Table of contents (This file) Chapter-2: Introduction MC_V0_B000_Intro_01TXT The nature of mathematics of cement MC_V0_B000_Intro_02TXT Mathematics of cement as against non-mathematical approach MC_V0_B000_Intro_03TXT Theoretical deductions versus empirical deductions. Chapter-3: Mathematical Models MC_V0_B000_Models_01TXT An introduction to mathematical model (MathCement model) MC_V0_B000_Models_02TXT Constituents of a mathematical model (MathCement model) MC_V0_B000_Models_03TXT Why create a mathematical model (MathCement model)? MathCement Preliminaries MC_V0_B000_TOC Topic: New table of contents including expanded tables of contents Mathcad Version:MC2001 Approach to MathCement Chapter-1: Preliminaries MC_V0_B000_Foreword_01TXT Foreword by Dr. A.K.Chatterjee MC_V0_B000_Preface_01TXT Preface - all editions MC_V0_B000_Ebook_01TXT About using electronic books MC_V0_B000_Readme_01TXT Read me MC_V0_B000_TOC_01TXT
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Table of contents (This file)
Chapter-2: Introduction
MC_V0_B000_Intro_01TXT The nature of mathematics of cement
MC_V0_B000_Intro_02TXT Mathematics of cement as against non-mathematical approach
MC_V0_B000_Intro_03TXT Theoretical deductions versus empirical deductions.
Chapter-3: Mathematical Models
MC_V0_B000_Models_01TXT An introduction to mathematical model (MathCement model)
MC_V0_B000_Models_02TXT Constituents of a mathematical model (MathCement model)
MC_V0_B000_Models_03TXT Why create a mathematical model (MathCement model)?
MathCement
Preliminaries
MC_V0_B000_TOC
Topic: New table of contents including expanded tables of contents
Mathcad Version:MC2001
Approach to MathCement
Chapter-1: Preliminaries
MC_V0_B000_Foreword_01TXT Foreword by Dr. A.K.Chatterjee
MC_V0_B000_Preface_01TXT Preface - all editions
MC_V0_B000_Ebook_01TXT About using electronic books
BIU MC_V0_B102_Review_2BIU Density, specific weight, specific volume and specific gravity
Mathematical techniques
MC_V1_B202_Crushing_3BIU Calculation of power for existing crusher at different capacities -application of Rittinger's law
MC_V1_B202_Crushing_3 Calculation of power for existing crusher at different capacities -application of Rittinger's law
BIU MC_V1_B202_Crushing_4BIU Calculation of power for existing crusher but to be used for different / changed material - application of Rittinger's law
MC_V1_B202_Crushing_4 Calculation of power for existing crusher but to be used for different / changed material - application of Rittinger's law
BIU MC_V1_B202_Crushing_5BIU Calculation of throughput for Roll Crusher
MC_V1_B202_Crushing_5 Calculation of throughput for Roll Crusher
BIU MC_V1_B202_Crushing_6BIU Double Rotor Hammer or Impact Crusher Calculations
MC_V1_B202_Crushing_6 Double Rotor Hammer or Impact Crusher Calculations
BIU MC_V5_B201_Fans_6BIU Calculation of pressure loss and flow through compressed air pipe lines
MC_V5_B201_Fans_6 Calculation of pressure loss and flow through compressed air pipe lines
BIU MC_V5_B201_Fans_7BIU Calculation of pipe diameter for flow through compressed air pipe lines
MC_V5_B201_Fans_7 Calculation of pipe diameter for flow through compressed air pipe lines
Production and distribution of compressed air
BIU MC_V5_B201_Fans_8BIU Calculation of frequency at which compressor would come on load as function of system demand.
MC_V5_B201_Fans_8 Calculation of frequency at which compressor would come on load as function of system demand.
BIU MC_V5_B201_Fans_9BIU Calculation of consumption of compressed
BIU MC_V5_B201_Fans_2BIU Compressed air pressure in a cylinder to deliver a desired clamping force
MC_V5_B201_Fans_2 Compressed air pressure in a cylinder to deliver a desired clamping force
BIU MC_V5_B201_Fans_3BIU Calculation of theoretical power to compress air polytropically
MC_V5_B201_Fans_3 Calculation of theoretical power to compress air polytropically
BIU MC_V5_B201_Fans_4BIU Calculation of theoretical power for isothermal compression of air
MC_V5_B201_Fans_4 Calculation of theoretical power for isothermal compression of air
BIU MC_V5_B201_Fans_5BIU Calculation of inter-cooler pressure for minimum power for two stage compression of air
MC_V5_B201_Fans_5 Calculation of inter-cooler pressure for minimum power for two stage compression of air
Calculation to find pressure range to control number of starts for compressor for given capacity of receiver based on operational demand of compressed air
MC_V5_B201_Fans_14
Calculation to find pressure range to control number of starts for compressor for given capacity of receiver based on operational demand of compressed air
MC_V5_B201_Fans_14BIUBIU
Calculation of number of starts for compressor for given capacity of receiver based on operational demand of compressed air
MC_V5_B201_Fans_13
Calculation of number of starts for compressor for given capacity of receiver based on operational demand of compressed air
MC_V5_B201_Fans_13BIUBIU
Calculation of capacity of receiver based on operational demand of compressed air
MC_V5_B201_Fans_12
Calculation of capacity of receiver based on operational demand of compressed air
BIU MC_V5_B201_Fans_9BIU Calculation of consumption of compressedair in operating a cylinder
MC_V5_B201_Fans_9 Calculation of consumption of compressedair in operating a cylinder
BIU MC_V5_B201_Fans_10BIU Calculation of reduction in consumption of compressed air in operating a cylinder with low pressure supply for retract stroke
MC_V5_B201_Fans_10 Calculation of reduction in consumption of compressed air in operating a cylinder with low pressure supply for retract stroke
BIU MC_V5_B201_Fans_11BIU Calculation of consumption of compressed air in operating a cylinder including effect of pressurising pipeline between valve and cylinder.
