New kVA rating with fan cooling (400 kVA)

Given

• Rating $S$ = 400 kVA; maximum efficiency at 80 % of full load
• Heat run at full load: $\theta_1 = 24$ °C after 1 h, $\theta_2 = 34$ °C after 2 h
• Fan cooling raises heat dissipation per unit area per °C ($\lambda$) by 15 %

Step 1: time constant and final temperature rise

With $a = e^{-1/\tau}$, $\theta_1 = \theta_m(1-a)$ and $\theta_2 = \theta_m(1-a^2)$:

$$\begin{aligned} 1 + a &= \frac{34}{24} = 1.4167 \;\Rightarrow\; a = 0.4167 \\ \tau &= \frac{1}{\ln(1/0.4167)} = \frac{1}{\ln 2.4} = 1.142\ \text{h} \approx 68.5\ \text{min} \\ \theta_m &= \frac{24}{1 - 0.4167} = 41.14\ ^\circ\text{C} \end{aligned}$$

Step 2: losses in terms of full-load copper loss $P_c$

At maximum efficiency, copper loss = iron loss:

$$\begin{aligned} P_i &= (0.80)^2 P_c = 0.6400\,P_c \\ \text{Full-load losses} &= P_i + P_c = 1.6400\,P_c \end{aligned}$$

The final rise is $\theta_m = \dfrac{\text{losses}}{S\lambda}$, so the losses allowed $\propto \lambda\,\theta_m$. At a new load of $x$ times full load, the losses are $P_i + x^2 P_c$ (iron loss is constant).

(i) Same final temperature rise ($41.14$ °C)

Losses allowed rise by 15 %:

$$\begin{aligned} 0.6400 + x^2 &= 1.15 \times 1.6400 = 1.8860 \\ x^2 &= 1.2460 \;\Rightarrow\; x = 1.1162 \\ \text{New rating} &= 1.1162 \times 400 = 446.5\ \text{kVA} \end{aligned}$$

(ii) Allowable temperature rise 50 °C

Losses allowed are now in the ratio $1.15 \times \dfrac{50}{41.14}$:

$$\begin{aligned} 0.6400 + x^2 &= 1.15 \times \frac{50}{41.14} \times 1.6400 = 2.2920 \\ x^2 &= 1.6520 \;\Rightarrow\; x = 1.2853 \\ \text{New rating} &= 1.2853 \times 400 = 514.1\ \text{kVA} \end{aligned}$$

Answer: time constant ≈ 1.14 h (68.5 min); final rise ≈ 41.1 °C; new rating (i) ≈ 446 kVA, (ii) ≈ 514 kVA.