June 2004

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PE Pyroprocessing Day 4Volatilisation Balance Calculation

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Exercise : Volatile Balance Calculation

Field work Mass determination Sampling

Calculations Mass balance and corrections Volatile balance and corrections Calculation of v and t coefficients

Interpretation

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Exercise : Field work

Stack gas& dust

Precipitator

Wastedust

Kilngas

Kilnfeed

Kiln load Clinker

Fuel

Sample

Flow measurement

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Exercise : Definitions Mass as received M

Mass as measured in the field (T/h)

Mass on dry basis M’ Mass without the moisture content (T/h)

Mass at LOI=0 M” Mass without the CO2 from the kiln feed (T/h)

Specific mass Mass of a component by unit mass of clinker (kg/kg ck or g/kg

ck)

1002%

1'OH

MM

100

%1'" 2CO

MM

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Exercise : Input data - Mass flow

Flow Moisture CO2

(T/h) (%) (% dry)

Kiln feed 48,0 0,5% 33,0%

Coal 4,5 4,0% -,-

Clinker 30,0 0,0% 0,0%

Waste dust 4,8 0,0% 28,0%

Stack dust 0,3 0,0% 30,0%

Kiln load 0,0% 30,0%

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Exercise : Input data - Elements SO3 K2O Na2O Cl

(% dry) (% dry) (% dry) (% dry)

Kiln feed 0,55% 1,00% 0,05% 0,008%

Coal ash -,- 2,00% 0,10% 0,050%

Clinker 0,50% 1,25% 0,25% 0,002%

Waste dust 4,75% 3,00% 0,25% 2,190%

Stack dust 21,25% 22,00% 3,50% 11,400%

Kiln load 2,20% 1,50% 0,10% 0,460%

Coal: 13% ash (as received)

4,1% S (as received)

Stack: 260,00 kg/h SO2

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Exercise: Calculation Mass Balance (1)

On dry basis (T/h):

input output

Kiln feed 47,76 Clinker 30,000

Coal 4,32 Waste dust 4,800

Stack dust 0,300

Stack gas 0,325

Total 52,08 Total 35,425

Input and output do not match. The difference is CO2

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Exercise: Calculation Mass Balance (2)On LOI-free basis (T/h):

The input and output are much closer but still differ by 0,987 T/h. A mass correction is therefore needed.

Note: Sulphur is considered on SO3 basis

Conversion: %SO3 = % S 80 / 32

%SO3 = % SO2 80 / 64

input output

Kiln feed 31,999 Clinker 30,000

Coal ash 0,562 Waste dust 3,456

Coal SO3 0,443 Stack dust 0,210

Stack gas 0,325

Total 33,004 Total 33,991

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Exercise : Calculation Mass Balance (3)Mass correction

Mc = M” k (Total output - Total input)

Weighting adjustment coefficients k

Kiln feed 1,00

Coal ash 0,00

Clinker 0,00

Waste dust 0,00

Stack dust 0,00

Stack gas 0,00

k 1,00

input output

Kiln feed 32,987 Clinker 30,000

Coal ash 0,562 Waste dust 3,456

Coal SO3 0,443 Stack dust 0,210

Stack gas 0,325

Total 33,991 Total 33,991

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Corrected mass flow (dry basis)

t/h kg/kg ck

Kiln feed 49,23 1,64

Coal 4,32 0,14

Clinker 30,00 1,00

Waste dust 4,80 0,16

Stack dust 0,30 0,01

Stack gas 0,33 0,01

Exercise : Calculation Mass Balance (4)

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Exercise : Volatile balance (1)

1 %K2O is percentage of ash

K2O balance

Corrected mass flows % K2O

t/h

Kiln feed 49,23 1,00%

Coal ash1 0,56 2,00%

Clinker 30,00 1,25%

Waste dust 4,80 3,00%

Stack dust 0,30 22,00%

Stack gas 0,33 0,00%

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Exercise: Volatile Balance (2)

It does not balance and differs by 81,43 kg/h! Let’s correct it!

K2O balance (kg/h)

input output

Kiln feed 492,34 Clinker 375,00

Coal ash 11,23 Waste dust 144,00

Stack dust 66,00

Stack gas 0,00

Total 503,57 Total 585,00

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Exercise : Volatile Balance (3)

K2O balance correctionMc = M” k (Total output - Total input)

Weighting adjustment coefficients k

Kiln feed 0,45

Coal ash 0,00

Clinker 0,45

Waste dust 0,10

Stack dust 0,00

Stack gas 0,00

k 1,00

input output

Kiln feed 528,98 Clinker 338,36

Coal ash 11,23 Waste dust 135,86

Stack dust 66,00

Stack gas 0,00

Total 540,21 Total 540,21

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Exercise: Volatile balance (4)

Corrected K2O composition

Corrected mass flows Corrected

t/h % K2O % K2O

Kiln feed 49,23 1,00% 1,07%

Coal ash 0,56 2,00%

Clinker 30,00 1,25% 1,13%

Waste dust 4,80 3,00% 2,83%

Stack dust 0,30 22,00%

Stack gas 0,33 0,00%

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Exercise : Kiln loadAn assumption has to be made on the kiln load mass flow rate

Typical values: 1.2 - 2.0 kg/kg ck

ASSUME: 1.6 kg/kg ck

Then

Kiln load = 1.6 30 = 48 T/h

Kiln load K2O = 720.0 kg/h

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Exercise : Volatile recirculation model

v = 1 - CK

t = K - M vK + F

Material Clinker

Losses FuelGas

Kiln load

trapping volatilization

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Exercise : CoefficientsK2O coefficients:

v = 1 - CK

= 1 - 338,36720,00

= 0,53

t = = 0,486720,00 – 528,98

0,53x720,00 + 11,6

201,86 392,88 11,23

191,02 381,64

t = 0,486 v = 0,53

528,98 720,00 338,36

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Coating regions

Heavy coating

Medium

coating

Low

coating