The Thermodynamics Inside HVAC and Climate-Control Systems
Walk into any large building on a humid afternoon and the air conditioning is doing two jobs at once, not one: it is cooling the air, and it is also drying it. A student who has only ever solved q = mcΔT problems will correctly work out the cooling part and completely miss the drying part — and in a humid climate the drying part can be the larger of the two loads. This article works through both calculations properly, shows why they have to be added rather than compared, and explains where the extra energy for the second one actually goes.
The formulas
Latent heat: Qlatent = ṁ · Δw · L
Total cooling load: Qtotal = Qsensible + Qlatent
What each term means
| Term | Meaning | Unit |
|---|---|---|
| ṁ | Mass flow rate of the air being conditioned | kg/s |
| cp | Specific heat capacity of (dry) air at constant pressure | ≈ 1.005 kJ kg⁻¹ K⁻¹ |
| ΔT | Temperature drop the air undergoes across the cooling coil — this alone is the "sensible" load, the part you can feel as a temperature change | K (or °C) |
| Δw | Drop in humidity ratio — mass of water vapour removed per kilogram of dry air | kg water / kg dry air |
| L | Latent heat of vaporisation of water — temperature-dependent, roughly 2 260 kJ/kg near 100 °C and closer to 2 450–2 500 kJ/kg near typical room and coil temperatures | kJ/kg |
Worked example 1 — the sensible load
An air-handling unit processes 2.00 kg/s of air, cooling it from 30 °C to 20 °C.
ΔT = 30 − 20 = 10 K (a temperature difference is the same in °C and K)
Qsensible = ṁ · cp · ΔT = 2.00 × 1.005 × 10
Qsensible = 20.1 kW
Worked example 2 — the latent load, and why it matters
The same air stream also has its humidity ratio reduced by 0.0030 kg water per kg dry air as it passes over the cold coil and moisture condenses out. Using L ≈ 2 450 kJ/kg for condensation near the coil's operating temperature:
Mass of water condensed per second = ṁ × Δw = 2.00 × 0.0030 = 0.0060 kg/s
Qlatent = 0.0060 × 2 450
Qlatent = 14.7 kW
Total cooling capacity actually required:
Qtotal = Qsensible + Qlatent = 20.1 + 14.7 = 34.8 kW
The latent portion here is 14.7 kW out of a 34.8 kW total — over 40% of the real cooling capacity the system needs to deliver, and it produces no temperature drop by itself. A design that only calculated the sensible 20.1 kW would leave the system badly undersized the moment humidity is added to the picture. This is exactly why sensible and latent heat are always added, never compared as if one could substitute for the other: cooling and drying are two physically distinct processes happening to the same air stream, one a temperature change and the other a phase change, and both consume real energy.
Where this is actually used
In a vapour-compression air-conditioning system, the cold evaporator coil does two things simultaneously by its very nature: air blown across a coil colder than the air's dew point both loses sensible heat (its temperature falls) and, once the coil surface drops below the dew point, has water vapour condense out of it directly onto the coil (the latent load). The ratio of sensible load to total load is called the sensible heat ratio, and a system's coil and fan are sized against the total load, not the sensible load alone — undersizing for the latent component is a recognised cause of a building that feels cool enough on a thermometer but still feels clammy, because the air was cooled without being adequately dried. The same two quantities, sensible and latent, are why HVAC engineers work with a psychrometric chart rather than a plain temperature scale: humidity ratio has to be tracked as its own variable alongside temperature at every stage of the process.
Common mistakes that cost marks
- Computing only q = mcΔT and calling it the cooling load. That number is the sensible load only. Whenever moisture is being removed, the latent load must be added separately — it is not folded into cp in any way.
- Using the specific heat of dry air when the air stream is humid. Moist air's effective specific heat is slightly higher than dry air's because water vapour itself has a higher specific heat; for careful work, humid specific heat (cp of dry air plus the vapour's contribution scaled by humidity ratio) should be used rather than the bare dry-air value.
- Treating L as a fixed constant regardless of temperature. Latent heat of vaporisation genuinely falls as temperature rises — using the familiar 2 260 kJ/kg (valid near 100 °C, the value most students memorise from Class 11–12 thermodynamics) for a process happening near room temperature or coil temperature overstates the true value by a meaningful margin; state which temperature the value applies to.
- Sign or direction errors on Δw. Cooling and dehumidifying removes moisture, so Δw is a drop in humidity ratio; heating and humidifying would be the reverse process, adding latent load in the other direction.
Exam relevance
| Exam | Typical use |
|---|---|
| IIT-JAM / CUET-PG Physical Chemistry | Sensible vs latent heat, energy balances involving phase change |
| GATE Chemistry / allied engineering | Applied thermodynamics, heat and mass balance calculations |
| CSIR-NET Physical Chemistry | First-law energy balances, latent heat of vaporisation problems |
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