Page 107 - Industrial Ventilation Design Guidebook

72 CHAPTER 4 PHYSICAL FUNDAMENTALS

where p t, is the partial pressure of water vapor in air and ph(T) is the sat-
urated water vapor pressure at temperature T. When 9 is 1 or 100%, we say
that the air is saturated.

Example 2a
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):
log/?£(20 °C) = 28.59051 - 8.2 log(20 + 273.16) + 0.0024804(20 + 273.16)
3142.31 _ , m
"20 + 273.16 ' b J
1>631
ph(2Q °C) = 10~ bar = 0.0234 bar
• Humidity x from Eq. (4.83):

x = 0.6220 - 1 0 °:°o 0234 = 0.01490 kg H 2 O/kg d.a.

• Densities p,, p h, and p from Eqs. (4.76), (4.78), and (4.79):
5
= 0.0234 • 1Q • 0.018053 = Q Q1731 -, 3
Ph
8.314-293.15 u.ui/31 kg/m
5
3
0 = (1.0-0.0234). IP -0.028964 = * UQ6 ke/m L fife)
Pt 1 16U6 kg/m
8.314-293.15 ' (~ x j
3
p = 1.1606 + 0.01731 = 1.178 kg/m
h k = 1.006 • 20 + 0.01490 • (2501 + 1.85 • 20) = 57.9 kj/kg d.a.
(b) Humid air, <p = 50%
p' h(2Q °C)= 0.0234 bar

p' h = 0.5 • 0.0234 = 0.0117 bar
x = 0.6220 • 1 0 °l°o Qi 17 = 0.00736 kg H 2O/k.g. d.a.
5
n - 0.0117-IP -0.018053 _ n nn oxr . ff/ms
ph 0 0865 kg/m
8.314-293.15 -°
5
n - (1.0-0.0117)-IP -0.028964 _ -, 174 ^ olrA _ ph _ 0.00863)
Pi &
~ 8.314 • 293.15 ~ ( ~ x ~ 0.00736 j
p = 1.174 + 0.00865 = 1.183 kg/m 3
h k = 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 pres-
sure of air is TT = 0.825 bar and the temperature of air is 20 °C.