MC_V5_B201_Fans_11 Calculation of consumption of compressed air in operating a cylinder including effect of pressurising pipeline between valve and cylinder.
BIU MC_V5_B201_Fans_12BIU
MC_V5_B201_Fans_18 Calculating of extend stroke time for a cylinder as a function of load to be moved considering cushioning effects.
Volume-22 : Business Mathematics
MC_V22_TOC Expanded Table of Contents - Vol.22
Book 200 Funds and assets creation
Book 201 MC_V22_B201_Interest_TOC Interest
BIU MC_V22_B201_Interest_1BIU Calculation of Effective Rate of Interest on an Investment when Nominal Rate of Interest is Known
MC_V22_B201_Interest_1 Calculation of Effective Rate of Interest on an Investment when Nominal Rate of Interest is Known
BIU MC_V5_B201_Fans_15BIU Estimation of leakage of compressed air , based on fall in pressure in ring main when the plant is shut down.
MC_V5_B201_Fans_15 Estimation of leakage of compressed air , based on fall in pressure in ring main when the plant is shut down.
BIU MC_V5_B201_Fans_16BIU Calculating compressed air cylinder bore to be able to develop a desired clamping force
MC_V5_B201_Fans_16 Calculating compressed air cylinder bore to be able to develop a desired clamping force
BIU MC_V5_B201_Fans_17BIU Calculating cylinder bore and compressed air consumption to move a load up an incline.
MC_V5_B201_Fans_17 Calculating cylinder bore and compressed air consumption to move a load up an incline.
BIU MC_V5_B201_Fans_18BIU Calculating of extend stroke time for a cylinder as a function of load to be moved considering cushioning effects.
Calculation of Rate of Nominal Interest per Year to Reach a Certain Future Value on a Fixed Deposit or Investment and Amount of Recurring Deposit Separately to Achieve a Desired Total Future Value.
MC_V22_B201_Interest_6
Calculation of Rate of Nominal Interest per Year to Reach a Certain Future Value on a Fixed Deposit or Investment and Amount of Recurring Deposit Separately to Achieve a Desired Total Future Value.
MC_V22_B201_Interest_6BIUBIU
Calculation of Rate of Nominal Interest per Year,to Reach a Certain Future Value on a Fixed Deposit or Investment and Recurring Deposit Separately or Together
MC_V22_B201_Interest_5
Calculation of Rate of Nominal Interest per Year,to Reach a Certain Future Value on a Fixed Deposit or Investment and Recurring Deposit Separately or Together
MC_V22_B201_Interest_5BIUBIU
Calculation of Future Value of a Fixed Deposit or Investment and Recurring Deposit Separately or Together
MC_V22_B201_Interest_4
Calculation of Future Value of a Fixed Deposit or Investment and Recurring Deposit Separately or Together
MC_V22_B201_Interest_4BIUBIU
Calculation of Rates of Interest on an Investment when Final Balance After the End of Deposit Period is Known
MC_V22_B201_Interest_3
Calculation of Rates of Interest on an Investment when Final Balance After the End of Deposit Period is Known
MC_V22_B201_Interest_3BIUBIU
Calculation of Nominal Rate of Interest on an Investment when Effective Rate of Interest is Known
MC_V22_B201_Interest_2
Calculation of Nominal Rate of Interest on an Investment when Effective Rate of Interest is Known
MC_V22_B201_Interest_2BIUBIU
BIU MC_V22_B202_Annuities_4BIU Calculation to find, present value of an annuity
MC_V22_B202_Annuities_4 Calculation to find, present value of an annuity
BIU MC_V22_B202_Annuities_5BIU Calculation to find, equated monthly instalment to repay a loan
MC_V22_B202_Annuities_5 Calculation to find, equated monthly instalment to repay a loan
Book 203 MC_V22_B203_Depreciation_TOC Depreciation
Text MC_V22_B203_Depreciation_1TXTIntroduction to Depreciation
Book 202 MC_V22_B202_Annuities_TOC Annuities
BIU MC_V22_B202_Annuities_1BIU Calculation of Payments for a Sinking Fund
MC_V22_B202_Annuities_1 Calculation of Payments for a Sinking Fund
BIU MC_V22_B202_Annuities_2BIU Calculation of Time Required for a Target Amount in Sinking Fund
MC_V22_B202_Annuities_2 Calculation of Time Required for a Target Amount in Sinking Fund
BIU MC_V22_B202_Annuities_3BIU Calculation to find, how much money will be in the Sinking Fund after a certain period.
