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

1.7.6 · D3 · Physics › Thermodynamics › Heat transfer — conduction (Fourier's law k), convection, ra

Yeh parent topic ke liye practice floor hai. Kuch bhi compute karne se pehle, aao hum har tarah ke question ko map karte hain jo teen heat-transfer laws tumhare saamne rakh sakti hain. Phir hum har box ka ek example solve karte hain taaki tum kabhi aisa case na dekho jo tune pehle na dekha ho.

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The scenario matrix

#	Cell (case class)	Which law	The trap it tests
A	Single slab, plain numbers	Conduction	°C-difference = K-difference
B	Slabs in series	Conduction	resistances add; find interface temperature
C	Slabs in parallel	Conduction	conductances add, not resistances
D	Degenerate: gradient $\to 0$	Conduction	$\Delta T=0\Rightarrow H=0$ (equilibrium)
E	Convection off a surface	Convection	$hA\Delta T$, linear in $\Delta T$
F	Radiation, surroundings negligible	Radiation	when may we drop $T_0^4$?
G	Radiation, surroundings NOT negligible	Radiation	full $T^4-T_0^4$, both in kelvin
H	Sign / net = 0 limiting case	Radiation	body in equilibrium with room, $P_{\text{net}}=0$
I	Real-world word problem	Combined	pick the dominant mode
J	Exam twist: unit sabotage	Radiation	°C used where kelvin required

Hum A–J neeche cover karenge. Prerequisites jo tum open rakhna chahoge: Thermal conductivity k, Newton's law of cooling, Black body radiation, Electrical resistance Ohm's law.

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Example 1 — Cell A: single slab, bare numbers

1. Fourier's law ka magnitude form likho. $H=kA\frac{\Delta T}{\Delta x}$ Yeh step kyun? Hume sirf flow ka size chahiye, isliye hum magnitude version use karte hain jahan $\Delta T=T_{\text{hot}}-T_{\text{cold}}$ — full law $H=-kAdT/dx$ ka minus sign pehle se hot-minus-cold lene se account ho gaya hai.

2. $\Delta T$ compute karo aur unit trick note karo. $\Delta T=22-2=20{(}^{\circ }\text{C or K, identical})$ Yeh step kyun? 20 °C ka difference exactly 20 K ke difference ke barabar hota hai — kyunki ek kelvin ek Celsius degree ke size ke barabar hota hai. Difference ke liye koi conversion nahi chahiye.

3. Values dalo. $H=0.6\times 6\times \frac{20}{0.20}=0.6\times 6\times 100=360\text{W}$

Verify: Units: $({\text{W m}}^{-1}{\text{K}}^{-1})({\text{m}}^2)(\text{K})/(\text{m})=\text{W}$ ✓. Magnitude 360 W — sau wala, jaise forecast kiya tha. Ek brick wall parent note ki 8000 W glass window se kaafi kam heat jaane deti hai, kyunki brick moti hai aur uska $k$ chhota hai. Sensible. ✓

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Example 2 — Cell B: slabs in series + interface temperature

1. Har thermal resistance $R=\Delta x/(kA)$ compute karo. $R_1=\frac{0.20}{0.6\times 6}=0.0556\text{K/W},R_2=\frac{0.05}{0.03\times 6}=0.2778\text{K/W}$ Yeh step kyun? Series mein same current $H$ dono slabs se guzarta hai (energy steady state mein kahi accumulate nahi ho sakti — dekho Second law of thermodynamics), isliye resistances add hoti hain, bilkul Electrical resistance Ohm's law mein ek current carry karne wale resistors ki tarah.

2. Resistances add karo, $H$ nikalo. $R_{\text{tot}}=0.0556+0.2778=0.3333\text{K/W},H=\frac{\Delta T}{R_{\text{tot}}}=\frac{20}{0.3333}=60\text{W}$ Yeh step kyun? $H=\Delta T/R_{\text{tot}}$ heat ke liye Ohm's law hai. Foam akele ne flow ko 360 W (Example 1) se 60 W tak cut kar diya.

3. Interface temperature $T_m$ nikalo. Same $H$ brick se flow karta hai, isliye brick ke across drop $H{R}_1$ hai: $\Delta T_{\text{brick}}=H{R}_1=60\times 0.0556=3.33\text{K}$ $T_m=T_{\text{inside}}-\Delta T_{\text{brick}}=22-3.33=18.67^{\circ }\text{C}$ Yeh step kyun? Figure dekho: 20 K drop ka zyada hissa (16.67 K) foam ke across padta hai, jo poor conductor hai. Temperature ki khadi cliff insulating layer mein hoti hai — insulation ka yahi poora point hai.

Verify: Foam ke across drop $=H{R}_2=60\times 0.2778=16.67$ K, aur $3.33+16.67=20$ K = total ✓. Interface $18.67^{\circ }$C par hai jo 2 aur 22 ke beech mein hai — physically valid ✓.

