# Vapor Pressure of Water and Ice and Calculation of Humid Air State Values

The partial pressure of water vapor in air cannot be higher than the vapor pressure of saturated water p{,(T) corresponding to air temperature T, If it were higher, condensation of water vapor would occur until the equilibrium state corresponding to the saturated vapor pressure w’as achieved.

Saturated water vapor pressure is most accurately found front vapor tables or can be approximated with the following equation:

Logpfc'(0) = 28.59051 -8.2 log(0 + 27.3.16)

(4.106)

+ 0.0024804(0 + 273.16) ■ 3142-31

0 + 273.16

The logarithm in Eq. (4.106) is Briggsian (a logarithm w’ith 10 as the base), pressure is in units of bar, and the temperature is in Celsius.

A simpler approximation for the pressure of saturated water vapor is

PhT) = p0exp( 11.78(T — 372.79)/(T- 43.15)), (4.107)

Where the constant p0 = 105 Pa and the temperature T is in degrees Kelvin.

When the temperature is under 0 °C, the saturation pressure ph, is calcu­lated using the vapor pressure of ice (ice turns into vapor directly, i. e., subli­mates) and we can use the following empirical formula:

Logpf,(0) = 10.5380997(4.108)

The logarithm in Eq. (4.108) is Briggsian, pressure has units of mbar, and the temperature is in Celsius.

For the vapor pressure of ice, the equation of Clapeyron can be obtained in the same way as for water:

DPh = Y(j)f^v’-)dT ’ ‘-4J09>

Where hj is the enthalpy of ice and Vj is the specific volume of ice.

The relative vapor pressure of air or the relative humidi ty is defined by the equation

Where ph is the partial pressure of water vapor in air and ph(T) is the sat­urated water vapor pressure at temperature T. When <p is 1 or 100%, we say that: the air is saturated.

Find the properties of saturated air and <p = 50% air, when the total pressure of air is p = 1.0 bar and the temperature of air is 20 °C.

(a) Saturated air, <p — 100%

• Pressure of saturated vapor p(,(20 °C) from Eq. (4.106):

Logp/'(20 °C) = 28.59051 -8.2 log(20+ 273.16) + 0.0024804(20+ 273.16)

-,3142-^1 = 1.631

20 + 273.16

P’h(20 °C) = 10_1-631bar = 0.0234 bar

• Humidity x from Eq. (4.83):

X = 0.6220 • 1 00i°0 Q234 = 0.01490 kg H20/kg d. a.

• Densities ph ph, and p from Eqs. (4.76), (4.78), and (4.79):

= 0.0234 • 105 • 0.018053 = Q 01731 k/m3 Pb 8.314-293.15 u. ui/JiRg/m

 Ph X

P. = (1-0-0-0234)-105-0.028964 = 1 UQ6 j. = p‘ 8.314 -293.15 i. iouц Kg/m

P = 1.1606 + 0.01731 = 1.178 kg/m3

Hk = 1.006 • 20 + 0.01490 ■ (2501 + 1.85 • 20) = 57.9 kj/kg d. a. (b) Humid air, <p = 50% p’h(20 °C)= 0.0234 bar p’h = 0.5 0.0234 = 0.0117 bar x = 0.6220 • i oi 17 = 0.00736 kg H20/k. g. d. a.

Ph=°-01Јi^r53=°-00865 k^m3

. _ (1.0-0.0117). 105- 0.028964 _ ^ _ 0h _ 0.00863′

Pt 8.314 -293.15 _ & (~ x ~ 0.00736^

P = 1.174 + 0.00865 = 1.183 kg/m3 h. = 1.006 • 20 + 0.00736 • (2501 + 1.85 • 20) = 38.8 kj/kg d. a.

Example 2b

Find the properties of saturated air and <p = 50% air, when the total pres­sure of air is 7t = 0.825 bar and the temperature of air is 20 °C.

(a) Saturated air, <p = 100%

P’h (20 °C) = 0.0234 bar

X = 0.6220 ■ Q.. g2jr^234 = °-01816 kg H^°/kS d-a —

Ph = 0.01731 kg/m3 y

 Ph X

D = (0-825- 0.0234) • 105 ■ 0.028964 = 0 953 . , 3 ‘

P‘ 8.314-293.15 w

/

P = 0.970 kg/m3 hk = 1.006 • 20 + 0.01816 • (2501 + 1.85 • 20) = 66.2 kj/kg d. a.

(b) Humid air, <p = 50%

 ,a.

P’h(20 °C)= 0.0234 bar pf, = 0.0117 bar x = 0.6220 • 82^5’Ffi 7 = °-00895 kS ^O/kg d Ph = 0.0865 kg/m3

N = (0-825 — 0.0117) • 105 • 0.028964 Pl 8.314-293.15

 Ph — 0.00863

0.967 kg/m3

X 0.00736^

Hk = 1.006 • 20 + 0.00895 • (2501 + 1.85 ■ 20) = 42.8 kj/kg d. a.

Comparing Examples 2a and 2b we notice that the total air pressure has effects on the humidity x, partial density of dry air p,, total pressure or pres­sure of humid air, and enthalpy bk. Knowing the total pressure is therefore es­sential in calculations of the thermodynamic properties of humid air.

Pressure and humidity have also an effect on the mass flows. We continue Examples 2a and 2b by calculating the dry air mass flow in a fan when the hu­mid air volume flow in the fan is 0.8 m3/s. According to Eq. (4.91) and the cal­culations above, we obtain

 (Ex. La) (Ex. Lb) (Ex. 2a) (Ex. 2b)

M, = piqv = 1.1606 • 0.8 = 0.928 kg d. a./s mh = 0.01490 • 0.928 = 0.01383 kg/s

Fflj = 1.174 ■ 0.8 = 0.939 kg d. a./s mb = 0.00736 • 0.939 = 0.00691 kg/s

M, = 0.953 ■ 0.8 = 0.762 kg d. a./s mh = 0.01816 • 0.762 = 0.01384 kg/s

Thj = 0.967 ■ 0.8 = 0.774 kg d. a./s mh = 0.00895 • 0.774 = 0.00693 kg/s

Thus the total mass flows m= (mt + rhf,) differ in different cases. Water va­por flow thb is obtained by multiplying the dry air mass flow by the corre­sponding humidity x (Eq. 4.93). As a basic quantity in humid air mass and energy balance calculations, we use dry air mass flow th„ and the effect of hu­midity on the energy balance is noted in the enthalpy hk (Eq. 4.87).