Heat dissipation and cooling solutions: Worked examples

6.4.5 · D3 · Hardware › Power, Thermal & Reliability › Heat dissipation and cooling solutions

Yeh page parent topic ka problem gym hai. Hum un do master formulas ko — jo tumne pehle se dekh rakhi hain — har us situation mein chalayenge jo ek exam ya real build mein aa sakti hai — including woh weird edge cases jahan arithmetic tumhe jhooth bolne ki koshish karti hai.

Kuch bhi compute karne se pehle, sirf do tools ko fresh se anchor karte hain, plain words mein.

Recall Do formulas, zero se dobara batate hain

Thermal resistance $R_{t h}$ ka matlab hai "ek watt heat push karne par cheez kitni degrees zyada garam ho jaati hai." Chhota number = heat ka aasaan raasta = achhi cooling. $R_{t h} = \frac{Δ T}{P} [°C per watt]$ Jis tarah se hum zyada use karte hain, rearrange karke: $Δ T = P \cdot R_{t h}$ (push ki gayi heat times resistance = temperature climb), bilkul electricity mein $V = I \cdot R$ ki tarah.

Series rule: jab heat ek layer se doosri, phir teesri se flow karti hai (chip → paste → heatsink → air), toh resistances add up hoti hain: $R_{t h, total} = R_{1} + R_{2} + \dots$ kyunki har layer apna temperature step add karti hai aur steps stack hoti jaati hain.

Neeche sab kuch sirf yahi do ideas hain, plus convection resistance $R_{t h, co n v} = \frac{1}{h \cdot A}$ (badi surface $A$ ya tez air $h$ ⇒ chhoti resistance).

The scenario matrix

Har heat problem is grid ke ek cell mein hoti hai. Humara kaam har row mein kam se kam ek worked example land karna hai.

#	Cell class	Kya tricky banata hai	Example jo isko hit karta hai
A	Forward — $P$ aur sab $R$ diye hain, temperature nikalo	plain series-add, phir $Δ T = P R$	Ex 1
B	Backward — temperature limit di hai, $R$ budget nikalo	fixed (upstream) part subtract karo	Ex 2
C	Convection design — $R = \frac{1}{h A}$ se $h$ ya area $A$ nikalo	product ko invert karo, units watch karo	Ex 3
D	Zero / degenerate input — $P = 0$ , ya ek "perfect" layer $R = 0$	formula phir bhi sense banana chahiye	Ex 4
E	Limiting behaviour — $h \to \infty$ , ya $A \to \infty$	resistance ek floor tak approach karti hai, zero tak nahi	Ex 5
F	Real-world word problem — liquid loop, transport vs surface	decide karo kaun si resistance dominate karti hai	Ex 6
G	Exam twist — units mismatch / trap (grams, kW, °C vs K)	pehle convert karo, warna answer 1000× galat hoga	Ex 7
H	Comparison / ratio — "kitni baar better?"	do resistances ka ratio, area cancel ho jaata hai	Ex 8

Har example neeche apne cell letter se tagged hai.

The worked examples

Cell A — plain forward calculation

Worked example Ex 1 (A): Poora chip-to-air path

Ek GPU $P = 220$ W dissipate karta hai. Path hai: junction-to-case $R_{1} = 0.15$ °C/W, thermal paste $R_{2} = 0.10$ °C/W, heatsink-to-air $R_{3} = 0.25$ °C/W. Ambient air $T_{a} = 22$ °C. Junction temperature $T_{j}$ nikalo.

Forecast: Padhne se pehle guess karo — kya $T_{j}$ 60 °C ke paas hoga, 100 °C ke paas, ya 130 °C se upar? Apna guess likho.

Step 1 — Series resistances add karo. $R_{t h, total} = 0.15 + 0.10 + 0.25 = 0.50 °C/W$ Yeh step kyun? Heat teeno layers se ek ke baad ek jaati hai, isliye har layer ka temperature step stack hota hai — series resistances add hoti hain.

Step 2 — Power ko temperature rise mein convert karo. $Δ T = P \cdot R_{t h, total} = 220 \times 0.50 = 110 °C$ Yeh step kyun? $R_{t h} = Δ T / P$ , isliye $Δ T = P R$ . Yeh batata hai ki hottest silicon air se kitna upar hai.

Step 3 — Ambient baseline add karo. $T_{j} = T_{a} + Δ T = 22 + 110 = 132 °C$ Yeh step kyun? $Δ T$ surrounding air ke relative hai; actual temperature usi baseline se measure hoti hai.

