Heat dissipation and cooling solutions: Worked examples

This page is the problem gym for the parent topic. We are going to take the two master formulas you already met and drive them through every kind of situation an exam or a real build can throw at you — including the weird edge cases where the arithmetic tries to lie to you.

Before we compute anything, let us re-anchor the only two tools we need, in plain words.

Recall The two formulas, restated from zero

Thermal resistance $R_{t h}$ is "how many degrees hotter something gets for each watt of heat you push through it." Small number = easy path for heat = good cooling. $R_{t h} = \frac{Δ T}{P} [°C per watt]$ Rearranged the way we mostly use it: $Δ T = P \cdot R_{t h}$ (heat pushed times resistance gives the temperature climb), exactly like $V = I \cdot R$ in electricity.

Series rule: when heat flows through one layer, then the next, then the next (chip → paste → heatsink → air), the resistances add up: $R_{t h, total} = R_{1} + R_{2} + \dots$ because each layer adds its own temperature step and the steps stack.

Everything below is just these two ideas, plus the convection resistance $R_{t h, co n v} = \frac{1}{h \cdot A}$ (bigger surface $A$ or faster air $h$ ⇒ smaller resistance).

The scenario matrix

Think of every heat problem as living in one cell of this grid. Our job is to land at least one worked example in each row.

#	Cell class	What makes it tricky	Example that hits it
A	Forward — given $P$ and all $R$ , find temperature	plain series-add, then $Δ T = P R$	Ex 1
B	Backward — given a temperature limit, find the $R$ budget	subtract the fixed (upstream) part	Ex 2
C	Convection design — find $h$ or area $A$ from $R = \frac{1}{h A}$	invert the product, watch units	Ex 3
D	Zero / degenerate input — $P = 0$ , or a "perfect" layer $R = 0$	formula must still make sense	Ex 4
E	Limiting behaviour — $h \to \infty$ , or $A \to \infty$	resistance approaches a floor, not zero	Ex 5
F	Real-world word problem — liquid loop, transport vs surface	pick which resistance dominates	Ex 6
G	Exam twist — units mismatch / trap (grams, kW, °C vs K)	convert first, or the answer is 1000× wrong	Ex 7
H	Comparison / ratio — "how many times better?"	ratio of two resistances, cancels area	Ex 8

Each example below is tagged with its cell letter.

The worked examples

Cell A — the plain forward calculation

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

A GPU dissipates $P = 220$ W. The path is: 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. Find the junction temperature $T_{j}$ .

Forecast: Guess before reading — will $T_{j}$ land near 60 °C, near 100 °C, or above 130 °C? Write your guess.

Step 1 — Add the series resistances. $R_{t h, total} = 0.15 + 0.10 + 0.25 = 0.50 °C/W$ Why this step? Heat goes through the three layers one after another, so each layer's temperature step stacks — series resistances add.

Step 2 — Convert power into a temperature rise. $Δ T = P \cdot R_{t h, total} = 220 \times 0.50 = 110 °C$ Why this step? $R_{t h} = Δ T / P$ , so $Δ T = P R$ . This is how far above the air the hottest silicon sits.

Step 3 — Add the ambient baseline. $T_{j} = T_{a} + Δ T = 22 + 110 = 132 °C$ Why this step? $Δ T$ is relative to the surrounding air; the actual temperature is measured from that baseline.

Verify: Units check — W × (°C/W) = °C. ✅ Sanity: 132 °C is above a typical 95–100 °C throttle limit, so this cooler is inadequate — a physically meaningful, believable result (see Thermal throttling).

Cell B — working backward to a resistance budget

Worked example Ex 2 (B): What heatsink do I need?

A CPU dissipates $P = 125$ W. Junction must stay below $T_{j, m a x} = 90$ °C. Ambient is $T_{a} = 35$ °C. The fixed upstream part (junction-to-case + paste) is $R_{up} = 0.30$ °C/W. What heatsink-to-air resistance $R_{h s}$ is allowed?

Forecast: Will the budget be tight (below 0.2 °C/W) or generous (above 0.5 °C/W)? Guess.

Step 1 — Find the whole temperature budget. $Δ T_{budget} = T_{j, m a x} - T_{a} = 90 - 35 = 55 °C$ Why? That is the maximum climb we are allowed from air to silicon.

Step 2 — Convert the budget into a total resistance ceiling. $R_{t h, total} = \frac{Δ T _{budget}}{P} = \frac{55}{125} = 0.44 °C/W$ Why? We flip $Δ T = P R$ into $R = Δ T / P$ — the largest resistance that still fits the budget.

