Question 6.AE.2: Determine the temperature rise △Th of a single-phase inducti......
Determine the temperature rise △T_h of a single-phase induction motor provided V_3=0.10 pu≡10%, T_{amb}=40 °C, and T_{rated}=100 °C. Assume k_{avg}=0.85 and \ell_{a v g}=1.40.
The rated temperature rise is △T_{rated}=T_{rated} – T_{amb}=T_{rise\ rat}=60 °C. For the harmonic with order h=3 and amplitude V_3=0.10 pu≡10% one obtains the weighted-harmonic factor by replacing V_{ph} / V_{p1} by 10 as follows \sum_{h=3} \frac{1}{(3)^{0.85}}(10)^{1.40}=9.88, resulting with Fig. 6.14 in the additional temperature increase due to the 3rd harmonic (average) \Delta T_{h=3}=11 \% \text { or } \Delta T_{h=3}=6.6^{\circ} \mathrm{C}.
Transformers, induction machines, and universal machines have different loss or temperature sensitivities with respect to voltage harmonics and, therefore, the additional temperature rises are different for all five types of devices. The additional temperature rise (loss) versus weighted-harmonic factor function indicates that the most sensitive components are single-phase induction machines, whereas the least sensitive devices are transformers with resistive load and universal machines. This is so because for any harmonic terminal voltage, a single-phase machine (and to some degree an unbalanced three-phase induction machine) develops forward-and backward-rotating harmonic fields and responds like being under short-circuit conditions due to the large slip s_h of the harmonic field with respect to the rotating rotor. In transformers the resistive and inductive loads can never have zero impedance due to the nature of the frequency dependency of such loads.
