DESIGN of Sodium Carbonate PRODUCTION PLANT Comprehensive Design Project

 

Chapter 8 

SAFETY MEASURES  

to an uncontaminated area, given mouth-to-mouth resuscitation and supplemental oxygen. Keep victim warm and quiet. Assure that mucus or vomited material does not obstruct the airway by positional drainage. Skin/eye contact:  • 

Skin contact cause irritant. 

• 

In case of contact, immediately flush skin with plenty of water. Cover the irritated skin with an emollient. Remove contaminated clothing and shoes. Cold water may be used. Wash clothing  before reuse. Thoroughly clean shoes before reuse. Get medical attention.

• 

Eye(s) contact cause irritant.

• 

Check for and remove any contact lenses. In case of contact, immediately flush eyes with  plenty of water for at least 15 minutes. Cold water may be used.

Ingestion:  • 

Do NOT induce vomiting unless directed to do so by medical personnel. Never give anything  by mouth to an unconscious person. Loosen tight clothing such as a collar, tie, belt or waistband. Get medical attention if symptoms appear.

Handling And Storage • 

Do not ingest. Do not breathe dust. Wear suitable protective clothing. In case of insufficient ventilation, wear suitable respiratory equipment. If ingested, seek medical advice immediately and show the container or the label. Avoid contact with skin and eyes. Keep away from incompatibles such as acids.

• 

Hygroscopic. Keep container tightly closed. Keep container in a cool, well-ventilated area. Do not store above 24°C (75.2°F). Hygroscopic

Stability And Reactivity • 

The product is stable.

Disposal Considerations • 

Waste must be disposed of in accordance with federal, state and local environmental control regulations.

Toxicological Information  • 

Routes of Entry: Inhalation. Ingestion.

• 

Acute oral toxicity (LD50): 4090 mg/kg [Rat].

• 

Acute toxicity of the dust (LC50): 1200 mg/m3 2 hours [Mouse].

• 

Effects on Humans: May cause damage to the following organs: upper respiratory tract, skin,

eyes. Hazardous in case of skin contact (irritant), of ingestion, of inhalation (lung irritant).

 

Chapter 8 

SAFETY MEASURES  

Permeable Exposure Limit • 

LDL (Lowest Published Lethal Dose) [Man] - Route: Oral; Dose: 714 mg/kg

Personal Protective Equipments • 

Splash goggles. Lab coat. Dust respirator. Be sure to use an approved/certified respirator or equivalent. Gloves.

• 

Personal Protection Protection in Case of a Large Spill: Splash goggles. Full suit. Dust respirator.

Boots. Gloves. A self contained breathing apparatus should be used to avoid inhalation of the  product. Suggested protective clothing might might not be sufficient; consult a specialist BEFORE handling this product.

 

Chapter 9

MATERIAL  BALANCE 

 

CHAPTER 09

MATERI MATE RIAL AL B AL A NC NCE E 

Material balances are based on the fundamental law of conservation of mass. In particular, chemical engineers are concerned with doing mass balances balances around c hemical processes. proc esses. Doing a ‘mass balance’ is similar in principle to accounting. In accounting, accountants do balances of what happens to a Company’s money. Chemical Chemical engi nee neers rs do a mass balance to account for what happens to each of the chemicals th at is used in a chemical process... pr ocess..... .. 

 

Chapter 9 

MATERIAL  BALANCE 

This chapter contains the material balance for the process we designed. There are number of assumptions and those are described in appropriate places. Some values are extracted from literature and those are mentioned in reference. This calculation based on 50Mg (tons) of product (Na 2 CO 3 ) output per day. All the rates are given in daily basis and haven’t mention in calculations. Most of the flow rates are calculated in mole. 9.1 Product Specification

Ingredients  Na 2 CO 3    Na 2 SO 3 Moisture and impurities

Percentage 99.5% 0.004% 0.495%

 

Table 9.1- Soda ash specification

Average molar weight of product

= (1063×0.995) + (183×0.005) + (14230.000042) = 105.566 g/mol

Production rate

= 50000kg/day

Production rate in kmol

= /  ./

= 473.6375kmo 473.6375kmol/day l/day Note: We used Microsoft Excel in our calculation and its goal seek function was used to do the calculation targeting above calculated flow rate by varying total NH 3  input to the NH 3 Absorption Unit. Microsoft Excel worksheet was attached at the end of the chapter.  

9.2 Components in Purified Brine

 NaCl  Na 2 SO 4 Ot Othe herr Na/C Na/Ca/M a/Mg g salts salts  

H2O Sa Salt lt (T (Tot otal al))

5.07 kmol/m3 0.05 kmol/m3 0. 0.00 009 9 kmol kmol/m /m3 45.42 kmol/m3 50 50.5 .549 49 kmol kmol/m /m3

 

 

 

 

Table 9.2- Purified brine specification

Density of brine

= 1122 kg/m3 

By calculating we found that amount of NaCl we need for our process is 1477.31kmol. Using above concentrations the Flow rate of purified brine was calculated. Flow rate of purified brine

=

.   ./

= 291.38m3 Flow rate of purified brine (mole)

= 291.38 m3 × 50.549 kmol/m3  = 14729.05kmo 14729.05kmoll

 

Chapter 9 

MATERIAL  BALANCE 

 Now using the flow rate and above compositions compositions of brine, = 291.38m3 × 0.05 kmol/m3 

Flow rate of Na 2 SO 4  (mole)

= 14.57kmol Flow rate of Other Na/Ca/Mg salts (mole)

= 291.38m3 × 0.009kmol/m3  = 2.62kmol = 291.38m3 × 45.42kmol/m3 

Flow rate of H 2 O (mole)

= 13234.56kmo 13234.56kmoll

9.3 NH 3 Absorption Unit

   NaCl  Na 2 SO 4

 

Other Na/Ca/Mg salts H2O  NH 3 OH

Brine

1477.31 14.57

kmol kmol

2.62 13057.79 176.77

kmol kmol kmol

Waste gas

 NH 3 CO 2

 

126.2654

kmole

29.67

kmole

 

  n   o    i    t   p   r    t   o   i   s   n    b   U

 

   A    3    H    N

Makeup ammonia + Gas from ammonia recovery unit

 

 

H2O  NH 3 OH  NaCl

13055.16 1136.389 1477.305

kmole kmole kmole

 Na 2 SO 4

14.56909

kmole

CO 2    NH 3  

32.28992 1085.883

 Ammoniated  Ammonia ted brine water 

Figure 9.1- NH 3 Absorption Unit

kmol kmol

 

