Review: Equilibrium Conversion XAe

heat→
increasing temp adds heat (product) & pushes rxn to left (lower conversion)

Makes sense from Le Chatelier’s principle

Heat is a reactant in an endothermic rxn→
increasing temp adds reactant (heat) & pushes rxn to right (higher conversion)

Clicker question material

maximize XA

From thermodynamics

XEB

T

600

500

350

0.15

0.33

0.75

High T0: moves XA,EB line to the right. Rxn reaches equilibrium fast, but low XA

Low T0 would give high XA,e but the specific reaction rate k is so small that most of the reactant passes through the reactor without reacting (never reach XA,e)

feed temp- reaction reaches equilibrium quickly but XA,e is low
Lower feed temp- higher XA,e but reaction rate is too slow to be practical
Cooling between reactors shifts XA,EB line to the left, increasing XA

to increase the conversion

Review: Endothermic Reactions

XEB

T

heating process

final conversion

Red lines are from the energy balance, slant backwards because ΔH°RX >0 for endothermic reaction

reactor- differential form of EB must be used
Multiple steady states: more than one set of conditions satisfies both the energy balance & mole balance

of T as a function of XA
Use k = Ae-E/RT to obtain k as a function of XA
Use stoichiometry to obtain –rA as a function of XA
Calculate:

may use numerical methods

reactor

Steady-State PFR/PBR w/ Heat Exchanger

Energy balance on small volume of SS PFR:

0

Heat exchange area per volume of reactor

Take limit as ΔV→∞:

Expand:

Plug in Q:

SS PFR, Ẇs=0

PFR energy balance is coupled to the PFR design eq, and PFR design eq is coupled to Arrhenius eq for k or Kequil

(these are the 3 equations that must be simultaneously solved)

Multiply Ua and (Ta-T) by
-1 (-1 x -1 = 1)

Switched sign & order in bracket

equations simultaneously with an ODE solver (Polymath)

If this were a gas phase rxn w/ pressure drop, change stoichiometry accordingly & include an equation for dΔP/dW

Tinside reactor)
Calculate k = Ae-E/RT where T was calculated in step a
Plug the k calculated in step b into the design equation to calculate VCSTR

Case 1: Given FA0, CA0, A, E, Cpi, H°I, and XA, calculate T & V

Solve TEB for T as a function of XA
Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT )
Plot XA,EB vs T & XA,MB vs T on the same graph. The intersection of these 2 lines is the conditions (T and XA) that satisfies the energy & mass balance

Case 2: Given FA0, CA0, A, E, Cpi, H°I, and V, calculate T & XA

XA,EB = conversion determined from the TEB equation
XA,MB = conversion determined using the design equation

XA

T

XA,EB

XA,MB

XA,exit

Texit

Intersection is T and XA that satisfies both equations

Isothermal CSTR: feed temp = temperature inside the CSTR

vs T
Intersections are the T and XA that satisfy both the mass balance and energy balance
Multiple sets of conditions are possible for the same rxn in the same reactor with the same inlet conditions!

Reactor must operate near one of these steady states- this requires knowledge of their stability!

XA,MB

ΔCP=0, first order kinetics, all feeds at the same temperature (Ti0=T0), constant Ta in jacket, and an overall heat transfer coefficient

Heat removed term ≡ R(T)

Heat generated term ≡ G(T)

A steady-state occurs when R(T) = G(T)

Bring terms that remove heat to other side of equation:

substitutions:

Heat removed:

Heat generated:

G(T)

When T0 increases, slope stays same & line shifts to right

R(T) line has slope of CP0(1+κ)

When κ increases from lowering FA0 or increasing heat exchange, slope and x-intercept moves

Ta

Ta>T0: x-intercept shifts right as κ↑

κ=0, then TC=T0

κ=∞, then TC=Ta

causes the reactor T to drift to a T between SS1 & SS2

R(T) > G(T) so T gradually falls to T=SS1

Suppose a disturbance causes the reactor T to drift to a T between SS2 & SS3

G(T) > R(T) so T gradually rises to T=SS3

Suppose a disturbance causes the reactor T to drop below SS1

G(T) > R(T) so T gradually rises to T=SS1

Suppose a disturbance causes the reactor T to rise above SS3

R(T) > G(T) so T gradually falls to T=SS3

SS1 and SS3 are locally stable (return to them after temp pulse)
SS2 is an unstable- do not return to SS2 if there is a temp pulse

Temperature

G(T) & R(T)

the steady state temperature (TS)
Notice that the number of steady state temperatures depends on T0

Increasing T above these TS cause a temperature jump to the higher TS

to jump to TS,cyan

Temperature Ignition-Extinction Curve

T0, entering temperature

Ts, steady-state temp

ignition temperature

extinction temperature

Upper steady state

Lower steady state

Plot TS vs T0

TS,upper

TS,lower

TS along dashed line are unstable

R(T), G(T)

T

Ignition temp: T where jump from TS,lower to TS,upper occurs

Slight decrease in T below TS,magenta causes reactor T to drop to TS,yellow

Extinction temp: T where drop from TS,upper to TS,lower occurs