The thermal conductivity of assembled armature laminations i

The thermal conductivity of assembled armature laminations is 20 times as great along the direction of laminations as in direction across the laminations. Calculate the loss that will be conducted across the laminations in a stack 40 mm thick and 6000 mm^2 in cross section with a difference of 20 degree C. Given that a difference of 5 degree C will cause 25 W to be conducted through a cross section of 2500 mm^2 in area and 20 mm thick measured along the laminations.

Solution

Thermal Conductivity = Q/t in joules/sec or watts = kA (T2 - T1)/d

where, k = thermal conductivity of material; A = Surface are; (T2 - T1) = Temperature Difference, d = thickness of the material.

Given that difference of 5 deg. Celcius (T2 - T1) cause 25 W (Q/t) to be conducted through a cross section of 2500 mm² in Area (A) and 20 mm thick (d) measured along the lamination.

So, Thermal Conductivity measured along the lamination = 25 W = k (2500)*(5) / 20

Therefore, k = (25 * 20)/ (2500 * 5) = 0.04

Thermal Conductivity along the laminations for stack 40 mm thick (d) and 6000 mm² in cross section area (A) with a difference of 20 deg. Celcius (T2 - T1)

Q/t = (0.04 * 6000 * 20)/ (40) = 120 W

Now, given that thermal conductivity along the laminations is 20 times as great than across the laminations.

So, Thermal Conductivity across the laminations = Thermal Condutivity along the laminations / 20

= 120/20 = 6 Watts.