Engineering task and calculation objective
The MIFL module calculates the adiabatic mixing of two humid air streams – a fundamental task in HVAC and drying technology. From the mass flow, temperature, total pressure and relative humidity of both streams, the module determines the state of the mixed air stream: mixing temperature, humidity ratio (moisture content), relative humidity, specific enthalpy and density of the humid air.
This calculation is needed wherever air streams are combined: for recirculated-air mixing in air-handling systems, for mixing fresh and exhaust air in dryers, for blending in humidified air, or for assessing whether a mixture enters the fog region and condensate forms. The evaluation follows the classic Mollier h-x methodology: the states are described via the partial pressure and saturation pressure of the water vapour, the humidity ratio x and the specific enthalpy h; the mixing point lies on the straight line connecting the initial states in the h-x diagram.
Anyone who wants to calculate the mixing temperature of humid air or determine the relative humidity after mixing thus obtains a closed-form balance calculation without graphical reading – including the split of the mass flows into dry air, vapour and, where applicable, liquid water.
Calculation workflow
- Determine the states of the individual streams: For both air streams, the saturation pressure and the partial pressure of the water are calculated from temperature, total pressure and relative humidity; from these follow the humidity ratio x (kg of water per kg of dry air), the humidity ratio at saturation, the specific enthalpy and the density of the humid air.
- Split the mass flows: Each humid air mass flow is split into its dry-air fraction and its water fraction (vapour, possibly liquid), because the mixing balances are referred to the dry air – it is the conserved quantity of the calculation.
- Set up the mixing balances: The dry-air mass balance, the water balance and the enthalpy balance yield the humidity ratio and specific enthalpy of the mixed stream; in the h-x diagram, this corresponds to the mixing point on the straight line between the two initial states, divided in the ratio of the dry-air mass flows.
- Evaluate mixing temperature and relative humidity: From the enthalpy and humidity ratio of the mixed stream, the mixing temperature is calculated; with the partial pressure of the water and the saturation pressure at the mixing temperature, the relative humidity of the mixture follows.
- Check the fog region: If the humidity ratio of the mixture exceeds the saturation humidity ratio at the mixing temperature, liquid water precipitates (fog); the water fraction is then split into saturated vapour and liquid water.
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| Molar mass of air | MGL | kg/kmol |
| Molar mass of water | MGW | kg/kmol |
| Heat of evaporation of water | Δhv | J/kg |
| Specific gas constant air | RL | J/(kg·K) |
| Specific gas constant water | RW | J/(kg·K) |
| Temperature | ϑ1 ϑ2 | °C |
| Temperature | ϑ1 ϑ2 | °C |
| Temperature of mixture | ϑM | °C |
| Total pressure | PGes1 PGes2 | Pa |
| Total pressure | PGes1 PGes2 | Pa |
| Total pressure of mixture | PGesM | Pa |
| Relative humidity | φ1 φ2 | % |
| Relative humidity | φ1 φ2 | % |
| Relative humidity | φM | % |
| Partial pressure water | P1 P2 | Pa |
| Partial pressure water | P1 P2 | Pa |
| Partial pressure of water | PM | Pa |
| Saturation pressure water | Ps1 Ps2 | Pa |
| Saturation pressure water | Ps1 Ps2 | Pa |
| Saturation pressure of water | PsM | Pa |
| Humidity level 1 | x1 x2 | kg/kg |
| Humidity level 1 | x1 x2 | kg/kg |
| Humidity level | xM | kg/kg |
| Humidity level at saturation | xs1 xs2 | kg/kg |
Worked example
In a ventilation system, 1.0 kg/s of humid air at 20 °C and 40 % relative humidity is mixed adiabatically with 0.5 kg/s of humid air at 40 °C and 80 % relative humidity. The total pressure is 1.0 bar. Find the humidity ratio, temperature and relative humidity of the mixture – a worked example of mixing two humid air streams.
Given values
| Mass flow humid air 1 | 1.0 kg/s |
| Temperature / rel. humidity stream 1 | 20 °C / 40 % |
| Mass flow humid air 2 | 0.5 kg/s |
| Temperature / rel. humidity stream 2 | 40 °C / 80 % |
| Total pressure p | 1.0 bar (100,000 Pa) |
Solution
Humidity ratios of the individual streams
Saturation pressures from the Magnus formula: ps(20 °C) ≈ 2,333 Pa, ps(40 °C) ≈ 7,367 Pa.
Partial pressures: pD1 = 0.40 · 2,333 = 933 Pa; pD2 = 0.80 · 7,367 = 5,894 Pa.
x = 0.622 · pD/(p − pD):
x1 = 0.622 · 933/99,067 = 0.00586 kg/kg; x2 = 0.622 · 5,894/94,106 = 0.0390 kg/kg.
Dry-air mass flows and enthalpies
ṁL1 = 1.0/(1 + x1) = 0.9942 kg/s; ṁL2 = 0.5/(1 + x2) = 0.4813 kg/s.
h = 1.005·t + x·(2,501 + 1.86·t) kJ/kg dry air:
h1 = 1.005 · 20 + 0.00586 · (2,501 + 37.2) = 35.0 kJ/kg
h2 = 1.005 · 40 + 0.0390 · (2,501 + 74.4) = 140.5 kJ/kg
Mixing balances
Water balance: xM = (ṁL1·x1 + ṁL2·x2)/(ṁL1 + ṁL2) = (0.9942 · 0.00586 + 0.4813 · 0.0390)/1.4754 = 0.0167 kg/kg.
Enthalpy balance: hM = (0.9942 · 35.0 + 0.4813 · 140.5)/1.4754 = 69.4 kJ/kg.
Mixing temperature and relative humidity
From hM = 1.005·tM + xM·(2,501 + 1.86·tM) it follows that:
tM = (69.4 − 0.0167 · 2,501)/(1.005 + 0.0167 · 1.86) = 26.8 °C.
Partial pressure: pDM = p·xM/(0.622 + xM) = 2,608 Pa; ps(26.8 °C) ≈ 3,512 Pa.
φM = 2,608/3,512 = 0.74 = 74 % — the mixture remains unsaturated, no fog forms.
Result
| Humidity ratio of the mixture xM | 0.0167 kg/kg dry air |
| Mixing temperature tM | 26.8 °C |
| Relative humidity φM | ≈ 74 % |
| Total mass flow | 1.5 kg/s humid air |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
Why are the balances referred to the dry-air mass and not to the humid air?
The dry-air mass remains constant during humidification, dehumidification and mixing – the total humid mass does not. This is why the humidity ratio x and the specific enthalpy h are consistently referred to 1 kg of dry air; only then do the balances become linear and the mixing point lies exactly on the straight line in the h-x diagram.
Can fog form when two unsaturated air streams are mixed?
Yes. The saturation line in the h-x diagram is curved; the mixing line between two states close to saturation can therefore cut into the fog region. A classic example is visible breath in winter: each stream on its own is unsaturated, yet the mixture is supersaturated and water condenses out.
Is the mixing temperature the mass-weighted average of the inlet temperatures?
Only approximately. Strictly, it is the specific enthalpy (referred to dry air) that is averaged, not the temperature. Since the heat capacity of humid air depends on the water content, the actual mixing temperature deviates from the simple temperature average – by several tenths of a degree up to a degree for strongly differing humidities.
What role does the total pressure play?
The humidity ratio depends directly on the total pressure via x = 0.622·pD/(p − pD): the same relative humidity means a higher moisture content at lower pressure. Mixing calculations for high-altitude locations or pressurised systems must therefore be carried out with the actual total pressure, not with standard pressure.