REPORT TO DEPARTMENT OF CHEMICAL ENGINEERING EGE UNIVERSITY FOR COURSE: CHE386 CONCEPTUAL DESIGN II
DESIGN REPORT I CSTR DESIGN FOR ETHYL ACETATE PRODUCTION
SUBMITTED TO Prof. Dr. Ferhan ATALAY
SUBMISSION DATE 08/03/10
GROUP 3 05070008901 05070008103 05070008849 05060008091 05060008017
Ürün ARDA Berna KAYA Demet ACARGİL M. Serkan ACARSER Tayfun EVCİL
SUMMARY
This report is about the production ethyl acetate by the esterification reaction of acetic acid and ethanol. Both components are in aqueous solution; the acetic acid is 96% pure and ethanol is 96.5% pure. The reactants are fed to a CSTR at 750 C. The products are also at 750C. In the CSTR calculation which is the main part of the report, six-blade turbine is chosen, and the motor power was calculated as 4.441 kW . A CSTR, most commonly, is heated by either a jacket or an internal coil. In jacket calculations heat transfer area was found as 6.36 m2, mass flow as 60.434 kg/s, T out as 197.94 0C , h i as 574.043, h o as 921.845 and U 0 as 347.32 W/m2K. In coil calculations heat transfer area was found as 3.14 m2, mass flow as 3.537 kg/s, T out as 164.866 0C, h i as 2001.16 , h o as 903.283 and U 0 as 818.318 W/m2K. Finally, the number of coils was calculated as 6. Necessary calculations for both jacket and coil were performed and the necessary comparisons were made in discussion part.
i
TABLE OF CONTENTS
Summary ....................................................................................................................... i 1.0 Introduction ........................................................................................................... 1 2.0 Results ..................................................................................................................... 3 3.0 Discussion ................................................................................................................ 8 4.0 Nomenclature ....................................................................................................... 11 5.0 References ............................................................................................................ 13 6.0 Appendix .............................................................................................................. 14
ii
1.0 INTRODUCTION Ethyl acetate (systematically, ethyl ethanoate, commonly abbreviated EtOAc or EA) is the organic compound with the formula CH 3 COOCH 2 CH 3 . This colorless liquid has a characteristic sweet smell (similar to pear drops) like certain glues or nail polish removers, in which it is used. Ethyl acetate is the ester of ethanol and acetic acid; it is manufactured on a large scale for use as a solvent. In 1985, about 400,000 tons were produced yearly in Japan, North America, and Europe combined.In 2004, an estimated 1.3M tons were produced worldwide. PRODUCTION Ethyl acetate is synthesized industrially mainly via the classic Fischer esterification reaction of ethanol and acetic acid. This mixture converts to the ester in about 65% yield at room temperature: CH 3 CH 2 OH + CH 3 COOH ⇌ CH 3 COOCH 2 CH 3 + H 2 O
The reaction can be accelerated by acid catalysis and the equilibrium can be shifted to the right by removal of water. It is also prepared industrially using the Tishchenko reaction, by combining two equivalents of acetaldehyde in the presence of an alkoxide catalyst: 2 CH 3 CHO → CH 3 COOCH 2 CH 3 By dehydrogenation of ethanol A specialized industrial route entails the catalytic dehydrogenation of ethanol. This method is less cost effective than the esterification but is applied with surplus ethanol in a chemical plant. Typically dehydrogenation is conducted with copper at an elevated temperature but below 250 °C. The copper may have its surface area increased by depositing it on zinc, promoting the growth