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Lecture18 Thursday 3/13/08
Solution to Tuesdays In-class Problem .User Friendly Energy Balance Derivations
Adiabatic (Tuesday’s lecture) .Heat Exchange Constant Ta .Heat Exchange Variable Ta Co-current .Heat Exchange Variable Ta Counter Current
Adiabatic Operation
Elementary liquid phase reaction carried out in a CSTR
The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.
(a) Assuming the reaction is irreversible, A B, (KC = 0) what reactor volume is necessary to achieve 80% conversion?
(b) If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume?(c) Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and
95% conversion. Compare with the CSTR volumes at these conversions(d) Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function
of temperature between 290K and 400K.
A B
FA0
FI
CSTR Adiabatic
Mole Balance
Rate Law
Stoichiometry
A B
T X
FA0 5mol
minT0 300K
FI 10molmin
HRx 20,000cal mol A (exothermic)
VFA0X
rA exit
rA k CA CB
KC
k k1e
E
R
1
T1 1
T
KC KC1 expHRx
R
1
T2 1
T
CA CA0 1 X
CB CA0X
Energy Balance – Adiabatic, CP = 0
T T0 HRx XiCPi
T0 HRx X
CPAICPI
T 300 20,000
164 2 18
X300 20,000
164 36X
T 300 100 X
Irreversible for Parts (a) through (c)
(a) Given X = 0.8, find T and V
rA kCA0 1 X (i.e., KC )
Calc KC
(If reversible)
Calc T Calc kŹ ŠrAGiven X Calc VCalc
VFA0X
rA exit
FA0X
kCA0 1 X
T 300 100 0.8 380K
k 0.1exp10,000
1.989
1
298 1
380
3.81
VFA0X
rA
5 0.8 3.81 2 1 0.8
2.82 dm3
(b)
Calc KC
(If reversible)
Calc X Calc kŹ ŠrAGiven T Calc VCalc
rA kCA0 1 X (Irreversible)
T 360K
X T 300
1000.6
k 1.83min 1
V5 0.6
1.83 2 0.4 2.05 dm3
(c) Levenspiel Plot
FA0
rA
FA0
kCA0 1 X
T 300 100X
Choose X Calc T Calc k Calc rACalc
FA0
rA
CSTR X = 0.6 T = 360
0
5
10
15
20
25
30
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
X
-Fa
0/R
a
CSTR 60%
CSTR X = 0.95 T = 395
PFR X = 0.6
PFR X = 0.95
Summary
CSTR X = 0.6 T = 360 V = 2.05 dm3
PFR X = 0.6 Texit = 360 V = 5.28 dm3
CSTR X = 0.95 T = 395 V = 7.59 dm3
PFR X = 0.95 Texit = 395 V = 6.62 dm3
(d) At Equilibrium
KC CBe
CAe
CA0Xe
CA0 1 Xe
Xe
1 Xe
Xe KC
1KC
KC KC2 expHR
R
1
T2 1
T
Choose T Calc KCCalc Xe , repeat
KC 1,000exp 20,000
1.987
1
290 1
T
T 290
T 330
T 350
T 370
T 390
KC 1,000
KC 14.9
KC 2.6
KC 0.95
KC 0.136
Xe 0.999
Xe 0.937
Xe 0.72
Xe 0.355
Xe 0.12
(e) Te = 358 Xe = 0.59
X CPA ICPI T T0
HRx
200
20,000 T 300
X 0.1 T 300
User Friendly Equations Relate T and X or Fi
User Friendly Equations Relate T and X or Fi
Heat Exchange
Elementary liquid phase reaction carried out in a PFR
The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.
Ta
AA B
Heat Exchange Fluid
Ý m C
FA0
FI
Rate Law (2)
(3)
(4)
Stoichiometry (5)
(6)
Parameters (7) – (15)
dXdV
rA FA0
rA k CA CB
KC
k k1 expER
1T1
1T
KC KC2 expHRx
R1T2
1T
CA CA0 1 X
CB CA0X
FA0 , k1, E, R, T1, KC2 , HRx , T2 , CA0
Mole Balance (1)
Energy Balance
Adiabatic and
(16A)
Additional Parameters (17A) & (17B)
Heat Exchange
(16B)
TT0 HRx XiCPi
CP 0
T0 , iCPiCPA
ICPI
dTdV
rA HRx Ua T Ta
FiCPi
FiCPiFA0 iCPi
CPX , if CP 0 then
dTdV
rA HRx Ua T Ta
FA0 iCPi
A. Constant Ta (17B) Ta = 300K
Additional Parameters (18B – (20B):
B. Variable Ta Co-Current
(17C)
C. Variable Ta Counter Current
(18C)
Guess Ta at V = 0 to match Ta = Tao at exit, i.e., V = Vf
Ta , iCPi, Ua
dTa
dV
Ua T Ta Ý m CPcool
, V0 Ta Tao
dTa
dV
Ua Ta T Ý m CPcool
V0 Ta ? Guess
More heat effects
T
Ta
V+ΔVV
mc
HC
FA, Fi V+ΔVV
Fi
T
Ta
Fi
TTAUQ a
Da
LD
V
LDA
WS
/44
turbine)(no 0
2
In - Out + Heat Added = 0
dV
dFH
dV
dHF
dV
HFd
TTUdV
HFd
VTTUHFHF
ii
ii
ii
aaii
aaVViiVii
0
0 VTTUQ aa
RiiaiiPiiii
aiii
Pii
RPiii
HHrHdV
dTCF
dV
HFd
rrdV
dFdV
dTC
dV
dHTTCHH
0
Pii
rg
Pii
aaaR
aaaRPii
aaaRiPi
CF
dV
dT
CF
TTUrH
dV
dT
TTUrHdV
dTCF
TTUrHdV
dTFC
0
Heat removed
Heat generated
XCCF
TTUrH
dV
dT
XCCFCXFCF
PPiiA
aaaR
PiPiiAPiiiAPii
0
00
Example: Constant Ta
Find conversion, Xeq and T as a function of reactor volume
1) Mole balance
2) Rates
3) Stoich
4) Heat effects
V
X
V
T
V
rate
X
Xeq
0A
a
F
r
dV
dX
TTR
Hkk
TTR
Ekk
kCCkr
RCC
CBAa
11exp
11exp
22
11
XCC
XCC
AB
AA
0
0 1
PIIPAPii
C
Ceq
PPiiA
aaaR
CCC
k
kX
CCF
TTUrH
dV
dT
1
0 0
Parameters
a
IPIPAA
AaaC
R
rrate
CCC
FTUkk
TTREH
, , , ,
, , , , ,
, , , , ,
0
021
21
Example: Variable Ta CoCurrent
Coolant balance:
In - Out + Heat Added = 0
0 0 ,
0
0
aaPCC
aaa
aaC
C
aaVVCCVCC
TTVCm
TTU
dV
dT
TTUdV
dHm
TTVUHmHm
dV
dTC
dV
dH
TTCHH
aPC
C
raPCCC
0
All equations can be used from before except Ta parameter, use differential Ta instead, adding mC and CPC
Example: Variable Ta Counter - Current
In - Out + Heat Added = 0
All equations can be used from before except dTa/dV which must be changed to a negative to arrive at the correct integration we must guess the Ta value at V=0, integrate and see if Ta0 matches; if not, reguess the value for Ta at V=0
PCC
aaaaa
CC
aaVCCVVCC
Cm
TTU
dV
dTTTU
dV
dHm
TTVUHmHm
0
0