MC_V22_B202_Annuities_3 Calculation to find, how much money will be in the Sinking Fund after a certain period.
Volume-23 : New Calculations
MC_V23_TOC Expanded Table of Contents - Vol.23
Book 200 Miscellaneous
Book 201 MC_V23_B201_Hopper_TOC Hopper design
BIU MC_V23_B201_Hopper_1BIU Calculation of valley angle of a hopper with square and rectangular cross section, given the wall angles and vertical height of the pyramidal part.
MC_V23_B201_Hopper_1 Calculation of valley angle of a hopper with square and rectangular cross section, given the wall angles and vertical height of the pyramidal part.
BIU MC_V23_B201_Hopper_2BIU Calculation of wall angles and valley angles of a hopper with square and rectangular cross section, given the opening dimensions and height of the pyramidal part.
BIU MC_V22_B203_Depreciation_2BIU Depreciation Calculation of Assets by Straight-line Method
MC_V22_B203_Depreciation_2 Depreciation Calculation of Assets by Straight-line Method
BIU MC_V22_B203_Depreciation_3BIU Depreciation Calculation of Assets by Reducing Balance Method
BIU MC_V22_B203_Depreciation_4BIU Depreciation Calculation of Assets by Accelerated Reducing Balance Method
BIU MC_V22_B203_Depreciation_5BIU Depreciation Calculation of Assets by Sum-of-the-Years' -Digits Method
BIU MC_V22_B203_Depreciation_6BIU Depreciation Calculation by Units of Production Method
Calculation to determine number of refractory bricks required for a kiln section
MC_V23_B202_Refra_1
Calculation to determine number of refractory bricks required for a kiln section
MC_V23_B202_Refra_1BIUBIU
Refractory designMC_V23_B202_Refra_TOCBook 202
Calculation of maximum filling height of a silo so that the lateral pressure does not exceed given safe value
MC_V23_B201_Hopper_4
Calculation of maximum filling height of a silo so that the lateral pressure does not exceed given safe value
MC_V23_B201_Hopper_4BIU
Calculation of lateral pressure on silo wall due to stored bulk material as function of filling height and checking the possibility of increasing capacity by increase of height.
MC_V23_B201_Hopper_3
Calculation of lateral pressure on silo wall due to stored bulk material as function of filling height and checking the possibility of increasing capacity by increase of height.
MC_V23_B201_Hopper_3BIUBIU
Calculation of wall angles and valley angles of a hopper with square and rectangular cross section, given the opening dimensions and height of the pyramidal part.
BIU MC_V23_B205_Ducts_1BIU Load analysis of a multi-support saddle mounted tertiary air duct
MC_V23_B205_Ducts_1 Load analysis of a multi-support saddle mounted tertiary air duct
Book 203 MC_V23_B203_Kiln_TOC Kiln
BIU MC_V23_B203_Kiln_1BIU Calculation to determine the gap between tyre and supporting pads
MC_V23_B203_Kiln_1 Calculation to determine the gap between tyre and supporting pads
BIU MC_V23_B203_Kiln_2BIU Calculation of heat of reaction of clinker based on laboratory analysis -based on VDZ procedure.
MC_V23_B203_Kiln_2 Calculation of heat of reaction of clinker based on laboratory analysis -based on VDZ procedure.
Book 204 MC_V23_B204_Fuels_TOC Fuels & combustion
BIU
MC_V23_B207_Preblend_1BIU Calculation of longitudinal and linear blending beds .
MC_V23_B207_Preblend_1 Calculation of longitudinal and linear blending beds .
Book 208 MC_V23_B208_Matrl_TOC Materials properties
BIU MC_V23_B208_Matrl_1BIU Calculation of specific heats of various materials as a function of temperature
MC_V23_B208_Matrl_1 Calculation of specific heats of various materials as a function of temperature
BIU MC_V23_B208_Matrl_2BIU Calculation of specific heats of various gases as a function of temperature
MC_V23_B208_Matrl_2 Calculation of specific heats of various gases as a function of temperature
Book 206 MC_V23_B206_Dedust_TOC Dedusting systems
BIU MC_V23_B206_dedust_1BIU Calculation to check suitability of existing Glass Bag House after proposed upgradation of kiln.
MC_V23_B206_Dedust_1 Calculation to check suitability of existing Glass Bag House after proposed upgradation of kiln.
BIU MC_V23_B206_dedust_2BIU Calculate total pressure a fan must develop to move air in a dedusting duct fitted with suction hood, filter and nozzle at discharge end .
MC_V23_B206_dedust_2 Calculate total pressure a fan must develop to move air in a dedusting duct fitted with suction hood, filter and nozzle at discharge end .
Book 207 MC_V23_B207_Preblend_TOC Preblending systems
BIU
BIU V0_A001_TablesCharts_8BIU Calculation of Specific Heats of Gases and Materials as a Function of Temperature
BIU MC_V0_A003_Mensuration_VS_3BIU Right Circular Cylinder
BIU MC_V0_A003_Mensuration_VS_4BIU Pyramid
BIU MC_V0_A003_Mensuration_VS_5BIU Right Circular Cone
BIU
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MathCement
MathCement
Preliminaries
MC_V0_B000_Foreword_01TXT
Topic: Foreword by Dr. A.K.Chatterjee
About this topic
Foreword
It was about 40 years back that I had the first opportunities of peeping into cement plants and tasting the flavours of cement technology. Within the first few years of my professional exposure, I had realized how complex and dynamic the cement processes were. I understood that two kilns or two mills of the same design and dimension might not behave in identical manners even in comparable situations. A quantitative and numerical understanding of the processes involved was the only way to succeed in achieving the targeted results.
