Heat dissipation and cooling solutions: Exercises — Hinglish deep dive

6.4.5 · D4 · Hardware › Power, Thermal & Reliability › Heat dissipation and cooling solutions

Yeh page Heat dissipation and cooling solutions ke liye ek self-test ladder hai. Har problem apna answer ek collapsible Solution box ke andar chhupata hai — pehle khud try karo, phir reveal karo. Difficulty formula pehchanne se shuru hokar poora cooling system synthesise karne tak jaati hai.

Definition Woh symbols jo tum har jagah reuse karoge

$T_{amb}$ = ambient temperature: aaspaas ki hawa (ya room) ka temperature jisme cooler ultimately heat dump karta hai. Yeh woh baseline hai jiske upar sab kuch measure hota hai — ek chip kabhi apne ambient se thanda nahi ho sakta.
$T_{j}$ = junction temperature: chip ke andar hottest active silicon ka temperature (transistors jahaan rehte hain, woh "junction" hai). Yeh woh number hai jisko hum actually protect karna chahte hain — yeh ambient plus total temperature rise ke barabar hota hai: $T_{j} = T_{amb} + Δ T$ . Jab $T_{j}$ chip ki spec cross karta hai, tab Thermal throttling trigger hoti hai.
$k$ = thermal conductivity: ek material property jo batata hai ki heat ek solid se kitni aasani se guzar sakti hai. Units W/m·K (thickness ke har metre aur temperature difference ke har kelvin mein heat ke watts). Copper $k = 400$ , aluminium $k = 205$ , air $k \approx 0.025$ — metals air se ~10,000× behtar hain.
$h$ = heat-transfer coefficient: ek moving fluid (hawa, paani) surface se heat kitni aasani se uthata hai. Units W/m²·K (surface ke har square metre aur kelvin mein watts). $k$ ke unlike yeh ek pure material constant nahi hai — yeh flow speed aur geometry par depend karta hai, kyunki tez flow insulating still-fluid boundary layer ko얇a karta hai. Still air $h \approx 8$ , fan air $h \approx 60$ , water $h \approx 1000$ .

Do building-block resistances (dono parent note mein derive kiye gaye hain) exactly inhi symbols use karti hain:

Conduction (heat solid se guzarna): $R_{t h, cond} = \frac{d}{k A}$ — thickness $d$ divided by conductivity $k$ times area $A$ .
Convection (heat moving air/liquid se uthna): $R_{t h, conv} = \frac{1}{h A}$ — heat-transfer coefficient $h$ times area $A$ ka reciprocal.
Series mein resistances add hoti hain: $R_{t h, total} = R_{1} + R_{2} + \dots$ (kyunki heat ko ek-ek karke har stage cross karni padti hai, isliye unke "pushes" $Δ T_{i}$ stack hote hain).

Exercises se pehle, ek picture — yeh dekhne ke liye ki woh do formulas apni woh shapes kyun leti hain — yeh baaki page ki visual-first foundation hai:

Figure — Heat dissipation and cooling solutions

Left (conduction): heat ek solid slab se squeeze hoti hai. Mota slab (bada $d$ ) ek lambi tunnel hai, isliye resistance $d$ ke saath badhti hai (fraction ke upar). Chauda slab (bada $A$ ) zyada parallel lanes deta hai, isliye resistance $A$ ke saath ghatti hai; better material (bada $k$ ) bhi isse ghataata hai — isliye $R = d / (k A)$ . Right (convection): heat surface se moving air mein ek patli still-air boundary layer ke across nikalti hai. Zyada surface $A$ ya tez flow (bada $h$ ) dono help karte hain, aur raaste mein koi "mota" cheez nahi hoti, isliye $R = 1/ (h A)$ — koi $d$ nazar nahi aata.

Ab woh chain jo exercises actually baar baar solve karti hain:

Is chain ko dekho: left par red CHIP heat source hai (power $P$ , "current"). Teeno colored boxes Exercise 2.1 se series resistances $R_{1}, R_{2}, R_{3}$ hain — junction-to-case, paste, heatsink-to-air. Right par blue AIR box $T_{amb}$ hai, woh floor jiske upar sab kuch baitha hai. Grey arrows dikhate hain ki heat ek-ek karke har box se guzarti hai: yahi "ek-ek karke" wajah hai ki resistances add hoti hain. Temperature har box mein $Δ T_{i} = P R_{i}$ badhti hai, aur neeche ke labels dikhate hain ki woh drops 90 °C rise aur 115 °C junction temperature mein sum hote hain jo tum Exercise 2.1 mein compute karoge.

