Q4Electrical Machine Design

Question

Q.4. A copper bar 12 mm in diameter is insulated with micanite tube which fits tightly around the bar and into the rotor slot of an induction motor. The micanite tube is 1.5 mm thick and its thermal resistivity is 8 ohm-meter. Calculate the loss that will pass from copper bar to iron if a temperature difference of 25°C is maintained between them. The length of bar is 0.2 m.

Answer 1

This is a cylindrical (radial) heat conduction problem, analogous to calculating the resistance of a cylindrical shell in electrical circuit theory, using the thermal-electrical analogy where thermal resistivity (ρ, in ohm-meter, using the electrical-analogy convention where 1/thermal-conductivity is treated as a thermal resistivity) plays the role of electrical resistivity.

Given data: copper bar diameter = 12 mm, so inner radius r1 = 6 mm = 0.006 m; micanite tube thickness = 1.5 mm, so outer radius r2 = 6+1.5 = 7.5 mm = 0.0075 m; thermal resistivity ρ = 8 Ω-m (i.e., ohm-meter in the thermal-electrical analogy, equivalent to °C·m/W); length of bar l = 0.2 m; temperature difference ΔT = 25°C.

Thermal resistance of the cylindrical shell: using the standard formula for radial conduction through a cylindrical shell (directly analogous to the resistance of a cylindrical resistor with current flowing radially):

R_{t h} = \frac{ρ}{2 π l} ln (\frac{r _{2}}{r _{1}})

R_{t h} = \frac{8}{2 π ( 0.2 )} ln (\frac{7.5}{6}) = \frac{8}{1.2566} \times ln (1.25) = 6.366 \times 0.2231 = 1.4206 Ω_{t h}

Heat loss: applying the thermal-electrical analogy (Q = ΔT/Rth, analogous to I = V/R):

Q = \frac{Δ T}{R _{t h}} = \frac{25}{1.4206} \approx 17.6 W

This heat loss of approximately 17.6 W represents the rate at which heat generated in the copper bar (by I²R loss) can be conducted radially outward through the micanite insulating tube to the surrounding iron (rotor slot wall) for the given 25°C temperature difference; this calculation is a standard step in verifying that a proposed conductor/insulation configuration in a rotor slot can adequately dissipate the expected copper loss without the insulation exceeding its permissible temperature limit, directly linking the electromagnetic design (which determines the copper loss) to the thermal design (which must safely remove that heat).

This same cylindrical-shell conduction formula applies generally to any insulated round conductor problem in machine design, such as heat flow through slot insulation surrounding a stator or rotor conductor, and is a standard analytical tool used throughout the thermal verification stage of electrical machine design to confirm that a proposed conductor/insulation arrangement can safely carry its rated current without the insulation exceeding its class-rated temperature limit.

Q4Electrical Machine Design

Question

Answer