Chapter 9 

MATERIAL  BALANCE 

Possible reactions inside the NH 3 Absorption Unit, NaCl (  aq aq)  + H 2 O (l )   + NH 3(g)   ↔ NaCl (  aq aq)   + NH 4 OH ( aq  aq)   +

2-

CO 2 (  aq aq)  + H 2 O (l )   → CO 3 2+

Ca

----------- (1) 

(  aq aq)  +

2H

----------- (2)

(  aq aq)

2(  aq aq)  +

2+

(  aq aq)  → CaCO 3 ↓(  ss)  

CO 3

----------- (3) 

-

Mg

  → Mg(OH)

  + 2 OH (  aq aq)

----------- (4)  2 ↓ (  ss)

(  aq aq)

Gaseous streams at output of NH 3 Absorption Unit,  NH 3  Total NH 3   (as ammonia gas and ammonium hydroxide) input to the NH 3 adsorption unit is 1262.654kmol. Gas stream of the output of NH3 Absorption Unit is assumed to have gas NH 3 . And the amount is equal to 10% of total NH 3  enter to the unit.  NH 3  output of NH 3 Absorption Unit

= 1262.654 kmol × 10% = 126.25kmol

CO 2 Assume reacted amount of carbon dioxide with salt is equal (mostly salt contains Ca 2+, Mg2+ which react 1:1 with CO 3 2+) to the salt amount (mole). Carbon dioxide output

= Input – Reacted with salt = 32.28992 kmol - 2.62kmol = 29.67kmol

Liquid streams at output of NH 3 Absorption Unit,  NH 3  Ammonia exits as ammonium hydroxide in liquid stream from the system. As mentioned above input of total NH 3  is 1262.654kmol. Ammonium Hydroxide amount

= 1262.654 - 126.25 kmol = 1136.389 kmol

 NaCl  No reaction on NaCl, So Input is equal equal to output. Output

= 1477.31 kmol

 

Chapter 9 

MATERIAL  BALANCE 

 Na 2 SO 4   No reaction on Na2 SO 4 , So Input is equal to output. Output

= 14.57 kmol

Other Na/Ca/Mg salts Here we assume 100% precipitation of salts of Na, Ca and Mg except Na 2 CO 3  and CaCO 3 . And equal amount of moles of water removes with the salt solid from the NH 3  Absorption Unit. This salt and water removes from the system at this unit. And it is assume that Na 2 SO 4  is 100% soluble in this condition. Salt precipitation

= 2.62kmol

H 2 O (Water) Water removed with the salt

= 2.62kmol (See Other Na/Ca/Mg salts)

Amount of water enter to the unit

= 13051.47kmol

Amount of water exit from the unit = Water enter to the unit - Water removed with with the salt = 13048.85kmol

9.4 Air Mixture

Makeup Ammonia

 NH 3

 

6.410432

CO 2    NH 3

32.28992 1085.883

kmol kmol

 Air mixture To NH3 absorption unit

kmole

Gas from ammonia recovery unit CO 2  NH 3

 

32.28992 1079.472

kmole kmole

Figure 9.2- Air mixture before NH 3 Absorption Unit

Air mixture has two input streams; both get mixed and go to the NH 3 Absorption Unit. CO 2  No two inputs of carbon dioxide dioxide and output stream equal to the input. 32.28 32.28992kmol. 992kmol.  NH 4  Output Ammonia Amount

= From Ammonia recovery unit + Ammonia Makeup = 1079.472 + 6.410432 = 1085.883kmol

 

 

Chapter 9 

MATERIAL  BALANCE 

9.5 Gas Washing Tower with Purified Brine

Purified brine

 

Waste gas

CO 2  N 2  NH 3

 

 

423.78 1604.452 6.313271

kmole kmole kmole

   NaCl  Na2SO4 Other Na/Ca/Mg salts H2O

1477.31 14.57 2.62 13234.56

kmole kmole kmole kmole

 NH 3 CO 2

56.81944 394.1129

 

 

kmole kmole

Waste gas from Carbonator unit

Gas washing with  purified brine Brine

Waste gas from NH 3 adsorption unit

 NH 3  NaCl  Na 2 SO 4

 

Other Na/Ca/Mg salts H2O  NH 3 OH

1477.31 14.57

kmol kmol

2.62 13057.79 176.77

kmol kmol kmol

CO 2

 

126.2654

kmole

29.67

kmole

 

Figure 9.3- Gas washing tower with purified brine

Output gas stream, CO 2  Assume carbon dioxide does not dissolve in brine. Output carbon dioxide amount

= From carbonator + From NH 3 Absorption Unit = 394.1129 + 29.67 = 423.78kmol

 N 2  Assume Nitrogen does not dissolve in brine. Output Nitrogen amount

= From carbonator + From NH 3 Absorption Unit = 1604.452 + 0 = 1604.452kmol

 

Chapter 9 

MATERIAL  BALANCE 

9.6 Carbonator Unit

 

 NH 3 CO 2  N 2

NH3 , CO CO2, 2, N2 Mixt ure

 

57 394 1,604

 

 

kmole kmole kmole

 Am mo ni ated b ri ne

H2 O  NH 4 OH  NaCl  Na 2 SO 4

13054 1136 1477 15

kmole kmole kmole kmole

 

 

Carbonation tower

CO2, N2 mixture Cooling System

 N 2 CO 2 H2 O

1,604 1,474 305

 

 

H2 O  NaHCO 3  NaCl  NH 4 Cl  NH 4 HCO 3  Na 2 SO 4  

 

13,359

kmole

1,047 430 1,047 32 15

kmole kmole kmole kmole kmole

kmole kmole kmole

NaHCO3 NaH CO3 Solutio n

Figure 9.4- Carbonator Unit

Possible reactions inside the Carbonator Unit,  NH 4 OH (  aq  g)  + H 2 O (l )   aq)   ↔ NH 3( g

----------- (5)  ----------- (6) 

2 NH 4 OH (  aq  g)  ↔ (NH 4 ) 2 CO 3(  aq aq)   + H 2 O (l )   aq)   + CO 2( g  g)  + H 2 O (l )  ↔ 2 NH 4 HCO 3(  aq aq)   aq)   + CO 2( g (NH 4 ) 2 CO 3(  aq

----------- (7)

2 NH 4 HCO 3(  aq aq)   ↓  + 2 NH 4 Cl (  aq aq)   + 2 NaCl (  aq aq)   ↔ 2 NaHCO 3(  aq aq)  

----------- (8) 

