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Published equation contexts

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

Why this formula appears here

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

Equation 8 · Datacenters & Infrastructure

The Physical Plant: Power, Cooling, and Networks in an AI Datacenter

This equation gives an approximation: it relates the quantities while allowing an approximation.

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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