Ampacity at new ambient and max temperature

Ampacity follows from the heat balance of the conductor. Neglecting solar heating, and with the resistance taken at the maximum temperature as constant:

$$I^2R = h\,A_s(\theta_c - \theta_a) \quad\Rightarrow\quad I \propto \sqrt{\theta_c - \theta_a}$$

New current-carrying capacity

Original: $\theta_c - \theta_a = 80 - 35 = 45\ °\text{C}$ at $I_1 = 400$ A.

New: $\theta_c - \theta_a = 75 - 40 = 35\ °\text{C}$.

$$\begin{aligned} I_2 &= I_1\sqrt{\frac{35}{45}} \\ &= 400 \times 0.8819 \\ &= 352.8\ \text{A} \end{aligned}$$

Conductor temperature at 320 A

The temperature rise is proportional to $I^2$. Using the new rating (35 °C rise at 352.8 A) with ambient 40 °C:

$$\begin{aligned} \theta_c - 40 &= 35\left(\frac{320}{352.77}\right)^2 = 45\left(\frac{320}{400}\right)^2 = 28.8\ °\text{C} \\ \theta_c &= 40 + 28.8 = 68.8\ °\text{C} \end{aligned}$$

Answer: ampacity = 352.8 A (about 353 A); conductor temperature at 320 A = 68.8 °C (below the 75 °C limit).

The small change of resistance with temperature and the solar heat gain are neglected.