Heat dissipation and cooling solutions: Question bank

6.4.5 · D5 · Hardware › Power, Thermal & Reliability › Heat dissipation and cooling solutions

Traps se pehle, kuch words hain jo tumhe bilkul seedha pata hone chahiye, kyunki neeche ke har item ka aadhar inhi par hai:

Definition The anchors

Temperature difference $Δ T$ = ek path ke hot side aur cold side ke beech ka gap, $Δ T = T_{hot} - T_{cold}$ , $^{\circ} C$ mein measure kiya jaata hai (ya K — ek degree ka size dono mein same hota hai). Heat sirf tab flow karti hai jab yeh gap non-zero ho, hot se cold ki taraf. Figure 1 (left) mein yeh upar se neeche tak ki drop ki height hai.
Heat flow $P$ = jis rate par heat energy ek path se move karti hai, watts (W) mein. Yeh plumbing analogy mein "current" hai. (Kuch textbooks is same quantity ko $Q$ likhte hain; is page par hum $P$ use karte hain throughout, parent note se match karne ke liye.)
Thermal resistance $R_{t h}$ = heat ko flow karne mein kitni mushkil hoti hai, $^{\circ} C/W$ mein. Yeh ek path ka " $Δ T$ per watt" cost hai: isme se $P$ watts push karo aur yeh $Δ T = P \cdot R_{t h}$ ka gap khol deta hai. Low hona achha hai.
Thermal conductivity $k$ = ek material property (copper vs air), $W/m \cdot K$ mein. Yeh is baat se fix hoti hai ki cheez kis cheez se bani hai.
Heat transfer coefficient $h$ = ek moving fluid kisi surface se heat kitni achhi tarah strip karta hai, $W/m^{2} \cdot K$ mein. Yeh flow par depend karta hai, sirf fluid par nahi.
Surface (or cross-section) area $A$ = us face ka area jisme se heat flow hoti hai ya strip hoti hai, $m^{2}$ mein. Bada $A$ = heat ke liye zyada parallel paths, isliye lower resistance. Yahi $A$ hai jo $R_{co n v} = 1/ (h \cdot A)$ aur $Δ T = P d / (k \cdot A)$ ke andar hota hai.

Trap fuel: log $k$ (ek material fact) aur $h$ (ek flow fact) ko swap kar dete hain jaise yeh ek hi knob ho. Yeh hain nahi.

Teen figures is page ke har answer ko anchor karte hain. Captions batati hain ki kaun sa item kaun si picture ki taraf point karta hai.

Figure 1. Left: water-tower analogy — heat flow $P$ current hai, $Δ T$ drop ki height hai, $R_{t h}$ pipe ki narrowness hai. Right: kyun series stages add karte hain — har stage total drop ka ek hissa consume karta hai, aur drops upar se neeche stack hote hain, bilkul circuit mein resistors ki tarah.

Figure 2. Convection boundary layer. Hot surface par almost-still air ka ek patla blanket chipa rehta hai aur use insulate karta hai — yahi blanket wajah hai ki still air ka $h$ chhota (~8) kyun hota hai. Fan us blanket ko patla kar deta hai, isliye heat us se zyada tez cross karti hai aur $h$ ~60 tak chadh jaata hai. Neeche ka material kabhi nahi badla; sirf flow badli.

Figure 3. Fins kyun aur patla thermal paste kyun. Left: ek solid block thodi si air ko touch karta hai; fins usi metal ko bahut zyada surface area $A$ mein failaate hain, aur $R_{co n v} = 1/ (h \cdot A)$ ghatता hai jaise $A$ badhta hai. Right: Fourier's law $Δ T = P d / (k \cdot A)$ ka matlab hai ki ek layer ke across temperature drop uski thickness $d$ ke saath badhti hai — isliye thick paste layer ek bada, unwanted drop banata hai thin layer ke muqable mein.

True or false — justify

Copper always cools better than aluminium because its conductivity is higher

False — copper ka

k

~2× higher hai, lekin conductivity sirf metal ke andar help karti hai; final bottleneck usually fins se convection hoti hai (

1/ (h \cdot A)

), aur copper ka extra weight chhote/kam fins force kar sakta hai, jo kabhi kabhi race haar jaata hai.

A bigger heatsink base always lowers junction temperature

False — base thickness

d

badhane se through-slab drop

Δ T = P d / (k \cdot A)

badhti hai; bada base tabhi help kar sakta hai jab extra material heat ko sideways zyada fins tak spread kare (heat-spreading), aur useful spreading distance ke baad extra mass sirf delay add karta hai, steady-state cooling nahi.

Radiation can be ignored for a normal CPU cooler

True typical conditions mein — room emissivity ke paas ek metal cooler ke liye, ordinary areas aur ~100 °C se kam surface temperatures par, radiation usually total ka kuch percent hi hoti hai; yeh universal constant nahi hai, lekin normal PC conditions mein conduction plus convection essentially saari heat carry karte hain, aur radiation sirf vacuum (space) mein dominate karti hai jahan convection impossible hai.

