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Equation 8 · Part 1 · The Physical Plant: Power, Cooling, and Networks in an AI Datacenter

Symbol ρ

ρcp≈4180 kJ/(m3⋅K)\rho c_p \approx 4180\ \mathrm{kJ/(m^3{\cdot}K)}
ρ\rho

What this part means

ρ is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

Its job in the formula

ρ is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

The passage around this formula

The following is my own worked calculation, using standard property values rather than any cited source. At around 27 °C and atmospheric pressure, air has ρ\rho cpc_p ≈\approx 1.17\ kJ/(m3⋅K)\mathrm{kJ/(m^3{\cdot}K)} ; liquid water has ρ\rho cpc_p ≈\approx 4180\ kJ/(m3⋅K)\mathrm{kJ/(m^3{\cdot}K)} . The ratio is roughly 3,600 to one. Rejecting 100 kW at a 15 K rise therefore requires about 5.7\ m3/s\mathrm{m^3/s} of air — near 12,000 cubic feet per minute, through a single rack aperture — or about 1.6\ L/s\mathrm{L/s} of water, which is a garden hose. There is no engineering cleverness that closes a factor of 3,600; the only levers are a larger Δ\Delta T or a larger duct, and both run out.

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Sources cited in the article section

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