Industrial Chemical Processes - Esercizi svolti

OUT IN

●​ = 3, 76 · ( 56907 − 728 ) = 213970

2

●​ = 0, 5 · ( 59914 − 732 ) = 29957

2

●​ = 1 · ( − 393510 + 92466 − 912

) =− 301956

2

●​ = 177517 + 25277 − 316977 =− 58029

Since H is higher than H this means that the adiabatic temperature is between 1750 and

OUT IN

−96479−

(

−114183

)

2000 K: = 1750 + ( 2000 − 1750 ) = 1828

−58029−

(

−114183

)

EX 40 - ADIABATIC TEMPERATURE, EXCESS of AIR

Calculate the theoretical combustion temperature for a gas containing 10% of CO, 10% of

CH and 80% of N , which is burnt with 130% excess of air. All gases are initially preheated

4 2

to 500 K. Consider 25°C as the reference temperature and 100 moles as the basis for

calculation.

To complete combust 10 moles of CO we need 5 moles of oxygen, for 10 moles of CH we

4

need 20 moles of oxygen so the total is 25 moles. The fuel is burnt with a 130% of excess air

therefore . The corresponding nitrogen is

= 25 + 1, 3 · 25 = 57, 5

,

2 but there are other 80 moles in the fuel so

57, 5 · 3, 76 = 216, 2 .

= 216, 2 + 80 = 296, 2

,

2

In the outlet there will be:

●​ for the combustion of CO: 10 moles of CO 2

●​ for the combustion of CH : 10 moles of CO and 20 moles of water → 20 mol of total

4 2

CO 2

●​ of unreacted oxygen

1, 3 · 25 = 32, 5

●​ of nitrogen

296, 2

The inlet enthalpies are:

●​ →

= 0 + 6644 − 728 = 5916 = 296, 2 · 5916 = 1752

2 2

●​ →

= 0 + 6811 − 732 = 6079 = 57, 5 · 6079 = 349

2 2

●​ →

=− 110520 + 6652 − 728 =− 104

= 10 · ( − 104 ) =− 1040

●​ →

=− 74850 + 9100 − 879 =− 66 = 10 · ( − 66 ) =− 660

4 4

●​ = 1752 + 349 − 1040 − 660 = 390

Let’s consider a possible outlet temperature of 1500 K:

●​ →

= 0 + 39145 − 728 = 38 = 296, 2 · 38 = 11256

2 2

●​ →

= 0 + 41337 − 732 = 41 = 32, 5 · 41 = 1333

2 2

●​ →

=− 393510 + 62676 − 912 =− 332

2

= 20 · ( − 332 ) =− 6640

2

●​ →

=− 241830 + 48848 − 837 =− 187

2

= 20 · ( − 187 ) =− 3740

2

●​ = 11256 + 1333 − 6640 − 3740 = 2200

Since H is higher that H the adiabatic temperature will be lower, so I try with 1200 K:

1500 K IN

●​ →

= 0 + 28819 − 728 = 28 = 296, 2 · 28 = 8302

2 2

●​ →

= 0 + 30492 − 732 = 30 = 32, 5 · 30 = 975

2 2

●​ →

=− 393510 + 45404 − 912 =− 349

2

= 20 · ( − 349 ) =− 6980

2

●​ →

=− 241830 + 35312 − 837 =− 207

2

= 20 · ( − 207 ) =− 4140

2

●​ = 8302 + 975 − 6980 − 4140 =− 1843

The adiabatic temperature is therefore computed with an interpolation:

390−

(

−1843

)

= 1200 + ( 1500 − 1200 ) = 1365

2200−

(

−1843

)

EX 41 - DEW POINT, AIR FUEL RATIO

Propane is burnt with an weight air-fuel ratio of 25. The system is working in a steady state.

Consider all gases in ideal conditions. We have to calculate the temperature at which the

liquid water just begins to form in the products (dew point).

The reactants are feeded at 25°C and 1 atm.

The combustion reaction is .

+ 5 → 3 + 4

3 8 2 2 2

( ) 4,76·5·29

The stoichiometric weight ratio is

= = 15, 7

1·44

ℎ

1

The operating ratio tells that for every kg of fuel (which correspond to mol) there

= 22, 7

44

25

are 25 kg of air (corresponding to of air).

