1. Strawberries contain about 15 % Solids and 85 % water.
To make a strawberry jam, crushed
      strawberries and sugar are mixed in a 45:55 ratio, and the mixture is heated to evaporate water
      until the residue contain one-third water by mass. Calculate how many grams of strawberries are
      needed to make a kilogram of jam.
                                             F2, sugar                  W, H2O
     F1, crushed strawberries
15 % Solids                          Mixer                      Evaporator                 P, Jam
85 % water                                                                        1/3 water by mass
                                                          = 
       Required: g of F1 per kg P
       Basis: 1 kg of P (1000 g)
       Solid balance @ whole system
       0.15 F1 + F2 =               iiiL
               
       F1 =   
                 F2
       F1 = 485.83 g of strawberries
   2. Fresh feed stream flowing at 100 kg/h contains 20 wt% KNO3 and the balance water. The fresh
      feed stream is combined with a recycle stream and is fed to an evaporator. The concentrated
      liquid solution leave the evaporator containing 50 wt % KNO3 is fed to a crystallizer. The crystals
      obtained from the crystallizer is wet and contains 4 wt % water. The liquid from the crystallizer
      constitute the recycle stream and contains 0.6 kg KNO3 per 1 kg water. Calculate:
      a) Weight of product crystal collected
      b) Weight of water removed in the evaporator.
      c) Weight of the recycled stream
F = 100 kg/h              W, H2O
20 wt % KNO3
80 wt% water                           50 % wt% KNO3
                     Evaporator                           Crystallizer
                                           C
                                                             P       KNO3
                                                                         4 % H2O
                                       0.6 Kg KNO3 / Kg H2O % H2O
Required: C, W, R
       Basis: 1 hr operation
       KNO3 balance @ whole system
               0.2(100) = (1 – 0.04)(P )
               P = 20.83 kg
       OMB @ whole system
               100 = W + 20.83
               W = 79.17 kg
   @ crystallizer
          OMB: C            = 20.83 + R
                                                           ith
           KNO3 balance:        0.5(C ) = 0.96(20.83) +     th
                                                               R
                    R = 76.68 kg
3. A stream of air at 40 oC and 1 atm has a dew point of 20 oC. Determine the following:
       a. Molar humidity
       b. Absolute humidity
       c. Relative Humidity
       d. % Humidity
       e. % (mole) water vapor
                                                             tᄀᄀ
                    ln (Pv) = 18.3036 -
                                                     ᄀt
           at 40 oC, Pv = 54.75 mm Hg
           at 20 oC, Pv = 17.19 mm Hg
           Pp = Pv at 20 oC
                        t
           Hm =    hi       t
                                = 0.023
                                 th
           Habs = Hm(              tᶈ
                                        L = 0.023( L = 0.014
                    t
           RH =    t 
                        x   100 = 31.40 %
                        t t.h2t                     iti
           %H =      t th th2 thᶈ
                                        x 100 =       t       x 100 = 29.62 %
                                                    hi  t 
                        t
           % H2O =      hi
                                 x 100 = 2.26 %
4. Dry air containing at 90 oC and 1 atm is humidified by spraying water and gas mixture leaves the
   humidifier saturated 60 oC and 1 atm. Calculate the moles of water added per 100 moles of dry air
   entering.
           Dry air                                                      Saturated air
           90 oC, 1 atm                Humidifier                60 oC, 1 atm
             E                                                          L
   Required: W/100 mole dry air
           Hm E = 0
   Using Antoine eq. At 60 oC, Pv = 148.39 mm Hg
                             t
           Hm L =     hi          t
                                      = 0.243
                                                t               ii t    tᶈ         t t
            ii t      tᶈ
                             = (0.243 – 0) t           tᶈ
                                                            x   ii t    tᶈ
                                                                             =   ii t    tᶈ
5. Contaminated air contains 3 mol% acetone and 2 mol% water vapor. The air was sent to an
   absorber to completely remove the acetone using water as absorber. The air leaving the absorber
   contains 0.5 mol % water and the liquid leaving the system contains 19 wt% acetone and 81 wt %
   water. Calculate the weight of water used per 100 mol of contaminated air
Contaminated air   E
      3 mol % C3H6O
      2 mol % H2O                                  air,     L
                                                          0.5 mol % H2O
                              Absorber                    99.5 mol % dry air
         R
        19 wt % C3H6O                          S          H2O
        81 wt % water
Required: S / 100 mols of E
Basis: 100 mols of E
Dry air balance:      (1 – 0.03 - 0.02)(100) = 0.995(L)
                              L = 95.5 mol
Acetone recovered: 0.03(100) = 3 mols
        0.19(R ) = 3(58)
        R = 915.8 g
        Water balance
        S = 0.81(915.8) + 0.005(95.5)(18)
        S = 750.4 g