of snowflake, fractal like structures (dendrites). Surface area can be again increased by deposition onto a zeolite, typically ZSM-5. Traces of rare earth and alkali metals are beneficial to the process. Byproducts of the dehydrogenation include diethyl ether, which is thought to primarily arise due to aluminum sites in the catalyst; acetaldehyde and its aldol products; higher esters; and ketones. Separations of the byproducts is complicated by the fact that ethanol forms an azeotrope with water, as does ethyl acetate with ethanol and water, and methyl ethyl ketone (MEK, which forms from 2-butanol) with both ethanol and ethyl acetate. These azeotropes are "broken" by pressure swing distillation or membrane distillation. USES Ethyl acetate is primarily used as a solvent and diluent, being favored because of its low cost, low toxicity, and agreeable odor. For example, it is commonly used to clean circuit boards and in some nail varnish removers (acetone and acetonitrile are also used). Coffee beans and tea leaves are decaffeinated with this solvent. It is also used in paints as an activator or hardener.Ethyl acetate is present in confectionery, perfumes, and fruits. In perfumes, it evaporates quickly, leaving but the scent of the perfume on the skin. -1-
LABORATORY USES In the laboratory, mixtures containing ethyl acetate are commonly used in column chromatography and extractions. Ethyl acetate is rarely selected as a reaction solvent because it is prone to hydrolysis and transesterification. In organic chemistry, especially in experiment, since ethyl acetate is very volatile and with low boiling point, it can be removed by compressed air in a hot water bath. OCCURRENCE IN WINES Ethyl acetate is the most common ester in wine, being the product of the most common volatile organic acid — acetic acid, and the ethyl alcohol generated during the fermentation. The aroma of ethyl acetate is most vivid in younger wines and contributes towards the general perception of "fruitiness" in the wine. Sensitivity varies, with most people having a perception threshold around 120 mg/L. Excessive amounts of ethyl acetate are considered a wine fault. Exposure to oxygen can exacerbate the fault due to the oxidation of ethanol to acetaldehyde, which leaves the wine with a sharp vinegar-like taste. OTHER USES In the field of entomology, ethyl acetate is an effective asphyxiant for use in insect collecting and study. In a killing jar charged with ethyl acetate, the vapors will kill the collected (usually adult) insect quickly without destroying it. Because it is not hygroscopic, ethyl acetate also keeps the insect soft enough to allow proper mounting suitable for a collection.
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2.0 RESULTS
Table 1. Assuming and reference data of coil, jacket and oil
Coil Assuming Data
Jacket Ref. Data
Assuming Data
Ref. Data
υ coil [m/s]
3
k [W/mK]
0.1105
υ jacket [m/s]
0.6
Cp oil [J/kg.K]
2204
T in [oC]
200
Cp [J/kg.K]
2147.4983
T in [oC]
200
ρ oil [kg/m3]
897.6
d o [in]
1.9
μ [Pa.s]
0.86*10-3
t shell [cm]
1.5
k oil [W/m.K]
0.109
d i [in]
1.61
ρ [kg/m3]
910.2406
t jacket [cm]
2.3
μ oil [Pa.s]
0.73*10-3
Table 2. Agitator calculation Agitator Calculation Results D tank [m]
H tank [m]
d ag [m]
E [m]
Re
Np [from fig.]
P [kW]
1.5
1.95
0.5
4.5
1*106
7
1.537
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P act [kW] P motor [kW] 3.074
4.441
Table 3. Properties of the components Properties
AcA
EtOH
EtAc
H2O
Cp [50°C] [J/kgK]
2160
2670
2020
4180
Cp [75°C] [J/kgK]
2280
2960
2120
4190
ΔH f [J/mol]
-486180
-277630
-463200
285840
Table 4. Density Correlations for Components (75°C)
Density Correlations for Components (75°C) T
MW
C1
C2
C3
C4
ρ (kmol/m3)
ρ (kg/m3)
AcA
348
60.0520
1.449
0.25892
591.95
0.2529
16.4726
989.2111
EtOH
348
46.0680
1.629
0.27469
