Fundamentals
Air is seldom dry; it normally contains varying amounts of moisture. Humid air is a mixture of dry air and water vapor. The term dry air denotes the mixture of all gases present in air (nitrogen, oxygen, carbon monoxide, and inert gases), except water vapor.
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TABLE 4.4 Analysis of Air at Sea Level
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Source: R. B. Keey, Drying Principles and Practice, 1972 |
TABLE 4.5 Minor Constituents of Dry Air
Parts per million
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Gas Symbol Molecular weight Volume Weight
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The molar mass of dry air is dependent on the consistency of air, but for standard air it is M( = 0.028964 kg/mol. In practical calculations we may use the rough value of 0.0290 kg/mol. The molar mass of water vapor is Mh = 0.0180153 kg/mol and the rough value used in practical calculations is 0.0180 kg/mol.
According to the state equation of ideal gas, the partial density of dry air in humid air is
P, = ^, (4-76)
Where pj is the partial pressure of dry air and R is the gas constant. According to present knowledge, the best value for the gas constant is
R = (8.31441 ±0.00026)
In practical calculations we use the value of R = 8.314 J/mol K.
The partial density of water vapor in humid air is
Ph = Pjj^ , (4.77)
Where ph is the partial pressure of water vapor.
The density of humid air is the sum of the partial densities of dry air and water vapor,
P = P, + Ph (4.78)
And the total pressure of humid air is a sum of the partial pressures of dry air and water vapor:
P = Pi + Ph (4.79)
The mass of dry air in a volume V is denoted as mt (p; = m,/V;) and the mass of water vapor in V is mh (ph — mh/Vh). Humidity of air, meaning the ratio of water vapor mass to dry air mass, is defined as
X= ^. (4.81)
Nt; ■ ‘
Using partial pressures, this definition can be written as
X = ^. (4.82)
Humidity x is thus a dimensionless number, but sometimes a “dimension” is added to it, as a reminder of its definition. We can, for instance, write x ~ 0.05 or x = 0.05 kg H20/kg d. a., where d. a. stands for dry air.
According to Eqs. (4.76), (4.78), and (4.80), humidity can be written as
X = 0.6220^ = 0.6220-%- . (4.83)
Mi Pi pi p-ph 1
By solving the partial pressure of water vapor from the equation above, we receive
K-M&rr,- m4]
We denote again the mass of dry air in a volume V as w, and the mass of water vapor as mh. When humid air is treated as an ideal mixture of two components, dry air and water vapor, the enthalpy of this mixture is
H = mjhj + mhhh, (4.85)
Where hj is the specific enthalpy of dry air (J/kg) and bh is the specific enthalpy of water vapor.
Technical calculations dealing with humid air are reasonable to solve with dry air mass flow rates, because these remain constant in spite of changes in the amount of water vapor in the air. For that reason a definition for enthalpy,
L’4-86)
Which is the enthalpy of humid air divided by the dry air mass, is made. The dimension of enthalpy hk is J/kg, but it is often written as J/kg d. a. as a reminder that the total enthalpy of the mixture is calculated in terms of a kilogram of dry air. Combining Eqs. (4.85) and (4.86), we get
Mjhk = m, hj+ mhbh,
And using Eq. (4.81) we have
Bk = hj + xhh. (4.87)
In calculations with humid air, when the pressure is not high (usually the atmospheric pressure of 1 bar), water vapor and dry air can be handled as an ideal gas, as we have already done in Eqs. (4.76) and (4.78). For ideal gases the specific enthalpy is just a function of temperature:
Hi = h,(T)
And
Bh = hh(T).
When 0 °C dry air is chosen as the zero point of dry air enthalpy, and 0 °C water as the zero point of water vapor enthalpy, the enthalpies of dry air and water vapor can be calculated from the equations
B, cn=‘ cpt(T)dT (4.SS)
J 273.1.5K
Bh(T) — I bo + f cph{T)dT (4.89)
J 273.15K ‘
Where cpi(T) is the specific heat of dry air (J/kg K), cph(T) is the specific heat of water vapor, and lho is the heat of vaporization at 0 °C. Its numerical value is
Lh0 = 2501 kj/kg
Specific heats cpi and cpb are somewhat dependent on temperature. In the temperature range of -10 to +40 °C, their average values are
Cpi = 1.006 kj/kg °C cph = 1.85 kj/kg °C.
At the temperature of +50 °C, their values are cpi = 1.008 kl/kg °C and cph = 1.87 kj/kg °C. ‘
Using numerical values mentioned above, the enthalpy of humid air bk (Eq. (4.87)) can be written as
Hk = 1.0060 + *(2501 + 1.850) kj/kg, (4.90a)
Where 0 is the temperature in Celsius. Equation (4.90a) can also be written as
B = cpi0 + x{lho + cph6) (4.90b)
We denoted the mass of dry air in a volume V as that is, p, = m/ V’!5 and the mass of water vapor in V as mh, that is, ph = mf/Vf,. In practical calculations we usually handle volume flow g„(nvVs) instead of volume V. For instance, the value of volume flow is known in the suction inlet of a fan when the operating point of the fan is defined. Volume flow qv, expressing the total air flow or the combined volume flow of water vapor and dry air, is not constant in various parts of the duct, because the pressure and temperature can vary. Therefore in technical calculations dealing with humid air, material flows are treated as mass flows. Also, while the humidity can vary, the basic quantity is dry air mass flow w,(kg d. a./s). If, for instance, we know the volume flow q,. of a fan, the dry air mass flow through the fan is
Riti = piqv (4.91)
Where p{ is the partial pressure of dry air in the suction inlet of the fan, in the same place where the total volume flow V is defined. Accordingly, the water vapor flow mh (kg H20/s) along the volume flow is
= Phqvi (4.92)
Where ph is the partial pressure of water vapor.
Due to the definition of humidity (4.87), on the basis of the Eqs. (4.91) and (4.92),
Ritf, = xm,, (4.93)
When an energy balance is written, an enthalpy flow of humid air H is needed. This can be written according to the Eqs. (4.85) and (4.93) as
H = thjkj + mhbh = + xbh)
Or briefly, using definition (4.86),
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(4.94) |
H = tnjbk.
In energy balance calculations, which we will handle later on, we use Eq. (4.94). Example I
A fan takes +15 °C humid air at 0.5 m3/s. What is the dry air mass flow rhj taken by the fan when the outdoor humidity is x = 0.009 and the outdoor pressure is p = 1.0 bar?
Partial pressure of water vapor is calculated from Eq. (4.84):
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Dry air partial pressure is
Pi = P~Ph = 1-0-0.01426 = 0.986 bar = 0.986 • 105 Pa. Partial density of dry air is calculated from Eq. (4.76):
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Dry air mass flow is calculated from Eq. (4.91):
Riii = 1.194 • 0.5 = 0.597 kg d. a./s.
Water vapor flow through the fan is
Thh = xtht = 0.009 • 0.597 = 0.005373 kg HzO/s.
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