tics of Cement

8/4/2019 tics of Cement

1/163

Not to be copied in any form or distribute

Property of:Softideas Pvt. Ltd., 5- Sinbai Niwas, GoModel Colony, Pune-411016


2/163

Chapter 5 Raw meal

Chapter 6 Pre heating

Chapter 7 Clinkerisation

Chapter 8 Clinker cooling

Chapter 9 Cement grinding

Chapter 10 Cement storage

Chapter 11 Cement despatch

Chapter 12 Cement quality check

Chapter 13 Raw materials

Chapter 14 Cement properties

Book II - Process Calculations of Support Systems

MathCement2000

Introduction

File Name :TOC_friedrich_v8

Topic: Table of Contents

To go to cover page please click here

About using Mathcad Electronic Books

Read me

Not to be copied in any form or distributed or resold

Property of:

Softideas Pvt. Ltd., 5- Sinbai Niwas, Gokhale Nagar Road,Model Colony, Pune-411016

About MathCement2000

Preface to the first edition

Book I - Process Calculations of Plant Sections

Chapter1

Quarry

Chapter 2 Crushing

Chapter 3 Pre blending

Chapter 4 Raw grinding


3/163

Useful Tables - Earth Sciences

Chapter 3 Useful Tables - Areas and Perimeters

Chapter 4 Useful Tables - Volumes and Surfaces

Book IV - Popups

Chapter 1 Cement Chemistry

Chapter 2 Cement Glossary

Chapter 3 Cement Kiln Types

Chapter 4 Cement Properties

Chapter 5 Cement Quality Tests

Chapter 6 Pressure

Chapter 7 Psychrometry

Chapter 8 General Topics

Chapter 1 Site Conditions

Chapter 2 Dedusting Systems

Chapter 3 Laboratory Investigations

Chapter 4 Fuels and Fuel Systems

Chapter 5 Conveying Systems

Chapter 6 Fans and Blowers

Chapter 7 Raw-mix Design

Chapter 8 Plant Water

Chapter 9 Insulation

Book III - Useful Tables "Copyright 1999 by MathSoft, Inc. Reprinted bypermission of MathSoft, Inc."

Chapter 1 Useful Tables - Basic Sciences

Chapter 2


4/163

Book V - Charts and Tables

Chapter 1 Steam Tables

Chapter 2 Humidity Chart

Chapter 3 Separation Degree

Chapter 4 Coal Reactivity

Chapter 5 Conversion Tables

Chapter 6 Fineness


Property of:Softideas Pvt. Ltd., 5- Sinbai Niwas, Gokhale Nagar Road,Model Colony, Pune-411016


5/163

MathCement 2000

Introduction

File Name : Intro_2_v8

Topic: About Using Mathcad Electronic Books

Notes on Navigating and Searching theElectronic Book

To browse through the book, follow the hyperlinks (boldunderlined text) within the files or the b lue arrow buttons onthe Electronic Book toolbar. They will bring you to the next orprevious topic. The Start Page button (house) on the toolbar

brings you back to the main Table of Contents. The Searchbutton (binoculars) opens a searchable word index that willlink to the selected section.

The top left icon in each document will bring you back to thefirst page in that section or from that file to the main Table ofContents.

Changing Equations and Formulasin an Electronic Book

The advantage of a Mathcad Electronic Book is that allequations are live. You can change the values of variables,constants, or other expressions in an Electronic Book to testdifferent results. You can also enter your own equations andaccess any of the commands on menubar.

Copying from an Electronic Book

To copy material from an Electronic Book, just drag yourmouse across the regions until you see a dotted linesurrounding the material you want, then drag those regionsto your main Worksheet window. You can [Shift] [Click] onindividual regions to select or deselect them. Be careful toinclude any necessary variable definitions.

To copy the entire file to your Worksheet window, use theCopy or Save As buttons in the toolbar.


6/163

Saving Annotations in an Electronic Book

You can save your own edits or annotations to the ElectronicBook. If you close the book, your edits will not be saved.

However, checking Annotate Book under the Book menu allowsyou to save some or all annotations. When you close the bookyou will be prompted to save all or some of your changes. Thenext time you open the book, you will see your annotations ingreen. You can always return to the original version of the book.

Units

By default, Mathcad uses units from the SI unit system(International System of Units). Mathcad offers several choices of

unit systems: SI, CGS, MKS, U.S. customary units, or no unitsystem. Results will be displayed in the unit system chosen by theauthor of each Mathcad Electronic Book. Please check the MathcadUser's Guidefor more information on using and converting units inMathcad. If you change unit systems, results will be changed to the newsystem, but any units in the problem or in text regions will remain as the

author entered them.


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We welcome your valuable comments which will help us to continuously improve this booin future.

The book is open to accept limited number of advertisement for future editions. Pleasecontact us for details.

We apologise to any one whose name has not been mentioned by oversight.

We also gratefully acknowledge encouragement and advices given to us by manypeople from the Cement Industry and particularly Mr. P.B.Kulakarni, Mr. M.M.Hirpara ofGujarat Ambuja Cements Ltd., Mr. Sunil S. Potdar of Gammametrics India,Mr.I.C.Ahuja of ACC

We gratefully acknowledge encouragement and support provided to us by seniorofficials in ACC, notably by Dr. A.K.Chatterjee, Mr. T.N.Tiwari, Mr.R.Vasudevan, Mr. R.Sachan, Mr. Umesh Pratap and many others.

We acknowledge references to various books,literature,journals for source materials andsome of them are:Cement Engineers Handbook by Otto Labahn and B.KolhaasCement Data Book by W.H.DudaIndustrielle Waermetechnik by J.StepanekProcess Technology of Cement Manufacturing by VDZ

Mathcad ,Electronic Handbook, Electronic Book are registered trademarks of MathsoftInc., Cambridge, USA.

MathCement 2000 is trademark of Softideas Pvt. Ltd.

This book has an Identification code : MC2000-sample-1 and should not be copied anddistributed .

MathCement 2000 is an intellectual property, conceptualised, developed and producedby collective efforts of members of Softideas Pvt. Ltd, Pune, India.

All other highlight colours are for special attention

Input variables have yellow highlighted backgroundImportant Colour Codes :

I D: MC2000-sample-1

Topic: Read me


Introduction

MathCement 2000


8/163

MathCement 2000

Introduction


Topic: About MathCement 2000

What is MathCement 2000 ?

Hello! welcome to the world of Mathematics of Cement -that'swhat MathCement 2000 all about.It is the first time that such abook, using live math, is presented to the Cement industry.Thefirst edition is being taken out in the year 2000.

This book is a mathematical walk through -Mathematics ofCement.The purpose:

To present a ready reconner to the cement engineers fortheir day to day calculations.To remove the drudgery of repeatitive calculations.Freedom from dependency on a computer programmer.To provide ready to ready to use sets of calculations forvarious aspects fo plant design and operations.Justchange the input data to suit your specific problems andyou have the step by step detailed calculations in seconds.

It's self documentingand therefore easy for scrutiny by thirdparty for checking and certification.


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MathCement 2000

Introduction


Topic: Preface to the First Edition

I D: MC2000-sample-1

Preface to the First Edition

MathCement 2000 is an intellectual property, conceptualised, developed and producedby collective efforts of members of Softideas Pvt. Ltd., Pune, India. With the advances inTechnology, electronic documentation is becoming more and more popular in manyindustry segments. We in Softideas realised the need for an Electronic Handbook forcement engineers in the plants. Mathcad from Mathsoft Inc. provides an ideal platform forsuch a development specially with its feature of Live Maths.

With Live Maths feature , calculations can be presented using normal mathematicalnotations and programme is generated by Mathcad automatically in the background.

We, at Softideas also realised that the engineers are normally busy with day to dayproduction related problems in the plants, and it is not practical to expect them toresearch and develop such a book. What was needed was that somebody developed itfor them in the final usable form. That's what we have done.

We know that there are plenty of scope for improvement and we expect users feed backto enable us come up with more comprehensive and improved design of this book.

Bahar TepanDirector, Softideas Pvt. Ltd.,

Saumitra PalDirector, Softideas Pvt. Ltd.,

Pune : 1st. August, 2001


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Book-I Chapter 1

Quarry Section

File Name : 1_1_quarry

Topic: Expanded Table Of Contents

1_1_quarry_1 Quarry Deposit Estimation

1_1_quarry_2 Quarry - Excavator

1_1_quarry_3 Quarry Dumper Calculations


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%CL2 15:=CL2 Limestone component (Gr. 2)

%CL1 70:=CL1 Limestone component (Gr. 1)

(Insert appropriate values to suit your plant condition so that total of allcomponents is 100%)

Raw material components

Step :-2

To find deposits required for individual components

TonnesDq 4.752 107

=

Dq L Trd Qd K1:=

Dq Total deposits of Raw material

Calculation

daysTrd 330:=

Book-I Chapter 1

Quarry Section

File Name : 1_1_quarry_1

Topic: Quarry Deposit Estimation

Estimation of raw material deposits in terms of no. of years of plant operation.

For a Cement Plant, life cycle is an important cosideration for plant design. It iscustomary to expect the plant to be operative for 30 years or more with maximum designcapacity utilisation for clinker production. Estimation of raw material deposits required,can be done in 2 steps as elaborated below.

