Lecture 9 Thermodynamics of humid air Part 1

Areas where thermodynamics of humid air is needed HVAC technology Drying processes Cooling towers and scrubbers (heat recovery from humid gas) Heat exhangers of humid air and flue gases

Components of dry and humid air Dry air Humid air O2 N2 CO2 Ar Ne He H2O Mv = 0.018015kg/mol 0.018kg/mol v = vapor O2 N2 CO2 Ar Ne He Mda = 0.02896kg/mol 0.029kg/mol da = dry air A good approximation xO2 = 21 Vol-% (mol-%) xN2 = 79% Vol-% (mol-%)

Basic definitions Both dry air and vapor are ideal gases Absolute humidity (kgv/kgda) Total pressure ptot = pda + pv Total density tot = da + v Note that the absolute humidity can also be

Basic definitions The dependence between vapor pressure and absolute humidity => => pda = ptot – pv Mv/Mda = 0.02896/0.018 =0.622

Basic definitions Relative humidity where is the saturated vapor pressure. The influence of dry air is negligible on the saturated vapor pressure p’v(T) => we can use the same values as in the steam tables. An approximation equation for the saturated vapor pressure , where T = temperature [K] In some references, the relative humidity means: f = x/x’ THIS DEFINITION IS NOT USED IN THIS COURSE

The influence of dry air on the saturated vapor pressure System 1 System 2 Saturated air vapor + dry air (T, p = pda + pv) What is the influence of pda on pv when T is constant? Vapor (T, p = pv0) Liquid (T, pv0 ) Liquid (T, p) At equilibrium At equilibrium pv0 represents the vapor pressure when dry air does not exists = pv’(T). pv represents the vapor pressure when dry air exists.

The influence of dry air on the saturated vapor pressure

The influence of the air on the saturated vapor pressure where vliq is in unit m3/mol Usually vliq is given in unit m3/kg => v’liq is the specific volume of liquid water in unit m3/kg

The influence of dry air on the saturated vapor pressure, example Initial values t = 20oC, pda = 9.97104 Pa v’liq = 0.001 m3/kg MH2O = 0.018 kg/mol For pure vapor pv’(20oC) = 2337 Pa (tabulated value) For the mixture of vapor and dry air The difference is ca. 1.8 Pa => it is negligible If pda = 9.97105 Pa => pv(T,pda) = 2354 Pa => we still can assume that pv(T,pda)  pv’(T)

Enthalpy of humid air H = mdahda + mvhv = mdahda + xmdahv => hk = hda + xhv [kJ/kgda] Both gases are ideal gases => hk only depends on the temperature Zero/reference-points for dry air and water are: Dry air: dry air at 0oC Water: water in liquid form at 0oC cp values for dry air and vapor are usually given in unit kJ/(kgoC)

Enthalpy of humid air Substituting results of integration in the definition of the enthalpy => hk = cpdat + x(cpvt + 2501) [kJ/kgda] NOTE! Temperature t is given in unit oC. x is the absolute humidity and is given in unit kgv/kgda cpda and cpv represent the average heat capacities over the temperature range of 0…t oC. Both heat capacities are given in units kJ/(kgoC) or kJ/(kgK). In most calculations, we only use the symbol h (not hk) for the humid air, if there is no risk for any misunderstanding.

Example 1 The air temperature is 27oC, relative humidity 18% and total pressure 99800Pa. What is the absolute humidity, enthalpy and density of the air? pv’(27oC) = 3564 Pa hk = cpdat + x(cpvt + 2501) = 1.0027 + 0.004 (1.86 27 + 2501) = 37.3 kJ/kgda

Mollier diagram, psychrometric chart, i,x diagram (Salin Soininen perspective) f = x/x’ f IS NOT THE RELATIVE HUMIDITY Saturation curve

Mollier diagram

Dew point temperature (kastepiste) What is the dew point temperature of the air? pv’(20oC) = 2337 Pa pv =  pv’(20oC) = 0.42337 = 935 Pa Find a temperature where pv’(t) = 935 Pa From the steam table pv’(6oC) = 934.9 Pa => tdp  6oC Air conditions t = 20oC = 40% ptot = 100kPa

Example 2 What is the dew point of the air? Air conditions t = 23oC x = 0.007kg/kgda p = 99.7 kPa Find a temperature where pv’(t) = 1110 Pa pv’(8oC) = 1073Pa pv’(9oC) = 1148Pa => tdp  8.5oC