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• Zakiyah Kamto Irfin

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In the combustion of heptane, CO2 is produced. Assume that you want to produce 500 kg of dry ice per hour and that 50% of the CO2 can be converted into dry ice, as shown in Fig. E1.27 How may kilograms of heptane must be burned per hour? CO2 Gas (50%)

Other Products

CO2 Solid (50%)

C7H16 Gas Reactor

500 kg/hr Diketahui C7H16 combustion --> C7H16 direaksikan dng O2 menghasilkan CO2 pers reaksi --> C7H16 stoikiometri

+

O2

1

-->

CO2

11

+ 7

8

Ditanya C7H16 burned per day ?

kg

Jawab Basis

1 hr atau 500 kg dry ice

BM CO2 BM C7H16

44 100

CO2 total terproduksi = 500 kg dry ice =

x

1 0.5

x

1 7

kg CO2 kg dry ice

1000 kg CO2

Mol CO2 yang dihasilkan = 1000 kg CO2 44 kg/kmol CO2 = 22.72727 kmol CO2 Mol C7H16 yang dibutuhkan = 22.72727 kmol CO2 =

3.246753 kmol C7H16

H2O

kmol C7H16 kmol CO2

Massa C7H16 yang dibutuhkan = 3.246753 kmol C7H16 324.6753 kg C7H16

x

100 kg/kmol C7H16

A liquified mixture of n-butane, n-pentane and n-hexane has the following composition in percent. n - C4H10 50 n - C5H12 30 n - C6H14 20 Calculate the weight fraction, mol fraction and mol percent of each component and also the average molecular weight of the mixture.

Diketahui Komposisi camp n-butane, n-pentane dan n-hexane n - C4H10 50% n - C5H12 30% n - C6H14 20% Ditanya a. weight fraction b. mol fraction c. average MW Jawab

fraksi berat fraksi mol BM rata2

BASIS Kompone n n - C4H10 n - C5H12 n - C6H14

avg MW = total mass total mole

100 kg

kg 50 30 20 100

=

BM C

weight fraction BM 0.5 0.3 0.2 1

12 BM H mol mol fraction 58 0.862069 0.570418 72 0.416667 0.275702 86 0.232558 0.15388 1.511294 1

100 kg 1.511294 kg/kgmol

=

66.16847

1

Tips for Solving Problems 1. Write the Given and Find statements 2. Draw a picture of the system labeling all Inputs and Outputs for each component. 3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values. 4. Determine whether the problem is a batch process or rate process. Determine whether components are genera 5. Write the conservation of mass for each component and for the entire system. 6. Use algebra to solve for the unknowns. 7. Check your work.

1 GIVEN: 10 gallon fish tank Initial concentration: 2 percent salt Final concentration: 3.5 percent salt

salt water initial concentration final concentration

10 2% 3.5%

FIND: Amount of salt to be added 2 DRAW A PICTURE Dry salt Final salt water

Initial salt water Mixer

3 ASSIGN VARIABLES Dry salt (DS) X lbm dry salt Initial salt water (ISW)

Final salt water (FSW)

Mixer

10 gal salt water 2% salt 98% H2O

Y lbm dry salt 3.5% salt 96.5% H2O

4 BATCH PROCESS / RATE PROCESS --> BATCH PROCESS 5 MASS BALANCE Total r H2O

10 62.3 0.1337

ISW gal lbm/ft3 ft3/gal

+

DS X

= lbm

FSW Y

83.2951 Salt

lbm

+

Csalt.ISW 0.02 83.2951 1.665902

CH2O.ISW

H2O 0.98

83.2951 81.6292

X

lbm

=

Y

+ + +

Csalt.DS 1X 1X

= = =

Csalt.FSW 0.035 0.035

+ + +

CH2O.DS

= = =

CH2O.FSW

0X 0X

0.965 0.965

6 from eq. 3 81.6292 = Y =

0.965 Y 84.58984 ???

from eq. 2 1.665902 X = =

+

1X 2.960644 1.294742 ???