Those days, I had the "Cement Engineers Handbook " by Otto Labahn as my bible for all numerical appreciation of the unit operations. Although the book provided the procedures of calculations for a range of problems, I still had to maintain my own diary to apply the procedures to the actual situation.
In the 70s, apart from the revised editions of the Hand Book by Otto Labahn, I had the added benefits of the "Cement-Data-Book" by Walter H.Duda, Cement Manufacturers Hand Book by Kurt E. Peray, etc. While Duda opened up a comprehensive coverage of numerical data, diagrams, tables and description of processes and machinery, Peray provided a host of engineering formulae that represented the basic tools for gaining a better understanding of the cement engineering and technology. Notwithstanding the availability of these publications, the cement plant engineers and chemists had to undertake the laborious tasks of making their own calculations with specific input of data. The books provided only the guidelines, procedures and approaches.
Quite contrary to the past trends, "Mathcement" is a serious endeavour to rescue the cement plant operational personnel from the above ennui. The modern computer facility and accompanying software have provided new opportunities of easy computing. M/s Softideas Pvt.Ltd. have relied on all the past hand books, or at least the majority of them, and superimposed "Mathcad" on them. The authors of "Mathcement" have tried to identify the specific computational requirements of the practicing cement engineers and more particularly of the personnel operating in the cement plants.
Hence,"Mathcement" does not suffer from the strains of exhaustiveness in coverage. On the contrary, it provides a sharp focus on practical and day-to-day needs of practicing professionals. The entire e-book has been designed in 4 parts; the first one covers essentially the unit operations of a plant starting from the limestone quarry to the cement dispatch as well as the essential requirements of quality checks and raw materials.
Page 1 of 2 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement
The second part of the book covers all the auxiliaries such as de-dusting, conveyors, fans and blowers, water, insulation, laboratory and plant site conditions. The third part of the e-book takes care of the limited number of physical, chemical and thermo-chemical tables that are essentially required for engineering calculations. The last, but not the least, is the fourth part of the e-book that takes care of refreshing the readers about some of the essential basics of cement chemistry.
On the whole, from the e-book "Mathcement", any practicing cement professional will find a way to solve his or her problem and would obtain the results by the touch of the keyboard only by putting in the input values specific to his or her requirements. I did not realize earlier that computations could be so pleasurable, so easy, so rapid and yet, so precise and useful.
I am sure, all the concerned professionals would derive immense benefit from this innovative e- book named "Mathcement".
Page 2 of 2 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement
MathCement
Introduction
MC_V0_B000_Intro_01TXT
Topic: The nature of mathematics of cement
About this topic
Mathematics of cement is not a distinct branch of cement technology in a way we understand in our day-to-day interaction.It is rather an approach to technical analysis in which scientists; engineers and technologists make use of mathematical symbols in the statement of a problem and then draws upon known mathematical theorems to support reasoning.The approach is thus applicable to specific subjects dealt in cement technology like crushing, grinding, pyro-processing and so on.
Every approach to analysis of a problem associated with design, operation or optimisation in relation to cement is necessarily exemplified by mathematics.We will conventionally use the term mathematics of cement to cover the entire spectrum of describing problems and solution techniques using mathematical techniques.The mathematical techniques that are used can stretch from simple geometry to matrix algebra, differential equations, programming etc.It is our intention to introduce various fundamental and advanced mathematical techniques to solve day-to-day technical and operational problems that are faced by cement technologists or scientists.
MathCement is a registered trade of Softideas Pvt. Ltd., trade name for the topics dealt under the realm of mathematics of cement.
References
Fri Sep 23 10:38:09 2005
D:\MathCement_demo_pdf\MC_V0_B000_Intro_01TXT.mcd
Page 1 of 1 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement
MathCement
Introduction
MC_V0_B000_Intro_02TXT
Topic: Mathematics of cement as against non-mathematical approach
About this topic
Mathematics of cement is basically an approach to analysis of cement technology and related problems.Therefore, it does not fundamentally differ from any non mathematical approach to such analysis.It should be noted that the fundamental purpose of any theoretical analysis, regardless of its approach, is to arrive at some conclusions from a given set of assumptions or data via a process of reasoning.The major difference between "mathematical approach" and "non-mathematical or literary approach" would lie in the fact that in the case of the former, the assumptions and conclusions are stated in mathematical symbols rather than in words and in mathematical equations rather than in explanatory text.Also, in case of mathematical approach, we use mathematical theorem,which exist in abundance to draw upon for reasoning in place of literary logic. Truly speaking, symbols and words are equivalent.You can see this in the fact that symbols are mostly expressed in words.It should, therefore, hardly matter which one is chosen over the other.But we all know that symbols are more convenient in deductive reasoning and hence they form the basic building blocks of mathematics.