Gaseous streams at output of Carbonator Unit,  NH 3  Ammonia occurs from decomposes of NH 4 OH in the carbonator column. We assumed the decomposed amount is equal to 5% of total NH 4 OH input to the Carbonator.  NH 3  output of carbonator Unit

= NH 4 OH input × 5% = 1136 × 5% = 126.25kmol

 

 

Chapter 9 

MATERIAL  BALANCE 

CO 2 Reacted amount (mole) of carbon dioxide in the column is equal to NaHCO 3 , NH 4 HCO 3 amount  produce. Carbone dioxide enters only from the gas stream from the gas cooler. Carbon dioxide output

= Input – Reacted for NaHCO 3 - Reacted for NH 4 HCO 3  = 1473.68 kmol - 1047.182 kmol - 32.387kmol = 394.1129 kmol

Liquid streams at output of NH 3 Absorption Unit,  NaHCO 3  The efficiency of the carbonator column was taken as 97% (generation of NaHCO 3  as a percentage to effective NH 4 OH input). Effective NH 4 OH amount (mole)

= NH 4 OH input – decomposes to NH 3 = 1136.388 kmol - 56.819 kmol = 1079.569kmo 1079.569kmoll

Generation of NaHCO 3 amount (mole)

= Effective NH 4 OH amount × 97% = 1079.569kmol× 97% = 1047.182kmol

 NH 4 HCO 3 This is an intermediate product in reaction chain and amount is inversely proportional to the efficiency. Generation of NaHCO 3 amount (mole)

= Effective NH 4 OH amount × 3% = 1079.569kmol× 3% = 32.387kmol

 NaCl Sodium spent to generate NaHCO 3 . Sodium Chloride output from the carbonator

= Sodium Chloride input - NaHCO 3  output = 1477.305kmol - 1047.182kmol 1047.182kmol = 430.1231kmol

 

Chapter 9 

MATERIAL  BALANCE 

 NH 4 Cl Ammonium chloride generated in the reaction is equal to NaHCO 3 generated in the reaction. Ammonium chloride output

= 1047.182kmol

 Na 2 SO 4   No reaction on Na2 SO 4 , So Input is equal to output.  Na 2 SO 4  output

= 14.57kmol

H 2 O (Water) Water amount exits from the unit is equal to the water enters to the system. Water exits from the unit

= Water from NH 3 Adsorption unit + Gas Cooler = 13054.34kmol + 304.9075kmol 304.9075kmol = 13359.25kmol

9.7 Filter H2O  NaHCO 3  NaCl  NH 4 Cl  NH 4 HCO 3  Na 2 SO 4

     

13,359 1,047 430 1,047 32 15

kmole kmole kmole kmole kmole kmole

F  r   o m    C   a r   b   o n  a  t    o r   U n i    t  

 

Residue solid

Filter 

H2 O  NaCl  NH 4 Cl

13,123.9 430.1 1,047.2

kmole kmole kmole

104.7 32.4 14.5

kmole kmole kmole

Permeate  

 NaHCO 3  NH 4 HCO 3  Na 2 SO 4  

 

 

 

 NaHCO 3 H2 O  Na 2 SO 4

942.464 235.3804 0.02

kmole kmole kmole

Figure 9.5- Filter

Product composition after the filter as follows,

 

Component  NaHCO 3 H2O  Na 2 SO 4  

Percentage 80% 19.98% 0.02%

Table 9.3- Residue solid composition

And the filter assumed to have 90% efficiency (based on NaHCO 3 ). As NaHCO 3 is our main product this calculation is based on NaHCO 3 .

 

Chapter 9 

9.7.1. 

MATERIAL  BALANCE 

Calculation for residue solid

 NaHCO 3   NaHCO 3 in residue solid

= Input to the filter × 90% = 1047.182kmol × 90% = 942.464kmol

Other components are calculated according to the above composition. H2O Water in the residue solid (wet)

=

.  .%   %

= 235.38kmol  Na 2 SO 4  Although Na 2 SO 4 is not a solid in this condition, considering End Product composition it assumed some amount of Na 2 SO 4  remain in the residue   solid. So Na 2 SO 4 amount remain in the residue  solid equal to the Na 2 SO 4 in the End Product. Calculations are shown in dryer calculations.  Na 2 SO 4  in the residue  solid 9.7.2. 

= 0.02kmol

Calculation for permeate

 NaHCO 3   NaHCO 3 in permeate

= Input to the filter × 10% = 1047.182kmol × 10% = 104.7kmol

H2O Water in permeate

= Water input – Water in filtered solid = 13359.25 - 235.38 = 13,123.9kmol

 Na 2 SO 4   Na 2 SO 4  in permeate

= 14.5kmol

 NaCl It assumed that total NaCl amount is filtrate through the filter. So amount is equal to input of the filter.  NaCl in permeate

= 430.1kmol

 

Chapter 9 

MATERIAL  BALANCE 

 NH 4 Cl It assumed that total NH 4 Cl amount is goes through the filter. So amount is equal to input of the filter.  NH 4 Cl in permeate

= 1,047.2kmol

 NH 4 HCO 3 It assumed that total NH 4 HCO 3 amount is goes through the filter. So amount is equal to input of the filter.  NH 4 HCO 3 in permeate

= 32.4kmol

9.8 Lime Kiln Hot Flue Gas

 N 2 CO 2

 

 

1604.45 1002.45

kmole kmole

  Heated CaCO3 Coke

Coal

426.50

kiln

CaCO 3

575.95

kmole

kmole

Heated Ari

O2  N 2

 

426.50 1604.45

 

kmole kmole

Figure 9.6- Lime Kiln

Calculation for the Lime Kiln has done considering material requirements (CO 2 , CaO) and energy requirements. Main base for the calculation was the CO 2  output from the Kiln.

