Design:
The motor controller will be screwed (M4) onto a heat sink which will be enclosed by a multi print 3-D printed (cracks sealed with JB weld) ASA enclosure with splash proof connectors and lid. The heat sink will need to be cut and drilled.
Motor Controller Datasheet:
https://docs.prohelion.com/Motor_Controllers/WaveSculptor22/User_Manual/Cooling.html
Heat Sink Selected:
https://www.digikey.com/en/products/detail/advanced-thermal-solutions-inc/ATS-EXL6-254-R0/5848412
Fans Selected:
Enclosure Connector:
https://amphenol-industrial.com/products/epower-lite-and-epower-lite-mini/
Calculations:
Ploss = ReqIo2 + (αIo + β)Vbus + C𝑓eqVbus2 (from datasheet)
Req= 1.0800E-2, 𝛼= 3.3450E-3, β= 1.8153E-2, C𝑓eq= 1.5625E-4
Vbus = 134.4V (battery 100% SOC), Io = 60/sqrt(2) A (rms value)
Ploss = 43.7757W (heat generated)
We can assume that all of the power loss will be converted to heat, therefore the heat generated will be considered 43.78W. For the sake of simplicity, we are also assuming all of this heat is being transferred to the heat sink attached to the motor controller. Also, the airflow will not be evenly distributed across each fin, but for the sake of simplicity we are assuming it is. (assume 36 fins)
Q = hAtotal(Tobject−Tambient) (Newton’s Law of Cooling)
Atotal = (((2 ⋅ fin height ⋅ length) ⋅ number of fins) + (length ⋅ width) (0.592m²)
Pr = 0.71 (Prandtl number for air)
Re = ρvL/μ (Reynold’s number)
v = Q / A (Air velocity caused by fans) (250.497m/s)
Q = Volumetric flow rate (1.26 m³/s), 1CFM=0.0283m3/s
A = Cross-sectional area through which the air flows (0.00503m²)
ρ is the air density (typically 1200 g/m³ at room temperature).
v is the airflow velocity (m/s), which can be estimated from the fan's specifications.
L is a characteristic length (0.250m)
μ is the dynamic viscosity of air (0.01918 g/ms)
Nu = 0.023 ⋅ Re0.8 ⋅ Pr0.33 ⋅ (L/D)0.5 (Nusselt’s number for airflow parallel to fins)
L = Length of the fin (in the direction of the flow, 0.250m)
D = Characteristic dimension (fin height, 0.027m)
Nu = hL/k
h: is the heat transfer coefficient
L: is the characteristic length (0.250m)
k: is the thermal conductivity air (0.026 W/mk)
h = (0.023 ⋅ Re0.8 ⋅ Pr0.33 ⋅ (L/D)0.5 ⋅ k)/L
Q = ((0.023 ⋅ Re0.8 ⋅ Pr0.33 ⋅ (L/D)0.5 ⋅ k)/L)Atotal(Tobject−Tambient)
Q = ((0.023 ⋅ ρvL/μ0.8 ⋅ Pr0.33 ⋅ (L/D)0.5 ⋅ k)/L)Atotal(Tobject−Tambient) (heat transferred
Q =