Verify: Units check — W × (°C/W) = °C. ✅ Sanity: 132 °C ek typical 95–100 °C throttle limit se upar hai, isliye yeh cooler inadequate hai — ek physically meaningful, believable result (dekho Thermal throttling).

Cell B — resistance budget ke liye backward kaam karna

Worked example Ex 2 (B): Mujhe kaisa heatsink chahiye?

Ek CPU $P = 125$ W dissipate karta hai. Junction $T_{j, m a x} = 90$ °C se neeche rehna chahiye. Ambient $T_{a} = 35$ °C hai. Fixed upstream part (junction-to-case + paste) $R_{up} = 0.30$ °C/W hai. Kaun sa heatsink-to-air resistance $R_{h s}$ allowed hai?

Forecast: Kya budget tight hoga (0.2 °C/W se neeche) ya generous (0.5 °C/W se upar)? Guess karo.

Step 1 — Poora temperature budget nikalo. $Δ T_{budget} = T_{j, m a x} - T_{a} = 90 - 35 = 55 °C$ Kyun? Yahi maximum climb hai jo air se silicon tak allowed hai.

Step 2 — Budget ko total resistance ceiling mein convert karo. $R_{t h, total} = \frac{Δ T _{budget}}{P} = \frac{55}{125} = 0.44 °C/W$ Kyun? Hum $Δ T = P R$ ko $R = Δ T / P$ mein flip karte hain — yahi sabse badi resistance hai jo budget mein fit hoti hai.

Step 3 — Jo part badal nahi sakta use subtract karo. $R_{h s} = R_{t h, total} - R_{up} = 0.44 - 0.30 = 0.14 °C/W$ Kyun? Junction aur paste fixed hain; sirf jo bacha hua hai woh heatsink ke liye available hai.

Verify: Plug back karo: total $= 0.30 + 0.14 = 0.44$ , $Δ T = 125 \times 0.44 = 55$ °C, $T_{j} = 35 + 55 = 90$ °C. ✅ Bilkul limit par. 0.14 °C/W heatsink ek serious tower/air setup hai — reality se match karta hai.

Cell C — convecting surface design karna

Worked example Ex 3 (C): Air diya hua hai, toh kitna fin area chahiye?

Tumhe $R_{h s} = 0.14$ °C/W hit karna hai (Ex 2 se). Forced air $h = 70$ W/m²·K deta hai. Tumhe kitna effective fin surface area $A$ chahiye?

Forecast: Kuch sau cm²? Kuch hazaar? Ek order of magnitude guess karo.

Step 1 — Convection resistance form yaad karo. $R_{h s} = \frac{1}{h \cdot A}$ Yeh tool kyun, Fourier's law kyun nahi? Fourier's law ( $Q = k A Δ T / d$ ) heat ke liye hai jo solid ke through jaati hai. Yahan heat fin se moving air mein ja rahi hai, isliye boundary-layer picture — Newton's law of cooling — sahi hai, aur uski resistance $1/ (h A)$ hai.

Step 2 — Area solve karo. $A = \frac{1}{h \cdot R _{h s}} = \frac{1}{70 \times 0.14} = \frac{1}{9.8} \approx 0.102 m^{2}$ Kyun? Hum product invert karte hain; do knobs hain air ( $h$ ) aur expose ki gayi metal ( $A$ ).

Step 3 — Interpret karo. $0.102$ m² $= 1020$ cm². Fins ka ek block ~1000 cm² surface ke saath — stacked-fin tower cooler ke liye bilkul normal hai.

Verify: Back-substitute karo: $R = 1/ (70 \times 0.102) = 1/7.14 = 0.140$ °C/W. ✅ Units: $\frac{1}{( W/m ^{2} K ) ( m ^{2} )} = K/W =$ °C/W. ✅

Cell D — zero aur degenerate inputs

Worked example Ex 4 (D): Idle chip aur "perfect" layer

Do sub-questions jo formulas ko unke edges par stress-test karti hain.

(a) Ek chip $P = 0$ W par idle hai, $R_{t h, total} = 0.5$ °C/W, ambient $25$ °C. $T_{j}$ nikalo.

Forecast: Kya yeh garam hoga?