Level 1 — Recognition

Exercise 1.1

Ek cooling path ka $R_{t h} = 0.6$ °C/W hai aur woh $P = 150$ W carry karta hai. Uske across temperature rise $Δ T$ kya hai?

Recall Solution

Seedha master law use karo — kuch rearrange karne ki zarurat nahi. $Δ T = P \times R_{t h} = 150 \times 0.6 = 90 °C$ Humne kya kiya: flow ko resistance se multiply kiya. Kyun: yeh Ohm's law $V = I R$ ka thermal version hai; $R_{t h}$ batata hai ki har watt aapko kitne °C "khareedta" hai. Answer: 90 °C.

Exercise 1.2

Copper ki thermal conductivity $k = 400$ W/m·K hai, aluminium ki $k = 205$ W/m·K. Same slab ke liye (same $d$ , same $A$ ), kis ki conduction resistance zyada kam hai, aur roughly kitne factor se?

Recall Solution

$R_{t h, cond} = \frac{d}{k A}$ . Jab $d$ aur $A$ fixed hain, resistance $k$ ke inversely proportional hai. Bada $k$ ⇒ chhota $R_{t h}$ . $\frac{R _{Al}}{R _{Cu}} = \frac{k _{Cu}}{k _{Al}} = \frac{400}{205} \approx 1.95$ Copper wins, roughly $1.95 \times$ (≈ 2×) zyada kam resistance ke saath. Yahi wajah hai ki copper high-end heatsink bases mein dikhta hai. Iss idea ka extreme version dekhne ke liye Heat pipes dekho. Answer: copper, ≈ 2× zyada kam.

Exercise 1.3

Har mechanism ko desktop CPU cooler mein uske dominant role se match karo: (a) chip die → heatsink base, (b) fins → surrounding air, (c) fins → distant wall.

Recall Solution

(a) chip die → heatsink base: conduction (heat solid metal se guzarti hai).
(b) fins → air: convection (moving air surfaces se heat uthata hai).
(c) fins → distant wall: radiation (electromagnetic — ~100 °C se neeche total ka 5% se kam, isliye yahaan negligible; $Q = ε σ A (T_{s}^{4} - T_{amb}^{4})$ se governed, upar footnote dekho). Answer: a=conduction, b=convection, c=radiation.

Level 2 — Application

Exercise 2.1

Ek CPU 150 W dissipate karta hai. Path hai junction-to-case $R_{1} = 0.2$ °C/W, thermal paste $R_{2} = 0.1$ °C/W, heatsink-to-air $R_{3} = 0.3$ °C/W. Ambient air 25 °C hai. Junction temperature $T_{j}$ nikalo.

Recall Solution

Step 1 — series resistances add karo. Heat ko chip package cross karna padta hai, phir paste, phir heatsink ek ke baad ek — koi stage skip nahi hoti. Jab resistors end-to-end baithe hain, har ek apna temperature drop contribute karta hai aur woh stack hote hain, isliye total resistance plain sum hai: $R_{t h, total} = 0.2 + 0.1 + 0.3 = 0.6 °C/W$ Step 2 — temperature rise. Master law $Δ T = P R_{t h}$ apply karo, kyunki exactly yahi batata hai ki total resistance is power par kitne °C khareedti hai: $Δ T = P \times R_{t h, total} = 150 \times 0.6 = 90 °C$ Step 3 — ambient add karo. Yeh 90 °C surrounding air ke upar ek rise hai, isliye actual junction temperature $T_{j}$ paane ke liye hum woh baseline $T_{amb}$ add karte hain jiske upar poora stack baitha hai: $T_{j} = T_{amb} + Δ T = 25 + 90 = 115 °C$ Answer: 115 °C. Isse Thermal Design Power (TDP) limits ke against compare karo — agar $T_{j}$ spec cross kare, toh Thermal throttling kick in karti hai.

Exercise 2.2

Ek heatsink ka effective surface area $A = 0.05$ m² hai. Natural convection mein $h = 8$ W/m²·K; fan ke saath $h = 60$ W/m²·K. Dono convective resistances aur improvement factor compute karo.

