Air Density Temperature Fan Capacity

Overview

Engineering reference data for Air Density Temperature Fan Capacity in gases and compressed air.

Key Formulas

Ideal Gas Law

$PV = nRT$

Pressure × Volume = moles × gas constant × temperature.

Boyle's Law

$P_1 V_1 = P_2 V_2$

At constant temperature.

Charles's Law

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

At constant pressure.

Variables

Key Operational Principles

• A fan is a constant volume device: at constant speed and dimensions, the volume flow rate remains unchanged regardless of temperature.
• However, air density varies with temperature (at constant pressure), so the mass flow rate changes:
  • Hotter air (lower density) → less mass transported.
  • Colder air (higher density) → more mass transported.
• Fan specifications are typically based on NTP conditions (20°C, 101.325 kPa, 1.204 kg/m³). Always check if a requirement is stated at NTP or actual operating conditions.

Practical Examples

Example 1: Required Fan Volume Capacity

A drying process requires 1 m³/s of air at NTP conditions (20°C). The fan will operate when the air is heated to 80°C. Calculate the required fan volume capacity at operating conditions.

Using the operating volume flow formula: $q_o = q_r \frac{273 + t_o}{273 + t_r}$ $q_o = (1\ \text{m}^3/\text{s}) \times \frac{273 + 80}{273 + 20} = 1.2\ \text{m}^3/\text{s}$

Conclusion: The fan must be selected for a volume flow of 1.2 m³/s (and the corresponding pressure loss).

Example 2: Fan with Hot Air

A fan delivers 10,000 m³/h of hot air at 60°C. The total pressure loss at this volume is estimated to be 500 Pa. A manufacturer's fan power consumption at these conditions is 2.5 kW.

1. Volume: The fan must handle the hot air volume directly: 10,000 m³/h.
2. Pressure: Using the pressure ratio chart (referenced in the source text), the pressure coefficient for 60°C is ~1.15. $P_{operating} = P_{system} \times 1.15 = 500\ \text{Pa} \times 1.15 = 575\ \text{Pa}$ Select a fan rated for 575 Pa.
3. Power: Using the power ratio chart, the power coefficient for 60°C is ~0.88. $Power_{adjusted} = Power_{catalog} \times 0.88 = 2.5\ \text{kW} \times 0.88 = 2.2\ \text{kW}$ The adjusted power consumption is 2.2 kW.

Example 3: Fan with Combustion Air

Transport 10,000 m³/h of standard air (20°C) at an operating temperature of 180°C. The required pressure at operating conditions is calculated using the pressure head formula (2).

Note: This example was initiated in the source text. Applying the formulas from the "Key Formulas" section will determine the necessary operating volume flow and the adjusted pressure the fan must develop.