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Q˙=m˙ cp ΔT\dot{Q} = \dot{m}\, c_p\, \Delta T

Why this formula appears here

The physical reason liquid cooling becomes necessary at all is straightforward: air has a volumetric heat capacity roughly one four-thousandth that of water, so once a rack’s heat density crosses roughly 20–30 kW, moving enough air through the rack to hold a safe temperature rise requires impractical airflow velocities and fan power. The heat-removal budget for a cold plate loop follows directly from the sensible-heat relation: Q˙=m˙ cp ΔT\dot{Q} = \dot{m}\, c_p\, \Delta T. where Q˙\dot{Q} is the heat to be removed (watts), m˙\dot{m} is the coolant mass flow rate, cpc_p is its specific heat, and Δ\Delta T is the temperature rise the coolant is allowed across the cold plate. This is the one relationship every cooling-loop…

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Q˙=m˙ cp ΔT\dot{Q} = \dot{m}\, c_p\, \Delta T

Equation 1 · Datacenters

How AI Datacenter Power and Cooling Actually Work

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

The physical reason liquid cooling becomes necessary at all is straightforward: air has a volumetric heat capacity roughly one four-thousandth that of water, so once a rack’s heat density crosses roughly 20–30 kW, moving enough air through the rack to hold a safe temperature rise requires impractical airflow velocities and fan power. The heat-removal budget for a cold plate loop follows directly from the sensible-heat relation: Q˙=m˙ cp ΔT\dot{Q} = \dot{m}\, c_p\, \Delta T. where Q˙\dot{Q} is the heat to be removed (watts), m˙\dot{m} is the coolant mass flow rate, cpc_p is its specific heat, and Δ\Delta T is the temperature rise the coolant is allowed across the cold plate. This is the one relationship every cooling-loop…

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