Possible reactions inside the Carbonator Unit,  CaCO 3( s s)  

 CaO (  ss)  + CO 2 ( g  g)  

→

C ( s s)  + O 2 (  gg) 

 CO 2 ( g  g)  

→

----------- (9)  ----------- (10) 

 

Chapter 9 

MATERIAL  BALANCE 

Gaseous products at outlet, CO 2 Considering above (9) and (10) reactions, Carbon dioxide output

= CaCO 3 reacted  + C reacted = 575.95kmol + 426.50kmol = 1002.45kmol

 N 2  Before calculating N 2 , it needs to calculate air consumption for the kiln. In literature we found that the kiln is supplied no excess air. C reacted

= 426.50kmol

According to the Stoichiometric  Stoichiometric relation, amount of O 2  consumed. O 2  required

= 426.50kmol

Assuming composition of air is O 2 – 21%, N2  – 79%  N 2  comes with O 2  

=

.  %   %

= 1604.45kmol Air amount

= O 2  required + N 2  comes with O 2  = 2030.95kmol

Solid products at outlet, It is assume that only solid product at outlet is CaO. Considering real situations it is possible to remain unreacted CaCO 3 in outlet but considering calculation complexity and effectiveness on the system we assumed it as 0kmol. Amount of CaO

= 575.95kmol

 

Chapter 9 

MATERIAL  BALANCE 

9.9 Slaker of lime

water coming from cooler CaO solid

17068 kmole/day

575.95 kmole/day

Hot water from cooler Cold CaO

Excess water

extra water

Ca(OH)2 Solution

1417.2 kmole/day

Slaker

Ca(OH)2 H2O weight of solution Temperature

576kmole/day 13,418 kmole/day 284,135 kg/day 60 0C

Vaporizes water

evaporated water vapors

1657.4kmole/day

Figure 9.7- Slaker of lime 

Possible reactions inside the Slaker Unit,  CaO ( s s)  + H 2 O (l )  

 Ca(OH) 2(  aq aq)  

----------- (11) 

→

Liquid products at outlet, Ca(OH) 2  Ca(OH) 2  amount

= Amount of CaO = 575.95kmol

H2O Water input amount

= Water from gas cooler - Excess water (remove) = 17068.07kmol - 1417.21kmol = 15650.85kmol

Water reacted with Ca(OH) 2  

= 575.95kmol

Water vaporized amount

= 1657.4kmol

Excess water

= Water input - Water reacted - Water vaporized

(see energy balance)

= 15650.85kmol - 575.95kmol - 1657.4kmol = 13417.507km 13417.507kmol ol

 

Chapter 9 

9.10 

MATERIAL  BALANCE 

Ammonia Recovery Unit

CO 2    NH 3

Cool gas (NH3)

 

 

32.2 32 .289 899 92 107 079 9.47 .472

kmol kmole/ e/d day kmole/ ole/da day y

Cool NH4Cl Solution Condensed water 

HE

 

H2 O

1061.655

kmole/day

 

H2O  NaCl  NH 4 Cl  NaHCO3  NH 4 HCO3  Na 2 SO4

Hot NH3 ,H2O mixture

H2 O CO 2    NH 3

 

 

 

1061.655 32.28992 1079.472

13,123.9 430.1 1,047.2 104.7 32.4 14.5

kmole/day kmole/day kmole/day kmole/day kmole/day kmole/day

kmole/day kmole/day kmole/day

 Amm on ia recovery unit

shaking of lime

Ca(OH) 2 H2 O

575.95 13417.51

 

 NaCl CaCl 2  NaHCO 3  

 NH 4 OH Ca(OH) 2  Na 2 SO 4 H2 O

 

 

 

Steam

kmole/day kmole/day

430.12 523.59 104.72

kmole/day kmole/day kmole/day

0.10 52.36 14.55 25,479.72

kmole/day kmole/day kmole/day kmole/day

Waste water 

Figure 9.8- Ammonia Recovery Unit  

Possible reactions inside the Carbonator Unit,  2NH 4 Cl (  aq aq)  + Ca(OH) 2 NH 4 HCO 3(  aq aq)   (13)

 2NH 3( g  g)  + CaCl 2(  aq aq) + H 2 O (l )  

→

 NH 3( g aq) + H 2 O (l )    g)  + CO 2(  aq

→

----------- (12)  -----------

 

Chapter 9 

MATERIAL  BALANCE 

Gaseous products at outlet, CO 2 Considering 99.7% efficiency for (13) reaction, Carbon dioxide output

= NH 4 HCO 3  × 99.7% = 32.4kmol × 99.7% = 32.289kmol

 NH 4  Considering 100% efficiency for (12) reaction and considering 99.7% efficiency for (13) reaction, Ammonia output

= NH 4 Cl × 100% + NH 4 HCO 3  × 99.7% = 1,047.2 × 100% + 32.4 × 99.7% = 1079.472kmol

Liquid products at outlet,  NaCl  NaCl involves with no reaction through the Ammonia Ammonia recovery column.  NaCl output

= 430.1kmol

 NaHCO3  NaHCO3 involves with no reaction through through the Ammonia recovery column.  NaHCO3 output

= 104.72kmol

 NH 4 HCO 3  Considering 99.7% efficiency for (13) reaction,  NH 4 HCO 3  output

= NH 4 HCO 3  × 0.3% = 32.4kmol × 0.3% = 0.10kmol

 Na 2 SO 4   Na 2 SO 4  involves with no reaction through the Ammonia recovery column.  Na 2 SO 4  output

= 14.55kmol

 

Chapter 9 

MATERIAL  BALANCE 

CaCl 2  According to the reaction (12). (Stoichiometric relation is 2:1) CaCl 2 at outlet

= NH 4 Cl ×

    

= 1,047.2 ×   = 523.59 kmol Ca(OH) 2  Ca(OH) 2 amount is 110% of the actual Ca(OH) 2   requirement CaCO 3   to the kiln is back calculated from here. Ca(OH) 2  at the outlet

= Ca(OH) 2  × 10% = 523.59 × 10% = 52.36kmol

H2O Water involve no reaction in the Ammonia recovery column and it is assume that no water vapor in gaseous output. Water output amount

= Water from slaker + Water from filter = 13417.5 kmol + 13,123.9 kmol = 26,541.38 kmol

 

Chapter 9 

9.11 

MATERIAL  BALANCE 

Gas Cooler

   N 2 CO 2 H2 O

(g)

 

1604.452381 kmole/day 1473.682222 kmole/day 304.9074568 kmole/day

Gas out

Water In

H2O

Cooler

3

300 m /day

16669 Kmole/day

Gas mixer

H 2 O condense  N 2 CO 2 H2 O  

(g)

1604.45238 kmole/day 1473.68222 kmole/day] 704.26438 kmole/day]

 

Hot Water Out

H2O

17068 Kmole/day

Figure 9.9- Gas Cooler  

Gaseous products at outlet, CO 2 Carbon dioxide output

= Carbon dioxide input = 1473.68kmol

 N 2   Nitrogen output

= Nitrogen input = 1604.45kmol

H2O The ga gaseo eou us ou outlet let aass ssu umed to to be be sat satu ura rate ted d wi with water ater at it its temper ature.  Water in gaseous outlet

= Dry air flow ×

     

= {(N2× 28) + (CO 2  × 44)} ×

.   

= {(1604.45 × 28) + (1473.68× 44)} × = 304.90kmol

 

399kmole/day

.   