Mathematics has the basic advantage of leading from explicit assumptions and precise reasoning at every stage of solution.Mathematical theorems normally have the "if-then" form such that to tap the "then" part, which is the result part of the theorem,one has to make sure that the "if " part, that is the condition part, conforms to the assumptions made.
Many solutions are well depicted by geometrical modeling.However, geometrical solutions have their limitations arising out of dimensional limitation.For instance if the number of variables are three, the graphical treatment of the solution would call for 3-d graph which is difficult to draw . However, if the number of variables happens to be four or more,making such higher dimensional graph becomes practically impossible.Thus mathematical equations are better suited to handle problems involving many variables.
We can, thus, summarise the advantage we get by adopting mathematical approach against literary approach as follows:1) Mathematical language of statement are more concise and precise2) Mathematics provide us with many techniques and theorems to adopt readily3) It ensures that all assumptions are made explicit4) It enables treatment of multi-variable situations (n-variables)
Questions are sometimes raised about how realistic mathematical solutions are?. We must remember that any theoretical approach to an analysis , mathematical or otherwise,falls within the domain of reality. Thus ,all such treatments are only extracts from real world.It is always unrealistic to the extent that it fails to take into consideration some of the complex real life influences.If we could cover real life situations in its entirety, we could then possibly play God.
D:\MathCement_demo_pdf\MC_V0_B000_Intro_02TXT.mcd
Page 1 of 2 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement
Finally, we may like to compare mathematical approach to a "mode of transportation" to be used in our journey from a start point to destination point at a fast speed.Assuming that you would like to go from home to a park which is situated 3 km away, you you normally pick your car as your mode of transportation so that you can reach your destination fast. You may also choose to walk the distance to make it a part of your aerobic excercise.Thus, to be able to reach your conclusions faster, as a theorist, you would prefer to ride the vehicle of ' appropriate mathematical techniques". It,however, goes without saying that you have to learn to drive this vehicle and some one has to build the vehicle.The skill you will acquire will serve you in good stead for a long time come and so it should be worth its while to learn them.We have, on our part, tried to build a good and friendly vehicle called "MathCement" covering mathematical approach to solving cement problems in technology and operations.
MathCement is a registered trade-mark of Softideas Pvt. Ltd., trade name for the topics dealt under the realm of mathematics of cement.
Fri Sep 23 12:59:07 PM 2005
D:\MathCement_demo_pdf\MC_V0_B000_Intro_02TXT.mcd
Page 2 of 2 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement
MathCement
Mathematical Models
MC_V0_B000_Models_03TXT
Topic: Why create a mathematical model (MathCement model)?
About this topicWe live our life in the real world in all its glory.It is the nature that we observe with awe and wonder at the marvel of creation.As human beings our thinking brains have always been inspired to continuously try and convert natural phenomena to the best of our advantage.For example,we discovered electricity by observing lightning and made it work for us in thousands of different ways.
There are people in this world for whom the existing solutions are not acceptable.So they look for newer solutions by first rejecting the existing solution.As the next step they create an approximation of the real life situation or object which can be experimented with.This representative of real life situation or object is a model.A model,thus, tries to replicate a real life situation or object as closely as possible.This model greatly facilitates experiment since it is much simpler to handle and manipulate without the danger of damaging or hurting the original real life situation.Thus scientists can make changes in the model or subject the model to different conditions and evaluate its effect .They then project, how it would behave when real life situation is subjected to similar changed conditions.As a next step, such findings need to be approved by the appropriate authority.Then, after the approval,they can finally introduce the changes in real life and see how best the predictions are met.
Thus, through a model, we see the future behaviour of a current real life situation under controlled conditions. This helps us to take a decision whether to adopt the new solution or not.
When the representative model is built with mathematical symbols and variables are related to each other by equations to closely represent the assumptions and constraints and applied with appropriate mathematical theorem leading to conclusions, we have created a mathematical model.
Now by varying the values of the variables we can foresee the resultant predictive behaviour of the model.Computer helps us further to arrive at the conclusions very fast. We can thus analyse the performance of the real life situations very quickly without disturbing the original object.
MathCement is a registered trade of Softideas Pvt. Ltd., trade name for the topics dealt under the realm of mathematics of cement.
Page 3 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
Top
Select dumper capacity to fall between Wmax and Wds
as calculated based on turn around timeWds 37.5 tonne=
Wmax 200 tonne=
WmaxHv
2BDLS⋅:=
Hopper should hold at least two dumper loads. So check for maximum •capacity of each dumper.
Hv 266.67 m3
=Hopper capacity feeding the crusher
Hv 266.67 m3
=
Hv if Hv1 Hv2≥ Hv1, Hv2,( ):=
Hv Selected hopper capacity - the higher value between Hv1 and Hv2
Hv2 266.67 m3
=
Hv2Qcr Tdtr⋅
BDLS:=
Hv2Hopper capacity -( volumetric) based on minimum turnaround time for dumper - that means the hopper should hold material for the time taken by a dumper
Hv1 200 m3
=
Hv1Qcr Ht⋅
BDLS:=
Hv1Hopper capacity -( volumetric) based on desired holding time
Now calculate the capacity of hopper feeding the crusher•
Page 6 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
DpbnNew product size of rock -volume-surface mean dia.
mnNew feed rate to be tried
PPower required by crusher for current output
mCurrent mass feed rate to the crusher
DpbProduct size of rock -volume-surface mean dia.
