23-Chem-A4 Chemical Reactor Engineering · May 2013
Question 3 of 5: Adiabatic Plug-Flow Reactor — Exit Temperature
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour, open-book exam; any non-communicating calculator permitted, Fogler’s Elements of Chemical Reaction Engineering allowed. Format: five questions, each 20 marks; any four constitute a complete paper (80 marks). All five are solved below for completeness. Per the paper’s instructions, all data are treated as exact and answers are given to three significant figures.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th ed., Prentice Hall) — rate laws, batch/PFR/CSTR design equations, Arrhenius temperature dependence, integral & differential data analysis; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactor comparison and the tanks-in-series model; supporting thermochemical and property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 3: Adiabatic Plug-Flow Reactor — Exit Temperature (20 marks)
Given. Adiabatic PFR (no heat loss, no shaft work). Exothermic reaction $-\Delta H_R=150$ kJ/mol; feed $T_0=600$ K; conversion $X=0.80$. Whole-stream $c_p=2.3$ kJ/kg·K and $M=18$ kg/kmol (steam properties).
Quantity
Value
$F_{A0}$ (reactant A)
0.0020 kmol/s
Steam carrier
0.020 kmol/s
$-\Delta H_R$
150 kJ/mol
$c_p$ (stream)
2.3 kJ/kg·K
Feed temperature $T_0$
600 K
Conversion $X$
0.80
Find. (a) the reactor exit temperature; (b) the qualitative effect of a 20% pressure rise.
Figure 4 — Adiabatic plug-flow reactor: reactant A in a steam carrier enters at 600 K; the exothermic reaction ($\Delta H_R=-150$ kJ/mol) raises the exit temperature at 80% conversion.
Approach. Apply the adiabatic energy balance (heat of reaction = sensible-heat gain of the stream) to get the exit temperature; recognise the rate data and pressure as irrelevant to the temperature at a fixed conversion.
Adiabatic energy balance sets the exit temperature. For an adiabatic
reactor with no shaft work, the enthalpy released by reaction is absorbed as sensible heat
by the flowing stream:
$$F_{A0}\,X\,(-\Delta H_R) = \dot m\,c_p\,(T-T_0).$$
The rate constant and activation energy do not appear—they size the reactor volume,
not the temperature reached at a given conversion.
Heat released by reaction. With $F_{A0}=0.0020$ kmol/s $=2.0$ mol/s,
$X=0.80$ and $-\Delta H_R=150$ kJ/mol:
$$\dot Q_{rxn}=F_{A0}X(-\Delta H_R)=(2.0)(0.80)(150)=240\ \text{kW}.$$
Thermal capacity of the stream. The molar mass and heat capacity of
the whole stream are taken as those of steam, so with total molar flow
$0.0020+0.020=0.022$ kmol/s and $M=18$ kg/kmol:
$$\dot m=(0.022)(18)=0.396\ \text{kg/s},\qquad \dot m\,c_p=(0.396)(2.3)=0.911\ \text{kW/K}.$$
Adiabatic temperature rise and exit temperature.
$$\Delta T=\frac{\dot Q_{rxn}}{\dot m\,c_p}=\frac{240}{0.911}=263.5\ \text{K}
\;\Rightarrow\; T=600+263.5=863.5\approx\boxed{864\ \text{K}}.$$
(b) Effect of a 20% pressure increase. A higher pressure raises the
reactant concentration and hence the reaction rate, so 80% conversion would be reached in
a smaller volume. But the adiabatic energy balance above contains no pressure term:
at the same conversion (80%) the released heat and the stream’s thermal capacity are
unchanged, so
$$\boxed{T_{exit}\ \text{is unchanged at}\ \approx864\ \text{K}.}$$
Pressure moves the reactor size, not the adiabatic temperature–conversion line.
