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Unformatted text preview: = 16,520 kJ/hK
(c) From Eq. (22), LW = T0∆Sirr = 298.15(16,520) = 4,925,000 kJ/h
(d) Combining Eqs. (3) and (4) of Table 2.1,
Wmin = QR 1 − T0
T
− QC 1 − 0 − LW
Tsteam
Tcw = 26,360, 000 1 − 298.15
298.15
− 27, 300, 000 1 −
− 4, 925, 000
373.15
298.15 = 373, 000 kJ/h
(e) From Eq. (5), Table 2.1, η = 373,000/(4,925,000 + 373,000) = 0.0704
= 7.04% Exercise 2.4
Subject: Secondlaw analysis for a membrane separation of a gas mixture
Given: Component flow rates in lbmol/h for the feed. Permeate of 95 mol% H2 and 5
mol% CH4. Separation factor, SP, for H2 relative to CH4, of 47. Phase condition;
temperature in oF; enthalpy, h, in Btu/lbmol; and entropy, s, in Btu/lbmoloR for the feed,
permeate, and retentate. Infinite heat sink temperature = T0 = 539.7oR.
Assumptions: Neglect heat transfer to or from the membrane.
Find: (a) Production of entropy, ∆Sirr ,Btu/hoR
(b) Lost work, LW, Btu/h
(c) Minimum work of separation, Wmin , Btu/h
(d) Secondlaw efficiency, η
Suggest other separation methods.
Analysis:
First compute the material balance to obtain the flow rates of the retentate
and permeate. From Eq. (14), for the separation factor, SP, using the subscript P for
permeate and R for retentate,
nH 2 / nH 2
P
R
47 =
(1)
nCH 4 / nCH 4
P R For 95 mol% H2 and 5 mol% CH4 in the permeate,
nH 2
P
0.95 =
nH 2 + nCH 4
P (2) P By component material balances for H2 and CH4 around the membrane separator.
nH 2 = 3,000 = nH 2 + nH 2
F
P
R
(3) and (4)
nCH 4 = 884 = nCH 4 + nCH 4
F P R Solving Eqs. (1), (2), (3), and (4) for the 4 unknowns,
nH 2 P = 2,699.9 lbmol / h, nH 2 R = 3001 lbmol / h, nCH 4
. P = 142.1 lbmol / h, nCH 4 Therefore, nP = 2,699.9 + 142.1 + 0 = 2,842 lbmol/h
nR = 300.1 + 741.9 + 120 = 1,162 lbmol/h
Also, nF = 2,842 +1,162 = 4,004 lbmol/h R = 741.9 lbmol / h Exercise 2.4 (continued)
Analysis: (continued)
(a) From Eq. (2), the entropy balance for the membrane
∆Sirr =
out ( ns ) − in ( ns ) = 1,162(2.742) + 2,842(4.222) − 4, 004(1.520)
= 9,099 Btu/h o R
(c) From Eq. (22), LW = T0∆Sirr = 536.7(9,099) = 4,883,000 Btu/h
(d) Combining Eqs. (3) and (4) of Table 2.1,
Wmin = −LW
= −4,883, 000 Btu/h
Note the negative value. The energy of separation is supplied from the high pressure of
the feed.
(e) Because the minimum work of separation is equal to the negative of the lost
work, Eq. (5), Table 2.1 does not apply. The efficiency can not be computed unless heat
transfer is taken into account to give a permeate temperature of 80oF.
Gas adsorption could also be used to make the separation. Exercise 2.5
Subject: Expressions for computing Kvalues.
Given: Seven common thermodynamic expressions for estimating Kvalues.
Find: Expression assumptions if not rigorous
Analysis: φiL
is a rigorous expression.
φiV
φ
(b) K i = iL is not rigorous. It assumes ideal solutions so that γ iL and γ iV are 1.0.
φiV
(c) K i = φiL is not rigorous. It assumes ideal solutions and an ideal gas,
such that φiV , γ iL , and γ iV are 1.0.
γφ
(d) K i = iL iL is a rigoro...
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 Spring '11
 Levicky
 The Land

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