Transformer heating and cooling time constants (25°C/37.5°C)

Given

• Temperature rise from cold: $\theta_1 = 25$ °C at $t = 1$ h, $\theta_2 = 37.5$ °C at $t = 2$ h
• On disconnection, the temperature falls from its final steady value to 40 °C in 2.5 h; ambient = 30 °C.

Heating: final rise and heating time constant

From cold, $\theta = \theta_m(1 - e^{-t/\tau_h})$. Let $a = e^{-1/\tau_h}$:

$$\begin{aligned} 25 &= \theta_m(1 - a) \\ 37.5 &= \theta_m(1 - a^2) = \theta_m(1-a)(1+a) \end{aligned}$$

Dividing:

$$\begin{aligned} 1 + a &= \frac{37.5}{25} = 1.5 \;\Rightarrow\; a = 0.5 \\ \theta_m &= \frac{25}{1 - 0.5} = 50\ ^\circ\text{C} \\ \tau_h &= \frac{1}{\ln(1/a)} = \frac{1}{\ln 2} = 1.443\ \text{h} \approx 86.6\ \text{min} \end{aligned}$$

Cooling time constant

Final steady temperature = 30 + 50 = 80 °C (rise 50 °C). After 2.5 h the temperature is 40 °C, i.e. a rise of 40 − 30 = 10 °C above ambient.

$$\begin{aligned} \theta &= \theta_i\, e^{-t/\tau_c} \\ 10 &= 50\, e^{-2.5/\tau_c} \\ \tau_c &= \frac{2.5}{\ln(50/10)} = \frac{2.5}{\ln 5} = \frac{2.5}{1.6094} = 1.553\ \text{h} \approx 93.2\ \text{min} \end{aligned}$$

Answer: final steady temperature rise = 50 °C; heating time constant = 1.443 h (86.6 min); cooling time constant = 1.553 h (93.2 min).

The cooling time constant is larger than the heating one, as expected, because heat dissipation is poorer when the transformer is not loaded (weaker oil circulation).