counter current multistage operations with reflux
DESCRIPTION
COUNTER CURRENT MULTISTAGE OPERATIONS WITH REFLUX. SINGLE-STAGE (FLASH) DISTILLATION. Distillate D, x D. UNECONOMICAL. IDEAL STAGE. Feed F, x F. Bottom Product B, x B. L 0. V 1. 1. V 2. L 1. 2. L 2. V 3. F, x F. 3. V 4. L 3. 4. L 4. V 5. 5. L 5. V 6. - PowerPoint PPT PresentationTRANSCRIPT
IDEAL STAGEFeedF, xF
DistillateD, xD
Bottom ProductB, xB
SINGLE-STAGE (FLASH) DISTILLATION
UNECONOMICAL
Simple counter-current flow cannot give as complete a separation as required
N = More concentrated L0
uneconomical x0 is fixed by other consideration
Where does L0 come from?
Multistage cascade with reflux at both ends for distillation
V1
-qC
m
p
LC
D
F
Enrichingsection
Strippingsection
F-1
F+1
N
1
B
C
S qS
L0
Condenser
Reboiler
NL
1NV
SL
F
ZERO REFLUX
• No liquid returned to stage 1• No condensation of V2 to
supply liquid leaving stage 1• The vapor leaving stage 1
would be the same quantity and composition of the vapor leaving stage 2.
• The vapor leaving stage 2 would be the same quantity and composition of the vapor leaving stage 3.
• Etc.
V1
-qCLC
D
F F
1
C
2
V2
V3
3
Multistage cascade with no liquid reflux
F
N-2
N-1
N
F
B
S qS
NL
1NV
0LS
1NL
2NL
If the vapor reflux were eliminated:
• No vapor returned to stage N• No vaporization of to
supply vapor leaving stage N• The liquid leaving stage N
would be the same quantity and composition of the liquid leaving stage N-1.
• The liquid leaving stage N-1 would be the same quantity and composition of the liquid leaving stage N-2.
• Etc.
1NL
Multistage cascade with no vapor reflux
A fractionating column by its inherent nature has two limits of operation based upon reflux ratio:
• Minimum reflux• Total reflux
MINIMUM REFLUX
D
B
F
L0 • There is insufficient liquid returned
to the column • There is only an infinitesimal
change in vapor and liquid compositions through the plates.
• Infinite number of plates would be needed.
• Actual operation of a column below or at minimum reflux is impossible.
Schematic representation of minimum reflux operation
TOTAL REFLUX
• All condensate is returned to the column
• It requires the least number of stages.
• Practically no overhead product and no bottom product can be made and no feed is introduced.
• It is possible to operate experimentally a fractionating column at total reflux when the system inventory is large and only very small samples of distillate and bottoms are removed.
D = 0
B = 0
F = 0
Schematic representation of total reflux operation
MINIMUM REFLUX VS TOTAL REFLUX
Large reflux ratio
Small reflux ratio
• More coolant• More heating medium• Greater operating cost
• Greater number of plates
• Greater investment cost
$/un
it pr
oduc
t
Number of stages
Nmin
Total cost
Operating cost
Equipment cost
Optimum designat minimum cost
Schematic relationship between reflux ratio and number of stages
Over-all:
Component i:
(1)
(2)
(3)
(4)
(5)
V1 = L0 + D
D01 ii0i1 xDxLyV
D01 iii xxy
11
0
VD1
VL
1DV
DL 10
MATERIAL BALANCE AROUND TOTAL CONDENSER
V1 H1 + qD = L0 h0 + D hD (6)
ENTHALPY BALANCE AROUND TOTAL CONDENSER
The total heat removed in the condenser can be expressed in terms of heat per unit mass of distillate stream times the mass of stream.
qD = D QD
V1 H1 + D QD = L0 h0 + D hD
V1 H1 + (V1 – L0) QD = L0 h0 + (V1 – L0) hD
(7)
(8)
(9)
Introducing eq. (1) into eq. (8) to eliminate D yields:
D0D100D0D111 hLhVhLQLQVHV
DD0
DD1
1
0
QhhQhH
VL
01
1DD0
hHHQh
DL
(10)
(11)
V1 H1 + (V1 – L0) QD = L0 h0 + (V1 – L0) hD
DD11DD00 QhHVQhhL
Introducing eq. (1) into eq. (8) to eliminate V1 yields:
MATERIAL BALANCE IN ENRICHING SECTION
V1, y1, H1
L1, x1, h1 L0
x0
h0
DxD
hD
qD
Vm+1
ym+1
Hm+1
Lm
xm
hm
F
Over-all:
Component i:
(12)
(13)
(14)
Vm+1 = Lm + D
Dm1m iimi1m xDxLyV
Dm1m im1mimi1m xLVxLyV
Dm
D1m
ii
ii
1m
m
xxxy
VL
Introducing eq. (12) into eq. (13) to eliminate D results in
D1mDm ii1miim xyVxxL
(15)m1m
1mD
ii
iim
xyyx
DL
Introducing eq. (12) into eq. (13) to eliminate Vm+1 results in:
Dm1m iimim xDxLyDL
1mDm1m iiiim yxDxyL
ENTHALPY BALANCE IN ENRICHING SECTION
(16)DmmD1m1m hDhLQDHV
Dm1mmmDm1m1m1m hLVhLQLVHV
Introducing eq. (12) into eq. (16) to eliminate D results in
DmD1mmmDmD1m1m1m hLhVhLQLQVHV
DD1m1mDDmm QhHVQhhL
DDm
DD1m
1m
m
QhhQhH
VL
(17)
m1m
D1mDm
hHQHh
DL
(18)
Introducing eq. (12) into eq. (16) to eliminate Vm+1 results in:
DmmD1mm hDhLQDHDL
DmmD1m1mm DhhLDQDHHL
D1mDm1mm QHhDhHL
Partial condenser
MATERIAL AND ENTHALPY BALANCES AROUND PARTIAL CONDENSER
V1, y1, H1
L1, x1, h1
L0
x0
h0
DyD
HD
qD
Over-all:
Component i:
(19)
(20)
(21)
V1 = L0 + D
D01 ii0i1 yDxLyV
0
DD01
i
iDiii x
yK;xxy
D01 i01i0i1 yLVxLyV
(22) D0
D01
D0
D1
ii
iii
ii
ii
1
0
K1xKxy
yxyy
VL
D1D0 ii1ii0 yyVyxL
ENTHALPY BALANCE:
V1 H1 + qD = L0 h0 + D HD (24)
The total heat removed in the condenser (qD) can be expressed in terms of heat per unit mass of distillate stream times the mass of stream.
