T_{0} = 290 K$ par hain, aur convection coefficient $h = 8 W m^{- 2} K^{- 1}$ hai. Kyunki panel hawa mein freely khada hai, dono faces se radiate aur convect hota hai. Total heat loss nikalo.

Recall Solution

Dono faces kyun: ek freestanding panel aage aur peeche se room ke samne exposed hai, isliye effective areas double ho jaate hain: $A_{rad} = A_{conv} = 2 A_{face} = 2 m^{2} .$ Dono channels kyun add karein: convection aur radiation parallel mein act karte hain — do independent channels jo same surface se heat drain karte hain — isliye unke currents add hote hain. Convection: $H_{c} = h A_{conv} (T_{s} - T_{0}) = 8 \times 2 \times (350 - 290) = 960 W$ . Radiation: $35 0^{4} = 1.50063 \times 1 0^{10}$ , $29 0^{4} = 7.0728 \times 1 0^{9}$ , difference $= 7.9335 \times 1 0^{9}$ . $H_{r} = 0.9 \times 5.67 \times 1 0^{- 8} \times 2 \times 7.9335 \times 1 0^{9} = 809.6 W$ $H_{total} = H_{c} + H_{r} = 960 + 809.6 \approx 1769.6 W$ Dono channels yahan matter karte hain — koi bhi negligible nahi hai, isliye "chhota term drop karo" shortcut nahi chalega. Sab terms positive hain: garam panel dono taraf se thande room ko heat lose karta hai.

Level 5 — Mastery

L5.1 — Temperature-drop bookkeeping ek window mein still-air film ke saath

Ek double-glazed unit hai: glass ( $k_{g} = 0.8$ , $Δ x_{g} = 0.004 m$ ) + air gap ( $k_{a} = 0.025$ , $Δ x_{a} = 0.012 m$ ) + glass ( $k_{g} = 0.8$ , $Δ x_{g} = 0.004 m$ ), sab ka area $A = 2 m^{2}$ hai. Andar $2 0^{\circ}$ C, bahar $0^{\circ}$ C. $H$ nikalo aur confirm karo ki air gap insulation karta hai.

Figure teen resistances ko blocks ki tarah draw karti hai jinki width unke resistance ke proportional hai: dono glass panes bahut patle hain, trapped-air block enormous hai. Visually, air gap stack ki almost poori "electrical" length hai — isliye use almost poori temperature drop khaani chahiye.

Recall Solution

KYA HAI: teen slabs series mein ⇒ teen resistances add karo. $R_{g} = \frac{0.004}{0.8 \times 2} = 0.0025 K/W (har glass pane)$ $R_{a} = \frac{0.012}{0.025 \times 2} = 0.24 K/W (air gap)$ $R_{tot} = 0.0025 + 0.24 + 0.0025 = 0.245 K/W$ $H = \frac{20}{0.245} \approx 81.6 W$ Parent note ke single 4 mm pane ( $8000 W$ ) se compare karo: air gap ne loss ~98% cut kar diya. Air gap dominant kyun hai: iska $R_{a} = 0.24$ har glass ke $R_{g} = 0.0025$ se ~96 guna zyada hai. Air gap ke across temperature drop: $Δ T_{a} = H R_{a} = 81.6 \times 0.24 = 19.6 K$ — almost poora $20 K$ trapped air mein baitha hai (isliye air still rehni chahiye, taaki convection use short-circuit na kar sake).

L5.2 — Design: ek radiator ko room ki cooling se match karo

Ek hot object ( $e = 1$ , radiating area $A_{rad} = 0.05 m^{2}$ ) ko $T_{0} = 300 K$ ke surroundings mein net $200 W$ radiate karna hai. Iske liye kitna surface temperature $T$ chahiye?

Recall Solution

Invert kyun karein: humein target heat current pata hai aur $T$ ke liye back solve karna hai — 4th-power law fourth root se invert hoti hai. $H = e σ A_{rad} (T^{4} - T_{0}^{4}) \Rightarrow T^{4} = T_{0}^{4} + \frac{H}{e σ A _{rad}}$ $\frac{H}{e σ A _{rad}} = \frac{200}{1 \times 5.67 \times 1 0 ^{- 8} \times 0.05} = 7.0547 \times 1 0^{10}$ $T^{4} = 8.1 \times 1 0^{9} + 7.0547 \times 1 0^{10} = 7.8647 \times 1 0^{10}$ $T = (7.8647 \times 1 0^{10})^{1/4} \approx 529.4 K$ Physics check: $T > T_{0}$ hona zaroori hai net outward radiation ke liye ( $H > 0$ , hot → cold; dekho Second law of thermodynamics).

Recall Quick self-check ladder

Ek-line reason ::: L1 mode = mechanism; L2 kelvin sirf powers ke liye; L3 series mein resistances add hoti hain; L4 dono faces + channels parallel mein add hote hain; L5 net law ko root ke andar $+ T_{0}^{4}$ ke saath invert karo.

Related deep threads: Newton's law of cooling, Black body radiation, Wien's displacement law, Planck's law, Greenhouse effect, Thermal conductivity k.