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In the Afternoon T = 90 o F RH = 80% P t =738 mmHg

Example 1. At night T = 68 o F P t =738 mmHg. In the Afternoon T = 90 o F RH = 80% P t =738 mmHg. Cooling. Q : How much water (in %) is condensed? Let air = 2 ; and water vapor=1. L lb. Cooling. Cooling. F lb. System is partially saturated Get p 1 * at 90 o F from the tables

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In the Afternoon T = 90 o F RH = 80% P t =738 mmHg

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  1. Example 1 At night T = 68oF Pt=738 mmHg In the Afternoon T = 90oF RH= 80% Pt=738 mmHg Cooling Q : How much water (in %) is condensed? Let air = 2; and water vapor=1

  2. L lb Cooling Cooling F lb System is partially saturated Get p1* at 90oF from the tables p1* (at 90oF) = 36.1 mmHg RH = 0.8 = p1/p1* = p1/36.1 Thus, p1= 28.9 mmHg Then p2= 738-28.9 = 709.1 Dew point is the temp at which p1* = p1 = 28.9 mmHg T dew point = 83oF Since T is < 83oF Therefore System is sat’d We can get p1* at 68oF from the tables p1* (at 68oF) = 17.5 mmHg = p1 Then p2= 738-17.5 = 727.5

  3. Basis : F = 100 lb Tot balance F = L + W dry air balance 100x0.9609 = P x 0.9765 + 0 Then, L = 98.40 lb and W = 1.60 lb % of water condensed = W/(F y1) = 1.6/3.91 x 100 = 41%

  4. Example 2 (Dehydration) By Absorption in silica gel you are able to remove all (0.93Kg) of the water from moist air at 15oC and 98.6 KPa. The same air measures 1000 m3 at 20oC and 108 KPa when dry. What was the RH of the moist air ? 1 = water vapor ; 2 = dry air

  5. Wet air dry air Silica Gel dry air H2O vapor 15oC Pt=98.6 Kpa Water Liquid 20oC n1= 0.0517 n2 = 44.35 0.93 Kg =.0517 kgmol 1000 m3 108 kPa y1=0.009673 y2 =1-y1 n=PV/RT = 44.35 kgmol p1* = 1.70 KPa (at 15oC) ( from steam tables) p1 = y1x Pt = 0.009673 x 98.6 = 0.1147 kPa %RH = p1/p1*x100 = 6.7%

  6. Example 3 ( Humidification) 1000 m3 of moist air at 101 KPa and 22 oC and with a dew point of 11oC enters a process. The air leaves the process at 98 KPa with a dew point of 58oC. How many Kg of water vapor are added to each Kg of wet air entering the process? 1 = water vapor ; 2 = dry air

  7. BASIS: 1000 m3 entering wet air at given T and P Therefore, nt = PV/RT = 41.19 Kgmol wet air in Material balances: F + W = P F y2 + W y2 = P y2 n2 = y2 x nt = 0.98703 x 41.19 = 40.66 kgmol = n2OUT (Tie comp.) THUS, nt.out= 40.66/.8047 = 49.87 kgmol  Water added = 49.87 - 41.19 = 8.68kgmol

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