Step :-1

To find total deposits required for raw material having different components

Qd Plant Capacity in terms of Clinker production Qd 3000:= Tonnes/day

K1 Factor for converting Clinker to Raw meal K1 1.6:=

L Plant life time expected L 30:= Years

Trd Plant running days in a year


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TonnesDqL5 2.376 106

=

DqL5 Dq

CL5

100:=

DqL5 Quarry deposit for Other Component

Other Component (example only - please note you have to use only as manycomponents as applicable for your plant totalling 100% )


=

DqL4

Dq CL4

100:=

DqL4 Quarry deposit for Clay


=

DqL3

Dq CL3

100:=

DqL3 Quarry deposit for Laterite


=

DqL2

Dq CL2

100:=

DqL2 Quarry deposit for Limestone Gr.2


=

DqL1

Dq CL1

100:=

DqL1 Quarry deposit for Limestone Gr.1

Calculation:

%CL5 5:=CL5 Other Component

%CL4 5:=CL4 Clay Component

%CL3 5:=CL3 Laterite Component


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m3Vd 125=

Vd

Hv

Nd:=

VdMaximum capacity of Dumpers

m3Hv 250=

Hv

Qcr Ht

BDLS 60:=

Hv Hopper capacity -( volumetric)

Calculation

Nd 2:=Nd Number of Dumpers

minutesHt 15:=Ht Hopper to hold material equivalentto crusher feed in terms of

minutes

Tonnes/m3BDLS 1.2:=BDLS Bulk density of Stone (limestone)

Tonnes/HrQcr 1200:=Qcr Crusher capacity

To find Dumper volumetric capacity

Step :-4 Check availability of Standard capacity Dumper close to thecalculated value

Step :-3 Determine maximum Dumper capacity. Feed hopper toreceive minimum 2 Dumper loads

Step :-2 Determine feed hopper capacity (volumetric)

Step :-1 Determine crusher capacity

Following steps can be followed to estimate the no. of dumpers required and their capacity.

Estimation of Number of Dumpers and Capacity.

Topic: Quarry Dumper Calculations

File Name : 1_1_quarry_3

Quarry Section

Book-I Chapter 1


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Book-I Chapter 2

Crushing Section

File Name : 1_2_crushing


1_2_crushing_1 Crusher and Auxiliaries Calculations


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K1 1.6:=

CL Total Limestone component as percentageof Raw mix =

CL 85:= %

Tcrw Number of days Crusher runs in a week Tcrw 6:= Days /Wee

Thd Number of hours the crusher system has to run per

day

Thd 12:= Hours/ Day

Ht Hopper to hold material equivalent to crusher

feed in terms of minutes

Ht 15:= minutes

Calculations

Qcr

Crusher capacity

RMw Raw material required per week

RMw Qdk K1 7:=

RMw 5.04 104

= Tonnes/Week

Book-I Chapter 2

Crushing Section

File Name : 1_2_crushing_1

Topic: Crusher and Auxiliaries Calculations

Calculation of Crusher Capacity for Raw material (Limestone)

To calculate the capacity of crusher following steps can be followed .

Step :- 1 Determine Kiln capacity

Step :-2 Determine weekly requirement of crushed Limestone material

Step :-3 Determine number of days Crusher has to run per week

Step :-4 Determine number of hours the crusher system has to run per day

Qdk Kiln Capacity of Clinker Qdk 4500:= Tonnes /Day

BDLS Bulk density of Stone (limestone) BDLS 1.2:= Tonnes./m

K1 Factor for converting Clinker to Raw meal


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T

HrQcrf 714=

Qcrf K2 Qcr:=

K2 1.2:=Over capacity factor K2

Qcr 595:=Crusher Capacity =Qcr T/Hr

Feeder capacity Qcrf T/Hr.

Capacity of Crusher feeding devices e.g Apron Feeder, Vibrating Screen etcshould have 20 to 30% over capacity.

Feeder for Crusher

m3Hv 123.958=

Hv

Qcr Ht

BDLS 60:=

Hv Crusher Hopper capacity -( volumetric)

Tonnes / HrsQcr 595=

Qcr

LSw

Tcrw Thd:=

Qcr Crusher Capacity required

Tonnes/WeekLSw 4.284 104

=

LSw CL

RMw

100

:=

LSw Limestone required per week


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Feeder capacity Qcrf T/Hr. Qcrf 714=T

Hr

Transportation from crusher

Crusher should be followed by a Plate conveyor or Impact conveyor of shortlength.This may be followed by Weigh Belt Scale to check crusher output.This isfollowed by Belt Conveyor transporting crushed material to the Mix Bed or RawMill Hopper as the case may be.The capacity of these Transport devices should be having 50% over capacity(over crusher capacity) This is in view of flushing from Crusher when a big stone iscrushed resulting in rush of material.

Capacity of Transporting equipment after crusher upto Mix Bed including Stackeris given by Qtcro T/Hr

Qtcro 1.5 Qcr:=

Qtcro 892.5=T

Hr


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Book-I Chapter 3

Preblending Section

File Name : 1_3_preblending


1_3_preblending_1 Preblending and Stockpile Equipment Calculations


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CL 85:= %

Qcr Crusher capacity Qcr 565:= Tonnes/ Hr

RMd Number of days of Raw Material stock required RMd 7:=

Calculation

Qsp Stock pile capacity in Tonnes.

RMw Raw material required per week

RMw Qdk K1 7:=

RMw

5.04 104

= Tonnes/Week

LSw Limestone required per week

LSw CL

RMw

100:=

LSw 4.284 104

= Tonnes/Week

Book-I Chapter 3

Preblending Section

File Name :

1_3_preblending_1

Topic: Preblending and Stockpile Equipment - Calculations

Calculation of Stockpile Capacity for Raw material (Limestone)

To calculate the capacity of stockpile following steps can be followed.

Step :- 1 Determine Kiln capacity

Step :- 2 Determine weekly requirement of crushed Limestone material

Qdk Kiln Capacity of Clinker Qdk 4500:= Tonnes /Day

BDLS Bulk density of Stone (limestone) BDLS 1.2:= Tonnes./m3

K1 Factor for converting Clinker to Raw meal K1 1.6:=

CL Total Limestone component as percentage

of Raw mix


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Qsp Stock pile capacity in Tonnes -Live Capacity

Qsp Qdk K1

RMd:=

Qsp 5.04 104

= Tonnes

Qstacker Stacker Capacity

Qstacker 1.5 Qcr:=

Qstacker 847.5= Tonnes/Hr

Stacker / Stacker Belt Capacity = Qstacker 847.5= Tonnes/Hr


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Calculation of Ball Mill Percentage Filling as perMeasurement

1_4_rawgrinding_7 Calculation of Grinding Ball Size

1_4_rawgrinding_8 Specific Heat of Raw Material as a Function ofTemperature

1_4_rawgrinding_9 Estimation of grindability based on the operatingparameters of Mill

1_4_rawgrinding_10 Calculation of Mill Output at Different Fineness of Product

1_4_rawgrinding_11 Calculation of Efficiency of Dynamic Air Separator

1_4_rawgrinding_12 Average Piece Weight ofGrinding Media

1_4_rawgrinding_13 Ball Mill Critical Speed and Ball Charge Calculations

1_4_rawgrinding_14 Verical Roller Mill -- Calculation of Power

Book-I Chapter 4

Raw Grinding Section

File Name : 1_4_rawgrinding


1_4_rawgrinding_1 Bond Work Index of the various material based onbond test mill result dry basis

1_4_rawgrinding_2 Calculation of power at Ball Mill shaft for Raw material ,based on the Bond's work index

1_4_rawgrinding_3 Raw Mill and Auxiliary Equipment Capacity Calculations

1_4_rawgrinding_4 Raw Material Drying - Estimation ofTotal Moisture to beEvaporated from Feed

1_4_rawgrinding_5 Raw Mill Heat Balance - Evaluation of Hot GasRequirement for Drying

1_4_rawgrinding_6


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1_4_rawgrinding_15n Radiation Loss in Raw Mill Heat Balance

Not to be copied in any form or distributed or resoldProperty of:Softideas Pvt. Ltd., 5- Sinbai Niwas, Gokhale Nagar Road,

Model Colony, Pune-411016


23/163

kwh/sh.t ( dry basis )Ws 11.261=

kwh/sh.t ( dry basis )Ws44.5

Sc0.23

Pr0.82

10

Pf

10

Ff

:=

Ws Bond index , kwh/sh.t dry basis

Calculation

Pr 1.36:=Pr Finished product per mill revolution , gm / rev

Sc 90:=Sc Screen size , ( Normally 90 size ) ,

Ff 2500:=Ff Feed size to test mill ( 80 % passing - Range 0-4 mm ) ,

Pf 60:=Pf Product fineness test result ( 80 % Passing - Range > 90 ) ,

The Bond index of various material based on the Dry basis -

Bond test mill result is calculated as given below.

Topic: Bond Work Index of the various material based on bond test mill resultdry basis

File Name : 1_4_rawgrinding_1


Book-I Chapter 4


24/163

Mf 8:= %

Mp Total product moisture)surface) Mp 2:= %

Fs Specific heat of raw material Fs 0.21:= kcal/kg.deg.c

Fa False air percentage Fa 10:= %

P Absorbed mill power P 3000:= kw

Eg Amount of dedusting gases Eg 150000:= m3/hr

Egt Dedusting gases temperature Egt 105:= deg.c

Egs Specific heat of dedusting gases Egs 0.31:= kcal/Nm3.deg.c

Ra Surface area for Radiation Losses Ra 185:= m2

Rf Radiation Loss Rf 50:= kcal/m2.deg.cdifference

Book-I Chapter 4



Topic: Raw Mill Heat Balance - Evaluation of Hot Gas Requirement for Drying

Hot gases required for drying of material in raw grinding in ball mill

The hot gases required for the drying of the feed moisture in the raw material whilegrinding in close circuit ball mill is calculated as below.

