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This leaves 4,000  634 = 3,366 lb/h of water and 1,000  500 = 500 lb/h of sulfate in the mother
liquor. The total flow rate of mother liquor is 3,366 + 500 = 3,866 lb/h. Thus, the concentration
of sulfate in the mother liquor = 500/3,866 x 100% = 12.9 wt% Na2SO4. From the diagram
below, the required temperature for this concentration at C in the mother liquor is 18oC.
500 Exercise 4.59
Subject: Dissolving crystals of Na2SO4 with water.
Given: 500 kg of Na 2SO 4 ⋅ 10H 2 O crystals and 500 kg of Na2SO4 crystals at 20oC.
Assumptions: Equilibrium according to phase diagram of Fig. 4.24.
Find: Amount of water to dissolve the crystals at 20oC.
Analysis: From Fig. 4.24, the solubility of Na2SO4 in water at 20oC is 15 wt% Na2SO4.
Molecular weight of Na2SO4 = 142. Molecular weight of Na 2SO 4 ⋅ 10H 2 O = 322.
Therefore, the kg of Na2SO4 in Na 2SO 4 ⋅ 10H 2 O = 500(142/322) = 220 kg.
The water in Na 2SO 4 ⋅ 10H 2 O = 500  220 = 280 kg.
Total Na2SO4 in the crystals = 500 + 220 = 720 kg.
Therefore, need a total of 720(85/15) = 4,080 kg water.
Additional water needed = 4,080  280 = 3,800 kg. Exercise 4.60
Subject: Adsorption of phenol (B) from an aqueous solution at 20oC with activated carbon.
Given: One liter of aqueous solution containing 0.01 mol phenol. Freundlich isotherm
equation for adsorption of phenol from aqueous solution by activated carbon at 20oC.
Assumptions: Attainment of equilibrium.
Find: Grams of activated carbon for (a) 75%, (b) 90%, and (c) 98% adsorption of phenol
*
0.233
Analysis: At equilibrium, Eq. (1) in Example 4.12 is: q B = 2.16cB
(1)
*
where, q B = mmol phenol adsorbed/g carbon
cB = mmol phenol in solution/liter solution
For each case, cB = 0.01(1,000 mmol/mol)(1fraction adsorbed)
mmol adsorbed = 0.01(1,000 mmol/mol)(1 liter)(fraction adsorbed)
*
g activated carbon needed = mmol adsorbed/ q B
Using these equations, the results are as follows:
*
Case
g activated
cB at equilib, mmol phenol
qB ,
adsorbed
carbon
mmol/L
mmol/g
(a) 75% adsorbed
2.5
7.5
2.67
2.81
(b) 90% adsorbed
1.0
9.0
2.16
4.17
(c) 98% adsorbed
0.2
9.8
1.48
6.62 Exercise 4.61
Subject: Adsorption of a colored substance (B) from an oil by clay particles at 25oC.
Given: Oil with a color index of 200 units/100 kg oil. Adsorption equilibrium data
Assumptions: One adsorption equilibrium contact.
Find: (a) Freundlich equation for the adsorption equilibrium data.
(b) kg clay to reduce color index to 20 units/100 kg oil for 500 kg of oil
*
1/
Analysis: (a) From Eq. (430), the Freundlich equation is q B = AcB n , where here,
*
q B = color units/100 kg clay and cB = color units/100 kg oil
By nonlinear regression of the data, A = 0.6733 and (1/n) = 0.5090
*
The linearized form of the Freundlich equation is log q B = log A + (1 / n) log cB
By linear regression of the data with the linearized form, A = 0.6853, (1/n) = 0.5050
The two results are close and both fit the data quite well. Use the nonlinear regression result:
*
0.509
qB = 0.6733cB (b) (1) (
Q = 500 kg oil
S = kg clay
cBF ) = 200 units/kg oil
Need, at equilibrium, cB = 20 units/kg oil
*
From Eq. (1), qB = 0.6733(20) 0.509 = 3.09 units/kg clay Apply Eq. (429 for a material balance on the color units:
*
q B = 3.09 = − Q
Q
500
500
(
cB + cBF ) = −
(20) + (200)
S
S
S
S Solving Eq. (2), S = 29,100 kg of clay (2) Exercise 4.62
Subject: Absorption o...
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 Spring '11
 Levicky
 The Land

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