Step 1. $Δ T = P \cdot R = 0 \times 0.5 = 0$ °C. Kyun? Koi heat flow nahi matlab koi temperature step nahi, chahe cooler kitna bhi bekar ho. Step 2. $T_{j} = 25 + 0 = 25$ °C. Chip ambient par baith jaata hai. Verify: Physically sense banata hai — ek unpowered chip apne surroundings se zyada garam nahi ho sakta. ✅

(b) Maan lo ek paste layer perfect ho, $R_{2} = 0$ °C/W, with $R_{1} = 0.2$ aur $R_{3} = 0.3$ . Total?

Step 1. $R_{t h, total} = 0.2 + 0 + 0.3 = 0.5$ °C/W. Kyun? Ek zero-resistance layer koi temperature step add nahi karti — yeh series sum se drop out ho jaati hai. Yeh ek bahut patli, bahut conductive interface ka degenerate limit hai (isliye liquid metal ko chase kiya jaata hai). Verify: Ek real paste sirf $0$ ke approach kar sakta hai, kabhi reach nahi kar sakta (air gaps hamesha kuch chhod jaate hain). Formula gracefully degrade karta hai. ✅

Cell E — limiting behaviour

Worked example Ex 5 (E): Agar tumhare paas infinite airflow hota to?

Ek cooler lo jiska total path ek solid part $R_{solid} = 0.12$ °C/W (junction + paste + fin conduction) aur ek convection part $R_{co n v} = \frac{1}{h A}$ mein split hota hai, $A = 0.05$ m² ke saath. $h$ ko $8$ (still air) se $\infty$ (imaginary hurricane) ki taraf push karo. $R_{t h, total}$ kya approach karta hai?

Forecast: Kya total resistance zero ho jaati hai, ya kisi floor par ruk jaati hai?

Step 1 — Total likho. $R_{t h, total} (h) = 0.12 + \frac{1}{h \times 0.05}$ Kyun? Solid path aur air path series mein hain (heat metal se phir air mein jaati hai), isliye add hoti hain.

Step 2 — Kuch $h$ values evaluate karo.

$h = 8$ : $R = 0.12 + \frac{1}{0.4} = 0.12 + 2.5 = 2.62$ °C/W
$h = 100$ : $R = 0.12 + \frac{1}{5} = 0.12 + 0.20 = 0.32$ °C/W
$h \to \infty$ : $\frac{1}{h A} \to 0$ , isliye $R \to 0.12$ °C/W

Yeh step kyun? Yeh dikhata hai ki convection term collapse ho raha hai jabki solid term stubbornly rehta hai.

Step 3 — Lesson. Chahe airflow kitna bhi violent ho, total 0.12 °C/W se neeche nahi ja sakta — solid conduction floor. Jab tumhare fins "airflow-limited" se kaafi aage ho jaate hain, bada fan kuch nahi khareedta; tumhe $R_{solid}$ par attack karna hoga (copper base, better paste, direct-contact heat pipes — dekho Heat pipes).

Verify: Jab $h \to \infty$ , $1/ (h A) \to 0$ monotonically, isliye $R$ upar se $0.12$ ki taraf decrease karta hai aur kabhi cross nahi karta. ✅ (Dekho figure.)

Figure — Heat dissipation and cooling solutions

Cell F — real-world word problem (liquid loop)

Worked example Ex 6 (F): Kya bada pump is AIO mein help karta hai?

Ek liquid loop mein hai: chip-to-water cold plate $R_{c p} = 0.05$ °C/W, aur radiator-to-air $R_{r a d} = 0.15$ °C/W. Ek salesman claim karta hai ki pump flow double karna (jo $R_{c p}$ ko thoda $0.04$ tak lower karta hai) worth it hai. Chip $200$ W dissipate karta hai, ambient $28$ °C. Pehle aur baad mein $T_{j}$ compare karo.

Forecast: Kya pump double karne se $T_{j}$ noticeably drop karega?

Step 1 — Total resistance, pehle. $R = 0.05 + 0.15 = 0.20 °C/W, T_{j} = 28 + 200 (0.20) = 68 °C$ Kyun? Cold-plate aur radiator series mein hain (heat paani ke saath ek se doosre tak ride karti hai).

Step 2 — Total resistance, baad mein. $R = 0.04 + 0.15 = 0.19 °C/W, T_{j} = 28 + 200 (0.19) = 66 °C$

Step 3 — Interpret karo kaun sa term rule karta hai. Change sirf $2$ °C hai, kyunki radiator ki air convection ( $0.15$ ) dominate karti hai, water contact ( $0.05$ ) nahi. Liquid cooling ka asli gift heat ko chip se door ek bade radiator tak transport karna hai — air ko final hand-off abhi bhi ordinary convection hai. Paise stronger pump se better, zyada radiator area/fans par lagana better hai.

