Chapter 5: Isothermal Reactor Design: Conversion
Chapter 5 Self Test
Zero-order Reaction in a CSTR and PFR
The zero order reaction
\(\text{A} + 2\text{B} \rightarrow \text{C}\)
is carried out in a CSTR and in a PFR. For an entering volumetric flowrate of 10 dm3/s at a
concentration of .5 mol/dm3, what is the space time for each reactor to achieve 90%
conversion.
Additional Information: kA = 0.01 mol/dm3•s
Hint 2: CSTR combined mole balance and rate law
Hint 1
Mole balance:
\(V = \frac{F_{A0} X}{-r_A}\)
Hint 2
Rate Law:
\(-r_A = k_A\)
Combine:
\(V = \frac{F_{A0} X}{k_A}\)
The zero order reaction
\(\text{A} + 2\text{B} \rightarrow \text{C}\)
is carried out in a CSTR and in a PFR. For an entering volumetric flowrate of 10 dm3/s at a concentration of .5 mol/dm3, what is the space time for each reactor to achieve 90% conversion.
Full Solution: CSTR
MOLE BALANCE:
\(V = \frac{F_{A0} X}{-r_A}\)
RATE LAW:
\(-r_A = k_A\)
\(\text{A} + 2\text{B} \rightarrow \text{C}\)
STOICHIOMETRY:
\(V = \frac{F_{A0} X}{k_A}\)
\(V = \frac{C \, v_0 \, X}{k_A}\)
COMBINE:
\(V = \frac{\left(0.5 \, \frac{\text{mol}}{\text{dm}^3}\right) \left(10 \, \frac{\text{dm}^3}{\text{s}}\right) (0.9)}{0.01 \, \frac{\text{mol}}{\text{dm}^3 \cdot \text{s}}} = 450 \, \text{dm}^3\)
\(\tau = \frac{V}{v_0} = \frac{450 \, \text{dm}^3}{10 \, \frac{\text{dm}^3}{\text{s}}} = 45 \, \text{s}\)
The zero order reaction
\(-r_A = k_A\)
\(\frac{r_B}{(1/2)} = \frac{r_A}{-1}\)
\(r_B = \frac{r_A}{2}\)
\(r_B = -\left(\frac{k_A}{2}\right)\)
\(k_B = \left(\frac{k_A}{2}\right)\)
is carried out in a CSTR and in a PFR. For an entering volumetric flowrate of 10 dm3/s at a concentration of .5 mol/dm3, what is the space time for each reactor to achieve 90% conversion.
Full Solution: PFR
Mole Balance:
\(V = F_{A0} \int_0^X \frac{dX}{-r_A}\)
\(-r_A = k_A\)
\(V = F_{A0} \int_0^X \frac{dX}{k_A} = \frac{F_{A0} X}{k_A}\)
\(V = \frac{F_{A0} X}{k_A}, \, \text{same as CSTR}\)
\(\tau = 45 \, \text{s}\)