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Example 3 — Cell C: slabs in parallel

1. Har path ko full $\Delta T$ milta hai; heat currents add hoti hain. $H_{\text{brick}}=\frac{15}{0.5}=30\text{W},H_{\text{glass}}=\frac{15}{0.1}=150\text{W}$ Yeh step kyun? Brick aur glass side by side baithte hain, dono same do temperatures ko bridge karte hain — yahi parallel ki definition hai. Parallel paths se guzarne wale currents add hote hain (load split karte hain), isliye hum yahan resistances add nahi karte.

2. Total heat current. $H_{\text{tot}}=30+150=180\text{W}$

3. (Combined resistance ke through cross-check — parallel mein conductances add hoti hain.) $\frac{1}{R_{\text{tot}}}=\frac{1}{0.5}+\frac{1}{0.1}=2+10=12\text{W/K},R_{\text{tot}}=0.0833\text{K/W},H=\frac{15}{0.0833}=180\text{W}$ Yeh step kyun? Rule confirm hota hai: parallel → conductances $1/R$ add hoti hain. Chhoti window (area ka 20%) 150/180 ≈ 83% loss carry karti hai — low-resistance path dominate karta hai.

Verify: Dono methods se 180 W ✓. Glass loss ka majority carry karti hai jaise forecast kiya tha ✓.

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Example 4 — Cell D: degenerate case, $\Delta T\to 0$

1. $\Delta T=0$ set karo. $H=kA\frac{\Delta T}{\Delta x}=0.6\times 6\times \frac{0}{0.20}=0\text{W}$ Yeh step kyun? Fourier's law $\Delta T$ mein linear hai. Temperature difference nahi matlab driving "push" nahi, isliye koi net heat flow nahi — wall thermal equilibrium mein hai, jo Second law of thermodynamics ke consistent hai: heat tabhi flow karta hai jab gradient ho.

Verify: $H=0$ jab $\Delta T=0$ — limiting/degenerate case well-behaved hai (undefined nahi, kyunki $\Delta x\ne 0$). ✓

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Example 5 — Cell E: convection off a hot plate

1. Newton's law of cooling apply karo. $H=hA(T_s-T_{\text{fluid}})$ Yeh step kyun? Convection heat ko is proportion mein le jaata hai ki surface temperature fluid se kitna upar hai — ek simple engineering model (dekho Newton's law of cooling). Wahan bhi $\Delta T$ °C mein = K mein.

2. Values dalo. $H=15\times 0.4\times (80-25)=15\times 0.4\times 55=330\text{W}$

Verify: Units $({\text{W m}}^{-2}{\text{K}}^{-1})({\text{m}}^2)(\text{K})=\text{W}$ ✓. 330 W — hundreds, jaise forecast kiya tha ✓.

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Example 6 — Cell F: radiation, surroundings negligible

1. Net radiation law likho. $P_{\text{net}}=e\sigma A\left(T^4-T_0^4\right),\sigma =5.67\times 10^{-8}$ Yeh step kyun? Ek body dono emit karti hai ($e\sigma A{T}^4$) aur room se absorb karti hai ($e\sigma A{T}_0^4$); net difference hota hai (Kirchhoff — same $e$ dono taraf).

2. Fourth powers evaluate karo. $T^4=1200^4=2.0736\times 10^{12},T_0^4=300^4=8.1\times 10^9$ Yeh step kyun? $T_0^4$, $T^4$ se $\approx 256\times$ chhota hai — lagbhag $0.4\%$. Ise drop karne se sirf $0.4\%$ error aata hai, isliye yeh yahan ek legitimate 80/20 shortcut hai.

3. Compute karo. $P_{\text{net}}=0.9\times 5.67\times 10^{-8}\times 4\times 10^{-4}\times (2.0736\times 10^{12}-8.1\times 10^9)$ $=0.9\times 5.67\times 10^{-8}\times 4\times 10^{-4}\times 2.0655\times 10^{12}\approx 42.2\text{W}$

Verify: $T_0^4$ bilkul drop karne par $42.3$ W milta hai — exact $42.2$ W se sirf ~0.2% alag, jo confirm karta hai ki shortcut safe hai jab $T\gg T_0$ ho ✓.

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Example 7 — Cell G: radiation with surroundings that matter

1. Dono fourth powers, dono kelvin mein. $T^4=340^4=1.33634\times 10^{10},T_0^4=290^4=7.0729\times 10^9$ Yeh step kyun? Jab surface aur surroundings close hain, $T_0^4$, $T^4$ ka bada fraction hota hai (yahan ~53%), isliye hum ise drop nahi kar sakte — aisa karna answer ko lagbhag double kar deta.