Step 3 — Subtract the part you cannot change. $R_{h s} = R_{t h, total} - R_{up} = 0.44 - 0.30 = 0.14 °C/W$ Why? The junction and paste are fixed; only what is left over is available to the heatsink.

Verify: Plug back: total $= 0.30 + 0.14 = 0.44$ , $Δ T = 125 \times 0.44 = 55$ °C, $T_{j} = 35 + 55 = 90$ °C. ✅ Exactly the limit. A 0.14 °C/W heatsink is a serious tower/air setup — matches reality.

Cell C — designing the convecting surface

Worked example Ex 3 (C): How much fin area, given the air?

You must hit $R_{h s} = 0.14$ °C/W (from Ex 2). Forced air gives $h = 70$ W/m²·K. What effective fin surface area $A$ do you need?

Forecast: A few hundred cm²? A few thousand? Guess an order of magnitude.

Step 1 — Recall the convection resistance form. $R_{h s} = \frac{1}{h \cdot A}$ Why this tool and not Fourier's law? Fourier's law ( $Q = k A Δ T / d$ ) is for heat through a solid. Here the heat is leaving the fin into moving air, so the boundary-layer picture — Newton's law of cooling — is the right one, and its resistance is $1/ (h A)$ .

Step 2 — Solve for area. $A = \frac{1}{h \cdot R _{h s}} = \frac{1}{70 \times 0.14} = \frac{1}{9.8} \approx 0.102 m^{2}$ Why? We invert the product; the two knobs are the air ( $h$ ) and the metal you expose ( $A$ ).

Step 3 — Interpret. $0.102$ m² $= 1020$ cm². A block of fins with ~1000 cm² of surface — perfectly normal for a stacked-fin tower cooler.

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

Cell D — zero and degenerate inputs

Worked example Ex 4 (D): The idle chip and the "perfect" layer

Two sub-questions that stress-test the formulas at their edges.

(a) A chip is idling at $P = 0$ W with $R_{t h, total} = 0.5$ °C/W, ambient $25$ °C. Find $T_{j}$ .

Forecast: Does it heat up at all?

Step 1. $Δ T = P \cdot R = 0 \times 0.5 = 0$ °C. Why? No heat flowing means no temperature step, no matter how bad the cooler is. Step 2. $T_{j} = 25 + 0 = 25$ °C. The chip sits at ambient. Verify: Makes physical sense — an unpowered chip cannot be hotter than its surroundings. ✅

(b) Suppose a paste layer were perfect, $R_{2} = 0$ °C/W, with $R_{1} = 0.2$ and $R_{3} = 0.3$ . Total?

Step 1. $R_{t h, total} = 0.2 + 0 + 0.3 = 0.5$ °C/W. Why? A zero-resistance layer adds no temperature step — it drops out of the series sum. This is the degenerate limit of a very thin, very conductive interface (why liquid metal is chased). Verify: A real paste can only approach $0$ , never reach it (air gaps always leave something). The formula degrades gracefully. ✅

Cell E — limiting behaviour

Worked example Ex 5 (E): What if you had infinite airflow?

Take a cooler whose total path splits into a solid part $R_{solid} = 0.12$ °C/W (junction + paste + fin conduction) and a convection part $R_{co n v} = \frac{1}{h A}$ with $A = 0.05$ m². Push $h$ from $8$ (still air) toward $\infty$ (imaginary hurricane). What does $R_{t h, total}$ approach?

Forecast: Does the total resistance go to zero, or stop at a floor?

Step 1 — Write the total. $R_{t h, total} (h) = 0.12 + \frac{1}{h \times 0.05}$ Why? Solid path and air path are in series (heat goes through metal then into air), so they add.

Step 2 — Evaluate a few $h$ values.

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

Why this step? It shows the convection term collapsing while the solid term stubbornly stays.

Step 3 — The lesson. No matter how violent the airflow, the total cannot drop below 0.12 °C/W — the solid conduction floor. Once your fins are far past "airflow-limited," a bigger fan buys nothing; you must attack $R_{solid}$ instead (copper base, better paste, direct-contact heat pipes — see Heat pipes).

Verify: As $h \to \infty$ , $1/ (h A) \to 0$ monotonically, so $R$ decreases toward $0.12$ from above and never crosses it. ✅ (See figure.)

Figure — Heat dissipation and cooling solutions

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

Worked example Ex 6 (F): Does a bigger pump help this AIO?

