Chemical Reaction Engineering - Exercises and Questions for exam

CHEMICAL REACTION

ENGINEERING 1

SUMMARY

NASA CEA EXERCISES......................................................................................................... 3

SELF EVALUATION 1............................................................................................................. 5

SELF EVALUATION 2........................................................................................................... 10

SELF EVALUATION 3........................................................................................................... 15

SELF EVALUATION 4........................................................................................................... 21

QUESTIONS.......................................................................................................................... 29

2

NASA CEA EXERCISES

IDENTIFICATION of THERMODYNAMIC PROPERTIES

of a GAS

Identify the thermodynamic properties of gaseous acetylene at atmospheric pressure and

200°C, 400°C and 600°C.

1.​ Tp problem (assigned temperature and pressure)

a.​ Enter the conditions under which you want to calculate the properties

2.​ Reactants

a.​ In this case the compound is one so indicating the properties per mole or mole

fraction does not change

b.​ Ident: Name

c.​ Name: C2H2, acetylene

d.​ Amount: for the determination of the thermodynamic properties of a pure

compound it does not matter (for example enter “1”)

3.​ Only

a.​ add “Acetylene” to display only the properties of acetylene otherwise, if for

thermodynamics the system was required to react it would also show the

properties of the other compounds

4.​ Activity: Execute CEA

METHANE COMBUSTION

Measure the equilibrium composition of the initial CH and O mixture at a temperature of

4 2

1200°C and 1 atm.

The initial moles correspond to the stoichiometric moles of the methane combustion reaction:

+ 2 → + 2

4 2 2 2

1.​ Tp problem

2.​ Reactants:

a.​ Ident: Name

b.​ Name: CH4, O2

c.​ Amount: 1.2

3.​ Activity: Execute CEA

CEA reports as a result: x =0,33 and x =0,66 and small residues of the radical OH and

CO2 H2O

molecular oxygen O .

2

The same result occurs if the same number of carbon, hydrogen and oxygen atoms are

inserted as reagents which correspond to 1 mole of CH and 2 moles of O :

4 2

●​ C: 1 mole

●​ H: 4 moles

●​ O: 4 moles

N-OCTANE COMBUSTION with a SUB QUANTITY of O

2

25

+ → 8 + 9

2

8 18 2 2 2 3

Consider the combustion reaction of n-octane, but in the case analyzed the oxygen supplied is

equivalent to 30% of the stoichiometric oxygen and is injected through air. The temperature

is 1000°C and the pressure 3 bar. 25

0,3·

The moles of air in the supply are therefore .

2

= = 17, 857

0,21

,

1.​ Tp problem

a.​ T: 1000

b.​ p: 3

2.​ Reactants (moles)

a.​ Ident: Name

b.​ Name: C8H18, Air

c.​ Amount: 1, 17,857

3.​ Activity: Execute CEA

The equilibrium composition will be composed largely of nitrogen (coming from the air) but

also contains CO, H .

2

DETERMINATION of the ADIABATIC TEMPERATURE

The "hp problem" is used as a problem if the reaction occurs only at a given pressure but the

temperature at which the reaction will occur is unknown (the inlet temperature of the

components may be known).

We want to study the combustion reaction of acetylene with stoichiometric quantities of the

reactants → 1 mole of acetylene and 2.5 moles of oxygen entering the reactor at 25°C.

1.​ Hp problem

a.​ Pressure: 1 atm

2.​ Reactants

a.​ Ident: Name

b.​ Name: C2H2, O2

c.​ Amount: 1, 2.5

d.​ Temp: 25°C, 25°C

3.​ Activity: Execute CEA

The adiabatic temperature (“T”) is 3341 K. 4

SELF EVALUATION 1

EX 1

Determine the heat capacity at constant pressure (kJ/kg K) of pure C4H6,butadiene at T =

36°C and P = 1.70 bar.

1.​ Problem: Tp problem

a.​ T=36°C

b.​ p=1,7 bar

2.​ Reactants

a.​ Ident: Name

b.​ Name: C4H6

c.​ Amount: 1

3.​ Only: C4H6

4.​ Activity: Execute CEA

The heat capacity at constant pressure is .