Ght

Hot gases temperature , Ght

280

:=deg.c

Ghs Specific heat of hot gases Ghs 0.34:= kcal/deg.c-Nm3

Rt Base temperature Rt 20:= deg.c

Amb Ambient air temperature Amb 35:= deg.c

Ambs Specific heat of ambient air Ambs 0.30:= kcal/Nm3.deg.c

Alt Altitude ( From mean sea level ) Alt 950:= m

Fq Fresh feed quantity Fq 100:= tph

Mf Total fresh feed moisture(surface)


25/163

Note :The latent heat of evaporation of water is 540 kcal/kg of water.

Kcal/hrHow 3.978 106

=

Kcal/hrHow W 540 Egt+ Amb( ):=

kg/hrW 6522=

kg/hrWFq 1000 Mf Mp( )

100 Mf:=

Moisture to be evaporated

Step :-4 Heat loss to evaporate moisture

Kcal/hrHor 7.862 105

=

Kcal/hrHor Ra Rf Egt Rt( ):=

Step :-3 Heat loss due to radiation

Kcal/hrHog 2.855 106

=

Kcal/hrHog Eg Egs Egt Rt( )273

Egt 273+

:=

Step :-2 Heat to dedusting gases

Note :The raw material temp. is normally less by 5 deg.c. than the exit gas temperature.

Kcal/hrHop 1.68 106

=

Kcal/hrHop Fq 1000 Fs Egt Rt 5( ):=

Step :-1 Heat to raw material

Heat output :

Gh Required hot gas quantity ( Nm3/hr )

RG Circulating air ( Nm3/hr )

Calculations


26/163

Note :If the heat input is less than the heat output then only we require hot gases.

Kcal/hrHi 2.83 106=

Kcal/hrHi Hif Hip+ Hia+:=

Total heat input

Kcal/hrHia 6.75 104=

Kcal/hrHiaEg Fa Ambs Amb Rt( )

100:=

Step :-3 Heat from false air

Kcal/hrHip 2.448 106=

Kcal/hrHip

P 816:=

Step :-2 Heat from grinding power

Kcal/hrHif 3.15 105=

Kcal/hrHif Fq 1000 Fs Amb Rt( ):=

Step :-1 Heat from fresh feed

Heat inputs :

Kcal/hrHo 9.682 106=

Kcal/hrHo Hop Hog+ Hor+ How+ Hoa+:=

Total heat loss

Kcal/hrHoa 3.825 105=

Kcal/hrHoa

Eg Fa Ambs Egt Rt( )

100:=

Step :-5 Heat loss to false air


27/163

Nm3/hrGh 77501=

Amount of hot gases required for Drying :

Nm3/hrRG 7745=

Nm3/hrRG if Mg Gh Mg Gh, 0,( ):=

Amount of recirculation gases :

Note "O.K."=

Note if Mg Gh "O.K.", "Change the hot gases temp. or the dedusting gases quantity",( ):=

Nm3/hrGh 77501=

Nm3/hrMg 85246=

Nm3/hrMg Eg273

Egt 273+

1.24 WFa

100Eg

:=

Note :If the dedusting gases quantity is less than the hot gases required then either

increase the hot gases temperature or increase the dedusting gases amount and run theheat balance again.

Nm3/hrGh 77501=

Nm3/hrGh if Hi Ho< Gh, 0,( ):=

Nm3/hrGh 7.75 104=

Nm3/hrGh root Ho Hi Gh Ghs Ght Rt( )+ Gh,:=

Nm3/hrGh 0:=

Let us solve for hot gases requirement.


28/163

radians 6.283=

radians 2acosH r

r

:=

Percentage filling :

mr 2=

mrDi

2:=

r Radius inside lining

F Percentage filling of grinding media

Calculations

h

H

r

Di

a

C.Gof Charge

h

H

r

Di

a

C.Gof Charge

mH 0:=H Empty measured height

mDi 4:=Di Mill inside diameter

To calculate the percentage of grinding media filling in the ball millbased on the empty measured height is as below.

Topic: Calculation of Ball Mill Percentage Filling as per Measurement

File Name : 1_4_rawgrinding_6r1


Book-I Chapter 4


29/163

F

1

2r2 r H r( ) sin

2

100

r2

:=

F 100= %

Notes :

The filling ratio for close circuit ball mill is in the range of 26-32 %The filling ratio for open circuit ball mill is in the range of 26-32 %For single chamber ball mill the filling ratio is in the range of 24-28 %


30/163

Cf 2.2=

Cf100

Ef

Pb

Fb:=

Circulation factor :

%Ef 57.1=

%Ef

Pb

Fb

Fb Cb

Pb Cb

100:=

Separator efficiency :

Cf Circulation Factor

Ef Separator Efficiency in %

Calculations :

%Cb 55:=Cb Separator coarse passing

%Pb 88:=Pb Separator product passing

%percent materialbelow cut size

Fb 70:=Fb Separator feed passing

micronsS 90:=S Partical cut off size

Calculation method of separator circulation factor and the separator efficiency based onthe percentage passing of feed, product and coarse at a perticular partical cut offsize areas given below.

Topic: Calculation of Efficiency of Dynamic Air Separator



Book-I Chapter 4


31/163

GIIiGrinding media loading , ton

BIiBall size , mm

SISpecific surface , m2/ton

GIIj

23.9

27.1

32.3

20.8

0

:=BIIj

60

50

40

30

25

:=WIAverage peice weight , gm/peice

j 0 4..:=

Chamber - II

GIiGrinding media loading , ton

BIiBall size , mm

SISpecific surface , m2/ton

GIi

32.5

29.0

26.5

23.9

16.4

0

0

0

:=BIi

100

90

80

70

60

50

40

30

:=WIAverage peice weight , gm/peice

i 0 7..:=

Chamber - I

The average peice weight of the grinding media is calculated as givenbelow.

Topic: Average Piece Weight ofGrinding Media



Book-I Chapter 4


32/163

Calculations

Peice weight of each ball : Surface area of each ball :

m2WIi

2 BI

1000

3

12

7.8 1000:= kg/peice SIi

2 BI

1000

2

8

:=

kg/peice m2

WIi

4.084

2.977

2.091

1.401

0.882

0.511

0.261

0.11

= SIi

7.854 103

6.362 103

5.027 103

3.848 103

2.827 10 3

1.963 103

1.257 103

7.069 104

=


33/163

Book-I Chapter 5

Raw Meal Section

File Name : 1_5_rawmeal


1_5_rawmeal_1 Calculation of Blending &Storage Silos and AuxiliaryEquipmentCapacity for Raw-meal Preparation

1_5_rawmeal_2 Calculation of Kiln Feed Raw-meal Systems

1_5_rawmeal_3n Calculation of Kiln Feed Raw-meal Quantity


Property of:Softideas Pvt. Ltd., 5- Sinbai Niwas, Gokhale Nagar Road,



34/163

Book-I Chapter 6

Preheating Section

File Name : 1_6_preheating


No files here currently


35/163

1_7_clinkerisation_8n Kiln Capacity check

1_7_clinkerisation_9n Degree of Decarbonation of Raw Meal -Definition

1_7_clinkerisation_10n Degree of Decarbonation of Raw Meal Calculation onOperating Data

1_7_clinkerisation_11n Reaction Enthalpy -Decarbonation and Clinkerisation

1_7_clinkerisation_12n Theoretical Heat of Clinker Formation - Calculation

1_7_clinkerisation_13n Calculation of Kiln Weight under Operating Conditions

1_7_clinkerisation_14n Calculation of Kiln Torque under Operating Conditions

1_7_clinkerisation_15n Calculation for Kiln Hydraulic Thruster

1_7_clinkerisation_16n Kiln Hydraulic Thruster - A Note

1_7_clinkerisation_17n Kiln Drive Specification

Book-I Chapter 7

Clinkerisation Section

File Name :

1_7_clinkerisation


1_7_clinkerisation_1 Calculation of Kiln and Auxiliary Equipment Capacities

1_7_clinkerisation_2 Leakages Through Kiln Air Seal

1_7_clinkerisation_3 Calculation of Specific Heats of Clinker as a Function ofClinker Temperature

1_7_clinkerisation_4 Kiln Retention Time and Related Parameters

1_7_clinkerisation_5 Kiln Sinter Zone Cooling Fans

1_7_clinkerisation_6 Kiln Nose Ring Cooling Fans

1_7_clinkerisation_7n Kiln Tyre / Shell Ovality


36/163

1_7_clinkerisation_18n Calculation of expansion of Kiln in hot condition

Calculation of Temerature Profile of Kiln Shell in hotcondition

1_7_clinkerisation_19n

1_7_clinkerisation_20n Calcination Function





37/163

mY 4.395:=Y Measured min. diameter

mX 4.405:=X Measured max. diameter

mD 4.4:=D Diameter of Tyre

Percentage ovality calculation

Oval shape is in Redhaving Major Dia.=Xand Minor Dia. = Y

Perfect Circle is inBlack of Dia.= D

DY

X

Ovality of Kiln shell beyond acceptable limit can effect life of Refractory lining in theKiln.Kiln shell gets rigidity only after a tyre has been mounted. However, in case of loosetyre there exists a gap between the shell and the tyre at the top of the shell.The ovality ofthe shell , therefore cannot be guaranteed.But it is important to machine the tyres

accurately with maximum ovality inside not to exceed 0.2%

Calculation for ovality of Kiln Tyres / Shells

Topic: Kiln Tyre / Shell Ovality

File Name : 1_7_clinkerisation_7n


Book-I Chapter 7


38/163

OP Percentage Ovality of Tyre ( % )

OPX Y

D

100:= %

%OP 0.227=

Tyre ovality should be practically zero when machined and maximum 0.2% on load(after erection)

During operation at a specific temperature the shell expand and fit exactly into theTyre and in such a state the shell ovality is then max 0.2% i.e same as that of tyre onload.

