1Coolant & duty
ρ=1,050, k=0.4, Pr=26.2
2Channel pathBuild one channel's path as an ordered list of straight sections (each with its own width, height and length) and 45°/90°/180° bends. The flow runs through them in order; use ▲▼ to reorder. Multiple parallel channels share the total flow.
3Base conduction (optional)
Results
Base / module temperature
69.7°C
at 250 W, 40°C inlet
Thermal resistance
0.1189K/W
module → inlet fluid
Total pressure drop
7.4kPa
0.074 bar · major+bends
Area-weighted HTC
876W/m²K
UA = 8.52 W/K
Coolant outlet
40.7°C
ΔT 0.7°C · Re 2,469–2,469
Per-section detail
#w×h (mm)Dhv (m/s)Reh (W/m²K)ΔP (kPa)
15×44.441.672,4698760.523
25×44.441.672,4698760.523
35×44.441.672,4698760.523
Reference & assumptions

New to Liquid Cold Plate Design? Read the guide: Cold Plate Design: Rectangular-Channel Heat Transfer and Pressure Drop — a plain-language explainer of the standard behind this calculator.

Rectangular-channel liquid cold plate, steady single-phase Newtonian flow. Each straight section is solved on its own hydraulic diameter Dh = 2·w·h/(w+h) and per-channel velocity. The Darcy friction factor uses the rectangular-duct Poiseuille number f·Re = 96·(1 − 1.3553α + 1.9467α² − …) for laminar flow (Shah & London, α = short/long side — 56.9 at a square, 96 at parallel plates, so a channel is not the 64/Re of a round pipe) and Swamee-Jain when turbulent, interpolated across the 2300–4000 transitional band. The heat-transfer coefficient h = Nu·k/Dh uses the constant-heat-flux (H1) rectangular-duct laminar Nusselt 8.235·(1 − 2.0421α + …) (3.61 at a square) and Dittus-Boelter when turbulent. Bend losses use ΣK·(ρv²/2) with representative sharp-milled-bend K (45°≈0.3, 90°≈1.1, 180°≈2.0), which vary with bend radius. Thermal resistance from the channel wall to the inlet fluid is R_conv = 1/UA (UA = Σ h·wetted-area over every section and channel) plus R_caloric = 1/(2·ṁ·cp); optional 1-D base conduction t/(k·A) adds the module path. The channel side and top walls are counted as fully-effective heat-transfer area (no fin-efficiency derating), which over-estimates UA for tall/thin channels — treat the thermal result as a first-order estimate. Fluid density and cp reuse this site's coolant presets; ν/k/Pr reuse the Heat Exchanger transport table. Verify against CFD or a bench test before committing a design.

Validated: the rectangular-duct correlations reproduce their Shah & London anchor values exactly — laminar f·Re = 56.92 and Nusselt = 3.61 for a square channel, and 96 / 8.235 for the parallel-plate limit. A hand-worked single 5×3 mm × 100 mm water channel at 1 L/min (Dh = 3.75 mm, v = 1.11 m/s, Re ≈ 4150 turbulent) reproduces h ≈ 6270 W/m²K, a 0.66 kPa section pressure drop, a 1.44 °C coolant temperature rise and a 180° bend loss of 1.23 kPa — all matching hand calculation.

Calculation steps
1. Channel hydraulics (per section)
Dh = 2·w·h/(w+h), v = (Q/N)/(w·h), Re = v·Dh/ν
3 channel(s), Q = 6 L/min
first section: Dh = 4.44 mm, v = 1.667 m/s, Re = 2,469 (transitional)
2. Heat-transfer coefficient (Shah-London / Dittus-Boelter)
laminar Nu = 8.235·(1 − 2.0421α + …) (rect. duct, H1); turbulent Nu = 0.023·Re^0.8·Pr^0.4; h = Nu·k/Dh
k = 0.4 W/m·K, Pr = 26.2
area-weighted h = 876 W/m²K, UA = 8.52 W/K
3. Pressure drop (Darcy-Weisbach + bends)
ΔP = Σ f·(L/Dh)·(ρv²/2) + Σ K_bend·(ρv²/2)
3 straight section(s), 2 bend(s)
major 1.57 + bends 5.83 = 7.4 kPa
4. Thermal resistance & temperatures
R_conv = 1/UA, R_caloric = 1/(2·ṁ·cp); T_base = T_in + Q·(R_conv+R_caloric)
Q = 250 W, T_in = 40°C
R = 0.1189 K/W → base/module 69.7°C, coolant out 40.7°C (ΔT 0.7°C)