514
0.23178
16.0271
738.3386
EtAc
348
88.1050
0.900
0.25856
523.3
0.278
9.4389
831.6123
Water
348
18.0150
-13.851
0.64038
-0.00191
1.8211E-06
54.4414
980.7616
-4-
Table 5. Viscosity Correlations for Components (75°C)
Viscosity Correlations for Components (75°C) T
C1
C2
C3
C4
C5
μ (Pa.s)
AcA
348
-9.0300
1212.300
-0.322
-
-
5.93*10-4
EtOH
348
7.8750
781.980
-3.0418
-
-
4.62*10-4
EtAc
348
14.3540
-154.600
-3.7887
-
-
2.58*10-4
Water
348
-52.8430
3703.600
5.866
-5.88E-29
10
3.81*10-4
Table 6. Properties of the components after mixing, at 75°C ρ,mix [kg/m3]
μ,mix [Pa.s]
k,mix [W/mK]
Cp,mix (Pa.s)
878.489
0.000429
0.15
2898.1
Table 7. Flow rates of components F AcA [kmol/min]
F EtOH [kmol/min]
F EtAc [kmol/min]
F Water [kmol/min]
F TOTAL [kmol/min]
0.636
0.745
0.453
0.54
2.374
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Table 8. Concentrations of components at the exit C AcA [kmol/m3]
C EtOH [kmol/m3]
C EtAc [kmol/m3]
C Water [kmol/m3]
10.411
11.959
7.273
8.698
-rAcA = 0.152 kmol/m3min V liquid = 2.983 m3 V tank= 3.43 m3
Table 9. Mole fraction of the components X AcA
X EtOH
X EtAc
X Water
0.27
0.31
0.19
0.23
Table 10. Weight fraction of the components X AcA
X EtOH
X EtAc
X Water
0.31
0.28
0.32
0.09
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Table 11. Results from calculations Coil
Jacket
Results
Results
m [kg/s] ΔT [0C] T out [0C] h i [W/m2.K] h ic [W/m2.K] h o [W/m2.K] U o [W/m2.K] D’ [m] Perimeter of one coil [m] A req [m2] # of coil H coil [m]
3.536 35.134 164.886 2001.1634 10266.838 903.282 818.318 1.1 3.45 3.0403 6 1
ΔH f [W] ΔH R [W] ΔH P [W] Q=ΔH rxn [W] D ji [m] D io [m] T out [0C] G [kg/m2s] d eq [m] h i [W/m2K] h o [W/m2K] U o [W/m2K] A req [m2] A o,cal [m2]
-7-
268321 -240722 246276 273875 1.530 1.576 197.944 538.834 0.0933 574.043 921.845 347.32 6.36 9.189
3.0 DISCUSSION A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics. The concentrations of reactants and products in an equilibrium mixture are determined by the analytical concentrations of the reagents (A and B or C and D) and the equilibrium constant K c . The magnitude of the equilibrium constant depends on the Gibbs free energy change for the reaction. So, when the free energy change is large (more than about 30 kJ mol-1), then the equilibrium constant is large (log K > 3) and the concentrations of the reactants at equilibrium are very small. Such a reaction is sometimes considered to be an irreversible reaction, although in reality small amounts of the reactants are still expected to be present in the reacting system. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, for example, as does carbon dioxide (volatile) in the reaction. In this case, the reversible reaction is between acetic acid and ethyl acetate, such that:
To make this endothermic reaction irreversible, the water must be removed from the system at all times. However, in CSTR, which has a closed top, this is not possible. So the reaction’s conversion becomes a percentage of the equilibrium reaction, which changes between 80-90% in ideal cases. 80% of the equilibrium conversion was assumed for this report. The reaction’s conversion depends on many factors, such as temperature, the weight percentage of the catalyst in reaction (the catalyst in this reaction is H 2 SO 4 , with a weight percentage of 1.91%), presence of inert in the system, purity of the components and whether the tank is perfectly mixed. In this report, the components were taken as aqueous solutions, with mole percentage of acetic acid being 96%, and that of ethanol’s percentage being 96.5%. These factors also affect the reactor tank’s volume, due to the CSTR design equation. With mole balance and CSTR design equation, the inlet and outlet molar rates were calculated, and the inlet water of the solutions was added to the effluent water.