Recall Solution

$R_{t h, conv} = \frac{1}{h A}$ use karo, kyunki fin surfaces convection se air mein heat shed karti hain aur yeh exactly uss stage ki resistance hai. Natural: $R = \frac{1}{8 \times 0.05} = \frac{1}{0.4} = 2.5 °C/W$ Forced: $R = \frac{1}{60 \times 0.05} = \frac{1}{3.0} = 0.333 °C/W$ Improvement: $\frac{2.5}{0.333} = 7.5 \times$ Itna bada kyun? $h$ 7.5× jump kiya kyunki fan ne still air ki insulating boundary layer얇i kar di. Kyunki $R \propto 1/ h$ , resistance same factor se giri. Answer: 2.5 → 0.33 °C/W, 7.5× improvement.

Exercise 2.3

Tumhe 80 W dissipate karna hai aur heatsink base 90 °C se zyada garam nahi hona chahiye, 30 °C ambient mein. Chip-to-base path already 0.25 °C/W hai. Kaun si heatsink-to-air resistance chahiye?

Recall Solution

Step 1 — temperature budget. Heat sirf difference par flow karti hai, isliye usable budget base cap minus ambient hai: poora path sirf $90 - 30 = 60$ °C rise kar sakta hai. Step 2 — total resistance allowed. Master law ko $R = Δ T / P$ mein rearrange karo, kyunki ab hum allowed rise aur power jaante hain aur woh resistance chahiye jo fit ho: $R_{t h, total} = \frac{Δ T}{P} = \frac{60}{80} = 0.75 °C/W$ Step 3 — upstream part subtract karo. Series resistances add hoti hain, isliye chip-to-base stage jo already kharch karta hai woh heatsink ke liye baaki chhod deta hai: $R_{t h, hs} = 0.75 - 0.25 = 0.5 °C/W$ Answer: 0.5 °C/W.

Level 3 — Analysis

Exercise 3.1

25 °C air mein ek 100 W chip ke liye do heatsink candidates hain:

A: aluminium, $R_{t h, hs} = 0.35$ °C/W, plus paste 0.10, plus junction-to-case 0.15.
B: copper, $R_{t h, hs} = 0.28$ °C/W, plus paste 0.10, plus junction-to-case 0.15.

Chip $T_{j} = 95$ °C se neeche rehna chahiye. Kaun sa cooler(s) pass karta hai, aur margin kya hai?

Recall Solution

Total resistances (series stages sum karo, kyunki heat ek-ek karke har stage cross karti hai): $R_{A} = 0.15 + 0.10 + 0.35 = 0.60 °C/W, R_{B} = 0.15 + 0.10 + 0.28 = 0.53 °C/W$ Junction temperatures ( $T_{j} = 25 + 100 R$ , yani ambient plus $P R$ ): $T_{j, A} = 25 + 100 (0.60) = 85 °C$ $T_{j, B} = 25 + 100 (0.53) = 78 °C$ Dono 95 °C se neeche hain, isliye dono pass karte hain. Margins: A ka $95 - 85 = 10$ °C; B ka $95 - 78 = 17$ °C. Answer: dono pass; A margin 10 °C, B margin 17 °C. B ka copper base 7 °C headroom deta hai — Overclocking ke liye useful.

Exercise 3.2

Tumhara fan-cooled heatsink 25 °C air mein 150 W chip ke liye $R_{t h} = 0.33$ °C/W deta hai. Tumhe $T_{j} \leq 80$ °C chahiye. Kya yeh pass karta hai? Agar nahi, toh $R_{t h}$ kitna kam hona chahiye?

Recall Solution

Current rise: $Δ T = 150 \times 0.33 = 49.5$ °C, toh $T_{j} = 25 + 49.5 = 74.5$ °C. Kyunki $74.5 \leq 80$ , yeh pass karta hai $80 - 74.5 = 5.5$ °C margin ke saath. (Agar fail hota, toh required resistance hoti $R = (80 - 25) /150 = 0.367$ °C/W — lekin hum pehle se woh beat kar rahe hain.) Answer: pass karta hai, 5.5 °C margin.