Liquid cooling wins because water conducts heat better than metal

False — water ka

k

metal se worse hai; liquid cooling transport par jeet jaata hai, yaani uski high heat capacity chip se heat ko ek bade radiator tak le jaati hai, surface heat transfer par nahi.

Series thermal resistances add, just like series electrical resistances

True — heat junction→paste→heatsink→air ek stage ke baad doosre se flow karti hai (Figure 1, right panel), isliye drops stack hote hain:

R_{t o t a l} = R_{1} + R_{2} + R_{3}

, bilkul electrical analogue

Δ T \leftrightarrow V

aur

P \leftrightarrow I

ke saath.

More thermal paste means a better thermal connection

False — paste sirf microscopic air gaps fill karne ke liye hoti hai; thick layer

R_{t h} = d / (k \cdot A)

badhati hai kyunki paste (

k \approx 3 - 12

) us metal se kaafi worse hai jise yeh alag karti hai. Patla behetar hai.

A fan raises the heat transfer coefficient

h

but leaves the fin material's

k

untouched

True — fan air ka slow boundary layer patla karta hai (Figure 2),

h

ko ~8 se ~60 tak boost karta hai; copper/aluminium ki apni conductivity ek material property hai jise airflow change nahi kar sakti.

Doubling fan RPM adds only a few decibels of noise

False — ek fan ke liye, radiated sound power speed ke saath bahut steeply badhti hai, roughly

\propto (RPM)^{5}

model ki jaati hai. Decibels mein yeh

10 lo g_{10} (2^{5}) \approx 15 dB

add hota hai jab RPM double ho jaata hai — kuch dB nahi, aur 15 dB ka jump "twice as loud" se kaafi zyada laghta hai.

Spot the error

R_{t h} = 1/ (h \cdot A)

, so I'll cut the heatsink area in half and add a stronger fan to compensate."

Trap:

A

aadha karne se

R_{co n v}

double ho jaata hai, aur sirf break even karne ke liye

h

ko double karna padega — lekin ek fan ka

h

saturate ho jaata hai, isliye tum usually haarte ho. Area ek sasta, reliable lever hai; ise trade mat karo.

"Ambient is 25 °C and

Δ T = 90

°C, so the junction sits at 90 °C."

Error:

Δ T

ambient ke upar ek rise hai, absolute temperature nahi. Junction

= T_{ambi e n t} + Δ T = 25 + 90 = 115^{\circ} C

; ambient add karna bhool jaana classic galti hai.

"Fourier's law

Δ T = P d / (k \cdot A)

says a thicker slab moves more heat."

Error: ek fixed heat flow

P

ke liye, thickness

d

temperature drop ke numerator mein hai, isliye ek thicker slab ek bada drop banata hai (worse cooling) — equivalently, ek fixed

Δ T

ke liye yeh kam heat move karta hai. Yahi wajah hai ki thermal paste patni honi chahiye, generous nahi.

"My CPU idles cool, so the cooler is over-specified and I can remove a fan."

Error: idle power, load power ka ek fraction hai; cooling ko peak dissipation handle karni hoti hai (Thermal Design Power (TDP) figure aur usse upar), warna chip load ke neeche Thermal Throttling hit karegi.

"Higher CFM fan → always lower temperature."

Error: CFM tabhi help karta hai jab tak heatsink surface saturate na ho jaaye — jab fins heat zyada tez hand off nahi kar sakti, extra airflow sirf noise add karta hai. Us point ke baad tumhe zyada area chahiye, zyada air nahi.

"Liquid metal TIM has

k \approx 70

, so I'll smear plenty on to be safe."

Error: liquid metal electrically conductive hai — jo excess spill ho jaata hai pins ya SMD components par woh unhe short-circuit kar deta hai. Iska saara advantage ek whisper-thin film se achieve hota hai.

"Blower (centrifugal) fans are worse than axial because they move less air."

Error: blowers raw flow ko static pressure se trade karte hain, jisse unhe dense fins aur cramped laptop channels se push karne ki capability milti hai jahan ek high-CFM axial fan resistance ke against stall kar jaata.

Why questions

Why does convection depend on airflow but conduction does not?

Conduction ek fixed solid ke andar hoti hai jiska

k

material se set hota hai; convection surface par ek stagnant fluid ka boundary layer par depend karti hai (Figure 2) — moving air us layer ko patla karti hai, isliye

h

(aur is tarah cooling) ek flow property hai.

Why is the natural-convection

h

(~8 W/m²·K) so much smaller than the forced value (~60)?

Still air mein single motion surface ke dwara warm hue air ka gentle rise hota hai, isliye insulating boundary layer mota rehta hai aur heat dhire dhire us se cross karti hai; ek fan air ko surface par metres per second ki speed se drive karta hai, us layer ko thin shear karta hai, aur patla blanket = heat ka tez crossing, isliye

h

lagbhag 7–8× chadh jaata hai — yeh measured, order-of-magnitude engineering values hain, exact constants nahi.

Why do heatsinks have fins instead of one solid block?