= 1, 5255 ·

EX 2

Evaluate the % conversion of CO2 (1-N_CO2_equil/ N_CO2_initial)*100 at equilibrium, in

producing CH4 by reacting a stoichiometric (for methanation) mixture of CO2 and H2, at

350°C and 1 bar?

The methanation is: .

+ 4 → + 2

2 2 4 2

The stoichiometric quantity of the reactants is: 5

●​ 1 mole for CO 2

●​ 4 moles for H 2

On the CEA:

1.​ Tp problem

2.​ Reactants

a.​ Ident: Name

b.​ Name: CO2, H2

c.​ Amount: 1, 4

3.​ Activity: Execute CEA

The expected mole fractions of the compound of equilibrium are:

To evaluate the conversion of CO , mass balances have to be solved.

2

C: → →

1 = 1 · 0, 28501 + 1 · 0, 00028 + 1 · 0, 02872 0, 31401 = 1

.

= 3, 185

The number of moles at equilibrium of CO is thus .

= · = 0, 0915

2 2 2

0,0915

The conversion is .

2

= 1 − = 1− = 0, 9085 = 90, 85%

1

2

2

EX 3 1

CH4 burns with a lack of air, of the stoichiometric required for total oxidation to CO2

3,8

and H2O. Calculate the equilibrium composition of the products at T = 1200°C and P = 3.50

atm as mole fraction (> 0.1%).

The combustion of methane is: .

+ 2 → + 2

4 2 2 2

2

The stoichiometric quantity of air is .

= = 9, 524

0,21

1

Instead, the feed is composed by of Air.

· 9, 524 = 2, 51

3,8

1.​ Tp problem

2.​ Reactants

a.​ Ident: Name

b.​ Name: CH4, Air

c.​ Amount, 1, 2.51

3.​ Activity: Execute CEA 6

Eliminating the compounds that have a molar fraction lower than 0,1%=0,001, the

composition of the equilibrium mixture is:

●​ Ar: 0,00472

●​ CO: 0,19891

●​ CO : 0,00181

2

●​ H : 0,39226

2

●​ H O: 0,00883

2

●​ N : 0,39330

2

EX 4

Determine the equilibrium temperature (K) reacting a mixture of C3H8 (3.9 moles) and CO

(3.0 moles) initially at T = 41°C and P = 7.10 bar in an adiabatic flow reactor.

1.​ Hp problem

a.​ p=7,1 bar

2.​ Reactants:

a.​ Ident: Name

b.​ Name: C3H8, CO

c.​ Amount: 3.9, 3

d.​ Temp: 41, 41

3.​ Activity: Execute

The temperature at which the reaction occurs is T=840,70 K. 7

EX 5

Determine the equilibrium temperature (K) reacting a mixture of CH4 (1.7 moles) and O2

(3.1 moles) initially at T = 65°C and P = 8.10 bar in an adiabatic flow reactor.

1.​ Hp problem

a.​ p=8,1 bar

2.​ Reactants

a.​ Ident: Name

b.​ Name: CH4, O2

c.​ Amount: 1.7 moles, 3.1 moles

d.​ Temp: 65°C, 65°C

3.​ Activity: Execute CEA

The adiabatic temperature at which the reaction occurs is 3331,51 K.

EX 6

C8H18 (isooctane) burns with a lack of air, 1/4.00 of the stoichiometric required for total

oxidation to CO2 and H2O. Calculate the equilibrium composition of the products at T =

1400°C and P = 4.60 atm as mole fraction (> 0.1%)

25

The combustion of the iso-octane is: + → 8 + 9

2

8 18 2 2 2

ℎ 1

25

The stoichiometric quantity of air is .

= · = 59, 52

0,21

2

59,52

The feed instead contains: .

= = 14, 88

4

1.​ Tp problem

a.​ T= 1400°C

b.​ p=4,6 atm

2.​ Reactants

a.​ Ident: Name

b.​ Name: C8H18, Air

c.​ Amount: 1, 14.88

3.​ Activity: Execute CEA 8

EX 7

Which species will be at equilibrium (at least 20 ppm) after mixing equal amounts of SiH4

and CO2 at -40°C and 1 bar?