-8-
Concentrations were calculated from the rate equation. The k values of the rate equation depend on temperature and the weight percentage of the catalyst in the system. Upon finding these k and concentration values the rate of acetic acid’s formation was calculated. After finding the rate, the reaction volume was calculated as 2.983 m3. The safety factor was added to the reaction volume and the tank volume was found as 3.43 m3. After finding the volume, the diameter and the height of the tank was calculated as 1.5 and 1.95 m respectively. The diameter of the agitator was calculated as 0.5 m. To heat the coil, many options are available. One can heat the components at a separate heat exchanger before feeding to the reactor, a jacket can be used, or a certain number of coils can be installed in the reactor. Choosing the heating fluid is very important, because choosing saturated steam (hence using steam generator) might cause additional expenses. Saturated steam is used in bigger industries in order to take advantage of its heating, and to generate electricity by means of steam turbines. Hot oil is a cheaper fluid to obtain, and for this reason hot oil was used as heating fluid. The velocity of the hot oil affects the heating of the reactor, as the velocity affects the Reynold’s number, the Nusselt number, and the convection heat coefficient as a result. The pipes and pumps can be picked to obtain the desired mass flow rate. The Nusselt number of the jacket and coils were calculated using the Chilton-Drew-Jebend’s correlation. The wall thickness of the jacket and reactor affects heat transfer because the mass flow rate, the heat transfer area and the temperature difference change. The jacket was calculated first, which can be seen in appendix. To calculate the heat requirement for this endothermic reaction, hypothetical steps formed such that the reactor’s temperature, which was 75°C, was virtually dropped down to 25°C; and the reaction take place in this temperature. The products and the remaining components are then heated up to 75°C. Since the temperature of the entering and exiting components were 75°C, the only remaining factor was ΔH rxn in energy balance equation. The formation enthalpies at 25°C were found from references. The velocity of hot oil was assumed as 0.6 m/s, and then the temperature difference was calculated between the entrance and the exit of the jacket as 2.05 0C. If a higher velocity had been assumed, the mass flow rate would increase, required heat transfer area decrease and the pressure drop would change, a pump and pipe with a bigger diameter would be needed and this would cause more expensive operations and the temperature difference will be affected. After finding the velocity and mass flow rate, using the correlations, h o was calculated as 921.845 W/m2K. The viscosity, specific heat and thermal conductivity of hot oil were taken at 200°C. With these data, h i was calculated as 574.043 W/m2K. After that U o was found as 347.32 W/m2K, and the area necessary for heating was calculated as 6.36 m2. The jacket satisfies the reactor. -9-