A liquid loop has: chip-to-water cold plate $R_{c p} = 0.05$ °C/W, and radiator-to-air $R_{r a d} = 0.15$ °C/W. A salesman claims doubling the pump flow (which slightly lowers $R_{c p}$ to $0.04$ ) is worth it. The chip dissipates $200$ W, ambient $28$ °C. Compare $T_{j}$ before and after.

Forecast: Will doubling the pump noticeably drop $T_{j}$ ?

Step 1 — Total resistance, before. $R = 0.05 + 0.15 = 0.20 °C/W, T_{j} = 28 + 200 (0.20) = 68 °C$ Why? Cold-plate and radiator are in series (heat rides the water from one to the other).

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

Step 3 — Interpret which term rules. The change is only $2$ °C, because the radiator's air convection ( $0.15$ ) dominates, not the water contact ( $0.05$ ). Liquid cooling's real gift is transporting heat to a big radiator far from the chip — the final hand-off to air is still ordinary convection. Money is better spent on more radiator area/fans than on a stronger pump.

Verify: $68 - 66 = 2$ °C, and the fixed $0.15$ radiator term is $\frac{0.15}{0.19} \approx 79%$ of the total — confirming it dominates. ✅

Cell G — the exam-trap: units

Worked example Ex 7 (G): The kilowatt / gram trap

A power module dissipates $P = 1.5$ kW through a cold plate. Water enters at $30$ °C and you are told the coolant absorbs the heat with a temperature rise you must find, given mass flow $\overset{m}{˙} = 0.05$ kg/s and water's specific heat $c_{p} = 4.18$ kJ/kg·K. Find the coolant's outlet temperature.

Forecast: People blurt "huge rise" because 1.5 kW sounds enormous. Guess first.

Step 1 — Convert to consistent units. $P = 1.5 kW = 1500 W$ ; $c_{p} = 4.18 kJ/kg\cdotK = 4180 J/kg\cdotK$ . Why? Mixing kW with J/kg·K gives an answer off by 1000. Convert first, always.

Step 2 — Use the flow-heating relation. $P = \overset{m}{˙} c_{p} Δ T \Rightarrow Δ T = \frac{P}{m ˙ c _{p}} = \frac{1500}{0.05 \times 4180}$ Why this tool? This is energy balance: the watts entering the water become $\overset{m}{˙} c_{p} Δ T$ of carried-away heat per second. It answers "how much does the fluid warm as it flows through?"

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

Step 4 — The lesson. Only a $7$ °C coolant rise from 1.5 kW — that is why water works: its huge $c_{p}$ (4× air) lets it swallow big power with a small temperature climb.

Verify: Units: $\frac{W}{( kg/s ) ( J/kg\cdotK )} = \frac{J/s}{J/(s\cdotK)} = K$ . ✅ If you had forgotten the kJ→J conversion you'd get $Δ T \approx 7180$ °C — instantly absurd, a built-in sanity alarm.

Cell H — ratio / comparison

Worked example Ex 8 (H): How many times better is the fan?

The same heatsink (same area $A$ ) runs first in still air $h_{1} = 8$ W/m²·K, then with a fan $h_{2} = 64$ W/m²·K. By what factor does thermal resistance improve?

Forecast: Guess the factor before doing algebra.

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

Step 2 — Take the ratio; watch $A$ cancel. $\frac{R _{1}}{R _{2}} = \frac{1/ ( h _{1} A )}{1/ ( h _{2} A )} = \frac{h _{2}}{h _{1}} = \frac{64}{8} = 8$ Why this step? Because the same heatsink is used, $A$ is identical and divides out — the improvement is purely the airflow ratio. No area number was even needed.

Step 3 — Interpret. An 8× drop in convection resistance from one fan. This is the single biggest lever in air cooling and the reason Case airflow and positive/negative pressure matters so much.

Verify: Ratio of resistances $= h_{2} / h_{1} = 8$ exactly; dimensionless, as a ratio must be. ✅

Recall Quick self-test

A 100 W chip, total path 0.4 °C/W, ambient 30 °C — junction temp? ::: $30 + 100 \times 0.4 = 70$ °C. Fan raises $h$ from 10 to 90 on the same heatsink — resistance improves by what factor? ::: $90/10 = 9 \times$ . Why can't a stronger pump cool much better in Ex 6? ::: The radiator-to-air convection dominates; water transport was never the bottleneck. A perfect (zero-resistance) TIM layer changes the series sum how? ::: Not at all — a zero term drops out.