 

Chapter 9 

MATERIAL  BALANCE 

Liquid products at outlet, H2O Water in liquid outlet

= From air mixture + Additional injection – In gaseous outlet = 1326.26 + 6,623 – 304.90 = 7644.77kmol

9.12 

Air Mixture (Before the Gas Cooler)  

 N 2 CO 2 H2 O

1604.452381 1473.682222 704.264384

 

Kmole/day Kmole/day   Kmole/day

CO 2 H2O

471.232 704.264384

 

Kmole/day Kmole/day

Air Mixer 

 N 2 CO 2  

 

1604.452381 1002.450222

Kmole/day Kmole/day

Figure 9.10- Air mixture before gas cooler  

Air mixture has two input streams; both get mixed and go to the Gas Cooler. CO 2 CO 2  in outlet

= From dryer + From CaCO 3  preheated = 471.232 + 1002.45 = 1473.68kmol

H2O Water in outlet

= From dryer + From CaCO 3  preheated = 704.26 + 0 = 704.26kmol

 N 2   Nitrogen output

= From dryer + From CaCO  preheated 3

= 0 + 1604.45 = 1604.45kmol

 

Chapter 9 

9.13 

MATERIAL  BALANCE 

Dryer

CO2 H2O  N2 CO2

1604.452381 1002.450222

Kmole/day Kmole/day

471.232 704.264384

Kmole/day Kmole/day

CO2, H2O Vapou Vapou r 

Flue gas out

NaCO3 NaC O3 Product NaHCO3 NaH CO3 Solutio n

 NaHCO3

942.464

Kmole/day

H2O

235.3804

Kmole/day

 Na2SO4 Flow rate

0.02 1177.844

Kmole/day kmole/day

 N2 CO2

Calcinations NaH NaHCO3 CO3

 Na 2 CO3 H2O  Na2SO4 Flue gas in 1000 C

1604.45 1002.45

Total

471.232

Kmole/day

2.348

Kmole/day

0.02

Kmole/day

473.6

Kmole/day

Kmole/day Kmole/day

Figure 9.11- Dryer  

Possible reactions inside the Carbonator Unit, 2NaHCO3 ( s s)  

 Na 2 CO 3 (  ss) + CO 2 ( g  g)  + H 2 O ( g  g)  

----------- (14)

→

Gaseous products at out let, Heat is supplies to the dryer through indirect contact of flue gas from the kiln. So composition of flue gas does not change by going through the dryer. In this calculation other gas outlet is considered. CO 2 CO 2  at outlet

= CO 2  generated by the reaction  

= NaHCO 3  amount reacted ×   = 471.232kmol H2O H 2 O at outlet

= H 2 O generated by the reaction + H 2 O in inlet - H 2 O in product = 471.232 + 235.38 - 2.348 = 704.26kmol

 

Chapter 9 

MATERIAL  BALANCE 

Solid products at out let,  Na 2 CO 3   Na 2 CO 3  in product

= Na 2 CO 3  generated by the reaction 

= NaHCO 3  amount reacted ×   

= 471.232kmol  Na 2 SO 4  Na 2 SO 4 amount in the product assume to be .02kmol. H2O H 2 O amount calculated assuming only material in the product is H 2 O except Na 2 CO 3  and Na 2 SO 4 .

 

Chapter 9 

9.14 

MATERIAL  BALANCE 

Material Flow Sheet

 

Chapter 10 

ENERGY BALANCE 

CHAPTER 10

ENER EN ERGY GY BA L A NC NCE E  

Energy is fundamental to the quality of our lives. Nowadays, we are totally dependent on an abundant and uninterrupted supply of energy for living and working. It is a key ingredient in all sectors of modern economies Energy supply must be sustainable and diverse.  And  A nd ener en erg gy need s to be us ed m or e efficiently........ 

 

Chapter 10 

ENERGY BALANCE 

10.1 Kiln Energy Balance  N 2 CO

 

1604.45

kmole/day

1002.45

kmole/day

2

Temperature 1000 C

CaCO 3 575.95 Temperature 566 C  

Coal 426.50 Temperature 30 C

kmole/day

O2

 

426.50

kmole/day

 N 2 1604.45 Temperature 900 C

kmole/day

 

Figure 10.1- Kiln 

Assumption • 

Limestone is mixture of several components CaCO 3 =78% SiO 2  =18.5% Al 2 O 3  =3.5% mole Percentage Mixture Cp is 136 J/moleK

• 

Heat loss from the kiln is considered as 15%

kmole/day

• 

This calculation is down by using trial and error method using excel

Gas phase enthalpy change were calculated using following equation 0

ΔH R 

 =   

C p = Specific heat capacity

C p is changed with Temperature like following equation

C p = a + b T – c /T2  Component Co2 O2  N2 CaO

a 10.34 8.27 6.5 10

b 0.00274 0.000258 0.001 0.05

c 195500 187700 0 108000

Table 10.1- a, b, c constant

   

Chapter 10 

ENERGY BALANCE 

Object

Fine the coke requirement for the kiln Energy balance  Enthalpy of inlet material material + Heat of combustion combustion = Enthalpy of outlet material + Dissociation Dissociation enthalpy of CaC CaCO3 O3 + Loss

Enthalpy of inlet material

Component O2  N 2 CaCO 3  

 

 

Flow rate (kmole/day) Enthalpy @ 900 C (J/mole) 426.50 45,385.0 1604.45 28,837.0 575.95 136.0 Total Enthalpy of inlet material (kJ/Day)

Enthalpy(kJ/Day) 19,356,702.5 46,267,593.3 78,329.2 65,702,625.0

Table 10.2- kiln inlet enthalpy

Enthalpy of outlet material Component CO 2 N2  

 

CaO

Flow rate (kmole/day) 1002.45 1604.45

Enthalpy @ 1000 C (J/mole) 51,060.0 32,231.0

575.95

73.6

Total Enthalpy of outlet material (kJ/Day) Table 10.3- kiln outlet enthalpy

Dissociation enthalpy of CaCO3

CaCO 3  supply rate per day =575.95 kmole/day

Enthalpy(kJ/Day) 51,185,108.3 51,713,104.7 42,380.3 102,940,593.4

CaCO 3s  → CaO s  + CO 2g  

( H) =177 =177 kJ kJ/m /mol olee

Heat absorbed for reaction=101,943,189.3 r eaction=101,943,189.33 3 kJ/Day

Heat of combustion

Calorific value of coke=32000 kJ/kg Assume kiln coke requirement is 426.5 kmole/day Coke supply rate per day =426.5 kmole/day Heat released due to combustion of coke=163776000 kJ/Day