The same steps were followed for the coil, only that the number of coils was additionally calculated as 6, with the coils being a certain distance such as 20 cm away from the tank wall. Using coil for this reactor is a better option, because coil has a smaller area which lessens the required amount of hot oil as 3.536 kg/s , therefore making the reaction costly efficient. In the agitation systems, we chose open turbine agitator with six-bladed impeller with four baffles. We chose our propeller speed as 2 rps, and the motor power needed was calculated as 4.441 kW. The safety factor and efficiency were also added for finding the actual power.
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4.0 NOMENCLATURE Fi :
Molar flowrate of ith component
X:
Conversion
T:
Temperature
Ci:
Concentration of ith component
[kmol/min] [oC] [kmol/m3]
D T : Tank diameter
[m]
H T : Height of tank
[m]
N:
Rotational speed
[rps]
N P : Power number E:
Distance between reactor bottom and impeller
[m]
D’:
Diameter of one coil
[m]
Po:
Operating power
[W]
Q:
Heat taken/given from the reactor
[W] [W/m2.K]
U 0 : Overall heat transfer coefficient D ji :
Inner diameter of jacket
[m]
D jo : Outer diameter of jacket
[m]
G:
[kg/m2.s]
Mass flux
Re: Reynolds number V tank : Volume of tank
[m3]
V liq : Volume of reaction mixture
[m3]
d ag : Agitator diameter
[m]
-r A : Rate of reaction with respect to component A k :
[kmol/m3min]
Specific reaction rate
m i : Mass flowrate of ith component
[kg/h]
do:
Outer diameter of coil
[m]
di:
Inner diameter of coil
[m]
hi:
Convective heat transfer coefficient inner fluid
[W/m2.K]
ho:
Convective heat transfer coefficient outer fluid
[W/m2.K]
- 11 -
Greeks
ρ:
Density
µ:
Viscosity
υi :
Volumetric flowrate of ith component
[kg/m3] [Pa.s] [m3/min]
η : Efficiency ΔH: Enthalpy of out/ in/ reaction
[W]
Subscripts i:
ith component
i:
Inner
o:
Outer
ag:
Agitator
- 12 -
5.0 REFERENCES
1. Richard M.Felder, Ronald W.Rousseau, 2000, Elementary Principles of Chemical Processes, 3rd Ed., John Wiley & Sons, Inc., USA. 2. Sümer Peker, Şerife Ş.Helvacı, 2003, Akışkanlar Mekaniği – Kavramlar, Problemler, Uygulamalar, 1st Ed., Literatür Yayıncılık, İstanbul. 3. Perry’s R. H., Chilton, 2008, Chemical Engineers’ Handbook, 8th Edition, Mc GrawHill Kokagusha, Tokyo. 4. Warren L. McCabe, Julian C.Smith, Peter Harriot, 1993, Unit Operations of Chemical Engineering, 5th Ed., McGraw-Hill, Singapore 5. Incropera, P.F., DeWitt, D.P., 2007, Fundamentals of Heat and Mass Transfer, 6th Ed., John Wiley & Sons, Inc., Canada. 6. J.M.Smith, H.C.Van Ness, M.M.Abbott, 2005, Introduction to Chemical Engineering Thermodynamics, 7th Ed., McGraw-Hill, Singapore 7. Octave Levenspiel, 1999, Chemical Reaction Engineering, 3rd Ed., John Wiley & Sons, Inc.,USA. 8. Atalay, F.S., 1994, "Kinetics of the Esterification Reaction Between Ethanol and Acetic Acid", Developments in Chemical Engineering and Mineral Processing, Vol.2, p.181-184. 9. http://www.processglobe.com/Liquid_Specific_Heat.aspx
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6.0 APPENDIX Calculation of the Volume of the CSTR
Feed Stream T=750C
Outlet Stream T=750C
Figure 1. A typical CSTR k1 CH 3 COOH + C2 H 5 OH CH 3 CO2 C2 H 5 + H 2 O k 2
+
A
B
k1 k
+
C
D
2
0 = = Treaction 75 C , θ B 1.1
k1
−3 = 22.63*10 1 min , k2 1.55*10−3 m3 kmol.min
= ; −rA 22.63*10−3 * C A0.5CB0.5 − 1.55*10−3 CC CD
= −rA k1 * C A0.5CB0.5 − k2 CC CD Name Acetic Acid
Composition A
Initial FA
Change − FA * x
FB
− FA * x
—
+ FA * x
0
Ethanol
B
0
Ethyl Acetate
Reference 8
C
0
0
0
Water
D
FD
0
FC = 21000 FC
+ FA * x 0
0
= FB FA (θ B − x ) 0
kg kmol = = kg min , FC 39.954 39.954 * 0.453 kmol min min 88.1 kg
= FC FA= * x ; 0.453 F= *0.416 , FA 1.09 kmol min A 0
0
0
FA == FA *0.96 ; 1.09 F= *0.96 ; FA 1.135 kmol min A 0, feed