If the ovalityof shell, specially under the tyre section is in the range of 0.4 to0.5%,then the refractory life is adversely affected.

In the manufacturing workshop the Kiln shell ovality should be measured after fixingthe spider rings. Otherwise the shell becomes oval under its own weight.

If the ovality of fabricated shells show an ovality of more than 0.2% even with thespiders mounted, it will be difficult to align the shells during erection prior to weldingof the shells at site.This is an important considerationwhy ovality of Kiln shells alsoshould be kept atminimum possible value.


39/163

Book-I Chapter 7



Topic: Kiln Capacity check

Calculation of capacity and loads for a given size of Kiln

QK Kiln Capacity evaluation ( T/Day )

Burner

D meters -Inside shell Dia.

d meters -Inside refrectory Dia.

D Inside ( shell ) diameter of the kiln D 4.8:= m

W Refractory thickness W 200:= mm


40/163

k.cal /m2.hrKTL

3.946 106

=

k.cal /m2.hrKTLK

d

2

4

:=

d 4.4=

md D 2W

1000:=

d Kiln diameter on refractory ( m )

k.cal. / hr.K 60 106

:=K Heat input to Kiln

Kiln thermal load is calculated based on total heat input to kiln on open cross sectionalarea.This should normally not exceed. 4.5 million k.cal. /m2/hr.but can go upto 6.0

million k.cal. /m2/hr.

KTL Kiln Thermal Load ( k.cal /m2.hr.)

RLD 14.583=

RLDL

D:=

RLD Kiln L / D ratio

Calculations :

% 3:= Kiln slope ( Range 3.0-4.0 )

deg. 35:= Angle of repose ( Clinker 35 deg. )

rpmN 3.5:=N Kiln speed

mL 70:=L Overall length of the kiln


41/163

t / day /m3QSP 4.607=

t / day /m3QSP QTH

Vi:=

QSP Specific output of kiln ( t / day /m3)

QTH 4.903 103=

t / day

t / dayQTH F1 D3.283 L

0.33:=

F1 7:=Output factor =F1 is a function of the type of kilnFor normal preheater kiln F1 = 3.0

For precalciner kiln F1 is 7 to 7.5

QTH Theoretical output of Kiln ( t/ day of clinker )

Vi 1.064 103

=m3

m3Vi

d2

4L:=

Vi Kiln volume on refractory (internal volume)( m3)


42/163

Book-I Chapter 7



Topic: Calculation for Kiln Hydraulic Thruster

For calculation of operating thrust of Kiln it is necessary to first determine the totaloperating weight of the kiln . Please refer to file no. 1_7_clinkerisation_13

For a more detailed note on Hydraulic Thruster please refer to file 1_7_clinkerisation_16

KILNF

TH

WK

TYRE

THRUST ROLLER

C.L of KILN

KILNSECTION

WK Total operating weight of Kiln WK 868:= MT

o Let the slope of Kiln be 2:=


43/163

It should be noted that the Thrust Roller on the upper side of the Tyre is to act only asstopper to prevent the kiln from riding out of the roller during its upward movement.

90.0

63.0

45.0

31.5

17.5 MT

Std. hyd. unit

Thrust rollers are fitted with standard hydraulic thrusting devices of standard ratings.A typical range is shown below

Normally Thrust Rollers are provided on Pier No. 1 only.However , it is possible toprovide two sets of Thrust Rollers in 1st. and 2nd. piers. Insuch a case the load isshared equally by each roller.

MTFTH 30.293=

MTFTH WK cos90

180

:=

FTH The thrust applied by kiln on its downhill movement due to the slope ( MT )

radian1 0.035=

180 0.035=1

180 :=

To convert slope from degree to radian = 1


44/163

S1 represents length upto end of first first section of kiln and S2 upto end of secondsection of kiln and so on

S represents length of cummulative sections of kiln

ALT

n:=L

i

LT

n:=i 1 n

..:=n 10

:=L

T70

:=

The shell temperatures for the sections are either measured or taken from experience

The kiln is theoretically divided into n equal section - normally 10 sections and isdenoted by A

Where L is kiln lengthLT L=At designed ambient temp.

Let total length of kiln at cold condition be LT m

L1

L2

L3

L

KILN

Support - 1 Support - 2Support -3

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

Calculation of expansion of Kiln in hot condition

Topic: Calculation of expansion of Kiln in hot condition



Book-I Chapter 7


45/163

S0 0:=

Si Li Si 1+:=

L

0

0

1

2

3

4

5

6

7

8

9

10

0

7

7

7

7

7

7

7

7

7

7

= S

0

0

1

2

3

4

5

6

7

8

9

10

0

7

14

21

28

35

42

49

56

63

70

=T

205

210

230

250

280

310

310

310

290

270

250

:=

ta 0:= ambient tempco eff. of linear expan = 0.01215 per deg. C

Ei 0.01215 Ti 1 ta( ):=

DX represents expansion of kilnsectionsE

0

0

1

2

3

4

5

67

8

9

10

0

2.491

2.551

2.794

3.037

3.402

3.7663.766

3.766

3.523

3.28

=

DXi Li Ei:= E1 2.491=

L1 13:= L2 38:= L3 58:=


46/163

0 10 20 30 40 50 60 70200

250

300

350

Kiln Shell Temp. Profile

Shell length from feed end - m

Shelltemperature-Deg.C

Ti

Si

DX10 22.963=Expansion of sec. 10DX5 23.814=Expansion of sec. 5





Expansion of each section is given by DXi in mm

The shell is equally divided into n equal sections (n is fixed at n =10 )

DX

0

0

1

2

3

4

5

6

7

8

9

10

0

17.435

17.86

19.561

21.262

23.814

26.365

26.365

26.365

24.665

22.963

=


47/163

L1

L2

L3

L

KILN

Support - 1 Support - 2Support -3

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

0 10 20 30 40 50 60 7015

20

25

30

Kiln Shell Expansion at each section

Shell lengths from feed end -m

Expansion-mm

DXi

Si


48/163

DXL2 110.14=

DXL2 if SS2( ) 5> DXL23, if SS1 4= DXL21, DXL22,( ),:=

DXL23

DX1

DX2+

DX3+

DX4+

DX5+

DX6+

ESS2

ML2+:=

DXL22 DX1 DX2+ DX3+ DX4+ DX5+ ESS2 ML2+:=

DXL21 DX1 DX2+ DX3+ DX4+ ESS2 ML2+:=

ML2 3=ML2 mod L2 A,( ):=

SS2 5=SS2 truncL2

A

:=

L2Full Shell sections upto

DXL1 32.38=


DXL13 DX1 DX2+ DX3+ ESS1 ML1+:=

DXL12 DX1 DX2+ ESS1 ML1+:=

DXL11 DX1 ESS1 ML1+:=

ESS1 ML1 14.944=ML1 6=ML1 mod L1 A,( ):=

ESS1 2.491=SS1 1=SS1 trunc

L1

A

:=

A 7=



49/163

DXL 226.658=DXL2 110.14=

DXL3 211.228=DXL1 32.38=

mmDXL 226.658=

DXL DX:=

DXL3 211.228=


DXL33 DX1 DX2+ DX3+ DX4+ DX5+ DX6+ DX7+ DX8+ DX9+ DX10+ ESS3 ML3+:=

DXL32 DX1 DX2+ DX3+ DX4+ DX5+ DX6+ DX7+ DX8+ DX9+ ESS3 ML3+:=

DXL31 DX1 DX2+ DX3+ DX4+ DX5+ DX6+ DX7+ DX8+ ESS3 ML3+:=

DXL2 110.14=ML3 2=ML3 mod L3 A,( ):=

SS3 8=SS3 trunc

L3

A

:=



50/163




Calculation of Grate Cooler Offset from Kiln axis1_8_clinkercooling_5n

Calculation of Grate Cooler Recuperation Efficiency -T.A.Duct Tapping from Kiln Hood

1_8_clinkercooling_4

Calculation of Grate Cooler Recuperation Efficiency -T.A.Duct Tapping from Cooler


Calculation of GrateCooler Drive Power1_8_clinkercooling_2

Calculation of Clinker Transport Equipment Capacity andClinker Stockpile



File Name :

1_8_clinkercooling

Clinker Cooling Section

Book-I Chapter 8


51/163

mm

N Max. grate speed ( Shaft speed , rpm ) N 22:= strokes/min

Calculations

F Cooler grate total drive force (Kilo Newton)

F Ga Df:= KN

KNF 345=

T Torque at ecenctric shaft (KN.m )

T FSl

2 1000:= KN.m

T 20.7

=KN.m

This torque has to be transmitted by Chain wheel

N Maximum RPM of driven sprocket ( strokes per minute )

Shaft power :

Book-I Chapter 8

Clinker Cooling Section

File Name :


Topic: Calculation of Grate Cooler Drive Power

Grate cooler motor power calculation :

The grate motor power is calculated as below. See Sketch alongside

Pm

Grate motor power Pm

kw

Ga Grate area Ga 30:= m2

Df Specific cooler grate drive force Df 11.5:= KN/m2

Sl Stroke length Sl 120:=


52/163

Ps2 N T

60:= KW

Ps 47.689= KW

Grate motor power :

