Energy Balance

ENERGY BALANCE

Reference temperature = 273 K

PRE-HEATER Inlet temperature of solution

=

25°C

Outlet temperature of solution

=

60°C

Qf

=

mCp dT

Qf

=

[(111.1*1.429)+(4.187*2.268)]*(298-273)

=

168.258

Qf

=

4206.4504 kJ/hr

Qp

=

168.258

=

10095.48 kJ/hr

Heat carried in by the feed stream

*

25

Heat carried out by the product stream

*

60

Heat to be supplied,

Qp - Qf =

Steam requirement

S

5889.03

=

( Qp - Qf )/ λs

(λs=2260 kJ/kg)

=

2.60576 kg/hr

Assuming 80% heat transfer efficiency in the equipments Actual steam requirement, S

Streams Reactant stream Product stream Steam Total

=

3.25 kg/hr

Energy inlet (kJ/hr) 4206.4504 5889.01 10095.46

Energy outlet (kJ/hr) 10095.48 10095.48

SULFONATOR

Qf

Qp

=

(235.8*1.17*30) + (10095.48) + [(228.2*1.3734)+(.437*4)] * 100

=

58128.25 kJ/hr

=

[(326.5*2.045) + (288.2*1.43136) + (4*.437) + (4.253*22.674)] *120

=

141524.46 kJ/hr

Heat of reaction: No. of moles of anthraquinone taking part in the reaction = 1.13365 kmol/hr



C14H8O2

+

H2SO4

A

+

H

(in the presence of M)

∆HR

=

∆Hf°

+

∆Hf°

=

(∆Hf° ASA + ∆Hf° W)

C14H8SO5 +

H2O

ASA

W

+

ʃ ∆CpdT -

(∆Hf° A + ∆Hf° H)

∆Cp

ʃ ∆CpdT ∆HR

Total heat

=

(-675320-285830) - (-58240-814000)

=

-88910 kJ/kmol

=

∆Cp Products

=

(588.96+76.554) – (297.723+140.04)

=

228.474

=

228.474 * 95

=

-88910 + 21705.03 =

=

-76186.914 kJ/hr

=

Qp – Qf + ∆HR

=

7209.296 kJ/hr

-

=

∆Cp Reactants

21705.03 kJ/kmol -67204.97 * 1.3365

Hence, heat should be supplied to get products at required temperature Steam required,S

=

7209.296/2260

=

3.19 kg/hr

Assuming around 80% heat transfer efficiency in the equipments Actual steam requirement, S

Streams Reactant stream Product stream Heat of reaction Steam Total

=

4 kg/hr

Energy inlet (kJ/hr) 58128.25

Energy outlet (kJ/hr) 141524.46

76186.914 7209.4 141524.5

141524.46

CRYSTALLIZER Qf

=

141524.46 + (15*4.187*25)

=

143094.6 kJ/hr

Qp

=

(326.5*2.045 + 288.2*1.311 + 4*.437 + 37.67*4.2) * 80

=

96448 kJ/hr

Heat to be removed =

Qp - Qf

=

-46646.6 kJ/hr

Brine water is to be supplied for this cooling purpose Brine is available @ 288 K Brine required

=

46646.6/(3.996*40)

=

179.6 kg/hr

Streams Reactant stream Product stream Brine Total

Energy inlet (kJ/hr) 143094.6 -

Energy outlet (kJ/hr) 96448 46646.6 143094.6

143094.6

Assuming 80% heat transfer efficiency, actual brine required = 225 kg/hr

PRECIPITATOR Qf

= [(326.5*2.045 + 10*4.2) * 80] + [(34.63*.677 + 253.97*4.187) * 25] = 84992.37 kJ/hr

Qp

= [ 151.53*3.5075 + (1.05*4.2)*280.97 + 192.63*2.045 ] * 85 = 183982.29 kJ/hr

Heat of reaction: C14H8SO5 + ASA

KCl

+

P



C14H7SO5K + PAS

+

HCl C

No. of moles of potassium anthraquinone-1-sulfonate formed = 0.4644 kmol

ʃ ∆CpdT

∆HR

=

∆Hf°

∆Hf°

=

(∆Hf° PAS + ∆Hf° C) -

=

(-1153981.62 – 167370) – ( -675320 – 420680)

