Question

9. A building has indoor air conditions of 70°F dry bulb temperature and 40% relative humidity. Determine the direction of water vapor flow under the following

Answer 1

Indoon condition = 70°F 90%. Relative humidity 70°F= 21:11 water rapoure flow from high vapour pressure Region to lower vapour pressure Region. saturated vapour pressure at 21:11% = 021523 kpa Relative humidity (9) is given 40% 4. Loe - 0.4 » Pu=(0.442.523) kpa = 1.0092 kPa This is the vapour pressure inside the Room » outside Condition 32°F 40% RH 32 8= 0°C Butureted rapour pressure=0.6113 kpa 8- Pet 20.4 Pu - (0.4*0.6113) = 0.24452 kpa e out side vapour pressure is less than the inside vapoare pressure, so vapour

will of low from room to outside. ② outside condition 90° F, 20% RH - 900f = 32.2.2°C saturated rapour pressure at this temporture = 4.859 kpa 8- Pi = 0.2 - Pu= (0.2 x 9.859) kpa = 0.971 kpa < 1.0092 kpa Here also outside vapour pressure is less than the inside vapour pressure (1.0892 Kpa). i varpour will flow from inside to outside. 10 2 X
saturated vapour pressure value has taken from steam table.