Pm 1.4 Ps:= KW

Pm 67= KW

Similarly calculate for all grates individually


53/163

F

Eccentric

o

Eccentric Drive

Gearbox output

s1/2

F

Eccentric

o

Eccentric Drive

Gearbox output

s1/2


54/163




Expected Power Draw of Cement Mill as a Function ofSpeed

1_9_cementgrinding_12n

Cement Mill Grinding Performance as a Function ofSurface Generated

1_9_cementgrinding_11n

Mill Radiation Loss1_9_cementgrinding_10n

Cement Mill Cooling Air1_9_cementgrinding_9n

Cement Mill Output at Different Finenesses1_9_cementgrinding_8n

Sizing of Close Circuit Cement Mill1_9_cementgrinding_7n

Calculation of Water Spray in the Mill for Cooling1_9_cementgrinding_6

Estimation of grindability of Clinker based on the operatiparameters

1_9_cementgrinding_5

Calculation of Cement Mill Output as a function ofGrinding Media Load


Cement Mill Heat Balance and Calculation of Hot Gas fDrying


Calculation of Cement Grindability based on Ziesel Valu1_9_cementgrinding_2

Calculation of Cement Mill and Auxiliary Equipment1_9_cementgrinding_1


File Name :

1_9_cementgrinding

Cement Grinding Section

Book-I Chapter 9


55/163

N 12:= rpm

F Percentage filling F 28:= %

G Grindability , Kwh/ton at 3000 blaine G 30:=

Kf K factor Kf 9.55:=

Calculations

The blaine correction factor :

Bf e

Pf 3000( )1000

0.49

:=

Bf 1.103=

The grindability at the required product fineness :

Gf G Bf:= Kwh/t .

Kwh/tGf 33.089=

Book-I Chapter 9


File Name : 1_9_cementgrinding_4

Topic: Calculation of Cement Mill Output as a function of Grinding Media Load

Production of the ball mill - Clinker grinding based on Grinding media loading

The theoretical production based on the grindability of the clinker is as givenbelow.

Pr

Production Pr

tph

Pf Product fineness Pf 3200:= blaine

Di Ball mill inside diameter Di 4:= m

L Ball mill effective length ( I + II Chamber ) L 14:= m

N Ball mill speed


56/163

The grinding media :

Gm Di

2 L 4.5 F

4 100:= ton

Gm 221.671= ton

The available mill shaft power :

PGm Kf Di( )

1.36:= kw

P 3113= kw

The production based on the grindability :

PrP

Gf:= tph

Pr 94= tph


57/163

tphCapnew 132=

Capnew roundCap

k0,

:=

k 0.907=

k 10

B2 B11000

0.213:=

Correction factor = k

Capnew To find what would be new capacity of mill ( tph )

Calculations :

cm2 /gmB2 2800:=B2 Desired fineness of cement acc. to Blaine

cm2 /gmB1 3000:=B1 Present fineness of cement acc. to Blaine

tphCap 120

:=Cap Present Mill output

For an operating mill capacity at a certain fineness is known . Following calculationmethod predicts expected outputs at other desired finenesses.

Calculation of Mill output at other finenesses

Topic: Cement Mill Output at Different Finenesses

File Name : 1_9_cementgrinding_8n


Book-I Chapter 9


58/163

Book-I Chapter 10

Cement Storage Section

File Name :

1_10_cementstorage


1_10_cementstorage_1 Calculation of Cement Transport and Silo and Auxiliar Equipment





59/163

Qsilo

Td Qcm Th( )

VBDc:=

Qsilo Silo capacity

Calculation

Tonnes/m3VBDc

1.2

:=VBD

cBulk density of cement for volume

HrsTh 16:=Th Mill running hours per day

daysTd 8:=Td Capacity in terms of number of days stock

To find capacity of Cement silo

Step :- 2

Tonnes / Hr.Qtr2 144=

Book-I Chapter 10

Cement Storage Section

File Name : 1_10_cementstorage_1

Topic: Calculation of Cement Transport and Silo and Auxiliary Equipment

Many times cement production can be enhanced by grinding coarser than originallycontemplated or by adding Pozzolanic material. Hence it is of great importance toforesee such possibilities at design stage and design the capacities for maximumanticipated capacity of the Mill

Step :- 1

To find the capacity of conveyor from Vibrating screen located at discharge of Cementmill

Transportation of cement from the mill after Vibrating Screen etc.should be designedfor 20 % over capacity w.r.t. mill capacity.

Qcm Max Capacity of Cement Mill Qcm 120:= Tonnes/ hr

Fod Over design factor Fod 1.2:=

Calculation

Qtr2 Conveyor capacity

Qtr2 Fod Qcm:=


60/163

Qsilo 12800= m3

Step :- 3

To find capacity of Silo Discharge Apparatus

Qp Demand of Packing m/c per hour Qp 100:= Tonnes/Hr

Over design factor (100 %) Fod 2:=

Calculation

Qd Silo Discharge Apparatus Tonnes/Hr

Qd Fod Qp:=

Qd 200= T/Hr

Note - There should be always a standby discharge apparatus


61/163

Book-I Chapter 11

Cement Despatch Section

File Name :

1_11_cmtdespatch


Currently no files here


62/163




Kiln Dust Loss in Terms of Kiln Feed Raw Meal-andApparant Degree of Calcination

1_12_quality_11n

To Calculate Quantity of CaO Required to Attain SpecificValue of Total Carbonate Content in Kiln Feed Raw Meal-

1_12_quality_10n

1_12_quality_9nTotal Carbonate Content in Kiln Feed Raw Meal-Calculation

1_12_quality_8nPercentage Liquid and Burnability Index Cement Clinker -Calculation

Complex Compounds of Cement Clinker - Calculation byBogue's Formulae

1_12_quality_7n

Burnability Factor of Kiln Feed Raw Meal - Calculation1_12_quality_6n

Hydraulic Ratio of Kiln Feed Raw Meal - Calculation1_12_quality_5n

Lime Saturation Factor Calculation1_12_quality_4n

Alumina Ratio and Requirement of Components1_12_quality_3n

Silica Ratio and Requirement of Components1_12_quality_2n

Loss on Ignition of Kiln Feed Raw Meal1_12_quality_1n

Topic: Expanded Table Of Quality Checks

File Name : 1_12_quality_n

Quality Checks

Book-I Chapter 12


63/163

LSFPWCaO

2.8 PWSiO2 1.65 PWAl2O3+ 0.35 PWFe2O3+:=

To find Lime Saturation Factor ( LSF ) When > 0.64

Condition 1:-

Check if the figure ARRM is > 0.64

ARRM 1.849=

ARRM

PWAl2O3

PWFe2O3:=

ARRM Alumina Ratio of Raw Meal

Calculation

%PWSiO2 22.3:=PWSiO2 Percentage of SiO2 in Kiln feed raw meal

%PWCaO 66.01:=PWCaO Percentage of CaO in Kiln feed raw meal

%PWAl2O3 5.4:=PWAl2O3 Percentage of Al2O3 in Kiln feed raw meal

%PWFe2O3

2.92

:=PW

Fe2O3Percentage of Fe2O3 in Kiln feed raw meal

Analysis on Loss Free Basis

Calculation of Lime Saturation Factor of Kiln feed raw meal ( LSF )

Topic: Lime Saturation Factor - Calculation

File Name : 1_12_quality_4n

Quality Checks

Book-I Chapter 12


64/163

Condition 2 :-

To find Lime Saturation Factor = LSF When < 0.64

LSFPWCaO

2.8 PWSiO2 1.1 PWAl2O3+ 0.7 PWFe2O3+:=

By using if statement we can evaluate LSF

LSF if ARRM 0.64>PWCaO

2.8 PWSiO2 1.65 PWAl2O3+0.35 PWFe2O3+

...,

PWCaO

2.8 PWSiO21.1 PWAl2O3+

...

0.7 PWFe2O3+

...,

:=

LSF 0.912=


65/163

PWMgO Percentage of MgO in Kiln feed raw meal PWMgO 1.365:= %

PWAlkalies Percentage of Alkalies in Kiln feed raw meal PWAlkalies 0.413:= %

PWSO3 Percentage of SO3 in Kiln feed raw meal PWSO3 0.338:= %

LOI Loss On Ignition LOI 35:= %

Calculation

TC Total Carbonate (Calculated from Analysis on Raw Basis)

TC 1.784 PWCaO 2.09 PWMgO+:=

TC 79.399= %

Note if the analysis is given in Loss Free Basis then multiply by conversion factor

=FLFtoR

FLFtoR100 LOI

100:=

Book-I Chapter 12

Quality Checks

File Name : 1_12_quality_9n

Topic: Total Carbonate Content in Kiln Feed Raw Meal- Calculation

Calculation of Total Carbonate Content in Kiln Feed Raw Meal

Analysis of Kiln Feed Raw Meal on Raw Basis

PWFe2O3 Percentage of Fe2O3 in Kiln feed raw meal PWFe2O3 1.898:= %

PWAl2O3 Percentage of Al2O3 in Kiln feed raw meal PWAl2O3 3.51:= %

PWCaO Percentage of CaO in Kiln feed raw meal PWCaO 42.907:= %

PWSiO2 Percentage of SiO2 in Kiln feed raw meal PWSiO2 14.495:= %


66/163

Conditional Calculation

Let us define Analysis Basis as = BASIS

If analysis is on Raw Basis then

If analysis is on Loss Free Basis then

Define BASIS

BASIS 1:=

TC if BASIS 0> 1.784 PWCaO 2.09 PWMgO+( ), FLFtoR 1.784 PWCaO2.09 PWMgO+

...