1.​ Tp problem

a.​ T=-40°C

b.​ p=1 bar

2.​ Reactants

a.​ Ident: Name

b.​ Name: SiH4, CO2

c.​ Amount: 1, 1

3.​ Activity: Execute CEA 9

SELF EVALUATION 2

EX 1

The global reaction A+B → (useful products) is carried out at constant T in a liquid phase.

0,4

Assume the reaction irreversible with a rate law . It is questioned whether

⎡ ⎤

= 0, 9

⎣ ⎦

·ℎ

it could be better (smaller reactors) to use a CSTR followed by a PFR of the same volume or

the reverse. The goal is a total conversion X = 90% feeding A in a concentration of .

6, 4

A

Determine the optimal total residence time [h] and reactors' sequence.

Since the reaction is occurring in a liquid phase, it can be assumed that the density is

constant, even more so temperature is constant and also velocity is constant.

The material balances are:

0,4

●​ PFR: = =− =− 0, 9

τ

∆ 0,4

●​ CSTR: = =− 0, 9

ϑ

If X =90% → → the volume is constant → →

,1 ,1

0, 9 = 1 − 0, 9 = 1 −

A

,0 ,0

→ .

,1

− 0, 1 =− = 0, 1 · = 0, 1 · 6, 4 = 0, 64

,1 ,0

,0

If the volume is the same for the two reactors, they have the same residence time.

For the PFR we can take the solution from the file “Simple solution”:

1

0,6 0,6

( )

⎡ ⎤

0

= − 0, 6 · 0, 9τ +

⎢ ⎥

⎣ ⎦

For the CSTR:

0

− 0,4 0 0,4

→

=− 0, 9 − =− ϑ · 0, 9

ϑ

With a trial and error procedure: θ

●​ if and CSTR → PFR: → not the right answer

θ = 2, 39 ℎ τ = = 1, 195 ℎ

2 1

0,6 0,6

( )

⎡ ⎤

0

○​ considering the last reactor (PFR): →

= − 0, 6 · 0, 9τ +

⎢ ⎥

⎣ ⎦

1 0,6

0,6 0,6

( ) ( )

⎡ ⎤

0 0

→

0, 64 = − 0, 6 · 0, 9 · 1, 195 + 0, 76 =− 0, 6453 +

⎢ ⎥

⎣ ⎦

0

→ = 1, 763

10

0 0,4

○​ from the CSTR: →

− =− ϑ · 0, 9

0,4

→ → the

1, 763 − 6, 4 =− 1, 195 · 0, 9 · 1, 763 − 4, 637 =− 1, 34

balance is not verified θ

●​ if and CSTR → PFR: → right answer

θ = 4, 55 ℎ τ = = 2, 275 ℎ

2

○​ considering the last reactor (PFR): →

1 0,6

0,6 0,6

( ) ( )

⎡ ⎤

0 0

→

0, 64 = − 0, 6 · 0, 9 · 2, 275 + 0, 76 =− 1, 2285 +

⎢ ⎥

⎣ ⎦

0

→ = 3, 144

0 0,4

○​ from the CSTR: →

− =− ϑ · 0, 9

0,4

→ → the

3, 144 − 6, 4 =− 2, 275 · 0, 9 · 3, 144 − 3, 256 =− 3, 237

balance verified → it may be the right answer

θ

●​ if and PFR → CSTR: → not the right answer

θ = 5, 62 ℎ τ = = 2, 81 ℎ

2 1

0,6

[ ]

0,6

○​ from the PFR: = − 0, 6 · 0, 9 · 2, 81 + (

6, 4

) = 2, 02

0,4

○​ from the CSTR: →

0, 64 − 2, 02 =− ϑ · 0, 9 · 0, 64 θ = 1, 88 ℎ

θ

●​ if and CSTR → PFR: → not the right answer

θ = 3, 03 ℎ τ = = 1, 515 ℎ

2

○​ considering the last reactor (PFR): →

1 0,6

0,6 0,6

( ) ( )

