New kVA rating with fan cooling (150 kVA)

Given

• 150 kVA; temperature rise from cold: 25 °C after 1 h, 37.5 °C after 2 h
• Fan improves heat dissipation by 20 %; maximum efficiency at 80 % of full load

Step 1: time constant and final rise

With $a = e^{-1/\tau}$:

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

Step 2: losses

At maximum efficiency (80 % load), iron loss = copper loss at that load:

$$\begin{aligned} P_i &= 0.8^2 P_c = 0.64\,P_c \\ \text{Full-load losses} &= 0.64P_c + P_c = 1.64\,P_c \end{aligned}$$

Step 3: new rating for the same final rise

Since $\theta_m = \dfrac{\text{losses}}{S\lambda}$, a 20 % higher $\lambda$ allows 20 % more losses for the same $\theta_m$. At $x$ times the old full load:

$$\begin{aligned} 0.64 + x^2 &= 1.2 \times 1.64 = 1.968 \\ x^2 &= 1.328 \;\Rightarrow\; x = 1.1524 \\ \text{New rating} &= 1.1524 \times 150 = 172.9\ \text{kVA} \end{aligned}$$

Answer: heating time constant = 1.443 h (86.6 min); final steady rise = 50 °C; new rating with fan ≈ 172.9 kVA.