Exercise 3.3

Ek cold plate paani ko $h_{w} = 1000$ W/m²·K se contact area $A_{w} = 0.004$ m² par touch karta hai. Radiator heat ko air mein dump karta hai; uska combined conductance $h_{a} A_{a} = 3.0$ W/K diya gaya hai — yeh simply radiator ke air-side heat-transfer coefficient $h_{a}$ aur uski total fin area $A_{a}$ ka product hai, ek number mein bundle kiya gaya kyunki $R = 1/ (h A)$ mein sirf product matter karta hai (ek bada radiator apni low resistance bade $A_{a}$ se paata hai, bade $h_{a}$ se nahi). Metal ke andar chhoti conduction ignore karke, do convective resistances aur total nikalo. Kaun dominant hai?

Recall Solution

Water side ( $R = 1/ (h_{w} A_{w})$ , convection resistance jahaan paani cold plate se heat uthata hai): $R_{w} = \frac{1}{h _{w} A _{w}} = \frac{1}{1000 \times 0.004} = \frac{1}{4} = 0.25 °C/W$ Air (radiator) side — yahaan hume pehle se product $h_{a} A_{a} = 3.0$ W/K diya gaya hai, isliye: $R_{a} = \frac{1}{h _{a} A _{a}} = \frac{1}{3.0} = 0.333 °C/W$ Total (series, kyunki heat pehle water stage phir radiator stage cross karti hai ek-ek karke): $R_{t h} = 0.25 + 0.333 = 0.583 °C/W$ Air side dominant hai (0.333 > 0.25). Yeh liquid cooling ki key insight hai: paani heat grab karne mein superb hai, lekin tumhe abhi bhi use radiator par air ko wapas dena padta hai — wahi bottleneck hai. Radiator ko feed karne ke liye Case airflow and positive/negative pressure dekho. Answer: $R_{w} = 0.25$ , $R_{a} = 0.33$ , total $0.58$ °C/W; air side dominant hai.

Level 4 — Synthesis

Exercise 4.1

Design check. Ek 200 W CPU ko 35 °C room mein $T_{j} \leq 90$ °C par rehna chahiye. Fixed path: junction-to-case 0.10, TIM 0.05. Tumhe ek heatsink choose karna hai. Available heatsink area 0.06 m² hai. Airflow ko kaun sa heat-transfer coefficient $h$ deliver karna chahiye?

Recall Solution

Step 1 — total budget: $Δ T = 90 - 35 = 55$ °C, toh $R_{t h, total} = \frac{55}{200} = 0.275 °C/W$ Step 2 — fixed upstream subtract karo: $R_{t h, hs} = 0.275 - 0.10 - 0.05 = 0.125 °C/W$ Step 3 — convection ke liye $h$ solve karo: $R_{t h, hs} = \frac{1}{h A} \Rightarrow h = \frac{1}{R _{t h, hs} A}$ $h = \frac{1}{0.125 \times 0.06} = \frac{1}{0.0075} = 133.3 W/m^{2} \cdotK$ Interpretation: $h \approx 133$ forced-air ke typical 60–100 se kaafi upar hai, isliye ek single stock fan nahi chalega — tumhe strong forced air chahiye (dense fins + high-static-pressure fan) ya phir liquid cooling par jaao. Answer: $h \approx 133$ W/m²·K.

Exercise 4.2

Paani heat ko ek flowing coolant ki tarah carry karta hai. Uska temperature rise $P = \overset{m}{˙} c_{p} Δ T_{water}$ se obey karta hai, jahaan $\overset{m}{˙}$ mass flow rate (kg/s) hai aur $c_{p} = 4180$ J/kg·K. Ek pump 1.5 L/min paani (density 1000 kg/m³) ek loop mein 250 W absorb karte hue move karta hai. Cold plate se guzarte waqt paani kitna garam hota hai?

Recall Solution

Step 1 — mass flow rate. 1.5 L/min ko SI mein convert karo: $1.5$ L/min $= 1.5 \times 1 0^{- 3}$ m³ per 60 s $= 2.5 \times 1 0^{- 5}$ m³/s. Density $1000$ kg/m³ se multiply karo: $\overset{m}{˙} = 1000 \times 2.5 \times 1 0^{- 5} = 0.025 kg/s$ Step 2 — $Δ T_{water}$ ke liye rearrange karo. $P = \overset{m}{˙} c_{p} Δ T_{water}$ se, dono sides ko $\overset{m}{˙} c_{p}$ se divide karo: $Δ T_{water} = \frac{P}{m ˙ c _{p}} = \frac{250}{0.025 \times 4180} = \frac{250}{104.5} = 2.39 °C$ Itna tiny kyun? Paani ki badi heat capacity ( $c_{p} = 4180$ J/kg·K) matlab hai ki chhota temperature climb bahut saari energy absorb karta hai — yahi liquid transport ki puri appeal hai. Coolant barely warm hota hai, toh poora loop near-uniform temperature par rehta hai aur radiator ek bade, dur ke surface par heat shed kar sakta hai. Answer: ≈ 2.4 °C rise.

