Unformatted text preview: ponent flow rates, ni , of feed and product 1, in kmol/h. Phase condition;
temperature in K; enthalpy, h, in kJ/kmol; and entropy, s, in kJ/kmolK for feed, product
1, and product 2. Infinite heat sink temperature = T0 = 298.15 K.
Find: Minimum work of separation, Wmin , in kJ/h
Analysis: From Eq. (4), Table 2.1,
Wmin = nb −
out From Eq. 21, nb
in b = h − T0 s For the feed stream (in), n = 30 + 200 + 370 + 350 + 50 = 1,000 kmol/h
b = 19,480  (298.15)(36.64) = 8,556 kJ/kmol
For product 1 (out), n = 30 + 192 + 4 + 0 + 0 = 226 kmol/h
b = 25,040  (298.15)(33.13) = 15,162 kJ/kmol
For product 2 (out), n = nfeed  nproduct 1 = 1,000  226 = 774 kmol/h
b = 25,640  (298.15)(54.84) = 9,289 kJ/kmol
From Eq. (4), Table 2.1,
Wmin = 226(15,162) + 774(9,289)  1,000(8,556) = 2,060 kJ/h Exercise 2.2
Subject: Minimum work for separating a mixture of ethylbenzene and xylene isomers.
Given: Component flow rates, ni , of feed ,in lbmol/h. Component split fractions for
three products, Phase condition; temperature in oF; enthalpy, h, in Btu/lbmol; and
entropy, s, in Btu/lbmoloR for feed and three products. Infinite heat sink temperature =
T0 = 560oR.
Find: Minimum work of separation, Wmin , in kJ/h
Analysis: From Eq. (4), Table 2.1,
Wmin = nb −
out From Eq. 21, nb
in b = h − T0 s For the feed stream (in), n = 150 +190 + 430 + 230 = 1,000 lbmol/h
b = 29,290  (560)(15.32) = 20,710 Btu/lbmol
For product 1 (out), using Eq. (12),
n = 150(0.96) + 190 (0.005) + 430(0.004) = 146.7 lbmol/h
b = 29,750  (560)(12.47) = 22,767 Btu/lbmol
For product 2 (out), using Eq. (12),
n = 150(0.04) + 190(0.99) + 430(0.99) + 230(0.015) = 623.3
lbmol/h
b = 29,550  (560)(13.60) = 21,934 Btu/lbmol
For product 3 (out), by total material balance,
n = 1,000  146.7  623.3 = 230 lbmol/h
b = 28,320  (560)(14.68) = 20,099 Btu/lbmol
From Eq. (4), Table 2.1,
Wmin = 146.7(22,767) + 623.3(21,934) + 230(20,099)  1,000(20,710) = 924,200 Btu/h Exercise 2.3
Subject: Secondlaw analysis of a distillation column
Given: Component flow rates, ni , from Table 1.5 for feed, distillate, and bottoms in
kmol/h for column C3 in Figure 1.9. Condenser duty, QC ,= 27,300,00 kJ/h. Phase
condition; temperature in K; enthalpy, h, in kJ/kmol; and entropy, s, in kJ/kmolK for
feed, distillate and bottoms. Infinite heat sink temperature = T0 = 298.15 K. Condenser
cooling water at 25oC = 298.15 K and reboiler steam at 100oC = 373.15 K.
Assumptions: Neglect shaft work associated with column reflux pump.
Find: (a)
(b)
(c)
(d)
(e) Reboiler duty, QR , kJ/h
Production of entropy, ∆Sirr , kJ/hK
Lost work, LW, kJ/h
Minimum work of separation, Wmin , kJ/h
Secondlaw efficiency, η Analysis: (a) From Eq. (1), the energy balance for column C3, QR = QC − nh +
in nh
out = 27,300,000  445.5(17,000) + 175.5(13,420) + 270(15,840)
= 26,360,000 kJ/h
(b) From Eq. (2), the entropy balance for column C3,
∆Sirr = ns +
out Q
−
Ts ns +
in Q
Ts = 175.5(5.87) + 270(21.22) + 27,300,000
26, 360, 000
− 445.5(25.05) −
298.15
373.15...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details