,

:=

TC 79.399=


67/163

Book-I Chapter 13

Raw Materials

File Name :

1_13_rawmaterials_n

Topic: Expanded Table Of Raw Materials

1_13_rawmaterials_1n Raw Materials for Cement Production





68/163

The final clinker is the product after combustion, during which process, the rawmaterials lose gaseous components . Thus the resultant balance material analysis isreported without possibility of any further loss. This report is termed as loss freebasis.By predicting the anticipated losses , it is possible to report the analysis of rawmaterial s in Loss Free Basis.

Raw materials as analysed in Laboratory are reported as Raw Basis

Tetracalcium alumino ferrite is represented by C4AF

Tricalcium aluminate is represented by C3A

Dicalcium silicate is represented by C2S

Tricalciun silicate is represented by C3S

Complex compounds of Cement are represented by :

Rest components represented by R

Fe2O3 represented by F

Al2O3 represented by A

SiO2 represented by S

CaO represented by C

The principle oxides that constitute cement clinker are:

Alumina : principal source of Al2O3

Laterite : principal source of Fe2O3

Sand stone : principal source of SiO2

Clay : principal source of SiO2

Lime stone : principal source of CaCO3 and hence CaO

Principle Raw Materials :

Variable Definitions for Raw materials section

Topic: Raw Materials for Cement Production

File Name : 1_13_rawmaterials_1n

Raw Materials

Book-I Chapter 13


69/163

F2 F3 FASH

And So On

LOI[ ] LOI1 LOI2 LOI3 LOIASH

When the total of all components including LOI is 100% then the analysis is reported inRaw Basis

When the total of all components excluding LOI is 100% then the analysis is reported inLoss Free Basis.

To convert from one basis to another following steps can be followed

Step :- 1 To convert to Loss Free Basis from Raw Basis

OI

OR

100 LOI100:=

OR%

Or

LetFRtoLF

100

100 LOI:=

LOI

Then

O1 FRtoLF OR:= FRtoLF

OI represents percentage oxide value in Loss Free Basis

OR represents percentage oxide value in Raw Basis

Since the Raw meal is a mix of two or more raw materials, the analysis is indexed toindicate the belongingness of particular values as follows.

raw material -1 raw material -2 raw material -3 Ash

CaO[ ] C1 C2 C3 CASH

SiO2[ ] S1 S2 S3 SASH

Al2 O3[ ] A1 A2 A3 AASH

Fe2 O3[ ] F1


70/163

%

C Percentage of CaO in Kiln feed raw meal C 66.01:= %

S Percentage of SiO2 in Kiln feed raw meal S 22.3:= %

MGO Percentage of MgO in Kiln feed raw meal MGO 2.1:= %

AL Percentage of Alkalies in Kiln feed raw meal AL 0.635:= %

REST Percentage of Rest oxides in Kiln feed raw meal REST 3.2:= %

LOI Percentage of Loss On Ignition in Kiln feed raw meal LOI 35:= %

FLFtoR100 LOI

100:=

LOI represents percentage lOSS ON IGNITION in Raw Basis

FRtoLF is common factor for conversion fron Raw To Loss Free basis

Step :- 2 To convert from Loss Free Basis to Raw Basis

OR100 LOI( ) OI

100:=

LOI%

Or

Let

FLFtoR100 LOI

100:=

LOI

Then

OR FLFtoR OI:= FLFtoR

OI represents percentage oxide value in Loss Free Basis

OR represents percentage oxide value in Raw Basis

LOI represents percentage lOSS ON IGNITION in Raw Basis

FLFtoR is common factor for conversion fron Raw To Loss Free basis

Analysis on Loss Free Basis

F Percentage of Fe2O3 in Kiln feed raw meal F 2.92:= %

A Percentage of Al2O3 in Kiln feed raw meal A 5.4:=


71/163

RESTR Percentage of Rest oxides in Kiln feed raw meal RESTR FLFtoR REST:= %

FR FLFtoR F:= FR 1.898=

AR 3.51=AR FLFtoR A:=

CR FLFtoR C:= CR 42.907=

SR FLFtoR S:= SR 14.495=

MGOR FLFtoR MGO:= MGOR 1.365=

ALR FLFtoR AL:= ALR 0.413=

RESTR FLFtoR REST:= RESTR 2.08=

LOI Percentage of Loss On Ignition in Kiln feed raw meal LOI 35:= %

FLFtoR 0.65=

Analysis on Raw Basis

FR Percentage of Fe2O3 in Kiln feed raw meal FR FLFtoR F:= %

AR Percentage of Al2O3 in Kiln feed raw meal AR FLFtoR A:= %

CR Percentage of CaO in Kiln feed raw meal CR FLFtoR C:= %

SR Percentage of SiO2 in Kiln feed raw meal SR FLFtoR S:= %

MGOR Percentage of MgO in Kiln feed raw meal MGOR FLFtoR MGO:= %

ALR Percentage of Alkalies in Kiln feed raw meal ALR FLFtoR AL:= %


72/163

Book-I Chapter 14

Cement

File Name : 1_14_cement_n

Topic: Expanded Table Of Cement

1_14_cement_1n Evaluation of combined water in cement





73/163

Corresponds to co-efficients when cement is hydratedfor 13 year

a13 data2 :=

a6.5 data1 :=

Corresponds to co-efficients when cement is hydratedfor 6.5 year

Corresponds to co-efficients when cement is hydratedfor 1 year

a1 data0 :=

data0 1 2

0

1

0.23 0.23 0.23

0.17 0.18 0.2

:=

Co- efficients are given by the following table ( acc. to Kantro and Copeland)

r a1 C3 S[ ] a2 C2 S[ ]+ a3 C3 A[ ]+ a4 C4 A F[ ]+=

let us call this ratio = r so that

WC

Ca1 C3 S[ ] a2 C2 S[ ]+ a3 C3 A[ ]+ a4 C4 A F[ ]+=

then WC/ C ratio is given by the following formula

Let the quantity of cement be C gms

Let combined water in set and hardened cement be WC gm

Water that is held in the pores of the solid paste is free water or capillary water.

The water that is held at the surface of the cement by surface tensionof gelparticles is called adsorbed water or gel water.

After hydration of cement water that is chemically combined with the compoundsand form part of the solid is called combined water.There also exists adsorbedwater and free water .

Topic: Evaluation of Combined Water in Cement

File Name :

1_14_cement_1n

Cement

Book-I Chapter 14


74/163

WC 2.479 gm=

WC r C:=

C 10gm:=Let quantity of cement be

This would mean that after 13 years, the cement has combined water of 24.789 %of its weight .

a3 0.109=r 24.789 %=

a2 0.522=a

0.23

0.196

0.522

0.109

= a10.196=r

0.248=a0 0.23=

r a0

C3 S[ ]

100 a1

C2 S[ ]

100+ a2

C3 A[ ]

100+ a3

C4 A F[ ]

100+:=

a if t 1= a1, if t 13= a13, a6.5,( ),( ):=

C4 A F[ ] 9:=i 0 3..:=

C3 A[ ] 13:=

C2 S[ ] 27:=

C3 S[ ] 51:=and let cement constituents be in %

t 13:=let the time period of hydration be = t yearsValid for t equal to 1,6.5 and 13 only

a13

0.23

0.196

0.522

0.109

=a6.5

0.234

0.178

0.504

0.158

=a1

0.228

0.168

0.429

0.132

=


75/163

Book-II Chapter 1

Plant Site Condition

File Name :

2_1_sitecondition


2_1_sitecondition_1 The Barometric pressure at site based on the altitudeand temperautre.





76/163

Bp 635=mm of Hg

mm of HgBp

P273

273 Amb+

:=

Barametric pressure at ambient temperature

P 716=mm of Hg

mm of HgP760

Ef:=

Site pressure at zero deg.c. temperature

Ef 1.061=

Ef1

1 0.0065Alt

288

5.255:=

Elevation factor

Calculations:

deg.cAmb 35:=Amb Ambient temperature

mAlt 500:=Alt Altitude ( Above mean sea level )

mm of HgBpBp Barometric pressure

The Baraometric pressure based on the altitude of the site and the ambient temperature iscalculated as given below.

Topic: The Barometric pressure at site based on the altitude and temperautre.

File Name :

2_1_sitecondition_1

Plant Site Condition

Book-II Chapter 1


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Chimney Design2_2_dedusting_13n

Electrostatic Precipitator- To calculate Migration Velocity2_2_dedusting_12n

Electrostatic Precipitator- Performance Evaluation2_2_dedusting_11n

Electrostatic Precipitator- Specific Collection Area2_2_dedusting_10n

Electrostatic Precipitator- Deutsch Efficiency Formula2_2_dedusting_9n

Sizing of Gas Conditioning Tower2_2_dedusting_8n

GCT and ESP -Calculation of Gas Volume2_2_dedusting_7

Glass Bag House -Calculation of Gas Volume as Functioof temperature.

Book-II Chapter 2

Dedusting Systems

File Name : 2_2_dedusting


2_2_dedusting_1 Estimation of Vent Air Volume

2_2_dedusting_2_1r K- Factors

2_2_dedusting_3 Calculation of Pressure Losses in Ductings

2_2_dedusting_4 Design Parameters for Design of Duct Systems

2_2_dedusting_5 Duct Balancing

2_2_dedusting_6


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Book-II Chapter 2

Dedusting Systems

File Name :

2_2_dedusting_1

Topic: Estimation of Vent Air Volume

ESTIMATION OF VENT AIR VOLUMES

Estimation of vent air volume is the first step in engineering a dust collection system. Theair volume to be vented form any machinery hood or transfer point depends on manyfactors such as the physical nature of the material handled, air currents and surgescreated by the movement of machinery parts and materials, the type of enclosure

provided. Therefore, certain accepted standards which give rule-of-thumb methods haveto be followed. The following recommended procedures extracted from:

Industrial Ventilation-Published by American Conference of GovernmentIndustrial Hygienists, U.S.A.

could be used as guideline in estimation of air volumes.

