Mixing of Air Streams

Mixing of air streams at different states is commonly encountered in many processes, including in air conditioning. Depending upon the state of the individual streams, the mixing process can take place with or without condensation of moisture.

Mixing Without Condensation: Figure 8.14 shows an adiabatic mixing of two moist air streams during which no condensation of moisture takes place. As shown in the figure, when two air streams at state points 1 and 2 mix, the resulting mixture condition 3 can be obtained from mass and energy balance.

Figure 8.14 Mixing of Two Air Streams Without Condensation

From the mass balance of dry air and water vapour:

 

m_a1ω₁ + m_a2ω₂ = (m_a1 + m_a2) ω₃

 

From energy balance:

 

m_a1h₁ + m_a2h₂ = (m_a1 + m_a2) h₃

 

From the above equations, it can be observed that the final enthalpy and humidity ratio of mixture are weighted averages of inlet enthalpies and humidity ratios. A generally valid approximation is that the final temperature of the mixture is the weighted average of the inlet temperatures. With this approximation, the point on the psychrometric chart representing the mixture lies on a straight line connecting the two inlet states. Hence, the ratio of distances on the line, i.e., (1 − 3)/(2 − 3) is equal to the ratio of flow rates m_a2/m_a1. The resulting error (due to the assumption that the humid specific heats being constant) is usually less than 1%.

Mixing with Condensation: As shown in Figure 8.15, when very cold and dry air mixes with warm air at high relative humidity, the resulting mixture condition may lie in the two-phase region, as a result there will be condensation of water vapour and some amount of water will leave the system as liquid water. Due to this, the humidity ratio of the resulting mixture (point 3) will be less than that at point 4. Corresponding to this will be an increase in temperature of air due to the release of latent heat of condensation. This process rarely occurs in an air conditioning system, but this is the phenomenon which results in the formation of fog or frost (if the mixture temperature is below 0°C). This happens in winter when the cold air near the earth mixes with the humid and warm air, which develops towards the evening or after rains.

Figure 8.15 Mixing of Two Air Streams with Condensation

Example 8.3: A glass of water is mixed with ice. The moisture from the air begins to condense on the surface of glass when temperature of water comes 10°C. If the room temperature and pressure are 20°C and 1.01325 bar, determine the partial pressure and mass of water vapour in grams per kg of dry air.

Solution:

From steam table:

Example 8.4: The pressure and temperature of mixture of air and water vapour at 1 bar and 20°C. The dew point temperature of the mixture is 14°C. Find partial pressure of water vapour in mixture, relative humidity, specific humidity, enthalpy of mixture per kg of dry air and sp. volume of mixture per kg of dry air.

Solution:

From steam table:

Enthalpy of water vapour, h_s = c_p × DBT + h_fg = 4.18 × 20 + 2,454.1 (from steam table) = 2,537.7 kJ/kg

Enthalpy of mixture per kg of dry air, h_mix = h_a + ω × h_s = 1 × 20 + 0.0099 × 2,537.7 = 45.12 kJ/kg of dry air

Specific volume of mixture is equal to the volume of 1 kg of dry air,

Example 8.5: 50 m³ of air at 20°C DBT and 15°C WBT are mixed with 20 m³ of dry air at 30°C DBT and 20°C WBT. Determine the DBT and WBT of the mixture.

Solution:

From psychrometric chart, at 20°C DBT and 15°C WBT, the psychrometric property values can be given as

Similarly from psychrometric chart, at 30°C DBT and 20°C WBT, the psychrometric property values can be given as

Example 8.6: The DBT and RH of air are 40°C and 70%, respectively. The atmospheric pressure is 1 bar. Determine the specific humidity and vapour pressure of moisture in air. If 8 g of water vapour is removed from the air and temperature is reduced to 30°C, find the relative humidity and DPT.

Solution:

DPT (saturation temperature at 0.0398 bar) = 29°C

Example 8.7: 100 m³ of air per minute at 30°C and 60% is cooled to 22°C DBT (sensible cooling). Find the heat removed from air, relative humidity of cooled air and WBT of cooled air (take air pressure = 1 bar).

Solution: