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MoistAirTab™ Example 5 - Humidification

Example 5

Steam (or liquid water) injection is frequently used in air conditioning for humidification. Moist air at 20°C db and 8°C wb is being humidified by injecting saturated steam at 110°C to a final stream with a dew point of 13°C. The rate of dry air flow is 2 kg/s. What is the final db of the moist air and what is the steam injection rate?

Solution

The solution is presented below as a table generated from Excel. The gray shaded areas represent the input data.

Key concepts:.
1. Adiabatic mixing assumes that no net gain or loss of total energy in the mixing process
2. The dew point temperature specifies the moisture content in the final mixture.
Overall Enthalpy balance	m_airh₁ + m_w h_w = m_airh₂ (1)
Water balance	m_air W₁ + m_w = m_air W₂
or	m_w = m_air (W₂ - W₁) (2)
From (1) and (2)	(h₂-h₁)/(W₂-W₁) = h_w
or	h₂= h_w (W₂-W₁) + h₁ (2a)

	Units	Moist Air	Steam	Outlet Mixture
Pressure	bar	1.01325	1.01325	1.01325
Dry air flowrate	kg_{dry air}/s	2		2
Dry bulb temperature	^oC	20	110	21.14
Wet bulb temperature	^oC	8		16.08
Dew Point temperature	^oC	-8.72		13
Humidity ratio	kg_w/kg_{dry air}	0.0018		0.0094
Enthalpy of moist air	kJ/kg_{dry air}	24.69	2690.95	45.06	solved by equation (2a)
Relative humidity	%	12%		60%
Steam flowrate	kg/s		0.0151		solved by equation (2)

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MoistAirTab™ Example 5 - Humidification

Example 5

Solution