Level 5 — Mastery

Exercise 5.1

Full-system reliability trade-off. Ek server CPU (TDP 165 W) 25 °C ambient mein run karta hai. Do designs:

Design X: total $R_{t h} = 0.42$ °C/W (quiet fan).
Design Y: total $R_{t h} = 0.30$ °C/W (louder fan).

Engineering ka ek well-known rule of thumb kehta hai ki component lifetime roughly har 10 °C operating temperature rise par half ho jaata hai. Design X ki junction temperature ko reference maanke, Design Y ki relative lifetime estimate karo.

Recall Solution

Step 1 — junction temperatures. $T_{j, X} = 25 + 165 (0.42) = 25 + 69.3 = 94.3 °C$ $T_{j, Y} = 25 + 165 (0.30) = 25 + 49.5 = 74.5 °C$ Step 2 — Y vs X ka temperature drop: $Δ = 94.3 - 74.5 = 19.8 °C cooler$ Step 3 — halving rule apply karo. Lifetime har 10 °C thande hone par $2$ se multiply hoti hai: $Lifetime ratio = 2^{Δ/10} = 2^{19.8/10} = 2^{1.98} \approx 3.94$ Interpretation: Design Y, thoda noisy hone ke bawajood, expected service life mein roughly 4× zyada tikta hai. Yeh cooling aur Reliability and MTBF ke beech quantitative link hai — thanda silicon bahut kam fail hota hai. Noise ki cost ek bada reliability win kharidti hai. Answer: Design Y ≈ Design X ki 3.9× lifetime.

Exercise 5.2

Capstone. Tumhare paas ek machine hai jo load mein thermal-throttle karti hai. Measured: 180 W dissipation, ambient 28 °C, junction 100 °C par throttle karta hai. Current stack: junction-to-case 0.12, old dried paste 0.30, heatsink-to-air 0.35. (a) Confirm karo ki yeh throttle karta hai. (b) Tum paste replace karte ho (0.30 → 0.06) aur better fan add karte ho jo heatsink resistance half karta hai (0.35 → 0.175). Kya throttling rukti hai, aur naya margin kya hai?

Recall Solution

(a) Current state. $R_{old} = 0.12 + 0.30 + 0.35 = 0.77 °C/W$ $T_{j} = 28 + 180 (0.77) = 28 + 138.6 = 166.6 °C$ Yeh 100 °C throttle point se kaafi upar hai — haan, yeh hard throttle karta hai (aur chip ko likely damage kar raha hai; Thermal throttling dekho). (b) Fixes ke baad. $R_{new} = 0.12 + 0.06 + 0.175 = 0.355 °C/W$ $T_{j} = 28 + 180 (0.355) = 28 + 63.9 = 91.9 °C$ Kyunki $91.9 < 100$ , throttling rukti hai, $100 - 91.9 = 8.1$ °C ke margin ke saath. Lesson: dried paste akele $180 \times 0.30 = 54$ °C rise contribute karta tha — ek single degraded interface poore system ko cook kar sakta hai. TIM koi detail nahi hai. Baki thermal picture ke liye PCB thermal vias aur Power supply efficiency dekho. Answer: (a) 166.6 °C par throttle karta hai; (b) fix ho gaya, $T_{j} = 91.9$ °C, 8.1 °C margin.

Self-check clozes

Power, resistance aur temperature rise ko link karne wala master law

Δ T = P \times R_{t h}

Conduction resistance formula

R_{t h, cond} = d / (k A)

Convection resistance formula

R_{t h, conv} = 1/ (h A)

Series thermal resistances aise combine hoti hain

adding them

Rise se junction temperature paane ke liye tumhe

ambient temperature add karni padti hai

Liquid cooling mein usual bottleneck stage hoti hai

radiator (air side), paani nahi

Lifetime rule of thumb

roughly halves per 10 °C hotter

°C aur K in formulas mein interchangeable kyun hain

ek 1 °C step equals a 1 K step (sirf differences appear hote hain)