Table of Contents
- Specific heat of air, cp: about 1.006 kJ/kg.K (or 1006 J/kg.K). Density of air, rho: about 1.2 kg/m3 at 20 degC and 1 atm. Together they give a volumetric heat capacity of roughly 1200 J/m3.K - the energy stored in each cubic metre of air per degree of temperature.
- For any recovery duty the sensible power is: Q = rho x volumetric_flow x cp x temperature_difference Example: 1 m3/s of air cooled by 10 K recovers about 1.2 x 1 x 1006 x 10 ≈ 12,100 W, or 12.1 kW. Multiply by operating hours and effectiveness to get daily energy, the input to any payback calculation.
- Density falls as air heats and rises with pressure, so high-temperature or high-altitude duties shift the numbers. The epsilon-NTU method folds these into the capacity rates, but the underlying cp and rho are where every estimate starts.
- Do I use cp of dry or moist air? Use moist-air cp (~1.01) for ventilation work; the difference is small but real at high humidity. Why is volumetric capacity useful? Because fans and ducts are sized by volume flow, not mass flow, so Q per m3/s per K is the quick design figure. Need help specifying the right air-to-air heat recovery for your project? Contact the QIYU engineering team - WhatsApp +86 157 5335 5505, or email kuns913@gmail.com. Frequently Asked Questions What specific heat of air do recovery calculations use? About 1.006 kJ per kg per K for dry air at room conditions; with moisture present you use the humid-air value, but 1.006 is the standard sensible-heat constant. Why does air density matter in recovery? Heat carried equals density times specific heat times flow times temperature difference; density converts volumetric flow to mass flow, and mass flow is what actually carries the energy across the core.
- Specific heat of air, cp: about 1.006 kJ/kg.K (or 1006 J/kg.K).
- Density of air, rho: about 1.2 kg/m3 at 20 degC and 1 atm.
Together they give a volumetric heat capacity of roughly 1200 J/m3.K - the energy stored in each cubic metre of air per degree of temperature.
For any recovery duty the sensible power is:
Q = rho x volumetric_flow x cp x temperature_difference
Example: 1 m3/s of air cooled by 10 K recovers about 1.2 x 1 x 1006 x 10 ≈ 12,100 W, or 12.1 kW. Multiply by operating hours and effectiveness to get daily energy, the input to any payback calculation.
Density falls as air heats and rises with pressure, so high-temperature or high-altitude duties shift the numbers. The epsilon-NTU method folds these into the capacity rates, but the underlying cp and rho are where every estimate starts.
Do I use cp of dry or moist air? Use moist-air cp (~1.01) for ventilation work; the difference is small but real at high humidity.
Why is volumetric capacity useful? Because fans and ducts are sized by volume flow, not mass flow, so Q per m3/s per K is the quick design figure.
Need help specifying the right air-to-air heat recovery for your project? Contact the QIYU engineering team - WhatsApp +86 157 5335 5505, or email kuns913@gmail.com.
Frequently Asked Questions
What specific heat of air do recovery calculations use?
About 1.006 kJ per kg per K for dry air at room conditions; with moisture present you use the humid-air value, but 1.006 is the standard sensible-heat constant.
Why does air density matter in recovery?
Heat carried equals density times specific heat times flow times temperature difference; density converts volumetric flow to mass flow, and mass flow is what actually carries the energy across the core.