It may be noted here, for guidance, that when the venting volume is expressed in cubicmeters, this refers to actual operating conditions at site under consideration. however, ifthe volume is expressed in NM3, then this volume is to be corrected for actual operatingtemperature and altitude above mean sea level.

1. CLOSED TOP BINS:

Volume = K x volume of entering material.The value of K ranges form 1 (for large bins, low feed rates, coarse feed) to

50 (small bins, high feed rate, fine material).

The minimum rate is 3600 M3/Hr per M2 of bin opening.

2. BUCKET ELEVATORS:

1800 M3/Hr/Sq. Metre of elevator casing section. For elevators over 10 M height,connect exhaust from both top and bottom.

3. BELT CONVEYORS:

3600 M3/Hr per sq. metre of openings in enclosure, but minimum

1950 M3/Hr/metre belt width for belt speeds under 1 M/sec.

2800 M3/Hr/metre belt width for belt speeds over 1M/sec.Additional volume for greater than 1 metre fall and dusty material:

1200 M3/Hr for belt width 300 mm to 1000 mm

1700 M3/Hr for belt width above 1000 mm


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4. FLAT DECK SCREEN:

3600 M3/Hr/M2 through hood openings, but minimum.

900 M3/Hr/M2 screen area.

5. CYLINDRICAL SCREEN:

1800 M3/Hr/M2 circular cross section of screen, but minimum.

7200 M3/Hr/M2 of enclosure opening.

6. AIRVEYOR & PNEUMATIC SCREW PUMP SYSTEMS:

Multiply free air volume of the system as follows:For conveying distances upto 150 M --- 1.5 timesFor conveying distances over 150 M --- 1.75 times

7. PNEUMATIC GRAVITY CONVEYOR & BLENDING SYSTEMS:

1.0 times free air volume of the system.

8. SCREW CONVEYORS:

If dust tight, ventilate feed point only. Use 1100 M3/Hr per metre of nominal conveyordiameter corrected for height of fall of feed.

9. CRUSHERS:

At feed end: Minimum of 2700 M3/Hr/M2 of feed hood opening.At discharge: Measure air current surges and allow 50% excess.

10. ROTARY CRUSHER:These act as fans. Restrict ingress of air at feed point. Measure air quantity atdischarge and add 50%. (Crusher manufacturer's recommendations to be called for).

11. GRINDING MILLS:

Use the larger of the following:A. 0.5 M/sec to 0.6 M/sec through the cross section area of the mill.B. 1 M/sec through each aperture.

C. 8.5 M3/Hr/HP of Raw Mill motor rating.

D. 13 M3/Hr/HP of Cement Mill motor rating.(N.B. These figures are for operating temperatures of the Mills).


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12. MECHANICAL AIR SEPERATORS:

Where cooling or drying is carried out in Separator, use outlet volumes from detailed

calculations. Maximum air volumes at 88 oC are indicated below for Air Separators:

Separator Size *Volume at Sea Level4.267 M dia. 31,300 M3/Hr

4.877 M dia. 44,000 M3/Hr

5.486 M dia. 59,500 M3/Hr

6.096 M dia. 74,000 M3/Hr

6.706 M dia. 83,300 M3/Hr

7.315 M dia. 92,700 M3/Hr* For any other temperature or altitude, correct volume by applying densitycorrection factor.

13. GRATE OPENINGS:

A. (Floor or bench dumping - not enclosed)

2700 to 3600 M3/Hr/M2 of grate area.B. (Enclosed three sides and top)

2700 M3/Hr/M2 of net face area.

14. 60 T/Hr CEMENT PACKER :

i) Packer Guard --- 100 M3/Min.

ii) Top of Packer --- 15 M3/Min.

iii) Control Screen --- 20 M3/Min.


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Book-II Chapter 2

Dedusting Systems

File Name : 2_2_dedusting_5

Topic: Duct Balancing

BALANCING OF DUCT SYSTEM RESISTANCE

1. In a ducting network having branch entries or exits, it is essential to balance systemresistance at every joint (i.e., where a branching takes place), in order to obtain desiredflow rate in each branch.

2. In the typical network shown below, balancingis to be done as follows:

10

1

2

11

12

3

4

5

6

7

8

9

Dedusting Network

13

10

1

2

11

12

3

4

5

6

7

8

9

Dedusting Network

13


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4. Find resistance Wg in 11_13&12_13 say R1 & R2 respectively (mm of Wg)

3. Select Main & Branch - 11_13 Main, 12_13 Branch entry

2. Select velocities in ducts 11,13&12,13 (m/sec)

1. Find volume flow rate - m3 at temp. Deg. C, at point 11 & 12 (m3/sec, 0C)

3. Actual Balancing: Typical at junction 13

Fan SP = R6 + Outlet VP at '6'

Total system resistance = R6=R8=R9

7:8 & 7:97 (R7 + R7,9)/ R9(R7 + R7,8)/ R8 =

(R5 + R5,7)/ R7(R5 + R5,6)/ R6 =5:6 & 5:7

5R3 + R3,4 + R4,5 = R5

R310:3 & (R2+2:3)3

R21:2 & (R13+13:2)2

R1311:13 & 12:1313

ResistanceBalance Resistance betweenJunction Reference


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5. Find which is higher resistance - say R1 > R2

6. Find percentage of balance = ((R1 - R2)/ R1) x 100 %

7. Item 6 should be < 20%

8. If item 6 > 20% redo, line with less resistance 12,13. Increase velocity, to find R12 , 13> R2 say, now R12_13 = R3

9. Recheck % balance, i.e., find which resistance is greater between R3 & R1say R3 > R1Therefore, % balance = ((R3 - R1)/R3) x 100% < 20% O.K.

10. Now between R3 & R1 which are within 20% of balance, R3 > R1

11. Now governing resistance is R3 = R13 which is resistance in line 12_13

12. Re-evaluate volume flow in line 11_13 for this governing pressureVol 11_13 = Qfinal = Qestimated x square root (R3/R1)

where R3 and R1 are in mm of Wg.

Qestimated = Original estimated volume flow (m3/sec)

Qfinal = actual volume flow or corrected (m3/sec)

13. Now we know volume flow at 13_2 which is Q11_13 + Q12_13=Q13_2If there is appreciable temp difference between two air current, find volume of mix andtemp.

14. Select velocity in 13_2, which should be velocity in 11_13 & 12_13

15. Find resistance in 13_2 (mm of Wg)

16. Add R13 to item 15 to get resistance in point 2 = R 13 + R13_2


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17. Find VP in line 13_2 (velocity will depend upon standard pipe)

VP = V2 x Pair/2g mm of Wg (for metric)

18. Find equivalent VP in terms of velocity in line 11_13&12_13, say, line 1&2respectivelytherefore, equivalent VP = VP1_2

=(Q1 + Q2)2 x Pair / (A1 + A2)

2 x2g mm of Wg (for metric system)

For metric system: Pair = Kg/m3

Q1 = m3/sec in Branch 1

Q2 = m3/sec in Branch 2

A1 = m2 - area or Branch 1

A2 = m2 - area or Branch 2

g = 9.81 m/s2 - Accelaration due to gravity

19. Find difference between VP in line 13_2 and equivalent VPi.e., (VP13_2 - VP1_2) < 2.5 mm of Wg

20. If fig in 19 above is > 2.5 mm of Wg then add this to resistance of line 13_2, to getresistance at point 2therefore, Pressure at point 2 = R13 + R13_2 + (VP13_2 - VP1_2) mm of Wg

This resistance is added to indicate requirement of extra pressure for accelaration tohigher velocity, required in the line, after a branch has been connected.

21. Proceed for balancing at joint 2 as described for joint 13 and so on, till completenetwork is balanced.

Abbreviations used:R1_2 - means resistance between point 1 & 2

R2 - means resistance / static pressure at point 2

VP - Velocity Pressure


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5) The gas velocity in the ESP = v m/sec. is constant

4) Under the influence of Electric field, the charged dust particles move towards thecollecting electrodes at uniform velocity = W cms. / sec.

3) Influence of Electric Wind is negligible.

2) Dust collected on the plates are not reentrained into the gas stream.

1) Dust particles are uniformly distributed over the entire cross sectional area of ESP

We shall assume the following:

0.998=

1RC

RO:=

Then the efficiency of dust collection

gms. / m3RC 0.1:=RC Dust concentration at ESP outlet

gms. / m3RO 60:=RO Dust concentration at ESP inlet

Efficiency of an ESP is defined by its capacity to remove dust from the carrying gases. Ifthe gas volume flow rate is constant, the dust concentration at the inlet and at the out letcan be denoted by the following variables.

