Unformatted text preview: Liquidliquid extraction of uranyl nitrate (UN) from water (W) with
tributyl phosphate (TBP)
Given: 100 grams of 20 wt% UN in water. Thus, FW = 80 g .
Assumptions: Equilibrium batch contacts. W and TBP are mutually insoluble.
Y
'
.
(1)
Distribution coefficient is given in the form of Eq. (515): K DUN = UN = 55
X UN
Find: Grams of TBP solvent needed to extract 90% of the UN by the following
schemes:
(a) One stage
(b) Two crosscurrent stages with 50% of solvent to each stage.
(c) Two countercurrent stages.
(d) Infinite number of crosscurrent stages.
(e) Infinite number of countercurrent stages.
Analysis: Let: '
K = K DUN X (i ) = X UN in aqueous liquid leaving Stage i Y (i ) = YUN in TBP liquid leaving Stage i
From Eq. (514), the extraction factor = E = KS/FW = 5.5S/80 = 0.0688S For 90% extraction, X(N)/X(F) = 1  0.90 = 0.10.
(a) For a single stage, using Eq. (513),
X (1)
1
1
= 010 =
.
=
(F)
X
1 + E 1 + 0.0688S (2) Solving Eq. (2), S = 130.8 g of TBP.
(b) For two crosscurrent stages with equal solvent additions, where S = total solvent, Eq.
X (2)
1
1
(521) applies:
= 010 =
.
=
(3)
2
2
(F)
X
1+ E / 2
1 + 0.0688S / 2
Solving Eq. (3), S = 62.9 g of TBP. Analysis (continued): Exercise 5.9 (continued) (c) For two countercurrent stages, Eq. (528) applies:
X (2)
1
1
= 010 =
.
=
(F)
2
X
1+ E + E
1 + 0.0688S + 0.0688S 2 (4) Solving Eq. (4), S = 36.9 g of TBP.
(d) For an infinite number of crosscurrent stages, Eq. (523) applies: X (∞)
1
1
= 010 =
.
=
(F)
X
exp E exp(0.0688S ) (5) Solving Eq. (5), S = 33.5 g of TBP.
(e) For an infinite number of countercurrent stages, if E < 1, Eq. (532) applies:
X ( ∞ ) / X ( F ) = 0.10 = 1 − E = 1 − 0.0688S
Solving, S = 13.1 g of TBP Exercise 5.10
Subject:
Liquidliquid extraction of uranyl nitrate (UN) from water (W) with
tributyl phosphate (TBP)
Given: 2 kg of 20 wt% UN in water. Thus, FW =1.6 kg. S = 0.5 kg of TBP.
Assumptions: Equilibrium batch contacts. W and TBP are mutually insoluble.
Y
'
.
(1)
Distribution coefficient is given in the form of Eq. (515): K DUN = UN = 55
X UN
Find: Percent recovery of UN for:
(a) Twostage batch extraction.
(b) Three batch extractions.
(c) Twostage cocurrent extraction.
(d) Threestage countercurrent extraction.
(e) Infinitestage countercurrent extraction.
(f) Infinitestage crosscurrent extraction.
Analysis: Let: '
K = K DUN X (i ) = X UN in aqueous liquid leaving Stage i
Y (i ) = YUN in TBP liquid leaving Stage i From Eq. (514), the extraction factor = E = KS/FW = 5.5(0.5)/1.6 = 1.72
(a) For a single stage, using Eq. (513),
X (1)
1
1
=
=
= 0.368
(F)
X
1 + E 1 + 1.72 (2) % extraction = (1  0.368) x 100% = 63.2%
(b) Three batch extractions is equivalent to a crosscurrent arrangement with equal
solvent additions, where S = total solvent. Eq. (521) applies: X ( 3)
1
1
=
=
= 0.257
3
3
(F)
X
1+ E / 3
1 + 172 / 3
. (3) % extraction = (1  0.257) x 100% = 74.3%
(c) Twostage cocurrent is same as one stage. Therefore, % extraction = 63.2%. Analysis (continued): Exercise 5.10 (continued) (d) For threestage countercurrent extraction, Eq. (528) applies:
X ( 3)
1
1
=
=
= 0.093
(F)
2
3
X
1+ E + E + E...
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 Spring '11
 Levicky
 The Land

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