Chapter 6: Isothermal Reactor Design: Molar Flow Rates
Gas Phase PFR
Problem:
The elementary irreversible gas phase reaction 2A -> B is carried out isothermally with no pressure drop in a PFR.
Given:
Use measures other than conversion to plot molar flow rates of A and B down the reactor.
Hint 1: What are the mole balances on A and B and the rate law?
mole balances on A ?
(a) $\frac{dF_{A}}{dV} = 2r_{A}$
(b) $\frac{dF_{A}}{dV} = -2r_{A}$
(c) $\frac{dF_{A}}{dV} = -r_{A}$
(d) $\frac{dF_{A}}{dV} = r_{A}$
mole balances on B ?
(a) $\frac{dF_{B}}{dV} = r_{A}$
(b) $\frac{dF_{B}}{dV} = -2r_{A}$
(c) $\frac{dF_{B}}{dV} = \frac{-r_{A}}{2}$
(d) $\frac{dF_{B}}{dV} = \frac{r_{A}}{2}$
rate law ?
(a) $r_{A} = -k{C_{A}}^2$
(b) $r_{A} = k{C_{A}}^2$
(c) $r_{A} = kC_{A}$
Hint 2: What are the concentrations of A and B?
(a) $C_{A} = C_{T0}\frac{F_{A}}{F_{B}}p\frac{T_{0}}{T}$
(b) $C_{A} = C_{T0}\frac{F_{A}}{F_{A}+F_{B}}p\frac{T_{0}}{T}$
Hint 3: What is the rate of formation of B?
(a) $r_{B} = \frac{1}{2}C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$
(b) $r_{B} = -2C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$
(c) $r_{B} = -\frac{1}{2}C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$
Hint 4: What are the combined mole balance, rate law and stoichiometry?
(a) $\frac{dF_{A}}{dV} = k_{A}C_{T0}\frac{F_{A}}{F_{A}+F_{B}}$
(b) $\frac{dF_{A}}{dV} = -k_{A}C_{T0}\frac{F_{A}}{F_{A}+F_{B}}$
Hint 5: What is the polymath program?
Hint 1 - Mole Balances and Rate Law
Mole Balances
Rate Law
Hint 2 - Concentrations
Stoichiometry
For Isothermal and No DP
Hint 3 - Rate of Formation of B
$$ \textrm{For } A + \frac{b}{a}B \rightarrow \frac{c}{a}C + \frac{d}{a}D $$
$$ \frac{r_{A}}{-a} = \frac{r_{B}}{-b} = \frac{r_{C}}{c} = \frac{r_{D}}{d} $$
$$ \textrm{For } 2A \rightarrow B $$
$$ \frac{r_{A}}{(-2)} = \frac{r_{B}}{(1)} $$
$$ r_{B} = \frac{1}{2}(-r_{A}) = \frac{k_{A}}{2}{C_{A}}^2 $$
Hint 4 - Combined Mole Balances, Rate Law and Stoichiometry
Combine
Hint 5 - Polymath Code
[Note: To=227C=500K]
Use Polymath to solve
One can back calculate X
See the Polymath Solution Below
POLYMATH 5.0 Results 01-19-2001 Calculated values of the DEQ variables
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = -ka*Cto^2*(Fa/Ft)^2 [2] d(Fb)/d(V) = 0.5*ka*Cto^2*(Fa/Ft)^2 Explicit equations as entered by the user [1] Cto = 0.2 [2] ka = 10 [3] Ft = Fa+Fb [4] x = (5-Fa)/5
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Full Solution
Mole Balances
Rate Law
Stoichiometry
For 2A->B
For Isothermal and No DP
Combine
[Note: To=227C=500K]
Use Polymath to solve
One can back calculate X
See the Polymath Solution Below
POLYMATH 5.0 Results 01-19-2001 Calculated values of the DEQ variables
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = -ka*Cto^2*(Fa/Ft)^2 [2] d(Fb)/d(V) = 0.5*ka*Cto^2*(Fa/Ft)^2 Explicit equations as entered by the user [1] Cto = 0.2 [2] ka = 10 [3] Ft = Fa+Fb [4] x = (5-Fa)/5
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