= 1.665902

0.035 Y

7 CHECK from eq.1 83.2951 lbm 83.2951 lbm

+ +

X lbm 1.294742 lbm

= =

Y 84.58984

eters and list their values. e whether components are generated or consumed.

gallon

lbm

lbm

eq (1)

Y Y

eq (2)

Y Y

eq (3)

lbm lbm

Diketahui massa CO2 padat yang dihasilkan prosentase CO2 padaat/CO2 semua pers reaksi C7H16 + 1

500 kg/hr 50% O2

-->

CO2

11

+ 7

Ditanya massa C7H16 yang dibutuhkan

Jawab

Basis 1 jam 1 mol CO2 padat yang dihasilkan = 500 = 11.36364 mol 44 2 mol CO2 keseluruhan yang dihasilkan = 11.36364 = 22.72727 mol 50% 3 mol C7H16 yang dibutuhkan --> berdasar persamaan stoikiometri C7H16 stoikiometri mol

+

O2

-->

CO2

+

1

11

7

3.246753

35.71429

22.72727

4 massa C7H16 yang dibutuhkan = 3.246753 x

100

H2O 8

H2O 8 25.97403

Tips for Solving Problems 1. Write the Given and Find statements 2. Draw a picture of the system labeling all Inputs and Outputs for each component. 3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values. 4. Determine whether the problem is a batch process or rate process. Determine whether components are genera 5. Write the conservation of mass for each component and for the entire system. 6. Use algebra to solve for the unknowns. 7. Check your work.

1,2,3

Given Distilat (D): 85% Etanol 15% H2O

1 kg Feed (F): 35% Etanol 65% H2O

Waste(W): 5% Etanol 95% H2O FIND

Hitunglah massa distilat per kg limbah!

4 BATCH PROCESS 5 MASS BALANCE TOTAL F

= 1=

ETHANOL

Xf . F

H2O

(1-Xf) . F = 0.65 =

6 SOLUTIONS

= 0.35 =

D D

+ +

W W

Xd . D 0.85 D

+ +

Xw . W 0.05 w

(1-Xd) . D + 0.15 D +

(1-Xw ). W 0.95 w

A

(EQ2 & EQ3) 0.85 0.15

A-1

1.1875 -0.1875

X

0.05 0.95

B

0.35 0.65

-0.0625 1.0625

0.375 0.625

D W

7 CHECK F

= 0.35 0.65 1

D

+ 0.31875 0.05625 0.375

w 0.03125 0.59375 0.625

eters and list their values. e whether components are generated or consumed.

Distilat (D): 85% Etanol 15% H2O

Waste(W): 5% Etanol 95% H2O

EQ1

EQ2

EQ3

A

A-1

X = A-1.B

2 4

3 -3

B

0.166667 0.166667 0.222222 -0.11111

8 -2

0.25 0.35 0.4

0.166667 0.166667 0.222222 -0.11111

-3.4 2.85

1 2

1.55 0.6 0.35 0.05

A

0.8 1.6 0 0.2 0.6049 -0.15314 0.967841 -0.51608

0 0 0.21 0.79

0 -1 -0.5 0

-1 0 -1 0

B

1.2 0 4.9 93.9

X

0.30245 -0.6049 0.160796 -0.07657 0.153139 1.225115 -0.51608 -0.96784 0.257274 0.24196 -0.48392 0.128637

12.86064 115.6049 20.57703 9.088515

NG EA W DI FG 12.86064 115.6049 20.57703 9.088515 C H2 O2 N2

10.28851 0 20.57703 0 0 24.27703 2.572129 91.32787

CO2 teradsorb kebutuhan O2 ekses air

88.34%

20.57703 3.7 17.98%

0 -20.577 -10.2885 0

-9.08851 0 -9.08851 0

100

1.2 3.55E-15 4.9 93.9 100

NG EA W DI

0.35 0.2 0.45

0.55 0.4 0.05

4.6 -4.15

0.6 1.85

0.55

-1.45

0.1 0.3 0.6

0.3 0.3 0.4

A natural gas analyzes CH4, 80.0 percent and N2, 20.0 percent. It is burned under a boiler and most of the CO2 is scrubbed out of the flue gas for the production of dry ice. The exit gas from the scrubber analyzes CO2, 1.2 percent; O2, 4.9 percent; and N2, 93.9 percent. Calculate the (a) Percentage of the CO2 absorbed. (b) Percent excess air used.

1,2,3 GIVEN Dry Ice (DI) CO2

Flue Gas (FG)

Natural Gas (NG) BURNER CH4 N2

80% 20%

CO2 O2 N2

Air (A)

1.2% 4.9% 93.9%

O2 N2 FIND % CO2 ABSORBED % EXCES AIR 4 BATCH PROCESS 5 BASIS NM TOTAL NM.KOMP

6 INVERSE A A^-1

MULT

100 kgmol FG NG C H O N

A

- DI

0 0 0.42 1.58

0 -2 -1 0

0.6049 0.151225 -0.15314 -0.03828 0.967841 -0.25804 -0.51608 0.12098

-0.30245 0.07657 -0.48392 -0.24196

0.080398 0.612557 0.128637 0.064319

12.13476 115.7887 19.41562 8.50781

NG A W DI

7 CH4 IN NATURAL GAS 9.71 kgmol CO2 ABSORBED

REACTION

-W

0.80 3.20 0 0.40

CH4

FG 1.2 0 12.2 187.8

CEK NM TOTAL NG+A=W+DI+FG 127.9234 127.9234

CO2 IN DRY ICE 8.50781 kgmol

8.51 kgmol 87.63882 % +

2 O2

-->

CO2

O2 REQUIRED = =

2* 19.42 kgmol

=

48.63124 kgmol

= =

29.22 kgmol 150.4748 %

O2 IN AIR (A)