 

Chapter 10 

ENERGY BALANCE 

Heat loss to Environment

15% generated heat is Lost to the environment Heat loss = heat of combustion

percentage

=24566400 kJ/day

Substituting this value in to the energy balance equation satisfied both side. Coke requirement is satisfies

10.2 Energy Balance for Air Preheated CaO

575.95 kmole/day

 

Hot CaO 1000 C

Hot Air 900 C

Cool air 25 C

Air flow rate

 

O2  N 2

 

Cool CaO TC

2030.95 kmole /day 426.50 kmole/day 1604.45 kmole/day

Figure 10.2- Air preheated

Assumption •  Pressure is constant inside unit. Enthalpy change can calculated as follows ΔH c

0

C p = a + b T – c /T2

 =

a, b, c constants are gave in upper part of this chapter • 

This calculation is done by using trial and error method method •  Heat loss to the surrounding is 11% of the

Air enthalpy change Air mixture temperature is change from 25 C to 900 C

Components  N 2 O2  

 

Enthalpy Change (kJ/kmole) 26974.06 31969.88 Total Enthalpy

Flow rate kmole/day 1604.45 426.50

Enthalpy Change (kJ/day) 43278598.8 13635151.83 56,913,751

Table 10.4- Air enthalpy change

 

Chapter 10 

ENERGY BALANCE 

Enthalpy Change of CaO

Hot CaO is cooled from 1000 C to 154 C Component CaO

Enthalpy Change (kJ/kmole) 111218.35

mole Flow rate(kmole/day) 575.95 Total Enthalpy

Enthalpy Change (kJ/day) 64056233.52 64,056,234

Table 10.5- CaO enthalpy change

Heat loss

Heat loss to the environment is 11% of the heat gain by cooling of the CaO Heat loss to the surrounding

= 7,117,359.28 kJ/day

 Enthalpy Change of CaO = Enthalp Enthalpyy Change of C CaO aO + Heat loss

This equation is satisfies. So air is heated up to 900 C before feed to the kiln.

10.3 Calcinations of Crude Bicarbonate Objective • 

To determine the flue gas out T from the calcinations.

Assumption

Pressure is constant inside unit. Enthalpy change can calculated as follows 0

2

ΔH c

 =

C p = a + b T – c /T

a, b, c constants are gave in upper part of this chapter • 

10 % Heat is lost from indirect dryer which is supply from flue gas

• 

Specific heat capacity of pure substance is get from parry hand book

• 

 NaHCO3

96kJ/kmole K

H2O

75.6kJ/kmole K

 Na2CO3

121.38kJ/kmole K

Kiln internal pressure 1.5bar  

 

Chapter 10

ENERGY BALANCE 

CO 2 H2O Temperature

471.232 704.264384 200

 

 N 2 CO 2

1604.452381 1002.450222

kmole/day kmole/day

622

C

Temperature

 

kmole/day kmole/day C

  CO2, H2O H2O Vapou r 

Flue gas out

NaCO3 NaC O3 Produc t NaHCO3 Solution

 NaHCO3

942.464

kmole/day

H2 O

235.3804

kmole/day

 Na 2 SO 4 Flow rate

0.02 1177.844

kmole/day kmole/day

Temperature

30

   

 Na 2 CO 3 H2O  Na 2 SO 4 Total Temperature  

Calcinations NaH NaHCO3 CO3

 

Flue gas in

C

471.232 2.348 0.02 473.6 200

kmole/day kmole/day kmole/day kmole/day C

1000 C

 N 2 CO 2 Temparature  

 

1604.45 1002.45 1000

kmole/day kmole/day C

Figure 10.3- Dryer

Inlet Enthalpy of the NaHCO 3  solution

Heat Capacity of mixture of NaHCO 3

= C p   NaHCO3  mole fraction + C p   H2O   mole

fraction

Enthalpy of inlet mixture

= 91.9 kJ/kmole

= C p mixture Flow rate

= 32,806,204.21 32,806,204.21kJ/Day kJ/Day Outlet Enthalpy of the Na 2 CO 3

Enthalpy of Out let Material of Na 2 CO 3 = C p Na2CO3 Flow rate =27,190,673.66kJ/Day

Outlet CO 2 , H 2 O mixture Enthalpy

During thermal decomposition of the NaHCO 3 , H 2 O & CO 2  mixture is relished .Assume the temperature is 200 C

H 2 O g enthalpy from steam table

H 2 O g  Tota Totall en ener erg gy

= 2876kJ/kg

= 2876 876 18 704 704.2 =36,455,025.6 kJ/kmole

 

Chapter 10 

ENERGY BALANCE 

CO 2  Enthalpy

=45.2kJ/kmole

Total energy CO 2  

=45.2 =4

kJ/day

=21,298.24 kJ/day Total Enthalpy of the Exit mixture

=36,479,658.89kJ/Day =36,479,658.89kJ/Day

Flue Gas Enthalpy Change

Mixture of flue gas is cooled 1000 C to T C If Temperature of flue gas out is 622 C CO 2  enthalpy change = = 19,457,719.97 kJ/day

 N 2  Enthalpy

= =24,535,963.06kJ/day

Total enthalpy change = 43,993,683.03 43,993,683.03 kJ/Day

Decomposition NaHCO3 s  Enthalpy Change NaHCO3 s  

Total energy absorb

Na 2 CO 3   s + H 2 O g  + CO2g

∆  .  / /   

= 9.24 = 8,708,367.36k 8,708,367.36kJ/Day J/Day

kJ/day

Heat loss

Heat loss from indirect dryer dryer 10 % supply from flue gas =4399368.303kJ/Day =4399368.303kJ/Day

Energy equation

Enthalpy change of flue gas = Hot product out (NaCO 3(s) ) + CO 2 , H 2 O mixture out + Reaction energy + heat loss to environment -Enthalpy of NaHCO 3  in

Substituting the all calculated value to this equation is satisfies. This calculation is down by using trial and error method. So flue gas exit temperature is 622 C

 

Chapter 10 

ENERGY BALANCE 

10.4 CaCO 3  Preheated

Objective •  To determine the CaCO3 feed T Assumption • 

ΔH c

0

Pressure is constant inside unit. Enthalpy change can calculated as follows

 =

C p = a + b T – c /T2

a, b, c constants are gave in upper part of this chapter • 

• 

10% heat loss during preheating

 No Cp Change in limestone with increasing T @ Cp

 N 2 CO 2

 

 N 2

136 J/mole

 

 

CO 2

1604.452 kmole/day 1002.45 kmole/day

 

