Unformatted text preview: F
2 2 nF
nP2 (16) The final equations are (6) combined with (12) and (14) through (16).
(c) From Eq. (13),
yA P nP1
1 yA P nP2 =9 (17) 2 and, yBP nP1
1 yBP nP2
2 = 1
9 (18) Combining (17) and (18), with component and total material balances around
the separator gives the following equations that can be used with Eq. (6): yA P = 0.9 yA F
1 nF
nP1 (19) Exercise 2.9 (continued)
Analysis: (c) (continued)
yA P = 01yA F
.
2 nF
nP2 (20) nP1 = nF 0.8 yA F + 01
. (21) nP2 = nF 0.9 − 0.8 yA F (22) (d) From Eq. (15), SPA,B = SFA / SFB
= 361
1 − SFA / 1 − SFB (23) SFA = 0.95 is given. Combining this with Eq. (23), gives SFB = 0.05.
This part then proceeds as in part (b) to give: nP1 = nF 0.90 yA F + 0.05 (24) nP2 = nF − nP1 or nP2 = nF 0.95 − 0.90 yA F (25)
yA P = 0.95 yA F
1 yA P = 0.05 yA F
2 nF
nP2 nF
nP1 (26) (27) Equations (24) through (27) are combined with Eq. (6).
A spreadsheet can be used to compute the dimensionless minimum work, Wmin
RT0nF
with the following results for Parts (a) through (d): Exercise 2.9 (continued)
From the plot on the previous page, it is seen that the dimensionless minimum work is
very sensitive to the feed mole fraction and to the product purities.
From the derivation of the minimum work equations, it is seen that they are independent
of the separation method and only depend on thermodynamics.
To prove that the largest value of Wmin occurs for a feed with equimolar quantities of A
and B, consider the case of a perfect separation, part (a), as given by Eq. (11).
Let W = Wmin/nFRT0 and y = yA F . Then, the derivative of W with respect to y
is, dW
y
−1
= − ln y + − ln(1 − y ) + (1 − y )
dy
y
1− y
= − [ ln y − ln(1 − y ) ]
For min/max, set the derivative to zero and solve for y.
Therefore,
dW
= 0 = − [ ln y − ln(1 − y ) ]
dy
Solving, y = 1 − y or y = 0.5. This is an equimolar feed.
Furthermore, it gives a maximum value of W . Exercise 2.10
Subject: Relative volatility of the isopentanenormal pentane system
Given: Experimental data for relative volatility 125250oF. Vapor pressure constants.
Find: Relative volatilities from Raoult's law over the same temperature range and
compare them to the experimental values.
Analysis: Combining Eq. (5), Table 2.3, for Raoult's law, with Eq. (221) for α .
s
PiC
α iC5 ,nC5 = s 5
(1)
PnC5
Only the first three constants of the extended Antoine equation are given. (T=K, P= kPa)
2345.09
s
PiC5 = exp 13.6106 −
T − 40.2128
(2), (3)
2554.60
s
PnC5 = exp 13.9778 −
T − 36.2529
Using a spreadsheet to calculate the relative volatility from Eqs. (1), (2), and (3):
s
T, F
T, K
Raoult's law
Expt. α
PiC5 , kPa
PnsC5 , kPa
α
125
1.26
325
214
167
1.28
150
1.23
339
314
251
1.25
352
446
364
1.22
175
1.21
200
1.18
366
614
512
1.20
225
1.16
380
823
700
1.18
250
1.14
394
1079
834
1.16
The Raoult's law values are within 2% of the experimental values. Exercise 2.11
Subject: Condenser duty of a vacuum distillation column separating ethyl benzene (EB)
and styrene (S).
Given: Phase condition, temperature, pressure, flow rate, and compositions for streams
entering and leaving a condenser, which produces subcoole...
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 Spring '11
 Levicky
 The Land

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