Question

1) An air-water vapor mixture at 95°F has a relative humidity of 45%. Calculate the partial pressure of the water vapor and its dew-point temperature. (Hint: Psychrometric chart is really useful for this.)

Answer 1

Le VW Given, Day bulb temperature, Ti = 95°F Relative humidity, d = 45% Td = 95°F TJ = 35°C (In degrees celcius) using "psychometric Chart". At, Td = 35°c, d = 45% → specific humidity (CW) is obtained by extending Intersection point of & Toed in longitudinal direction anned WIDE mann

CScanne - 0.84m®/kg ant or ßatutation temperature, LALA 0.86m®/kg 550 NyDNO Relative Rumidity 90% 85% 80° 175% A K roK-0.4 M 185% K K WM160PKV OK 45% 1354W 35 0.88m®/kg tatky 12Enthalpy deviation kung or P -0.6 km 40 43 0.9mⓇ/kg of dry air- 0.002 0.004 0.006 0.008 0.010 0.012 10.014 10.016 0.018 0.020 0.022 0.024 Specific humidity, kJ/kg of dry air ---

enthalpy Lokallalelelal wet bulb Tw) - w (Specipic humidity | dew.jpoient d 35°C Day bulb temperature co = 0.016 kg/1g of day air but, w = [0.622 Po 7 o I Pa-Pre Scanned with can scariner

here, Pg= Vapour pressure Pa = Atmospheric pressure Pa = 1.0132 bar From 0.016 = 0.622 x Pze 1.0132-Pre 0.016 (1.0132-Pre) = 0.622x Pro 0.01621-0.016 Pre = 0.622 Pre • Pre = 0.0254 bar ] cscan

Dew point temperature (Ten) from psyclorometic chart, Tdow = 2100