1604.452 kmole/day 1002.45 kmole/day

CaCO 3 Temperature  

575.95 kmole/day 566 C

Figure 10.4- Cyclone 

Energy Equation

 Heat IN = Heat Out + Loss Enthalpy Change of Flue Gas = Enthalpy Change CaCo3 + Losses Flue Gas Enthalpy Change mixture of flue gas is cooled 622C to 210 C

CO 2  en enth thalp alpy y  N 2  Enth Enthalp alpy y Total enthalpy change

20 20,9 ,913 13,1 ,120 20.6 .69 9 kJ kJ/D /Day ay 26 26,1 ,177 77,9 ,959 59.3 .30 0 kJ kJ/D /Day ay 47,091,079.98 kJ/Day

Table 10.6- flue gas enthalpy change

 

Chapter 10 

ENERGY BALANCE 

Enthalpy Change Of CaCO3

42,376,113.55 42,376,113.55 kJ/Day

Heat loss 10% enthalpy change of the flue gas

Loss of heat

4,709,108.00 kJ/Day

Energy balance equation is satisfied using these values. Using cyclone CaCO 3  is heated up to 566C. This Calculation is done by using trial and error method.

10.5 Air Mixer Energy Balance Objective • 

To fine the gas mixture temperature Assumption  • 

ΔH c

0

Pressure is constant inside unit. Enthalpy change can calculated as follows

 =

C p = a + b T – c /T2

a, b, c constants are gave in upper part of this chapter • 

 Nearly 4% of heat is lost to the environment

• 

Kiln internal pressure 1.5bar

• 

Air Mixture pressure is 1bar

 N 2 CO 2 H2 O Temperature Stream 3  

 

1604.452381 1473.682222 704.264384 T

kmole/day kmole/day kmole/day C

 

CO 2 H2O Temperature Stream 2

471.232 704.264384 200

 

kmole/day kmole/day C

Air Mixer 

 

 N 2

1604.452381

kmole/day

CO 2 Temperature Stream 1

1002.450222 210

kmole/day C

Figure 10.5- Air mixture before gas cooler  

 

Chapter 10 

ENERGY BALANCE 

Stream 2 carrying energy is calculated in the calcinations energy balance section

Stream 2 energy

= 36,479,658.89 kJ/day

Stream 1 energy

= CO 2  energy + N 2  energy

CO 2  energy change =  N 2  energy change = Temperature of the mixture is 200 C

Steam 1 energy = 92,634.44 kJ/kmole

Heat loss to the surrounding surrounding = (92,634.4 (92,634.44+36, 4+36,479, 479,658. 658.89) 89) .4 =1,462,891.73 kJ/day

Steam 3 energy calculation If out let temperature is 140 C, Water vapor enthalpy is = 2756.8 kJ/kg Energy of water vapor

=

kJ/day

34,947,288.97 kJ/day

Gas carrying energy can calculated same as steam 1.if out let temperature is 140 C Gas carrying energy =110,493.69 kJ/day

Total energy of steam stea m 3 = 36,520,674.39 kJ/day

Energy equation

Stream 1 energy + Stream 2 energy = Stream 3 energy + loss

Substituting values to the energy equation is satisfies so out let temperature is 140 C

 

Chapter 10 

ENERGY BALANCE 

10.6 Heat Balance for Gas Cooler Object •

  To Find the Cooling water demand required for gas cooler. Assumption • 

Cooling water system used is ones through system having 30 0C temperature of inlet water.  

• 

Efficiency of the gas cooler is 95%.

• 

Out let gas stream is saturated with water vapor and quantity of energy carries with the steam is neglected.

 

 N 2

1604.452381 kmole/day

 

CO 2 H2O g

1473.682222 kmole/day 304.9074568 kmole/day 40 0C

Temperature

Gas out

H2O Temperature

 N 2

 

1604.45238 kmole/day

CO 2

1473.68222 kmole/day]

H2O g

704.26438

Temperature

 

Water In

kmole/day] 140 0C

Gas Ga s mixer mi xer

Cooler

300 m3/day 16669 kmole/day 30 0C

H 2 O condense

 

399 kmole/day

Hot Water Out

H2O

17068 kmole/day

Temperature

50

0

C

Figure 10.6- Gas cooler  

Heat given by the hot stream = Heat taken by the cold stream

Enthalpy change 

0

ΔH R 

=

Δ 

C p = a + b T – c /T2 0

ΔH R 

= a   ΔT  + b   ΔT2 – c /ΔT 

 

Chapter 10 

ENERGY BALANCE 

Component CO 2 O2  

 

 N 2

a b 10.34 0.00274 8.27 0.000258

6.5

 

c 195500 187700

0.001

Table 10.7- Soda ash specification

Total energy of the inlet gas mixture

0

= 35,057,783 kJ/day

Total energy of the out let gas mixture = 102,867 102,867 kJ/day Enthalpy change due to condensate of the of the water vapor = 9,751,817 kJ/day Total enthalpy change in gas stream = Total energy of the inlet gas mixture - Total energy of the out let gas mixture - Enthalpy change due to condensate of the of the water vapor = (35,057,783-10 (35,057,783-102,867-9,751,81 2,867-9,751,817) 7) kJ/day = 25203099 kJ/day Cooling water flow rate required = m kg Total Tot al enth enthalp alpy y chan change ge in gas stream stream = m C water

ΔT

= 25203099

Cooling water flow rate required required = m = 16669 kg/day = 300m3/day

Condense water and heated cooling water is mixed and then supply hole quantity in to slaker to make Ca(OH) 2   solution.

10.7 Slaking of Lime

Objective • 

Find the quantity of evaporated water vapor from the slaker.

Assumption • 

15% brix, Ca(OH) 2  mixture is required for the process.

• 

Heat loss to the surrounding is 10 %from the energy due to the reaction enthalpy.