DpaFeed size of rock -volume-surface mean dia.
List of parameters used
User defined units Top
An existing crusher which crushes rocks to reduce volume-surface mean diameter of feed to a smaller dia. product.These data being known from the operations log of the crusher, also the corresponding output and power requirement figures are known.Find the new power requirement for a new capacity at a different product dia.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This is worksheet can be used to supplement calculation of power required to crush material in an existing crusher at different outputs by using Rittinger's law.
About this topic
Topic: Calculation of power for existing crusher at different capacities -application of Rittinger's law
Page 3 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
Roll crusher
Top
A roll crusher of given diameters of rolls are set at a certain distance apart to achieve a certain nip angle.Find the maximum size of particle that can be fed.Also , knowing the width of the working face of the rolls, and density of feed material , find the throughput when the crusher runs at a certain speed. Consider real throughput as a percentage of theoretical throughput.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This is worksheet can be used to calculate the throughput in a Roll Crusher.
Page 5 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Roll crusher
Top
A roll crusher of given diameters of rolls are set at a certain distance apart to achieve a certain nip angle.Find the maximum size of particle that can be fed.Also , knowing the width of the working face of the rolls, and density of feed material , find the throughput when the crusher runs at a certain speed. Consider real throughput as a percentage of theoretical throughput.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This is worksheet can be used to calculate the throughput in a Roll Crusher.
About this topic
Topic: Calculation of throughput for Roll Crusher
MC_V1_B202_Crushing_5
Crushing section
MathCement 2000
D:\MathCement_demo_pdf\MC_V1_B202_Crushing_5.mcd
Page 1 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
R
ωAngular speed of the rolls
Linear velocity of rolls v
Actual feed rate (mass flow) Mva
Input datar 500:= mmRadius of rolls of the crusherwr 0.4:= mWidth of roll face
Distance between the roll surfaces S 12.5:= mm
Angle of nip θ 31:= deg
Actual throughput capacity as percent of theoretical throughput
q 12:= %
User defined units
Hz1
s:=
List of parameters used
rRadius of rolls of the crusher
wrWidth of roll face
Distance between the roll surfaces S
Angle of nip θ
Actual throughput capacity as percent of theoretical throughput
q
Roll running frequency f
ρFeed material density
Theoretical feed rate (volumetric) Fvth
Actual feed rate (volumetric) Fva
Half the distance between the roll surfaces d
Radius of maximum size of feed particle
D:\MathCement_demo_pdf\MC_V1_B202_Crushing_5.mcd
Page 2 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
therefore :
Rr d+
cos α( ) r−:=
R 25.36= mm
Radius of maximum size of feed particle
R 25.36= mm
Now calculate cross sectional area of •material flow through the rolls
As
As S 103
⋅ wr⋅:=
As 5000= mm2
Cross sectional area of material flow through the rolls As 5000= mm
2
Angular speed of the rolls ω
f 2:= HzRoll running frequency
Feed material density ρ 2500:= kg/m3
Top
Calculation algorithm
Find the radius of maximum size of feed •particle
R
Half the distance between the roll surfaces d
dS
2:= d 6.25= mm
Radius of maximum size of feed particle R
αθ
2
π
180⋅:=
cos α( ) r d+
r R+=
D:\MathCement_demo_pdf\MC_V1_B202_Crushing_5.mcd
Page 3 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Page 4 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Feed material density
Derived data
Radius of maximum size of feed particle R 25.36= mm
ω 12.57= rad/sAngular speed of the rolls
Linear velocity of rolls v 6.28= m/s
Actual feed rate (mass flow) Mva 9.42= kg/s
Top
Observation
References 1)Chapter on "Size Reduction" - Unit Operation of Chemical Engineering2)Hallmark Technologies, Pune
Fri Sep 23 1:45:57 PM 2005
Results Given data
r 500= mmRadius of rolls of the crusherwr 0.4= mWidth of roll face
Distance between the roll surfaces S 12.5= mm
Angle of nip θ 31= deg
Actual throughput capacity as percent of theoretical throughput
q 12= %
Roll running frequency f 2= Hz
ρ 2500= kg/m3
D:\MathCement_demo_pdf\MC_V1_B202_Crushing_5.mcd
Page 5 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
G1Gas flow rate at point 1 =G1
Let us calculate leakage air between two ponts" 1 and 2 "in a system under suction.The gas flows from 1 to 2.
List of parameters used
degC 1:=Metric_ton MT:=
MT tonne:=
kwh 1kW hr⋅:=short_ton ton:=
µ 106−m:=
microns1
1000mm:=
User defined units
In a raw grinding mill measurement is carried out at inlet (point 1) and outlet (point 2).Gas flow rate at inlet is measured and gas analysis is done to find oxygen and carbon dioxide content at both points.Based on the given data find the rate of false air entry into the mill between inlet and outlet.
Statement of problem
False air entry into grinding mill system affects drying operation as well as power consumption.Excessive false air lowers the temperature of hot gases which affects heat transfer efficiency and also lowers the heat available for drying of raw material.This is because heat is required to go into the false air to raise its temperature to mill ext gas temperature.As also the mill fan has to handle higher quantity of gases, the power consumption goes up.As more gas volume has to pass through the mill system which results in additional pressure loss , also adds to additional power consumption for the mill fan. Controlling of false air to reasonable value is, therefore, very important.