= 862

29

The combustion of 22,7 mol of propane would lead to the formation of

of CO and of H O. The 862 mol of air are

3 · 22, 7 = 68, 1 4 · 22, 7 = 90, 8

2 2

constituted by of oxygen and of

826 · 0, 21 = 173, 5 826 · 0, 79 = 652, 5

nitrogen.

The nitrogen will remain in the outlet, while the oxygen in the outlet will be

.

173, 5 − 22, 7 · 5 = 60 90,8

The molar fraction of water is therefore: .

= = 0, 104

90,8+68,1+652,5+60

2

The partial pressure of water is therefore .

= = 10132, 5 = 10, 13

2 2

At this pressure, the saturation temperature is

10,13−9,59

= 45 + ( 50 − 45 ) = 46° = 320

12,35−9,59

EX 42 - MATERIAL BALANCE

Consider a process for the synthesis of butyraldehyde by hydroformylation of propylene:

.

= + + ↔

3 2 2 3 2 2

Specifically, consider a fresh feed stream containing 5% of inert (propane) and 95%

propylene. This feed is mixed with 210 kmol/h of carbon monoxide and an equal amount of

hydrogen (syngas). Under reactor conditions, propane is an inert and it is too expensive to

separate it from propylene, so a purge stream is preferred. The conversion per passage of

propylene in the reactor is 30%.

The productivity of butanal is 180 kmol/h. The overall conversion is 90%. It is required to

calculate:

1.​ the propylene flow rate fed to the process

2.​ the recycle flow rate

3.​ the purge composition

4.​ the total purge to recycle ratio for propylene

5.​ the composition of the feed to the reactor

Assuming that the phase separator is perfectly separating the three streams, with a material

3 8

balance around it we can find that .

= = 180 ℎ

8 2

Using the conversion of propylene in the reactor, we have that →

=

8

2 . The moles of propylene in the outlet of the reactor are therefore

= = 600

ℎ

3 2 2 .

= − = 420 ℎ

4 3

With the material balance around the phase separator we obtain that .

= = 420 ℎ

Let’s write some molar balances: 0 6 6

−

●​ total conversion: (1)

= = 1− = 0, 9

0 0

4 5 6

●​ balance around the recycle splitter: (2)

= +

2 5 0

●​ balance around the mixer: (3)

= +

so we have 3 equations in 3 unknowns.

6

6 0

From (1): → .

1 − = 0, 9 = 0, 1

0

4 5 6 4 5 0 5 4 0

From (2): → →

= + = + 0, 1 = − 0, 1

2 5 0 2 4 0 0 0

From (3): → → →

= + = − 0, 1 + 600 = 420 + 0, 9

0

600−420

Request 1: .

= = 200 ℎ

0,9

5 4 0

= − 0, 1 = 420 − 0, 1 · 200 = 400 ℎ

6 0 .

= 0, 1 = 20 ℎ

The feed is composed by 95% of propylene and 5% of propane: →

0, 95: 200 = 0, 05:

0 .

= = 10, 53 ℎ

Since inert isn’t generated inside the process, if we took the molar balance around the entire

0 6

plant: .

= = 10, 53 ℎ

In the recycle splitter, the composition of the streams is the same.

6 20

Request 3: the composition in stream 6 (purge) is so it must be the

= = 65, 5%

20+10,53

5

same in stream 5. Since in stream 5 there are that correspond to 65,5% of the

= 400 ℎ

total, the molar flowrate of propane is: →

400: 0, 655 = : ( 1 − 0, 655

)

5 .

= = 210, 7 ℎ

5

Request 2: the recycle flowrate is therefore .

= 400 + 210, 7 = 610, 7 ℎ

20

Request 4: the purge to recycle ratio for propylene is .

= = 0, 05

400

2 5 0

The amount of propane in stream 2 is .

= + = 210, 7 + 10, 53 = 221, 2 ℎ

Request 5: The composition of the feed of the reactor is:

2 600

600

●​ = = = 48, 3%

1241,2

600+210+210+221,2

2 221,2

●​ = = 17, 8%

1241,2

2 210

●​ = = 16, 9%

1241,2

2 210

●​ = = 16, 9%

1241,2

2

EX 43 - MASS BALANCE, ENERGY BALANCE

Methanol is produced from CO and hydrogen according to the reaction:

CO + 2H → CH OH

2 3

∆ =− 90, 1

, 298

The catalysts currently used are based on Cu/ZnO/Al O which enable an extremely high

2 3

selectivity to be achieved.