⎡ ⎤

0 0

→

0, 64 = − 0, 6 · 0, 9 · 1, 515 + 0, 76 =− 0, 8181 +

⎢ ⎥

⎣ ⎦

0

→ = 2, 139

0 0,4

○​ from the CSTR: →

− =− ϑ · 0, 9

0,4

→ → the

2, 139 − 6, 4 =− 1, 515 · 0, 9 · 2, 139 − 4, 261 =− 1, 848

balance is not verified θ

●​ if and CSTR → PFR: → not the right answer

θ = 2, 56 ℎ τ = = 1, 28 ℎ

2

○​ considering the last reactor (PFR): →

1 0,6

0,6 0,6

( ) ( )

⎡ ⎤

0 0

→ →

0, 64 = − 0, 6 · 0, 9 · 1, 28 + 0, 76 =− 0, 691 +

⎢ ⎥

⎣ ⎦

0

= 1, 86

0 0,4

○​ from the CSTR: →

− =− ϑ · 0, 9

0,4

→ → the balance

1, 86 − 6, 4 =− 1, 28 · 0, 9 · 1, 86 − 4, 54 =− 1, 476

is not verified

It can be done also graphically.

EX 2 11

The material balance of a batch reactor is considering T and V constant.

=

∆ 1,2

Assuming that

≃ = =− 2

∆

We take into account two couple of concentration, with their time of sampling:

●​ and at 48°C → , ,

= [ 0 5. 5 ] = [

6 4. 98 ] ∆ = 5, 5 = 330 ∆ =− 1, 02

−

1,2

6+4,98 −1,02

→ →

= = 5, 49 =− 2 · 5, 49 − 0, 185 =− 2 · · 7, 72

1

2 5,5

−

→ 0, 012 =− · 1

●​ and at 63°C → ,

= [ 0 4 ] = [

6 4. 98 ] ∆ = 4 ∆ =− 1, 02

−

1,2

6+4,98 −1,02

→ →

= = 5, 49 =− 2 · 5, 49 − 0, 255 =− 2 · · 7, 72

2

2 4

−

→ 0, 0165 = · 2

We put the two equations in a system:

−

0, 012 = · 1

−

0, 0165 = · 2

−

0,012 0,012

From the first equation: and substituting it in the second

= 0, 0165 = · 2

−/ −

1

1

+

−

→ → →

0, 0165 = 0, 012 · 1, 375 =− +

2 2

2 1

( )

1 1 1,375

→

1, 375 = − = 1 1

−

1 2

1 2

0,012

And replacing it in the first equation: .

= = 15, 03

−/

EX 3 12

The general material balance for a irreversible reaction in a batch reactor is . In this

=

α

case the rate of production could have this shape: .

= =− 2

Considering that the sampling time is sufficiently small, the derivative could be considered as

a finite difference and the concentration in the rate of production is defined as the average of

0

∆ +

α α

the initial and final concentration: where .

=− 2 =

∆ 2

We consider the measurements: α

∆ α ( )

6,7+4,62

4,62−6,7

●​ and → →

= [ 0 14. 8 ] = [

6. 7 4. 62 ] =− 2 =− 2

14,8 2

∆

α 0,07025

→ →

− 0, 1405 =− 2 · 5, 66 = α

5,66 ∆ α

●​ and → →

= [ 14. 8 29, 6 ] = [

4, 62 3, 41 ] =− 2

∆

α

( ) α

3,41+4,62 0,0409

3,41−4,62 → → .

=− 2 − 0, 0818 =− 2 · 4, 015 = α

29,6−14,8 2 4,015

α α

0,0409 4,015

0,07025

Equaling the two equations: → →

= = 0, 5822 0, 7094 = 0, 5822

α

α α

5,66 4,015 5,66

→ .

α = 0, 5822 = 1, 575

0,7094 −3

0,07025

From the first equation: = = 4, 58 · 10 = 0, 00458

α

5,66 13

EX 4

In the case of a PFR, with an elementary and irreversible reaction occurring: =−

τ

=−

τ

From the stoichiometry and considering a constant volume → so the

= =

two equations are linearly dependent.

Assuming constant massive density of the mixture, we know that the velocity in the reactor is

constant, and so also the volumetric flowrate.

100

= = 1 = τ

100

The solution to the material balance is found in the file “Simple Solution” and is

From the equation, everythin