0, feed
0, feed
- 14 -
FC = FA * x 0
= FD FA (θ D − x )
1h 1000 kg ton 1 year 1 day * * * * year 365 day 24 h 60 min 1 ton
= x xe *0.8 = = 0.416 , xe 0.52 = ; x 0.52*0.8
0
Remaining = FA FA (1 − x )
0
θ= 1.1 = B
FB
FB= 1.09*1.1 = 1.199 kmol min
0
FA
0
0
FB = FA (θ B − x ) , FB = 1.09 (1.1 − 0.416 ) ; FB = 0.745 kmol min 0
FB
0
F= *0.965 , 1.199 F= *0.965 , FB 1.242 kmol min B B 0, feed
0, feed
FH O , from AcA = FA 2
0, feed
− FA = 1.135 − 1.09= 0.0454 kmol min 0
FH O , from EtOH = FB 2
0, feed
0, feed
− FB = 1.242 − 1.199= 0.0435 kmol min 0
FH O , feed = FH O , from AcA + FH O , from EtOH = 0.0454 + 0.0435 ; FD = FH O , feed = 0.0889 kmol min 2
2
2
0
2
FD = FD + FA x ; FD =+ 0.0889 1.09*0.416 ; FD = 0.542 kmol min 0
0
FA = FA (1 − x ) , FA = 1.09 (1 − 0.416 ) ; FA = 0.636 kmol min
See in Table 7
0
FTotal = FA + FB + FC + FD , FTotal = 0.636 + 0.745 + 0.453 + 0.542 ; FTotal = 2.376 kmo min l
Design Equaition of the CSTR ; V = CA = 0
FA 0 , CA =
ν0
θ B 1.1 =
FA
0
FA x 0 , (ν ν 0 in liquid phase ) = −rA
, CA =
FA * M w, A *1 ρ A
0
0
,
0
ρA
Reference 7
M w, A
FD 0.0889 0 = = 0.0815 FA 1.09
θD =
0
3
CA
0
1050 kg m ; C A 17.485 kmol m3 = 0 60.05 kg kmol
C= C A (1 − x= ) 17.485 (1 − 0.416 ) ; A
C= 10.211 kmol m3 A
CB C A (θ B −= x ) 17.485 (1.1 − 0.416 ) ; =
CB 11.959 kmol m3 =
0
0
CC C= x 17.485*0.416 ; = A 0
CC 7.273 kmol m =
3
CD C A (θ D += x ) 17.485 ( 0.0815 + 0.416 ) ; = CD 8.698 kmol m3 = 0
−rA
22.63*10−3 *10.2110.5 *11.9590.5 − 1.55*10−3 *7.273*8.698
−rA = 0.152 kmol m3 .min
- 15 -
See in Table 8
1.09*0.416 ; Vliquid 2.983 m3 = 0.152
Vliquid
D 3 = = = Assuming Vtank 1.15 Vliquid , H 1.3 D and d ag 1.15* 2.983 ; Vtank 3.43 m3 =
Vtank
π
= Vtank
π
See in Table 8
*1.3* D 3 = D 2 H ; 3.43 4 4 D = 1.497 m , D 1.5 m
; H 1.95 m = = H 1.3*1.5 = d ag
1.5 ; = d ag 0.5 m 3
E 1 = D 3
;
See in Table 2
= = 4.5 m E 1.5*3
Reference 4
Calculation of Motor Power
= = , Pact 2= = Assuming N 120 rpm 2rps= P , η 0.9 , ( safety factor ) 1.3
ρ mix Nd ag2 878.489* 2*0.52 = Rec = µmix 4.29*10−4
Re 1*106
From N P vs Rec figure N P is found as 7 ; NP =
P 7 = ρ mix * N 3 * d ag5 See in Table 2
P = ( 7 *878.489* 23 *0.55 ) P = 1537.35 W Pact 2*1537.35 = = 3074.7 W
= Pmotor
Pact * ( safety factor ) 3074.7 *1.3 = η 0.9
= = Pmotor 4441.23 W
4.441 kW
- 16 -
Design of Jacket by Heating Process
Feed Stream Tinitial=750C
Outlet Stream Tfinal=750C
Hot oil Tin=2000C
Hot oil Tout= ?
Figure 2. A typical CSTR with Jacket Total energy balance between reaction zone and jacket;
Q = ∆H out − ∆H in + ∆H rxn , ∆T = (75 − 75) = 0 So; ∆H out = ∆H in = 0 ⇒ Q= ∆H rxn
750C
Q
∆Hˆ R
C + D A + B
250C
750C
∆Hˆ P 0
25 C ∆Hˆ f Figure 3. Hypotetical step of ∆Hrxn
∆H rxn = ∆H R + ∆H P + ∆H f At 25 0 C ∆Hˆ f , A = −486180 J mol
∆Hˆ f , B = −277630 J mol
∆Hˆ f ,C = −463200 J mol
∆Hˆ f , D = −285840 J mol
∆Hˆ f = ∆Hˆ f , Product − ∆Hˆ f , Reactant J 1.09 kmol 103 mol 1 min * * * ∆Hˆ f = {[ −463200 − 285840] − [ −486180 − 277630]} mol 1 kmol 60 s min ∆H f = 268.321*103 W
CP , A
75 + 25 = 50 0 C 2 = 2160 J kg .K CP , B 2670 J kg .K
C P ,C
= 2020 J kg .K CP , D 4180 J kg .K
= At TAvg
- 17 -
See in Table 3
∆H R
J kg kmol *60.05 *1.09 2160 kg .K kmol min 1 min * ( 25 − 75 ) K * 60 s J kg kmol + 2670 * 46.07 *1.199 kg .K kmol min
∆H R = −240.722 *103 W 2160 + 2670 ∆H P = + 2020 + 4180
J kg kmol *60.05 *0.636 kg .K kmol min J kg kmol * 46.07 *0.745 kg .K kmol min 1 min * ( 75 − 25 ) K * 60 s J kg kmol *88.1 *0.453 kg .K kmol min J kg kmol *18 *0.542 kg .K kmol min
