DATACENTER TWIN LAB

Infrastructure Engineering · Lesson 20 of 24 · 25 minutes

Coordinate liquid cooling and thermal ride-through

Battery minutes do not create cooling minutes. Start with the manuscript’s mixed-cooling rack, then move to a separately labelled facility thermal model. The visual goal is to make both heat paths and their shared rejection boundary visible, so that the learner never mistakes a liquid-cooled label for complete removal of room heat.

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Learning objectives

Editorial derivation

Keep heat and electrical input separate

Cooling electricity is not thermal-removal capacity. For a sensible single-phase balance, heat rate = mass flow × specific heat × temperature rise, or density × volumetric flow × specific heat × rise. The illustrative 20 MW room balance requires about 1,381.98 m³/s air at density 1.2 kg/m³, specific heat 1,005 J/(kg K) and a 12 K rise. This aggregate balance does not specify a complete plant.

Illustrative assumption

Expose the heat that liquid cooling leaves behind

At a 100 kW rack and an illustrative 80% liquid capture, the liquid path removes 80 kW and the air path still handles 20 kW. With water specific heat 4,180 J/(kg K) and a 10 K rise, liquid flow is about 1.914 kg/s. Under the stated air assumptions, residual airflow is about 1.382 m³/s. Direct-to-chip, rear-door and immersion arrangements have different interfaces and auxiliary dependencies; do not merge them into one universal loop.

Illustrative assumption

Start the thermal clock

In a lumped loss-of-cooling model, temperature-rise rate is net heat input divided by effective thermal capacitance. With 150 MJ/K capacitance, 20 MW heat input, zero heat removal and a 5 K permissible rise, the interval is 37.5 s. Only thermally coupled mass on this time scale belongs in that capacitance, and local hot spots may reach limits sooner. A ten-minute UPS therefore does not establish ten minutes of thermal continuity.

Predict → run → explain

  1. Read the declared boundary and units. Predict the selected result before calculating.
  2. Run the starting case and inspect the ledger, assumptions and model limits.
  3. Apply the challenge, run again, and explain the difference. Export only the last completed inputs and results.

Liquid kW = rack kW × capture fraction; residual air kW = rack kW − liquid kW; water kg/s = liquid kW × 1,000/(water cp × water ΔK); residual air m³/s = residual air W/(density × cp × ΔK); rise K/s = net MW/capacitance MJ/K; lumped interval s = capacitance × allowed ΔK/net MW when net MW > 0.

Declared starting inputs

{
  "heatMW": 20,
  "removedMW": 0,
  "capacitanceMJK": 150,
  "deltaK": 5,
  "rackKW": 100,
  "captureFraction": 0.8,
  "waterDeltaK": 10,
  "waterCp": 4180,
  "airDensity": 1.2,
  "airCp": 1005,
  "airDeltaK": 12
}
Worked default — unrounded educational outputs
QuantityValueUnit
liquidKW80kW
residualAirKW20kW
waterKgS1.9138755980861244kg/s
residualAirM3S1.3819789939192924m³/s
temperatureRiseKS0.13333333333333333K/s
thermalLimitS37.5s or null for no positive net rise
statusfinite-lumped-limitthermal state

Challenge

How much room heat remains at the new rack boundary?

{
  "rackKW": 120,
  "captureFraction": 0.75
}

30 kW remains. This rack change does not silently change the separately declared 20 MW facility transient or its 37.5 s limit.

Worked challenge
QuantityValueUnit
liquidKW90kW
residualAirKW30kW
waterKgS2.15311004784689kg/s
residualAirM3S2.0729684908789388m³/s
temperatureRiseKS0.13333333333333333K/s
thermalLimitS37.5s or null for no positive net rise
statusfinite-lumped-limitthermal state

Two heat paths, two continuity clocks — static equivalent

  1. Reveal a rack cutaway, cold plates, residual air path and CDU.
  2. Split the 100 kW heat balance into labelled 80 kW liquid and 20 kW air.
  3. Move to the separately labelled 20 MW facility model and interrupt heat removal.
  4. Run a physical-time cursor against the 37.5 s thermal limit and a separately stated 600 s electrical interval.

Heat-path table, flow calculations and temperature-rise-versus-time table with explicit assumptions.

Control capture fraction, rack power and restore time separately. Pause/scrub the labelled transient; show instantaneous full restoration only as an added preview assumption.

Narrative steps are not physical seconds. Essential states and quantities remain available without animation.

Assessment and answer criteria

8.1 · Find water mass flow for 500 kW with an 8 K rise and specific heat 4,180 J/(kg K).

Read answer and marking criterion

14.95 kg/s.

Criterion: Convert kW to W.

Source: DCIE-R1 Appendix A, PDF pp.59-61

8.2 · Derive the lumped thermal interval and identify a non-conservative case.

Read answer and marking criterion

Integrate constant net heat input: t = C_eff ΔT / Q. Local concentration, poor mixing or overestimated coupled mass can shorten the true interval.

Criterion: State constant load, lost heat removal and spatial averaging.

Source: DCIE-R1 Appendix A, PDF pp.59-61

8.3 · Find residual room heat at 120 kW and 75% liquid capture.

Read answer and marking criterion

30 kW.

Criterion: Do not remove the residual air path from the diagram.

Source: DCIE-R1 Appendix A, PDF pp.59-61

Source and model boundaries

Data Centre Infrastructure Engineering, reference manuscript revision 1 (DCIE-R1), 11 October 2026, chapter 8, PDF pp.38–41. PDF page 1 includes the cover. manuscript-derived; not independent source verification.

Original educational adaptation. The private PDF and missing archive are not public downloads. Source statement, Editorial derivation, Illustrative assumption and Unresolved remain distinct.

Result schema: infrastructure-screen-1; content 2.0.0; model 1.0.0. Raw inputs, outputs, units, assumptions, source locators and canonical hashes. No supported in-browser Python verification protocol exists for this result family.