A drier must remove 200 Kg of H2O per hour from a certain material. Air at 22oC

F kgmol/hr --------->      --------------> P kgmol/h, air @ 72oC, rh 80%
air @ 22oC, rh 50%                            ^
                                                              | w= 200/18 kgmol/h H2O



Data for P* from table B.3 at page 638;

ToC                                                   P* mmHg                                          P* Kpa

22                                                      19.827                                                 2.64
72                                                      254.6                                                   33.94

Composition of F

PH2O = 0.5 (2.64) = 1.32 Kpa

Pair = 103 - 1.32 = 101.68 Kpa

Ptotal = PH2O  + Pair = 103.0 Kpa


Composition of P

PH2O = 0.8 (33.94) =  27.152 Kpa

Pair = 103 - 27.152 = 75.848 Kpa

Ptotal = PH2O  + Pair = 103.0 Kpa


Basis: W= 200 Kg H2O/h

Balances:

\(H_2O\ :\ \ \ F\ \left(\frac{1.32}{103}\right)\ +\ \frac{200}{18}\left(1.00\right)\ =\ P\ \left(\frac{27.152}{103}\right)\)

\(air:\ F\ \left(\frac{101.68}{103}\right)\ =\ P\ \left(\frac{75.848}{103}\right)\)

F = 32.63 Kmol/h

P = 43.74 Kmol/h

Kmol dry air used: \(F\ \left(\frac{101.68}{103}\right)\ =\ 32.63\ \left(\frac{101.68}{103}\right)\ =\ 32.21\ \frac{kmol}{h}\ \ \ \ \)

Kg of dry air used:  32.21 (29) = 934.079 kg/h