qD = D QD
V1 H1 + D QD = L0 h0 + D HD
(25)
(26)
V1 H1 + (V1 – L0) QD = L0 h0 + (V1 – L0) HD
DD11DD00 QHHVQHhL
DD0
DD1
1
0
QHhQHH
VL
(27)
Replacing D in equation (26) with (V1 – L0):
01
1DD0
hHHQH
DL
(28)
(L0 + D) H1 + D QD = L0 h0 + D HD
1DD010 HQHDhHL
Replacing V1 in equation (26) by (L0 + D):
MATERIAL BALANCE IN ENRICHING SECTION WITH PARTIAL CONDENSER
V1, y1, H1
L1, x1, h1
L0
x0
h0
DyD
hD
qD
Vm+1
ym+1
Hm+1
Lm
xm
hm
F
m
Over-all:
Component i:
(29)
(30)
(31)
Vm+1 = Lm + D
Dm1m iimi1m yDxLyV
Dm1m im1mimi1m yLVxLyV
Dm
D1m
ii
ii
1m
m
yxyy
VL
D is eliminated from equation (30) by substituting D with (Vm+1 – Lm):
D1mDm ii1miim yyVyxL
(32)m1m
1mD
ii
iim
xyyy
DL
Dm1m iimim yDxLyDL
Vm+1 is eliminated from equation (30) by substituting Vm+1 with Lm + D
1mDm1m iiiim yyDxyL
ENTHALPY BALANCE IN ENRICHING SECTION
V1, y1, H1
L1, x1, h1
L0
x0
h0
DxD
hD
qD
Vm+1
ym+1
Hm+1
Lm
xm
hm
F
(33)
(34)
DmmD1m1m HDhLQDHV
Dm1mmmDm1m1m1m HLVhLQLVHV
D is eliminated from equation (33) by substituting D with Vm+1 – Lm
DDm
DD1m
1m
m
QHhQHH
VL
DD1m1mDDmm QHHVQHhL
Vm+1 is eliminated from equation (33) by substituting Vm+1 with Lm + D
DmmD1mm HDhLQDHDL
m1m
D1mDm
hHQHH
DL
D1mDm1mm QHHDhHL
(35)
Over-all: BVL 1pp
Component i:B1pp ii1pip xByVxL
B1pp i1ppi1pip xVLyVxL
Bp
B1p
ii
ii
1p
p
xx
xy
VL
(36)
(37)
(38)
Replacing B in equation (37) with 1pp VL
B1pBp ii1piip xyVxxL
ENTHALPY BALANCE:
B1p1pBpp hBHVqhL
qB = B QB
B1p1pBpp hBHVQBhL
B1pp1p1pB1pppp hVLHVQVLhL
BB1p1pBBpp QhHVhQhL
BBp
BB1p
1p
p
hQhQhH
VL
(39)
(40)
(41)
Over-all: BVL 1NN
Component i:B1NN ii1NiN xByVxL
B1NN i1NNi1NiN xVLyVxL
BN
B1N
ii
ii
1N
N
xxxy
VL
(42)
(43)
(44)
Replacing B in equation (43) with 1NN VL
B1pBp ii1piip xyVxxL
MATERIAL BALANCE AROUND THE FEED PLATE
F = FV + FL
LV FFF HHh
LV FFF xyx
1pV
Hp+1
yp+1
pLhp
xp
Vm+1
Hm+1
ym+1
Lm
hm
xm
Over-all:
1mpm1pLV VLLVFF
Component i:
1mpm1pLV i1mipimi1pF,iLF,iV yVxLxLyVxFyF
1m1pV VVF
pmL LLF
(45)
(48)
(46)
(47)
1mp
m1pLV
i1mimL
imiV1mF,iLF,iV
yVxLF
xLyFVxFyF
(49)
V1pLp1p1mpm F,iiVF,iiLii1miim yyFxxFyyVxxL
pm
V1p
pm
Lp
pm
1p1m
ii1m
F,iiV
ii1m
F,iiL
ii
ii
1m
m
xxV
yyF
xxV
xxF
xx
yy
VL
If the feed is a saturated liquid, the last term in eq. (49) drops out.
If the feed is a saturated vapor, the middle term on the right side of eq. (49) drops out.
ENTHALPY BALANCE AROUND THE FEED PLATE
pp1m1mmm1p1pFLFV hLHVhLHVhFHFLV
pmL1m1mmm1pV1mFLFV hLFHVhLHFVhFHFLV
VL F1pVFpL1p1m1mpmm HHFhhFHHVhhL
pm
F1p
1m
V
pm
Fp
1m
L
pm
1p1m
1m
m
hhHH
VF
hhhh
VF
hhHH
VL VL
(50)
(51)
• If the feed is a saturated liquid, the last term in eq. (51) drops out.
• If the feed is a saturated vapor, the middle term on the right side of eq. (51) drops out.