R = 8.314 J mol^-1 K^-1 a) For a first order irreversible reaction A rightarrow
ID: 530153 • Letter: R
Question
R = 8.314 J mol^-1 K^-1 a) For a first order irreversible reaction A rightarrow B derive the mole balance equations for an isothermal Plug Flow Reactor (PFR) and an isothermal Continuous Stirred Tank Reactor (CSTR). b) Using the mole balance equation, calculate the volume required in a PFR in order to achieve 85 % conversion of A, when the kinetic constant is equal to k_t = 6.85 h^-1 and the volumetric flow rate through the reactor is V dot = 4 m^3 h^-1. c) Calculate the volume required in a CSTR which reaches the same conversion of 85 %, at the same volumetric flow rate and with the same kinetic constant. d) Calculate the conversion reached with a new PFR reactor having the same volume as the CSTR one calculated above. e) Plotting a qualitative diagram of the concentration profile inside each of the three reactors -CSTR with 85 % conversion, PFR-1 with 85 % conversion and PFR-2 with the same volume as the CSTR. Explain briefly and comment on the results.Explanation / Answer
Rate of moles of Ain = Rate of moles of A out+ Rate of loss of A due to chemical Reaction
For 1st order system this becomes
FAO= FA+ KCAV, V= volume of reactor, FA= molar flow rate of A, FAO = initial molar flow rate of A, K is rate constant, CA= concentration at any time, CAO= initial concentration. V= volume of reactor
FAO= FAO*(1-XA)+ KCAV
FAOXA= KCAV
CA= CAO*(1-XA)
FAOXA= KCAO*(1-XA)*V
Or XA/(1-XA)= KCAOV/FAO
T= space time = V/VO, VO= volumetric flow rate
T= V/VO= VCAO/FAO
Hence
XA/(1-XA)= KT-equation for 1st order reaction
Given K= 6.85/hr, Vo=4m3/hr, XA=0.85
Hence for a CSTR,6.85*V/4= 0.85/0.15, V= 3.1L
n case of ideal PFR, the concentration changes throughout the length of reactor. Hence a differential element is only to be considered.
Rate of moles of Ain = Rate of moles of A out+ Rate of loss of A due to chemical Reaction
For 1st order system this becomes
FA= FA+dFA+KCAdV
-dFA= KCAdV
FA= FAO*(1-XA)
-dFA= FAO*dXA
Hence
FAOdXA= KCAdV
Or
CA= CAO*(1-XA)
FAO*dXA= KCAO*(1-XA)dV
dXA/(1-XA)= KCAOdV/FAO
when integrated the equation becomes
-ln(1-XA)= KT,
Given XA=0.85
-ln(1-0.85)= 6.85*V/4
V= 1.89*4/6.85= 1.1 L
When the CFSTR volume is 3.1 L
-ln(1-XA)= 6.85*3.1/4= 5.31
XA= 1-0.005= 0.995
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