=

-225351.62 kJ/kmol

+

(∆Hf° ASA + ∆Hf° P )

∆Cp

ʃ ∆CpdT ∆HR

Total heat

=

∆Cp Products

=

(1144.357 + 25.8785) – (588.96 + 50.485)

=

530.82 * 60

=

31849.23 kJ/kmol

=

-225351.62 + 31849.23

=

-193502.39 * 0.4644

=

-89862.76 kJ/hr

=

Qp – Qf + ∆HR

=

9127.157 kJ/hr

Steam required=

9127/2260

-

=

∆Cp Reactants

4.038 kg/hr

Assuming 80% efficiency, actual steam requirement = 5 kg/hr

Streams Reactant stream Product stream Heat of reaction Steam Total

Energy inlet (kJ/hr) 84992.3725 89862.76 9127.01 183982.14


COOLER Inlet temperature

=

85°C

Outlet temperature =

25°C

Qf

=

(133.5*3.5075 + 2.5*2.045 + .28*.709) * 85

=

40252.79 kJ/hr

=

11839.05 kJ/hr

Heat released

=

-28413.74 kJ/hr

Brine required

=

28413.74/(3.996*10)

Qp

=

711.05 kg/hr

Assuming 80% efficiency, actual brine requirement = 888 kg/hr

Streams Reactant stream Product stream Brine Total

Energy inlet (kJ/hr) 40252.79 -

Energy outlet (kJ/hr) 11839.05 28413.74 40252.79

40252.79

NEUTRALIZER Qf

Qp

=

(11839.05 + 0.86*1.145*25)

=

11863.66 kJ/hr

=

(133.5*3.5075 + 0.45*.8645 + 4.326*2.7 + 0.14*4.187+.919*.35)*25

=

12030.7 kJ/hr

Heat of reaction: Reaction 1: 2 C14H8SO5

+



Na2CO3

ASA

S

2 C14H7SO5Na + NAS

CO2

+

O

ʃ ∆CpdT

∆HR

=

∆Hf°

∆Hf°

=

(∆Hf° NAS + ∆Hf° O + ∆Hf° W) - (∆Hf° ASA + ∆Hf° S)

=

[-2*1091200 – 393500 – 285830] – [-2*675320 – 1130770]

=

-380320 kJ/kmol

+

No. of moles of sodium carbonate reacted = 0.004355 kmol

ʃ ∆CpdT

=

∆HR1 = =

0 ∆Hf° -1656.3 kJ/hr

H2O W

Reaction 2: 2 HCl

+

C

∆Hf°



Na2CO3 S

2 NaCl

+

CO2

N

O

+

H2O W

=

(∆Hf° N + ∆Hf° O + ∆Hf° W) - (∆Hf° C + ∆Hf° S)

=

[-2*408760 – 393500 – 285830] – [-2*167370 – 1130770]

=

-31340 kJ/kmol

No. of moles of sodium carbonate reacted = 0.003776 kmol ∆HR2 =

-118.33 kJ/hr

∆HR

=

∆HR1 +

=

1774.63 kJ/hr

∆HR2

Heat to be removed + ∆HR

=

Qp - Qf

=

12030.7 – 11863.66 – 1774.63

=

-1607.6 kJ/hr

Brine required= =

1607.6/(3.996*10) 40.23 kg/hr

Assuming 80% efficiency, actual brine requirement = 50 kg/hr

Streams Reactant stream Product stream Heat of reaction Brine Total


Energy outlet (kJ/hr) 12030.7 1607.6 13638.3

RECYCLE HEAT BALANCES PAN DRYER Amount of solid to be dried

=

292.2 kg dry solid/hr

Inlet air temperature, Tg2

=

120°C

Outlet air temperature, Tg1

=

85°C

Inlet solids temperature, Ts1

=

85°C

Outlet solids temperature, Ts2

=

100°C

Assumptions:

@ 25°C, humidity of air is 50% (which is available from flue gas) Hence, absolute humidity of available air = 0.01 kgwater/kg dry air