246.276*103 W ∆H P = 273.875*103 W ∆H rxn = ( −240.722 + 246.276 + 268.321) *103 = Q = ∆H rxn = 273.875*103 W
See in Table 11
= Q m oil CP ,oil ∆T m oil ρ m oil ρ = π 2 A D jo − D 2ji ) ( 4 2 = = Assuming υ 0.6 m s , tshell 1.5*10−= m and t jacket 2.3*10−2 m
= , υ m ρυ A=
D ji = D + 2* tshell = 1.5 + 2*1.5*10−2
;
D ji = 1.53 m
D jo = D ji + 2* t jacket = 1.5 + 2* 2.3*10−2 ; D j 0 = 1.576 m
- 18 -
At Tin = 200 0 C 2204 = J kg .K , ρoil 897.6 kg m3
CP ,oil
m oil 897.6 = ; m oil 60.43 kg s π 2 2 (1.576 − 1.53 ) 4 = Q 60.43*= 2204* ∆T 273.875*103 0.6
∆T= 2.056 0 C ; ∆T= G =
( 200 − T )
Tout= 197.944 0 C
out
m oil 60.43 ; G 538.795 kg m 2 = = 2 2 2 2 0.785 ( D jo − D ji ) 0.785 (1.576 − 1.53 ) D − D ) (1.576 (= 2 jo
d eq =
2 ji
2
− 1.532 )
1.53
D ji
See in Table 11 d eq 0.093 m =
;
For calculation of the properties of hot oil average temperature must be used; 200 + 197.944 = TAvg = 198.972 0 C 2 At TAvg = 198.972 0 C 200 0 C
µoil
−3 = 0.73*10 Pa.s , koil 0.109= W m.K and CP ,oil 2204 J kg .K
C *µ = 0.027 * P ,oil oil koil
hi d eq koil
13
d eq * G * µoil
0.8
13
2204*0.73*10−3 0.093*538.795 hi 0.093 = 0.027 * * −3 0.109 0.109 0.73*10
0.8
hi = 574.482 W m 2 .K For the calculation of h o density, viscosity, specific heat capacity and thermal conductivity correlations were made for each component from References 3. Density Correlation; C
ρ= C
1 T 1+ 1− C3 2
C 4
for all components except water , For water ρ water =C1 + C2 T + C3 T 2 + C4 T 3
At 75 0 C
ρA
kmol 1.4486 = 16.4725 0.2529 348 m3 1+ 1− 591.95 0.25892
kmol 60.05 kg m kmol For other components the density correlation results are shown in Table 4. = ρ A 16.4725 * ; ρ A 989.211 kg m3 3
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ρ mix = ρ A * x A + ρ B * xB + ρC * xC + ρ D * xD 0.636 0.745 0.453 0.542 + 738.34* + 831.61* + 960.76* 2.378 2.378 2.378 2.378 3 = 878.489 kg m
ρ mix = 989.211* ρ mix
Viscosity Correlation;
C = µi exp C1 + 2 + C3 *ln T + C4 * T C5 T
(i
for each component , T [ K ])
At 75 0 C
1212.3
+ ( −0.322 ) *ln 348 + 0*3480 µ= exp −9.03 + A 348
µ A = 5.93*10−4 Pa.s For other components the viscosity correlation results are shown in Table 5. 13 = µmix
n
xµ , µ ∑= i =1
i
13 i
13 mix
0.27 * ( 5.93*10−4 ) + 0.31* ( 4.624*10−4 ) 13
13
+0.19* ( 2.58*10−4 ) + 0.23* ( 3.8*10−4 ) 13
13
µmix = 4.29*10−4 Pa.s Thermal Conductivity Calculation;
kmix = x A k A + xB k B + xC kC + xD k D kmix =0.27 *0.1714 + 0.31*0.12395 + 0.19*0.14229 + 0.23*0.16555 kmix = 0.15 W m.K Heat Capacity Calculation; 0.636*60.05 = 0.31 0.636*60.05 + 0.745* 46.07 + 0.453*88.1 + 0.542*18 Other values are shown in Table 9 xw, A
CP ,mix = CP , A * xw, A + CP , B * xw, B + CP ,C * xw,C + CP , D * xw, D CP ,mix = ( 2.28*0.31 + 2.96*0.28 + 2.12*0.32 + 4.19*0.09 ) *1000 CP ,mix = 2591.1 J kg .K
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See in Table 6
C *µ ho D = 0.55* P ,mix mix kmix kmix
0.25
d ag2 * N * ρ mix * µmix
23
2 120 0.25 rps *878.489 0.5 * ho *1.5 2591.1*0.000429 60 = 0.55* * 0.15 0.15 0.000429
23
ho = 921.84 W m 2 .K 1 1 1 D0 = + , U 0 ho hi Di
1.576 1 1 1 * = + U 0 921.84 574.482 1.53
Reference 5
U 0 = 342.32 W m 2 .K Q= U 0 A0 ∆Tln
∆T1 = = ∆T2
( 200 − 75)= 125 0 C 75 ) 122.944 0 C (197.944 −=
123.969 0 C ∆Tln = 273.875*103 = 342.32* A0,req *123.969 A0,req = 6.36 m 2
See in Table 11
π= A= DH π *1.5*1.95 0, calc A0,calc = 9.189 m 2
A 0,calc >A 0,req
So; the jacket satisfies our reactor design.
