Chapter2, Lesson D - Relative and Absolute Humidity of Air

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2D-5 :	Relative and Absolute Humidity of Air						4 pts

A small rigid cylindrical vessel contains 1 kg of humid air at 80^oC under a total pressure of 1 bar. The humid air inside the vessel contains 0.150 kg of water vapor. Determine the relative humidity of the air in the container.



Given:	T	80	^oC
	P_tot	1	bar =	100	kPa
	m_tot	1.000	kg wet air
	m_H2O	0.150	kg H₂O

Find:	h_r =	??? %

Assumption:		Air is a non-condensable gas.

Solution :	Let's begin with the definition of relative humidity:

							Eqn 1

	The vapor pressure is equal to the saturation pressure at the system temperature.
	We can find this in the saturation temperature section of the steam tables:
					P* (80^oC) =	47.39	kPa

	For an ideal gas:		P_H2O = y_H2O P_tot				Eqn 2

	Test if ideal:		Ideal Gas EOS :				Eqn 3

	Solve for the molar volume :						Eqn 4

					V	29.36	L/mole

	Therefore, since V > 20 L/mole, we can treat the wet gas as an ideal gas.

	For all gases, mole fraction is defined as :

							Eqn 5

	Where :			n_i = m_i / MW_i			Eqn 6

	Data:		MW_H20=	18.016	g H₂O / mol H₂O
			MW_BDA =	29.0	g bone-dry air / mol bone-dry air

	Since we know the mass of water and bone-dry air in the tank, as well as their molecular weights, we can calculate the number of moles of water and BDA in the tank using Eqn 6.


	Then, we can calculate the mole fraction of water in the gas in the tank, using Eqn 5. Next, we can use the given total pressure to calculate the partial pressure of water in the gas using Eqn 2.



	Here are the numerical results:			n_H2O	8.33	mol H₂O
				m_BDA	0.850	kg BDA
				n_BDA	29.31	mol BDA
				y_H2O	0.221	mol H₂O / mol wet gas
				P_H2O	22.12	kPa

	Finally, we can calculate the relative humidity using Eqn 1:
						h_r =	46.7%

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