Full-load heat run and 50% overload time

Heating from cold follows $\theta = \theta_m\left(1-e^{-t/\tau}\right)$. Final rise is proportional to total losses.

Time constant and final steady rise at full load

Let $x = e^{-1/\tau}$ ($\tau$ in hours).

$$\begin{aligned} 20 &= \theta_m(1-x), \qquad 32 = \theta_m(1-x^2) = \theta_m(1-x)(1+x)\\ \frac{32}{20} &= 1 + x \;\Rightarrow\; x = 0.6\\ \tau &= \frac{1}{\ln(1/0.6)} = 1.958\ \text{h}\\ \theta_m &= \frac{20}{1-0.6} = 50\ ^\circ\text{C} \end{aligned}$$

Losses

Full-load copper loss $P_c = 2P_i$.

• Full load: $P_i + 2P_i = 3P_i$
• 50% overload: $P_i + (1.5)^2(2P_i) = 5.5P_i$

Final rise on overload:

$$\theta_{mo} = 50 \times \frac{5.5}{3} = 91.67\ ^\circ\text{C}$$

Time to reach 50 °C on 50% overload

$$\begin{aligned} 50 &= 91.67\left(1-e^{-t/1.958}\right)\\ e^{-t/1.958} &= 1-\frac{50}{91.67} = 0.4545\\ t &= 1.958\ln\frac{1}{0.4545} = 1.543\ \text{h} \end{aligned}$$

Answer: final steady rise at full load = 50 °C ($\tau$ = 1.958 h); on 50% overload the transformer reaches 50 °C in about 1.54 h (≈ 92.6 min).