2. Subtract karo, phir multiply karo. $T^4-T_0^4=1.33634\times 10^{10}-7.0729\times 10^9=6.2905\times 10^9$ $P_{\text{net}}=0.85\times 5.67\times 10^{-8}\times 1.5\times 6.2905\times 10^9\approx 454.9\text{W}$

Verify: Units: $({\text{W m}}^{-2}{\text{K}}^{-4})({\text{m}}^2)({\text{K}}^4)=\text{W}$ ✓. Ek modest few-hundred watts, jaise forecast kiya tha — chhota 50 K gap net ko Example 6 ki blazing filament se kaafi neeche rakhta hai, chahe area bahut bada ho ✓.

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Example 8 — Cell H: limiting case, net radiation = 0

1. Net law apply karo $T=T_0$ ke saath. $P_{\text{net}}=e\sigma A(T^4-T_0^4)=e\sigma A(295^4-295^4)=0\text{W}$ Yeh step kyun? Statue tab emit aur absorb dono same rate par karti hai jab wo room ke temperature par ho. Wo abhi bhi radiation se glow karti hai — lekin net exchange zero hota hai. Isliye bodies ambient temperature reach karne ke baad cool hona band kar deti hain (Second law of thermodynamics phir: koi gradient nahi, koi net flow nahi).

Verify: $T^4-T_0^4=0\Rightarrow P_{\text{net}}=0$ regardless of $e$, $A$, $\sigma$ — equilibrium limit exact hai ✓.

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Example 9 — Cell I: real-world word problem (pick the dominant mode)

1. Pehle conduction aur convection rule out karo. Yeh step kyun? Vacuum mein koi matter nahi hota, isliye conduct karne ke liye kuch nahi aur convect karne ke liye koi fluid nahi — dono modes ko medium chahiye. Sirf radiation vacuum cross kar sakta hai. Isliye saari loss radiative hai.

2. Tiny emissivity ke saath radiative loss compute karo. $P=e\sigma A(T^4-T_0^4)$ $T^4=350^4=1.500625\times 10^{10},T_0^4=295^4=7.573502\times 10^9$ $T^4-T_0^4=7.432748\times 10^9$ $P=0.05\times 5.67\times 10^{-8}\times 0.06\times 7.432748\times 10^9\approx 1.27\text{W}$ Yeh step kyun? Silvering $e$ ko $0.05$ tak force kar deta hai, radiative loss ko ek black surface ke mukable ~20 guna slash kar deta hai. Vacuum ke saath jo baaki dono modes kill kar deta hai, coffee barely ek watt se zyada lose karti hai.

Verify: Just over 1 W — ek unsilvered, air-filled container se kaafi kam, jaise forecast kiya tha. Yahi thermos ka poora engineering point hai ✓.

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Example 10 — Cell J: exam twist, the °C sabotage

1. Galat calculation (°C, forbidden). $P_{\text{wrong}}=1\times 5.67\times 10^{-8}\times 0.1\times (127{)}^4=5.67\times 10^{-9}\times 2.601\times 10^8\approx 1.475\text{W}$ Yeh step kyun? Yahi trap hai: $T^4$ ek absolute quantity hai, isliye kelvin mandatory hai. Sirf ek difference $\Delta T$ °C mein reh sakta hai.

2. Sahi calculation (kelvin). $T=127+273=400\text{K},P_{\text{correct}}=5.67\times 10^{-9}\times 400^4=5.67\times 10^{-9}\times 2.56\times 10^{10}\approx 145.2\text{W}$ Yeh step kyun? Pehle convert karo, phir fourth power lo. Sahi radiated power galat wale se lagbhag 100 guna hai.

3. Ratio. $\frac{P_{\text{correct}}}{P_{\text{wrong}}}=\frac{400^4}{127^4}={\left(\frac{400}{127}\right)}^4\approx (3.15{)}^4\approx 98.4$

Verify: Galat answer ~98× chhota hai — ek catastrophic error, jo illustrate karta hai ki kelvin $T$ ki kisi bhi power mein non-negotiable kyun hai ✓.

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Recall Aage badhne se pehle self-test

Ek slab aur uski twin ko (a) series mein, (b) parallel mein same $\Delta T$ ke across rakha jaata hai. Kaun sa arrangement zyada total heat pass karta hai? ::: Parallel — resistances halve ho jaati hain (conductances add hoti hain), isliye current single slab ke mukable double hoti hai; series mein resistance double hoti hai, current halve hoti hai. Net radiation formula mein $T_0^4$ drop karna kab safe hai? ::: Jab $T\gg T_0$ ho, jaise $T=1200$ K vs $T_0=300$ K, jahan $T_0^4$, $T^4$ ka ~1% se kam ho. Ek body room temperature par — kya wo radiate kar rahi hai? Kya uska net radiation zero hai? ::: Haan wo radiate karti hai ($e\sigma A{T}^4$); lekin net zero hota hai kyunki wo equal-temperature surroundings se equal amount absorb karti hai.