Fins surface area

A

ko multiply karte hain jo air se exposed hoti hai (Figure 3, left), aur

R_{co n v} = 1/ (h \cdot A)

ghatता hai jaise

A

badhta hai; equal mass ka ek solid block kaafi kam air touch karta hai, isliye woh heat dhire dump karta hai bawajood same metal hold karne ke.

Why is the tiny junction-to-case resistance often the hardest to reduce?

Yeh chip package ke andar rehti hai, silicon aur lid geometry se fix hoti hai — tum ise khol nahi sakte; yeh ek floor set karti hai jo tum cross nahi kar sakte chahe tumhara external cooler kitna bhi achha ho.

Why does the parent compare

Δ T = P \cdot R_{t h}

to Ohm's law

V = I \cdot R

Kyunki maths identical hai: temperature difference "voltage" hai, heat flow

P

(watts) "current" hai, aur thermal resistance "resistance" hai — isliye circuits se saari series/parallel intuition seedhi carry over ho jaati hai.

Why does forced convection give roughly a 7–8× improvement over still air?

Kyunki

R_{t h} = 1/ (h \cdot A)

aur fan

h

ko ~8 se ~60 W/m²·K tak lift karta hai (lagbhag 7.5×) jabki

A

unchanged rehta hai — resistance exactly usi ratio se drop ho jaati hai.

Why is black anodizing on a heatsink mostly cosmetic for cooling?

Yeh radiation improve karta hai, lekin normal PC conditions mein 100 °C se neeche radiation sirf kuch percent heat removal hoti hai, isliye real benefit oxidation resistance aur looks hai — meaningful temperature drop nahi.

Edge cases

What cools an electronic device in the vacuum of space, where there is no air?

Koi fluid nahi hai isliye zero convection hai, isliye chain yeh hai: conduction heat ko board aur mounting straps ke through external radiator panels tak le jaati hai, jo phir use radiation (

T^{4}

term) ke roop mein thande space mein dump karte hain. Spacecraft heat pipes, loop heat pipes aur cold plates add karte hain taaki heat ko parts se un panels tak conduct karein — koi bhi cheez heat nahi hataati jab tak ki ek conducting path radiator tak ya radiator ki apni emission na ho.

At steady state with a perfectly matched cooler, what happens to the temperature over time?

Yeh badhna band ho jaati hai aur constant rehti hai: heat in equals heat out, isliye

Δ T

P \cdot R_{t h}

par settle ho jaata hai — "bucket" tabhi bhar ta hai jab tak hole utni tezi se paani drain kare jitni tezi se aata hai.

If the fan fails (h drops to the natural-convection value) on a chip designed for forced cooling, what happens?

R_{t h}

~7–8× jump karta hai, isliye

Δ T

same factor se jump karta hai; chip apni thermal limit ki taraf race karta hai aur ya toh hard throttle karta hai (dekho Thermal Throttling) ya khud ko protect karne ke liye shut down ho jaata hai.

What is the thermal effect of a zero-thickness (ideal) thermal paste layer?

Limit

d \to 0

mein, paste ki apni resistance

d / (k \cdot A) \to 0

— lekin air gaps fill karne ke liye tumhe abhi bhi kuch paste chahiye, isliye real optimum hai "jitna patla ho sake while still filling", literally zero nahi.

What happens to

R_{t h} = 1/ (h \cdot A)

as heatsink area grows without bound?

Yeh zero approach karta hai, isliye theory mein cooling perfect ho jaati hai — lekin practice mein fins crowd ho jaate hain, unke beech airflow stall ho jaata hai, aur effective

h

fall ho jaata hai, isliye infinite area diminishing, infinite nahi, returns deta hai.

Why does a thick heatsink base still help sometimes, despite the thickness penalty?

Kyunki ek base pure 1-D slab nahi hai: chhote chip footprint ke neeche enter hoti heat base ke through sideways spread hoti hai taaki chip se door fins tak pahunche. Ek thicker base is spreading resistance ko lower karta hai aur outer fins ko behtar feed karta hai — benefit tabhi tak rehta hai jab tak spreading gain through-thickness drop

P d / (k \cdot A)

se outweigh na ho.

If two chips share one heatsink, how does the total heat load change the base temperature?

Powers shared resistance se pehle add hoti hain:

Δ T = (P_{1} + P_{2}) \cdot R_{t h}

, isliye base akele kisi bhi chip se zyada hot run karta hai — yeh ek real trap hai jab ek CPU aur VRM ko ek cooler ke neeche combine kiya jaata hai.

Recall Quick self-test

The single knob a fan changes ::: Heat transfer coefficient $h$ (kabhi bhi material ka $k$ nahi). $Δ T = 90$ °C rise, ambient 25 °C → junction ::: $115^{\circ} C$ (rise plus ambient). Why thin thermal paste beats thick ::: $R_{t h} = d / (k \cdot A)$ thickness $d$ ke saath badhta hai; paste ek poor conductor hai, isliye iska kam hona behtar hai jab gaps fill ho jaayein. Doubling fan RPM adds how much noise (sound power $\propto RPM^{5}$ ) ::: Lagbhag $10 lo g_{10} (2^{5}) \approx 15$ dB.