1. Lecture 9 Thermodynamics of humid air; Part one

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

3. 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

4. Basic definitions p-V-T-relationships for dry air (da) and vapor (v)
=>
=>
pda = ptot – pv
Mv/Mda = /0.018 =0.622

5. 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

6. The influence of the dry air on the vapor pressure
System 1
System 2
What is the influence of
pda on pv when T is
constant?
Vapor (T, p = pv0)
Vapor + dry air
(T, p = pda + pv)
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.

7. The influence of the dry air on the vapor pressure

8. The influence of the dry air on the 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

9. The influence of the dry air on the vapor pressure, example
Initial values
t = 20oC, pda = 9.97104 Pa v’liq = m3/kg MH2O = 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)

10. 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)

11. Entalpy of humid air
Substituting results of integration in the definition of the enthalpy =>
hk = cpdat + x(cpvt ) [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.

12. 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?
Equations needed in calculations:
ptot = pda + pv
tot = da + v
hk = cpdat + x(cpvt )

13. 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 ) = 1.00 (1.86  ) = 37.3 kJ/kgda

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

15. Mollier diagram

16. 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) = Pa => tdp  6oC
Air conditions
t = 20oC
= 40%
ptot = 100kPa

17. 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