Follow the Heat

Read the explanation

The chip. Our example rack draws a steady 100 kW of electricity, the magazine’s example on page 18. All of it ends up as heat: energy is conserved, so the chips change its form, not its amount.

The cold plate. Cold plates sit on the hottest chips with liquid running through them and collect a share of the rack’s heat, the liquid capture fraction. With the magazine’s assumed 80 per cent, 80 kW passes into the liquid; the other 20 kW comes from parts the plates do not touch and goes into the room air (page 20). At 0 per cent the rack is cooled by air alone: all 100 kW goes into the room air. The share is an assumption, not a property of liquid cooling; real splits depend on the equipment.

The technology loop. The liquid carries its heat to the CDU and returns cooler. The flow it needs is the heat divided by (cp × the temperature rise); the loop liquid is treated as water, with rounded values cp = 4.18 kJ/(kg·K) and 1 kg per litre (page 23). For 80 kW at a 10 K rise that is 80 kW ÷ (4.18 kJ/(kg·K) × 10 K) = 1.91 kg/s, about 115 L/min; at a 5 K rise it is 3.83 kg/s, about 230 L/min. Halve the rise and the flow doubles; the heat does not change. A larger rise would need less water, but chips and other components have temperature limits, and the pumping it would save is not modelled here (page 23).

The CDU heat exchanger. Heat crosses the exchanger’s plates from the warmer technology loop to the cooler facility water. The two liquids stay separate: heat flows from warmer to cooler, through heat exchange rather than refrigeration. This model leaves out the CDU pump’s electricity.

The room air. The 20 kW in the room air has not gone anywhere. The air keeps carrying it until an air handler moves it into facility water, or another cooling system takes it. Residual air heat is part of the job, not a rounding error. In this model all 100 kW reaches heat rejection.

Heat rejection. A dry cooler uses fans to push the heat into outdoor air and consumes no water for cooling at the site, but it only works when the outdoor air is cooler than the water. A chiller with a dry cooler lets the heat leave even when the outdoor air is warmer than the water, but adds compressor work, W = Q ÷ COP: at the magazine’s illustrative COP of 5, cooling 100 kW takes 20 kW of work, so 120 kW leaves the condenser. An evaporative cooling tower rejects heat mainly by evaporating part of the circulating water: if all 100 kW left that way, up to about 150 L per hour would evaporate (100 kJ/s × 3,600 s/h ÷ 2,400 kJ/kg, with 1 kg per litre), about 1.5 m³ an hour for every MW. Real towers also shed some heat without evaporating, which means less evaporation, and lose extra water to blowdown and drift, which means more (page 58).

Outside. Everything ends up in the atmosphere: the 100 kW of IT heat, plus 20 kW of compressor work if a chiller is used. Pumps, fans and electrical losses are left out; each would add heat of its own.

What changed, and what did not. The electricity fixed the heat: 100 kW in means 100 kW of heat to move. The capture fraction chose the route, not the amount. The temperature rise changed the traffic: halve it and twice as much water has to flow for the same heat. A larger rise would need less water, but chips and other components have temperature limits, and the pumping it would save is not modelled here (page 23). Heat rejection decided the price of letting the heat go. A dry cooler consumes no water but needs outdoor air cooler than the water. A chiller lets the heat leave even when the air is warmer, but adds its own work, so more heat leaves than the chips made. A cooling tower pays in water: up to about 150 L per hour for this rack.