Use this worksheet to calculate possible false air entry raw grinding mill
About this topic
Topic: Calculation of false air coming into grinding mill system
Page 2 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
Carbon dioxide measured at point 2 =CO2st2
CO2st2 17.2%:=
Oxygen in air =O2air O2air 21%:=
Carbon dioxide in air =CO2air CO2air 0%:=
Calculation algorithm
Oxygen and Carbon dioxide in gases at various points in the system can be measured .We consider, for simplification, that air has 21% Oxygen and 0 % Carbon dioxide by volume. We can then calculate for difference in values of Oxygen and Carbo dioxide at different points and evaluate the percentage of false air with ref. to unmixed gases.For ease of understanding, we call the mixed gases as the mixture of gas and false air and gas as unmixed gas (without false air).Percentage of false air will be referred to as percentage of gas (unmixed gas) at first point (mill entry point for example).
As false air enters it dilutes the hot flue gas supplied to mill for drying purposes.So Oxygen level goes up and carbon dioxide level goes down.
False air entry between point 1 and point 2 as percentage of gas volume at point 1 is given by FAO!_2 based on Oxygen measurement
FAO1_2O2st2 O2st1−
O2air O2st2−:=
FAO1_2 13.04 %=
False air entry between point 1 and point 2 as percentage of gas volume at point 1 is given by FACO21_2 based on Carbon dioxide measurement
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
nNumber of measurement =n
G1Gas flow rate at point 1 =G1
Let us calculate leakage air between two ponts" 1 and 2 "in a system under suction.The gas flows from 1 to 2.
List of parameters used
User defined units
In a raw grinding mill measurement is carried out at inlet (point 1) and outlet (point 2).Gas flow rate at inlet is measured and gas analysis is done to find oxygen and carbon dioxide content at both points.Based on the given data find the rate of false air entry into the mill between inlet and outlet.
Statement of problem
False air entry into grinding mill system affects drying operation as well as power consumption.Excessive false air lowers the temperature of hot gases which affects heat transfer efficiency and also lowers the heat available for drying of raw material.This is because heat is required to go into the false air to raise its temperature to mill ext gas temperature.As also the mill fan has to handle higher quantity of gases, the power consumption goes up.As more gas volume has to pass through the mill system which results in additional pressure loss , also adds to additional power consumption for the mill fan. Controlling of false air to reasonable value is, therefore, very important.
Use this worksheet to calculate possible false air entry raw grinding mill
About this topic
Topic: Calculation of false air coming into grinding mill system
MC_V1_B204_Rawmill_16
Raw Mill Section
MathCement 2000
D:\MathCement_demo_pdf\MC_V1_B204_Rawmill_16.mcd
Page 1 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
Important: Please enter values as "0" if measument is not done Oxygen measured at point 1
=O2st1
O2st1 8.2:= %
Oxygen measured at point 2 =O2st2
O2st2 10:= %
Carbon dioxide measured at point 1 =CO2st1
CO2st1 19.7:= %
Carbon dioxide measured at point 2 =CO2st2
CO2st2 17.2:= %
Oxygen in air =O2air O2air 21:= %
Carbon dioxide in air =CO2air CO2air 0:= %
Oxygen measured at point 1 =O2st1 O2st1
Oxygen measured at point 2 =O2st2 O2st2
Carbon dioxide measured at point 1 =CO2st1
CO2st1
Carbon dioxide measured at point 2 =CO2st2
CO2st2
Oxygen in air =O2air O2air
Carbon dioxide in air =CO2air CO2air
Input data
Let us calculate leakage air between two ponts" 1 and 2 "in a system under suction.The gas flows from 1 to 2.
Gas flow rate at point 1 =G1 G1 30:= m3/s
Number of measurement =nn=2 if measurement is done for both Oxygen and carbon dioxiden=1 if measurement is done for only Oxygen or Carbon dioxide
Atleast one set of values must be available
n 1:=
D:\MathCement_demo_pdf\MC_V1_B204_Rawmill_16.mcd
Page 2 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
m3/sG2 39.27=
G2 G1FA1_2
100G1⋅+:=
Gas flow rate at point 2 including false air G2
%FA1_2 30.9=
FA1_2FAO1_2 FACO21_2+
n:=
Average of measurement FA1_2
%FACO21_2 14.53=
FACO21_2CO2st1 CO2st2−
CO2st2 CO2air−100⋅:=
False air entry between point 1 and point 2 as percentage of gas volume at point 1 is given by FACO21_2 based on Carbo dioxide measurement
%FAO1_2 16.36=
FAO1_2O2st2 O2st1−
O2air O2st2−100⋅:=
False air entry between point 1 and point 2 as percentage of gas volume at point 1 is given by FAO!_2 based on Oxygen measurement
As false air enters it dilutes the hot flue gas supplied to mill for drying purposes.So Oxygen level goes up and carbon dioxide level goes down.