=

Y2

The enthalpy of air entering the drier,HG2=

[(Cg + Y2Cv)(Tg2 – To) + λ Y2]

Specific heat of air,

Cg

=

1.005 kJ/kg°C

Specific heat of vapor,

Cv

=

1.884 kJ/kg°C

Enthalpy of vapor,

λ

=

2502.3 kJ/kg

HG2

=

147.882 kJ/kg

HG1

=

(1.005+1.884Y1)85 + 2502.3

=

85.425 + 2662.44 Y1

The enthalpy of the solid entering the drier, Hs1

=

Cp(Ts1 – To) + X1(Ts1 – To) Cw

=

(1.311+ 0.077*4.187) * 80 kJ/kg

=

130.7014 kJ/kg

The enthalpy of the solid leaving the drier, Hs2

=

1.369*100

=

136.9 kJ/kg

Assuming no heat loss due to radiation to the surroundings

Moisture balance: Ss X1

+

Gs Y2

=

Ss X2

Ss(X1 – X2)

=

Gs(Y1 + Y2)

292.2(0.077)

=

Gs(Y1 + 0.01)

Gs(Y1 – 0.01)

=

22.5

Gs HG2

=

Ss Hs2 +

+

Gs Y1

Enthalpy balance: Ss Hs1 + 292.2*130.7 +

Gs(147.882)

Gs(62.458 – 2662Y1)

Gs HG1 +Q

= 292.2(136.9) + Gs(85.424 + 2662Y1) + 0 =

1837.78

By solving, Gs

=

1724.13 kg dry air /hr

Y1

=

0.023

Heat source Ss Hs1 Gs HG2 Ss Hs2 Gs HG1 Total

Heat in kJ/hr 38190.54 254964.34

293154.88

Heat out kJ/hr

40002.18 253151.17 293153.35

COOLER Qf

Qp

Qp – Qf

=

(172.63*2.045 + 13*0.709 + 229.97*4.2) * 85

=

112890.1448 kJ/hr

=

(172.63*2.045 + 13*0.709 + 229.97*4.2) * 25

=

33203 kJ/hr

=

-79687.14 kJ/hr

Brine required= =

68533.856/(3.996*10) 1994.17 kg/hr

Streams Feed stream Product stream Brine Total

Energy inlet (kJ/hr) 112890.1448 112890.1448

Energy outlet (kJ/hr) 33203 79687.14 112890.14

NEUTRALIZER 2 Qp

Qp

=

(172.63*2.045 + 13*0.709 + 229.97*4.2 + (50.65*1.145)) * 25

=

34652.85 kJ/hr

=

(185.824*4.426 + 238.57*4.187 + 20.836*0.709 + 21.024*0.919)*25

=

46386.084 kJ/hr

Heat of reactions: Same reactions as the previous neutralizer occur ∆HR1

=

-380320 kJ/kmol

No. of moles of sodium carbonate reacted = 0.6 kmol ∆HR2

=

-31340 kJ/kmol

No. of moles of sodium carbonate reacted = 0.356 kmol Total heat of reactions=

Total heat

- 228192 - 11162.4

=

- 239354.4 kJ/hr

=

46386.084 – 34652.85 – 239354.4

Heat to be removed =

-227621.16 kJ/hr

Brine required

1.582 kg/s

Streams Reactant stream Product stream Heat of reaction

=

Energy inlet (kJ/hr) 34652.85 239354.4

Energy outlet (kJ/hr) 46386.084 -

Brine Total

274007.25

227621.16 274007.244

EVAPORATOR Qf

=

46386.084 kJ/hr

Evaporator operated @ 120 °C to completely vaporize water Qp

=

(185.824*4.426 + 20.836*0.709 + 21.024*0.919)*120 + 2260*238.57

=

641954.29 kJ/hr

Qp – Qf=

-595568.2125 kJ/hr

S

263.5 kg/hr

=

Assuming 80% efficiency in heat transfer equipments steam requirement is 330 kg/hr

Streams Feed stream Product stream Steam Total



Energy Balance

Recommend Documents