Chilled Water System Planner
Carry a cooling load through water flow, pipe losses, pump head and an explicit chiller COP scenario. Results are preliminary and use visible fluid-property assumptions.
Engineering example
For a 100 kW load and 5 K water ΔT, the default fluid properties yield about 4.78 L/s. A 100 mm, 50 m branch with K=8 is checked for velocity, pipe loss and pump head. At COP 5.5, the simplified plant input scenario is about 18.2 kW.
Use Q = ρ·cp·V·ΔT. Confirm mean-temperature fluid properties, fitting schedule, pump curve, chiller performance map, redundancy and controls before design selection.
Formulas, source and scope
Water flow V̇ = Q / (ρ·cp·ΔT) with cp = 4180 J/(kg·K). Pipe loss is delegated to the pipe engine and pump head/power to the pump engine. Plant input = Q / COP.
Formula source: site methodology / engineering assumption in this repository. No standard edition is pinned for this tool.
FAQ
What does the 100 kW example produce?
At ΔT = 5 K and ρ = 998 kg/m³ the engine flow is about 4.78 L/s. At COP 5.5 the plant input is about 18.2 kW. Pipe and pump figures follow the other two engines.
Are the fluid defaults a standard table?
No. Density 998 kg/m³ and viscosity 1.002 mPa·s are preliminary defaults in the engine. Verify them at the design mean temperature.
Estimate load → · Review pipe losses → · Review pump duty → · Operating-cost scenario → · Data-center cooling chain →