EXCESS O2

= -1 0 -2 0

9.71 kgmol

+

2 H2O

Tips for Solving Problems 1. Write the Given and Find statements 2. Draw a picture of the system labeling all Inputs and Outputs for each component. 3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values. 4. Determine whether the problem is a batch process or rate process. Determine whether components are generated or consumed. 5. Write the conservation of mass for each component and for the entire system. 6. Use algebra to solve for the unknowns. 7. Check your work.

eters and list their values. e whether components are generated or consumed.

A synthesis gas analyzing CO2, 4.5 percent; CO, 26 percent; H2, 13 percent; CH4, 0.5 percent; and N2, 56 percent, is burned in a furnace with 10 percent excess air. Calculate the Orsat analysis of the flue gas

1,2,3 GIVEN

Water (W) H2O

Flue Gas (FG)

Syn Gas (SG) BURNER CO2 CO H2 CH4 N2

4.5% 26% 13% 0.5% 56%

n CO2 n O2 n N2

Air (A) 10% EXCESS AIR O2 N2

? ? ?

21% 79%

FIND ORSAT ANALYSIS OF FLUE GAS 4 BATCH PROCESS / STEADY STATE 5 BASIS

100 kgmol SG

6 O2 REQUIRED CO

+

0.5 O2 --> 26

CH4

H2

TOTAL

+ 0.5

+ 0.5 O2 --> 13 6.5

+ 2 H2O 0.5

1

1 H2O 13

20.5 kgmol

O2 ENTER N2 ENTER

22.55 kgmol 22.55 *

7 CHECK NM ELEMEN

1 CO2 1

2.05 kgmol

NM TOTAL NM. KOMP

26

2 O2 -->

O2 EXCESS

EXCESS AIR (A)

1 CO2 13

=

22.55 +

IN CO2 CO H2 CH4 N2 H2O O2

4.5 26 13 0.5 140.831 0 22.55

-

79% = 21% 84.83095 =

OUT n CO2 n CO n H2 n CH4 n N2 n H2O n O2

84.83095 kgmol 107.381 kgmol

GEN + + + + + + +

CONS 26.5 0 0 0 0 14 0

ACC

-

0 26 13 0.5 0 0 20.5

0.5 = = 45.1 = =

31 28 80.1 281.6619

IN C H O N

4.5 + 26 + 9+ 281.6619

= = = = = = =

OUT 0 0 0 0 0 0 0

31 0 0 0 140.831 14 2.05 187.881 kgmol OUT

26 + 2 26 +

31 28 80.1 281.6619

ORSAT ANALYSIS 17.83 % 80.99 % 1.18 % 100.00 %

1,2,3 GIVEN Feed

Product

Reactor

C3H6 NH3 O2 N2

15% 7% 78%

C3H3N H2O N2 O2

? ? ? ?

FIND 1 2 3 4 4 BASIS

LR EXCESS REACTANT RATIO C3H3N/NH3, JIKA KONVERSI = 30% COMPOSITION PRODUCT FEED

5 FEED C3H6 NH3 O2 N2

100 kgmol MOL 42 17 32 14

15 7 78 0

C3H6 M R SR LR EXCESS

1 NH3 7 7 0

1 C3H3N 78 10.5 67.5

7 -7

NH3 C3H6 O2

114.2857 642.8571

KONVERSI

30%

C3H6 M R SR

1.5 O2

15 7 8

1 NH3 15 2.1 12.9

RATIO MOL C3H3N/NH3 =

1.5 O2 7 2.1 4.9

2.1 X 100 = 7

1 C3H3N 78 3.15 74.85

30 %

2.1 2.1

4 BASIS

FEED

5 FEED C3H6 NH3 O2 N2 3 H2O 21 -21

M R SR LR EXCESS

42 17 32 14

MASSA

mol 15 0.357143 7 0.411765 78 2.4375 0 0

C3H6 0.357143 0.357143 0

1 NH3 0.411765 0.357143 0.054622

C3H6 NH3 O2

3 H2O 6.3 6.3

100 kg

15.29412 355

1.5 O2 2.4375 0.535714 1.901786

1 C3H3N 0.357143 -0.35714

3 H2O 1.071429 -1.07143