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Maximum temperature of slot conductor

Short answer

Overhang temperature: theta_o = theta_a + p A(L+2l_o)/(2 l_o S_p lambda). Slot centre adds rise p rho_t L^2/8. Thus theta_max = theta_a + J^2 rho_e A(L+2l_o)/(2 l_o S_p lambda) + J^2 rho_e rho_t L^2/8.

If the ambient temperature is θa, determine the maximum temperature of conductor placed in slot if the slot insulation is comparatively very thick.

When the slot insulation is comparatively very thick, no heat flows from the conductor into the iron. All copper loss of the slot portion flows along the conductor to the overhang, and the overhang dissipates the whole loss of the conductor to the air. The maximum temperature is at the middle of the slot length.

Notation

  • JJ = current density, ρe\rho_e = electrical resistivity, so loss per unit volume p=J2ρep = J^2\rho_e
  • AA = conductor cross-section, SpS_p = perimeter of the overhang surface (W dissipated per unit length per °C is SpλS_p\lambda)
  • LL = slot (embedded) length, lol_o = overhang length at each end, ρt\rho_t = thermal resistivity of copper
  • λ\lambda = specific heat dissipation of the overhang surface (W/m²°C), θa\theta_a = ambient temperature
  l_o            L (slot)            l_o
 ~~~~~ |=====================| ~~~~~
 theta_o      theta_max        theta_o

Step 1: overhang temperature

The total loss of one conductor, pA(L+2lo)p A (L + 2l_o), is dissipated from the two overhang surfaces, of area 2loSp2 l_o S_p (overhang temperature taken as uniform):

pA(L+2lo)=2loSpλ(θo−θa)θo=θa+pA(L+2lo)2loSpλ\begin{aligned} p A (L + 2 l_o) &= 2 l_o S_p \lambda (\theta_o - \theta_a) \\ \theta_o &= \theta_a + \frac{p A (L + 2 l_o)}{2 l_o S_p \lambda} \end{aligned}

Step 2: rise of slot portion above the overhang

At distance xx from the slot centre, the heat crossing the conductor section is pAxp A x. So

dθdx=−ρt p x  ⇒  θ=θmax−ρtpx22\frac{d\theta}{dx} = -\rho_t\,p\,x \;\Rightarrow\; \theta = \theta_{max} - \frac{\rho_t p x^2}{2}

At x=L/2x = L/2, θ=θo\theta = \theta_o:

θmax−θo=p ρt L28\theta_{max} - \theta_o = \frac{p\,\rho_t\,L^2}{8}

Step 3: maximum conductor temperature

θmax=θa+J2ρeA(L+2lo)2loSpλ+J2ρe ρt L28\theta_{max} = \theta_a + \frac{J^2\rho_e A (L + 2 l_o)}{2 l_o S_p \lambda} + \frac{J^2\rho_e\,\rho_t\,L^2}{8}

The maximum temperature is the ambient, plus the rise of the overhang (set by its cooling), plus the parabolic rise along the slot. A long core or a high current density increases it sharply. This is why large machines use ventilating ducts, thin high-conductivity insulation or direct conductor cooling.

✓ Answer checked by Claude Sonnet 5.5

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