Verify: $68 - 66 = 2$ °C, aur fixed $0.15$ radiator term $\frac{0.15}{0.19} \approx 79%$ total ka hai — confirm karta hai ki yeh dominate karta hai. ✅

Cell G — exam-trap: units

Worked example Ex 7 (G): Kilowatt / gram trap

Ek power module $P = 1.5$ kW ek cold plate se dissipate karta hai. Paani $30$ °C par enter karta hai aur tumhe bataaya jaata hai coolant heat absorb karta hai — temperature rise tumhe nikalni hai, given mass flow $\overset{m}{˙} = 0.05$ kg/s aur water ki specific heat $c_{p} = 4.18$ kJ/kg·K. Coolant ki outlet temperature nikalo.

Forecast: Log blurt karte hain "bahut bada rise" kyunki 1.5 kW bada lagta hai. Pehle guess karo.

Step 1 — Consistent units mein convert karo. $P = 1.5 kW = 1500 W$ ; $c_{p} = 4.18 kJ/kg\cdotK = 4180 J/kg\cdotK$ . Kyun? kW ko J/kg·K ke saath mix karne se answer 1000 se galat aata hai. Pehle convert karo, hamesha.

Step 2 — Flow-heating relation use karo. $P = \overset{m}{˙} c_{p} Δ T \Rightarrow Δ T = \frac{P}{m ˙ c _{p}} = \frac{1500}{0.05 \times 4180}$ Yeh tool kyun? Yeh energy balance hai: paani mein jaane wale watts per second $\overset{m}{˙} c_{p} Δ T$ carried-away heat ban jaate hain. Yeh answer deta hai "fluid flow karte waqt kitna warm hota hai?"

Step 3 — Compute karo. $Δ T = \frac{1500}{209} \approx 7.18 °C, T_{out} = 30 + 7.18 = 37.2 °C$

Step 4 — Lesson. 1.5 kW se sirf $7$ °C coolant rise — isliye water kaam karta hai: iski huge $c_{p}$ (4× air) ise badi power ko chhote temperature climb ke saath swallow karne deti hai.

Verify: Units: $\frac{W}{( kg/s ) ( J/kg\cdotK )} = \frac{J/s}{J/(s\cdotK)} = K$ . ✅ Agar tumne kJ→J conversion bhool jaate toh $Δ T \approx 7180$ °C milta — turant absurd, ek built-in sanity alarm.

Cell H — ratio / comparison

Worked example Ex 8 (H): Fan kitni baar better hai?

Wahi heatsink (same area $A$ ) pehle still air $h_{1} = 8$ W/m²·K mein, phir fan ke saath $h_{2} = 64$ W/m²·K mein chalta hai. Thermal resistance kis factor se improve hoti hai?

Forecast: Algebra karne se pehle factor guess karo.

Step 1 — Dono resistances likho. $R_{1} = \frac{1}{h _{1} A}, R_{2} = \frac{1}{h _{2} A}$

Step 2 — Ratio lo; dekho $A$ cancel ho jaata hai. $\frac{R _{1}}{R _{2}} = \frac{1/ ( h _{1} A )}{1/ ( h _{2} A )} = \frac{h _{2}}{h _{1}} = \frac{64}{8} = 8$ Yeh step kyun? Kyunki wahi heatsink use ho raha hai, $A$ identical hai aur divide out ho jaata hai — improvement purely airflow ratio hai. Area number ki zaroorat bhi nahi thi.

Step 3 — Interpret karo. Ek fan se convection resistance mein 8× drop. Yeh air cooling mein sabse bada single lever hai aur isliye Case airflow and positive/negative pressure itna matter karta hai.

Verify: Resistances ka ratio $= h_{2} / h_{1} = 8$ exactly; dimensionless, jaise ek ratio hona chahiye. ✅

Recall Quick self-test

100 W chip, total path 0.4 °C/W, ambient 30 °C — junction temp? ::: $30 + 100 \times 0.4 = 70$ °C. Fan $h$ ko 10 se 90 tak same heatsink par raise karta hai — resistance kis factor se improve hoti hai? ::: $90/10 = 9 \times$ . Ex 6 mein stronger pump kyun bahut better cool nahi kar sakta? ::: Radiator-to-air convection dominate karti hai; water transport kabhi bottleneck tha hi nahi. Ek perfect (zero-resistance) TIM layer series sum ko kaise change karti hai? ::: Bilkul nahi — ek zero term drop out ho jaata hai.