---
title: "Ampacity at new ambient and max temperature"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:ampacity-change-new-temperatures"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/ampacity-change-new-temperatures
site: BCT Engineering
---

# Ampacity at new ambient and max temperature

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2080 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/ampacity-change-new-temperatures
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

A particular ACSR conductor has the current carrying capacity of 400 A at an ambient temperature of 35°C and maximum allowable temperature of 80°C. Compute the current carrying capacity of the same conductor if the ambient temperature and maximum allowable temperatures will be 40°C and 75°C respectively. Also compute the conductor temperature if the conductor carries the current of 320 A.

## Short answer

Using I ∝ √(θc – θa), new capacity = 400 √(35/45) = 352.8 A. For 320 A, temperature rise is 35×(320/352.8)² = 28.8 °C, so conductor temperature = 40 + 28.8 = 68.8 °C.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With solar heating neglected, ampacity is related to the temperatures by:
   A. I is proportional to (θc - θa)
   B. I is proportional to sqrt(θc - θa)
   C. I is proportional to (θc - θa)^2
   D. I is constant
   Answer: B. I^2 R = hAs(θc - θa). (Key point: Relation)

2. What is the new ampacity at 40 °C ambient and 75 °C maximum?
   A. 400 A
   B. About 352.8 A
   C. About 377 A
   D. About 320 A
   Answer: B. I2 = 400 sqrt(35/45) = 352.8 A. (Key point: New ampacity)

3. Temperature rise is proportional to:
   A. I
   B. I^2
   C. 1/I
   D. sqrt(I)
   Answer: B. Heat produced is I^2R. (Key point: Temperature rise)

4. The conductor temperature at 320 A with 40 °C ambient is about:
   A. 68.8 °C
   B. 75 °C
   C. 80 °C
   D. 59.5 °C
   Answer: A. Rise = 45(320/400)^2 = 28.8 °C; 40 + 28.8 = 68.8 °C. (Key point: Final temperature)

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