Quarter-load heat run and 30% overload time

Given

• Heat run at quarter load: rise 30 °C after 2 h, 40 °C after 3 h (from cold)
• Maximum efficiency at 70 % of full load
• Find the final steady rise at full load, and the time on 30 % overload (1.3 × full load) to reach that rise

Step 1: time constant and final rise at quarter load

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

$$\begin{aligned} 30 &= \theta_q(1 - a^2), \qquad 40 = \theta_q(1 - a^3) \\ \frac{1 - a^3}{1 - a^2} &= \frac{1 + a + a^2}{1 + a} = \frac{40}{30} \\ 3a^2 - a - 1 &= 0 \;\Rightarrow\; a = \frac{1 + \sqrt{13}}{6} = 0.7676 \\ \tau &= \frac{1}{\ln(1/0.7676)} = 3.781\ \text{h} \\ \theta_q &= \frac{30}{1 - 0.7676^2} = 73.03\ ^\circ\text{C} \end{aligned}$$

Step 2: losses

Maximum efficiency at 70 % load gives $P_i = 0.7^2 P_c = 0.49\,P_c$ ($P_c$ = full-load copper loss).

Step 3: final rise at full load

The final rise is proportional to losses (same cooling):

$$\theta_{FL} = 73.03 \times \frac{1.49}{0.5525} = 196.9\ ^\circ\text{C}$$

Step 4: time on 30 % overload

Final rise on overload: $\theta_{OL} = 73.03 \times \dfrac{2.18}{0.5525} = 288.1$ °C. Time (from cold, same $\tau$) to reach 196.9 °C:

$$\begin{aligned} 196.9 &= 288.1\left(1 - e^{-t/3.781}\right) \\ e^{-t/3.781} &= 1 - 0.6835 = 0.3165 \\ t &= 3.781 \times \ln\frac{1}{0.3165} = 4.35\ \text{h} \approx 261\ \text{min} \end{aligned}$$

Answer: heating time constant = 3.78 h; final steady rise at full load ≈ 197 °C; on 30 % overload the same rise is reached in ≈ 4.35 h (about 4 h 21 min).

(The data give an unusually high full-load rise. This follows directly from the figures given; the method is what matters.)