Heating constant and 1-hour rating (40 kVA)

Temperature rise above ambient (35 °C):

• after 1.5 h: $\theta_1 = 59 - 35 = 24\ ^\circ$C
• after 3 h: $\theta_2 = 71 - 35 = 36\ ^\circ$C

Heating from cold: $\theta = \theta_m(1-e^{-t/\tau})$.

i) Heating time constant

Let $x = e^{-1.5/\tau}$. Since 3 h = 2 × 1.5 h:

$$\begin{aligned} \frac{\theta_2}{\theta_1} &= \frac{1-x^2}{1-x} = 1 + x = \frac{36}{24} = 1.5 \;\Rightarrow\; x = 0.5\\ \tau &= \frac{1.5}{\ln 2} = 2.164\ \text{h} \end{aligned}$$

ii) Final steady temperature rise

$$\theta_m = \frac{24}{1-0.5} = 48\ ^\circ\text{C}$$

(Final oil temperature = 35 + 48 = 83 °C.)

iii) One-hour rating

The 1-hour rating is the load that, starting cold, brings the rise to 48 °C in exactly 1 hour. Let its final rise be $\theta_m'$:

$$\begin{aligned} 48 &= \theta_m'\left(1-e^{-1/2.164}\right) = \theta_m'(1-0.630) = 0.370\,\theta_m'\\ \theta_m' &= 129.72\ ^\circ\text{C} \end{aligned}$$

Final rise ∝ losses. With $P_c = 2P_i$ at full load, losses at load fraction $k$ are $P_i(1+2k^2)$, and at full load $3P_i$:

$$\begin{aligned} \frac{1+2k^2}{3} &= \frac{129.72}{48} = 2.702\\ k^2 &= \frac{3(2.702)-1}{2} = 3.554 \;\Rightarrow\; k = 1.885 \end{aligned}$$

One-hour rating $= 1.885 \times 40 = 75.4$ kVA.

Answer: $\tau$ = 2.164 h, final steady rise = 48 °C, 1-hour rating ≈ 75.4 kVA (1.885 × full load).