An appreciation of Deutsch Formula

This example is for understanding of Deutsch Formula and not for evaluation of ESPefficiency

Topic: Electrostatic Precipitator- Deutsch Efficiency Formula

File Name : 2_2_dedusting_9

Dedusting Systems

Book-II Chapter 2


86/163

m2df 0.02=

m2df 2 b dx:=

df Now the elemental collecting area ( m2 )

gms.g 0.09=

b . 2d . dx is equal to the volume of the inter electrode space of length dx -

considered in m3

g RX b 2 d dx:=

mdx 0.01:=dx Let us consider an inter electrodes spaceof length at a distance = x from inlet.

mb 1:=b Let us denote the electric field height

md 0.15:=d Let the distance between collecting anddischarge electrode be

gms. /m3RX 30:=RX Let dust conc. at distance x from inlet be

gms.gg The amount of dust particle at this space

RC

RO

d

x

dxb

L

W V

Collecting Electrode

Discharge Electrode


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RO

RC

g1

g

d 6.397=

Now let us put the limits of dust loading and time and integrating we get the following

k 0.035=

kdg

g:=

gms.dg 3.12 103=

dg dg:=

Since dust is removed from gas , we will assign negative (-) sign to the value of dg

dg 3.12 10 3= gms.

dg RX V:=

RX 30=

Since the dust content in this space is RX gms.dust settled =dg

m3V 1.04 104=

m3V 2 b dxW

100

dt:=

cm/secW 13:=W Migration velocity

dt .04:=quantity of dust = dg gms. settle on this elementalarea at a time dt sec.after a time tx ( the time totravel fron inlet to distance = x )

This means , all the dust particles with migration velocity of = W/100 m/sec., within adistance of W x dt/100 from the two surfaces of collecting area will settle down on theplates. Since the particles are assumed to be uniformly distributed, the gas volume which

contain all these dust particles = V m3

Let us now assume that a quantity of dust = dg gms. settle on this elemental area at atime dt after a time tx ( the time to travel fron inlet to distance = x )


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mg / m3RC 1000 RC:=

RC 0.000136=gms. / m3

RC RO e

W

100

t

d

:=

RC Dust conc. at outlet (gms. / m3)

t 15=sec

sectL

v:=

m/secv 0.8:=v Gas velocity

mL 12:=L Total length of ESP

Since the gas travels through the total active length of the ESP = L m in total time = tsec. at a velocity = v m/sec. we have:

RX RO e

Wtx

100 d

:=

Or

lnRX

RO

W

100 d

tx 0( ):=lnRX

RO

W

100 d

tx 0( ):=

Or

K1

0

tx

tW

100 d

d:=

tx 5:=time tx ( the time to travel fron inlet to distance = x )

K1

RO

RC

g1

g

d:=


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efficiency of collection

v velocity of gas in ( m/sec )

d Distance between Collecting and Discharge Electrodes in ( m )

L Total active length of ESP in ( m )

W Migration velocity in ( cm. / sec )

Where

0.9999977397=

exp1

100W

L

v d( )

2.26 106

=

e

W

100

L

v d

1 e

W

100

L

v d

:=

Since t= L/v we can write the expression as

1 e

W

100

t

d

:=

Or

1RO e

W

100

t

d

RO:=

Or

0.998=

1RC

RO:=

Or

mg / m3RC 0.136=


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Now please refer to evaluation of Sp. Collecting Area = f m2/m3/sec

f W

L

v d:=

So the Deutsche formula can be written as

1 e

W

100

L

v d

:=

Or

1 e

W f

100:=

Corrollaries

In the process of deriving the basic formula for ESP collection efficiency we can make thefollowing important observations.

1 e

W

100

L

v d

:=

1) For same efficiency and all other conditions of operations remaining same W/d is aconstant = K1 Or W= K1*d

i.e. W increases in direct linear proportion to d - the distance between collecting and

discharge electrodes.This would , however, mean higher voltage and rating ofTransformer / Rectifier sets

2 ) W / v is a costant i.e W increases in direct proportion of v - velocity of gases.But thevelocity cannot be increased very much as it has negative influence of re- entraining thedust particles.


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Volatility2_3_laboratory_11

Separation of ESP and Filter Dust2_3_laboratory_10

Burnability Investigation2_3_laboratory_9

Raw Mix Investgation2_3_laboratory_8

Fuel Investgation2_3_laboratory_7

Mineralogical Investigations2_3_laboratory_6

Chemical Tests2_3_laboratory_5

Tests Conducted at Physical Laboratory2_3_laboratory_4

Purposes of Tests2_3_laboratory_3

Coal Samples -A Point of View2_3_laboratory_2

Quantity of Raw Material Samples for Lab. Investigation2_3_laboratory_1


File Name : 2_3_laboratory

Laboratory Investigations

Book-II Chapter 3


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Book-II Chapter 3


File Name : 2_3_laboratory_5

Topic: Chemical Tests

Tests Conducted at Chemical Labs:For ease of understanding a sample Lab report has been referred to:

Summary report/observation:

1) It may be observed that Sulphate Resistant cement is defined as ASTM - Type V

where C3A 5% and (C4AF + 2C3A) 20%.

2) Alkali equillibrium factor Q is < 1 (Q=0.71). Hence no alkali by pass is required.If Chlorine content is > 0.04% then Alkali by pass is necessary (Cl - Free of Loss basis).Alkali & Sulphates combine to form alkali sulphates.when Q = 1 all alkalies combine with sulphates. (Equillibrium)

Q > 1 There is excess of SO3.

Q < 1 There is Alkali excess.

719462

80

22

3

ClOKONa

SO

Q ++=

where SO3, Na2O etc are % of these present in Clinker (including ash effect) on loss freebasis. 80, 62 etc. represent Molecular weight of SO3, Na2O etc.

3) Influence of fuel ash is significant and reduces Lime standard by 10 points, eg. from105 in Raw meal to 95 in clinker.

4) Burnability is normally poor with high Quartz content in Raw meal. But due to the fact

that most of the quartz is in size fraction < 90, the influence of same on Burnability isminimum.

5) Raw mix and ESP dust (size fraction 0 - 32) show similar chemical composition.

Individual Tests:All raw materials are assessed interms of colour, structure, hardness in mole's scale,moisture content on receipt and moisture absorption when kept under water.Under water storage of material reveals its properties with respect to flowability speciallyfor clay etc. Crushing strength is determined for crusher selection.


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Bulk density of material is determined after crushing the material in laboratory crusher.This value is used to determine storage value, eg. for stockpile value, Raw materialhopper value etc. The grain size analysis is done after crushing in the Laboratory

crusher to predict grain size distribution after crushing operation in industrial scale andthis information is used in crusher selection & design.

Chemical composition of Raw materials is determined by Atomic Absorption test. Thiscan also be done by X'ray analysis. Individual components are determined as under:

%SiO2 - By Atomic absorption test (AAS)

%Al2O3 - By Atomic absorption test (AAS)

%TiO2 - Calorimetric - By Photometer

%Fe2O3 - By Atomic absorption test (AAS)%Mn2O3 - By Atomic absorption test (AAS)

%CaO - Complexometric - Titration%MgO - By Atomic absorption test (AAS)%SO3 - (Total Sulphur) LECO

%P2O5 - Calorimetric - By Photometer

%Na2O - By Flame Photometer

%K2O

%Cl - By Titration


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Book-II Chapter 3


File Name : 2_3_laboratory_8

Topic: Raw Mix Investgation

Rawmix Design:One of the most important purpose of various tests is to determine a suitable Raw mixcapable of producing cement of desirable quality. We have already analysedcomposition of various raw materials and correctives as well as fuel ash. Followingparameters are fixed:

1) Consider coal ash absorption in clinker as 100%.2) Set value for lime standard at 95%. This is the most important set value fordetermining Raw mix composition so that Lime saturation factor in clinker is 95%.Normal range 92 - 97% in clinker.3) Second set point is for Silica ratio (2.4 to 2.8 in clinker).4)Third set point is for Alumina ratio (1.8 to 2.2 in clinker).For solving the equations, following rule may apply.For 2 component mix - 1 set point i.e., Lime saturation factor

3 component mix - 2 set point i.e., Lime saturation factor and Silica ratio4 component mix - 3 set point i.e., Lime saturation factor, Silica ratio, Alumina

ratio

Normal liquid phase - 22 - 28%If liquid phase is less, then cement is harder to burn, and produces SRC(Sulphateresistant cement). Various formulae for different ratios are indicated. If TiO2 and Mn2O3are separately determined, then TiO2 should be included in Al2O3 and Mn2O3 should beincluded in Fe2O3 for calculating the ratios.


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Reactivity of Coal as a Function of Fineness2_4_fuel_6n

Calculation of Flame Temperature2_4_fuel_5n

Calculation of Products of combustion2_4_fuel_4n

Calculation of Combustion air2_4_fuel_3n

Calculation of Coal Analysis and Heat Values of Coal2_4_fuel_2

Calculation of Capacities in Coal Preparation Section2_4_fuel_1


File Name : 2_4_fuel

Fuels and Fuel Systems

Book-II Chapter 4


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Book-II Chapter 4

Fuels and Fuel Systems

File Name : 2_4_fuels_2

Topic: Calculation of Coal Analysis and Heat Values Coal

Fuel Analysis and Calculation of Heat Values of Coal

This document covers fuel analysis and heat values calculation for solid fuel and liquidfuel namely, coal and fuel oil respectively.

If a fuel as received in the plant is analysed without any change in its state then it will be

termed as "Analysis (as received basis)".

Normally ultimate analysis of fuel on as received basis will be available from thelaboratory.

Solid fuel i.e., Coal, in many a case, is dried to a large extent before being fired into thefurnace. Dryers used are capable of drying the surface moisture of coal and the inherentmoisture remains in coal. This is because the temperature of coal is not allowed toexceed 80-100 0C (to avoid coking and preignition in the pipeline) and at thistemperature, with sufficient heat available only surface moisture can be dried where as todry inherent moisture, the coal should be heated to 108 0C. This is not done due toassociated 'Process problems'.

In any material's drying operation, inherent moisture cannot be dried at temperature

tics of Cement

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