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Design of Coil by Heating Process
Outlet Stream Tinitial=750C
Feed Stream Tinitial=750C
Hot Oil Tout= ? Hot Oil Tin=2000C
Figure 4. A typical CSTR with coil Assume υ 3= m s and Tin 200 0 C = di 0.0409 m , d o 0.0482 m = =
Reference 2
m oil ρ m oil ρ m 897.6 υ = = = = 3 π 2 π A do ) 0.04822 ( 4 4 m oil = 3.536 kg s Q m oil CP ,oil ∆T = 2204* ∆T 273.875*103 Q 3.536*= = ∆T= 35.134 0 C ; ∆T=
( 200 − T ) out
Tout= 164.866 0 C
200 + 164.866 = 182.433 0 C 2 W m.K , ρoil 910.248 kg m3 = 0.110
At TAvg = koil CP ,oil
See in Table 1
J kg .K , µoil 0.86*10−3 Pa.s = 2147.464
CP ,oil * µoil d *υ * ρoil hi d 0 = 0.023* i * koil koil µoil 0.8
0.4
0.8
0.0409*3*910.248 2147.464*0.86*10−3 hi 0.0482 = 0.023* * 0.110 0.110 0.86*10−3 hi = 2001.1634 W m 2 .K
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0.4
d hic =+ hi 1 3.5 o di hic = 10266.838 W
0.0482 2001.1634* 1 + 3.5 = 0.0409 m 2 .K
C *µ ho D = 0.87 * P ,mix mix kmix kmix
13
d ag2 * N * ρ mix * µmix
0.62
2 120 13 rps *878.489 0.5 * ho *1.5 2591.1*0.000429 60 = 0.87 * * 0.15 0.15 0.000429
0.62
ho = 903.282 W m 2 .K 1 1 1 d0 = + , U 0 ho hic di
0.0482 1 1 1 * = + U 0 903.282 10266.838 0.0409
Reference 2
U 0 = 818.318 W m 2 .K Q= U 0 A0 ∆Tln
∆T1 = = ∆T2
( 200 − 75)= 125 0 C 75 ) 89.866 0 C (164.866 −=
106.468 0 C ∆Tln = 273.875*103 = 818.318* A0,req *106.468 A0,req = 3.143 m 2 We put the coils 20 cm apart from the reactor walls ' 1.5 − 0.4 D= D − 2* 20*10−2 =
D' = 1.1 m
' π *1.1= 3.45 m Perimeter of one coil= P= π D=
See in Table 1
= A0,calc π= d 0 P * n π *0.0482*3.45* n A0,calc = 1.914* n if A0,calc = A0,req n = 6 coils If the space between the coils is 6 cm, the space from bottom of reactor is 45 cm, and the comparison of height of the tank with the height of the coil is shown below;
H coil = H bottom + d 0 * n + H SpacesBetweenCoil * ( n − 1) H coil =( 45*10−2 ) + ( 0.0482*6 ) + ( 6*10−2 * ( 6 − 1) ) H coil =1 m < H tank =1.95 m So; the coil satisfies our reactor design like mentioned in the discussion part.
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