Page Comparison

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Q = ((0.023 ⋅ ρvL/μ​^0.8 ⋅ P_r^0.33 ⋅ (L/D​)^0.5 ⋅ k)/L)A_total​(T_object−T_ambient) (heat transferred)

Q = 27.2885 ⋅ ΔT

Q = 8.9679 ⋅ ΔT

We can conclude that with the given cooling system, the temperature difference between the surface of the motor controller and the ambient temperature will need to be >5C to transfer the heat generated at peak power usage. Basically, it only cools the heat plate if we’re not just blowing hot air over hot metal.

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