The cable tray size is calculated using the total cable cross-sectional area divided by the allowable fill ratio, with tray width and depth derived from the required cross-sectional area.

Basic Formula

1. Calculate the area of a single cable: $$A_{\text{cable}}=\pi \times {\left(\frac{D}{2}\right)}^2$$ where $D$ is the cable outer diameter.
2. Calculate the total cable area: $$A_{\text{total}}=A_{\text{cable}}\times N$$ where $N$ is the number of cables.
3. Determine the required tray cross-sectional area: $$A_{\text{tray}}=\frac{A_{\text{total}}}{\text{Fill Ratio}}$$ The fill ratio is typically 40% for power cables and 50% for control or signal cables according to NEC and IEC standards, allowing for proper ventilation and heat dissipation.
4. Calculate tray dimensions: If the tray depth (height) is known, the required width can be calculated as: $$W_{\text{tray}}=\frac{A_{\text{tray}}}{H_{\text{tray}}}$$ Conversely, if the width is fixed, the required depth is: $$H_{\text{tray}}=\frac{A_{\text{tray}}}{W_{\text{tray}}}$$

Practical Considerations

• Void factor: Include 20–30% extra space for air gaps between cables and future expansion .
• Cable type: Power cables require more space for heat dissipation, while data or control cables may allow slightly higher fill ratios .
• Tray type and layout: Ladder, ventilated, or solid-bottom trays may affect airflow and allowable fill .
• Environmental factors: Temperature, humidity, and chemical exposure can influence tray sizing and material selection .
• Future expansion: Always plan for additional cables to avoid costly reinstallation .

Example Calculation

Suppose you have 20 cables, each 20 mm in diameter, and you want a tray with a 50% fill ratio:

1. Single cable area: $$A_{\text{cable}}=\pi \times (20/2{)}^2=314.16\text{ mm²}$$
2. Total cable area: $$A_{\text{total}}=314.16\times 20=6283.2\text{ mm²}$$
3. Required tray area: $$A_{\text{tray}}=6283.2/0.5=12566.4\text{ mm²}$$
4. If tray depth is 50 mm, required width: $$W_{\text{tray}}=12566.4/50\approx 251.3\text{mm}$$ Choose the nearest standard width, e.g., 300 mm . This method ensures compliance with standards, proper heat dissipation, and room for future cable additions.