Question 6.AE.4: Determine for an activation energy of E=0.51 eV the slope E/......

Power Quality in Power Systems and Electrical Machines [1073571]

Determine for an activation energy of E=0.51 eV the slope E/K and for the additional temperature rise $△T_h$ =3=6.6 °C (see Application Example 6.2 with $T_{amb}$ =40°, $T_2=T_{rat}$ =100 °C) the reduced lifetime of a single-phase induction motor.

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The slope $(E / K) \text { is } E / K=5.91 \cdot 10^3$ kelvin. Thus one obtains from $t_1=40 e^{-5.91 \cdot 10^3 \Delta T_h / 373\left(373+\Delta T_h\right)}$ ; with $△T_h$ =6.6 °C, a reduced lifetime of $\left.t_1\right|_{\Delta T_{h-3}=6.6^{\circ} \mathrm{C}}=30.37 \text { years. }$
From Application Examples 6.3 and 6.4 one concludes that lower values of the activation energy E result in lower decreases in lifetime.

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