 

Chapter 10 

ENERGY BALANCE 

water coming from cooler CaO solid Temperature

575.95 kmole/day

Temperature

154 0C

17068 kmole/day 50

0

C

 

Hot water from cooler

  Cold CaO

Excess water

extra water

 

temperature

Ca(OH)2 Solution

Slaker

Ca(OH) 2

576

H2O

13,418 kmole/day

weight of solution Temperature

kmole/day

  Vaporizes water

284,135 kg/day 0

60 C

evaporated water vapors

1657.4kmole/day

Figure 10.7- Slaker  

Energy with the inlet CaO stream

= m × C p × T

Component CaO

a 10

b 0.05

c 108000

Table 10.8- a, b, c constant for CaO

1417.2 kmole/day 50 0C

C p = a + b T – c /T2

Energy with the inlet CaO stream = (10+0.05 × (154+273) ×108,000/(154+273) 2) ×4.2 × 575.95 =

74,402.4 kJ/day

Amount of the input energy of the water from the cooler

= m × C water ×ΔT = 15,650.9×18×4.2× (273+50) = 382,175,186 kJ/day

Total Heat in = Energy with the inlet CaO stream + Amount of the input energy of the water from the cooler

Total Heat in = 74,402.4 + 382,175,186 = 382,249,589 kJ/day

 

Chapter 10 

ENERGY BALANCE 

Reaction Enthalpy:

 

CaO + H 2 O

Ca(OH) 2  

ΔH=

- 65 kJ/m kJ/mole ole

Heat generated due to reaction = 65 × 1000 × 576 kJ/day

Loss to surrounding

= 3,743,675 kJ/day

Cp of the Ca(OH) 2  mixture = 3.6183 kJ/kgK

Heat out with Ca(OH) 2 

= 3.6183 × 284135 × (273+60) 

= 342,353,062 kJ/day Energy with water vapor = Heat in+ Heat generated due to reaction - Heat out with Ca(OH) Ca(OH) 2 - loss.

Energy with water vapor  

+37,436,750 - 342,353,062 - 3,743,675 = 382,249,589 +37,436,750 = 73,589,615 kJ/day

Water vapor quantity

= 73,589,615/(4.2 × (100-50)+2256.7) = 29,833 kg/day =1,657.4 kmole/day

 

Chapter 10 

ENERGY BALANCE 

10.8 Recovery of Ammonia Column Energy Balance

  Boundary line

CO 2   NH 3 Temperature 

32. 3 2.28 2899 992 2 1079 1079.4 .472 72 T

km kmol ole/ e/da day y kmol kmole/ e/da day y C

Cool gas (NH3)

H2 O

1061.655

kmole/day Condensed water 

HE Cool NH4Cl Solution

H2 O CO 2    NH 3 Temperature  

1061.655 32.28992 1079.472 70

kmole/day kmole/day kmole/day C

Hot NH3 ,H2O mixture

H2 O  NaCl  NH 4 Cl  NaHCO3  NH 4 HCO 3  Na 2 SO 4 Temperature Flow rate molecular weight  

 

13,123.9 430.1 1,047.2 104.7 32.4 14.5 30 14,752.8 22.42532

kmole/day kmole/day kmole/day kmole/day kmole/day kmole/day C kmole/day kg/kmole

 Am mo ni a recovery uni t recovery

shaking of lime

Ca(OH) 2 H2 O Temperature  

575.95 13417.51 60

kmole/day kmole/day C

Steam

Bottom T low pressure steam

105 2

C bar

weight of solution Cp

284135.4 3.6183

Kg/day kJ/kgK Waste water 

 NaCl CaCl 2  NaHCO3  NH 4 OH Ca(OH) 2  Na 2 SO 4 H2O T Total Flow rate molecular weight  

 

 

430.12 523.59 104.72 0.10 52.36 14.55 25,479.72 105 26,605.16 20.84993

kmole/day kmole/day kmole/day kmole/day kmole/day kmole/day kmole/day C kmole/day kg/kmole

Figure 10.8- NH3 Recovery column 

Objective •  • 

To fine Quantity of steam consumption Fine out let temperature of the cool gas

 

Chapter 10 

ENERGY BALANCE 

10.8.1 Find Outlet Temperature of the Cool Gas Assumption • 

20 % Heat is lost to surrounding from HE •   NH 4 Cl Solution is heated from 30C to 50C • 

Pressure is constant inside unit. Enthalpy change can calculated as follows ΔH c

0

C p = a + b T – c /T2

 =

a, b, c constants are gave in upper part of this chapter • 

Column Bottom And top Temperatures are got from literature Top Temperature 70 C Bottom Temperature 105C

Energy equation Enthalpy change of cool steam + loss = Enthalpy change of hot stream Enthalpy change of hot stream

Hot gas mixture is cooled up to T. Assume T is 40 C  NH 3  & CO 2  mixture is cool from 70 C to 40 C this energy relies is = mole flow rate  NH 3  & CO 2 mixture Energy = 1,837,568 kJ/day Steam Steam Cond Conden ense se en ener ergy gy

= 1,06 1,061. 1.55 55 22 2257 57 18 kJ kJ/d /day ay = 43,130,800 kJ/day

Total Energy

= 44,968,368 kJ/day

Enthalpy change of cool steam 

Assume heat capacity of the NH 4 Cl mixture is same as water heat capacity because larger portion of mixture is include H 2 O. Energy grip by cool stream =

= =34,737,885 kJ/day

 

Chapter 10 

ENERGY BALANCE 

Heat loss to the surrounding

= total energy relies hot stream

0.2

= 8,993,674 kJ/day

Substituting this value to the energy equation it satisfies. So assume temperature is Correct

10.8.2 Fine Quantity of Steam Consumption Assumption

• 

Fully insulated the column no heat losses

•   NH4Cl

solution feed stream line & waste stream from the column have more water compare

with other substances. So Assume specific heat of solution is equal to the water specific heat • 

2 bar saturated steam is used for column heating

Apply energy balance for the system bounded by boundary line

Quantity of Energy supply to the column = Energy out from boundary- Energy in through boundary

Energy out from boundary 

Waste steam energy Waste steam is have column bottom T at 105 C, Cp same as water

Waste steam energy

= = 880,666,741.8 kJ/day

Heat loss from HE = 8,993,674 kJ/day

Heat carried from cool water =1061.655 =1061.655 175.2 175.2 18 =3,348,035.5 kJ/day Heat carried from cool NH 3 , CO 2  mixture = flow rate Cp

temperature

=12,696,681.8 kJ/day

 

Chapter 10 

ENERGY BALANCE 

Energy in through boundary 

Ca(OH) 2 mixture inlet energy

= =284135.4*3.6183*(60+273) =342,353,062 kJ/day

 NH 4 Cl solution inlet energy = = 22 22.42 14752.8 4.2 (273+30) = 421,023,166 kJ/day

Quantity of Energy supply to the column using steam

+8,993,674 +3,348,035.5 +12,696,681.8 +12,696,681.8 - 342,353,062 - 421,023,166 kJ/day = 880,666,741.8 +8,993,674 = 135,632,833.6 kJ/day

2 bar saturated steam enthalpy change = 2211.6 kJ/kg Steam consumption

=135,632,833.6/2211.6 =135,632,833.6/2211.6 =61,328 kg/day = 2.555330108 t/hour