Question: deriving deriving this CSTR Derivation without integrationCSTR solve 11-3 step by step in an external sheet 2. CSTR with heal exchanger, UA (TaT), and large

deriving
 deriving deriving this CSTR Derivation without integrationCSTR solve 11-3 step by
deriving this CSTR
step in an external sheet 2. CSTR with heal exchanger, UA (TaT),
Derivation without integrationCSTR
solve 11-3 step by step in an external sheet
and large coolant flow ratc. XEB=HRx(FA0UA(TTa))+1CP1(TT0)(T81 1. Adiabatic ( Q=0) CSTR, PFR,
Batch, or PBR. The rclationship between conv sion. XEB, and temperature for
Hs=0, constant CP1, and CP=0, is XEB=HRS1CP1(TT0)T=T0+CP1(HRx)X For an exothermic reaction (HRx)>0

2. CSTR with heal exchanger, UA (TaT), and large coolant flow ratc. XEB=HRx(FA0UA(TTa))+1CP1(TT0)(T81 1. Adiabatic ( Q=0) CSTR, PFR, Batch, or PBR. The rclationship between conv sion. XEB, and temperature for Hs=0, constant CP1, and CP=0, is XEB=HRS1CP1(TT0)T=T0+CP1(HRx)X For an exothermic reaction (HRx)>0 2. CSTR with heal exchanger, UA (TaT), and large coolant flow ratc. XEB=HRI(FA0UA(TT0))+1CP1(TT0) dWdT=FA0(()iCp+CM)paUa(TsT)riHRn(T) 3B. PFR in terms of converslon 3C. PBR in terms of molar flow rates dWdT=FsCrssCg(TsT)+riHns(T) 3D. PFR in terms of molar flow rates \[ \frac{d T}{d V}=\frac{\left.U_{a}\left(T_{2}-T ight)+r_{1} ight\lrcorner H_{\mathrm{z}_{1}}(T)}{\searrow F_{1} C_{r_{3}}} \] 4. Batch dtdT=N1CP1(r1V1)(Hn1)UA(TT2) 5. For Semibatch or unsteady CSTR dtdT=i=1nNiCPiQWii=1nFi0(CP1(TTi0)+1HR1(T)](rV)) 6. For multiple reactlons in a PFR dVdT=F1Cr1Ua(T1T)+r1,HR2 tsistA lial maite of Fuatipulating the tiletar balance Set. min 824 to 8.4 For an exteermic maction (Mim4}>0 Suc. 42 The Enerer thance 477 3. P7 WYPEs with heat rurhander. 3. Fint in luras af cumatilin (Ty-1.1.) 11. PFIl in ierme of qubersion TW-LE 3C. Pat in Acrine of mular fere ratrs 3D. PFR le terms of malur the rakis (T-1G) 4. Bateh ITt. HH3 1. For bemihatch or uncirade ChTIR Example 11-3 Adiabatic Liquid-Phase Isomerization of Normal Butane Normal butane, C4H10, is to be isomerized to isobutane in a plug-flow reactor. Isobutane is a valuable product that is used in the manufacture of gasoline additives. For example, isobutane can be further reacted to form iso-octane. The 2014 selling price of n-butane was $1.5/gal, while the trading price of isobutane was $1.75/gal This elementary reversible reaction is to be carried out adiabatically in the liquid phase under high pressure using essentially trace amounts of a liquid catalyst that gives a specific reaction rate of 31.1h1 at 360K. The feed enters at 330K. (a) Calculate the PFR volume necessary to process 100,000gal/ day (163kmol/h) at 70% conversion of a mixture 90mol%n-butane and 10mol%-pentane, which is considered an inert. (b) Plot and analyze X,XeT, and rA down the length of the reactor. (c) Calculate the CSTR volume for 40% conversion

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Chemical Engineering Questions!