Oxygen and Carbon dioxide in gases at various points in the system can be measured .We consider, for simplification, that air has 21% Oxygen and 0 % Carbon dioxide by volume. We can then calculate for difference in values of Oxygen and Carbo dioxide at different points and evaluate the percentage of false air with ref. to unmixed gases.For ease of understanding, we call the mixed gases as the mixture of gas and false air and gas as unmixed gas (without false air).Percentage of false air will be referred to as percentage of gas (unmixed gas) at first point (mill entry point for example).
Calculation algorithm
D:\MathCement_demo_pdf\MC_V1_B204_Rawmill_16.mcd
Page 3 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
Quantity of false air entry between point 1 and 2 =FA
FA G2 G1−:=
FA 9.27= m3/s
Results
Quantity of false air entry between point 1 and 2 =FA FA 9.27= m3/s
Observation
References
Fri Sep 23 2:02:24 PM 2005
D:\MathCement_demo_pdf\MC_V1_B204_Rawmill_16.mcd
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
CO2RM CO2RM CO2 content in Raw Meal
CO2CMCO2CM CO2 content in Partially Calcined Meal
MgCO3RMMgCO3RM Content of MgCO3 in 1 kg. Raw Meal
CaCO3RMCaCO3RM Content of CaCO3 in 1 kg. Raw Meal
DRecirculating dust = D kg/kg cl
List of parameters used
kgraw 1kg:=
kgcl 1kg:=
degC 1:=
User defined units
For given data of a kiln calculate apparent degree of calcination and actual degree of calcination
Statement of problem
Use this worksheet to calculate calculate degree of decarbonation of raw meal.
About this topic
Topic: Degree of Decarbonation of Raw Meal -Definition and calculation
Page 2 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
CO2RM
CD 88.22%=
A more refined definition can be as stated below.
CDR
1
1 CO2RM−
1
1 CO2CM−−
1
1 CO2RM−1−
:=
CDR 91.9 %=
By this method the value of CD is reported higher
Apparent degree of calcination CD:
Degree of calcination is worked out based on the ratio between loss of CO2 content from partially calcined raw Meal and total CO2 content in raw meal .It implies ,therefore, if the lower the value of balance CO2 content higher is the degree of calcination.
CO2 content in Partially Calcined Meal ,CO2CM as measured from samples collected from kiln inlet, is influenced by circulating dust at kiln inlet. This dust is practically 100 percent calcined and falsely reduces the value of CO2CM. This result shows higher degree of calcination as is termed as apparent degree of calcination CD .
To find the nett or actual degree of calcination without the influence of circulating dust, we employ correction factor:
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement 2000
Dedusting Systems
MC_V1_B302_Dedust_23
Topic: Sketches for K- Factors - set 3
About this topic
hv = velocity head
9
θ
K
ANGLE CIRCULAR RECTANGULAR
θ
10
20
30
45
60
CIRCULAR_K
0.025
0.102
0.218
0.435
0.652
RECTANGULAR_K
0.37
0.146
0.310
0.625
0.940
D:\MathCement_demo_pdf\MC_V1_B302_Dedust_23.mcd
Page 1 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
10
θ
BASED ON VELOCITY IN BRANCH ANGLE K VALUE
K
0.06
0.12
0.18
0.22
0.44
θ
10
20
30
45
60
11 PLATE/FLANGE
A B
A BCIRCULAR K = 0.87 0.48RECTANGULAR K = 1.25 0.70
D:\MathCement_demo_pdf\MC_V1_B302_Dedust_23.mcd
Page 2 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
12
DD
0.7D0.7D
K - REFERED TO VELOCITY AT D
WIRE GUARD 5 CM MESH K = 0.25
CONE WITHOUT GUARD CIRCULAR K RECTANGULAR K
θ
10
20
30
45
60
CIRCULAR_K
0.42
0.30
0.24
0.20
0.29
RECTANGULAR_K
0.53
0.38
0.31
0.29
0.39
Observation
References
Fri Sep 23 3:48:53 PM 2005
D:\MathCement_demo_pdf\MC_V1_B302_Dedust_23.mcd
Page 3 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
C4 Ca
SiO2 S Sm S1 S2 S3 S4 Sa
Al2O3 A Am A1 A2 A3 A4 Aa
Fe2O3 F Fm F1 F2 F3 F4 Fa
Statement of problem
In this example we determine the proportion of two raw material components only. LSF is fixed as set point.Also to be calculated are:Composition of raw mix and also the composition of clinker as loss free basis
User defined units
MathCement 2000
Laboratory Investigation and Raw Mix Design SystemsMC_V1_B303_Lab_25BIU
Topic: Raw-mix Design based on Lime Saturation Factor
About this topic
Use this worksheet to design a two component raw mix to attain a desired lime saturation factor in the raw mix
Raw mix design is a process of determining the quantitative proportions of the components of Raw mix ensuring that the clinker produced from such mix attain desired chemical and mineralogical composition.
This method is applicable to two raw material components, with the lime saturation factor selected for the clinker. To simplify the following calculations, symbols are used for the designation of the clinker components, the raw materials, and the coal ash; these symbols are placed in table below.
Calculation symbols for designation of clinker and raw material components
Compounds Clinker Raw mix Raw mat. Raw mat. Raw mat.Raw mat.CoalNo. 1 No. 2 No. 3 No. 4 ash
CaO C Cm C1 C2 C3
D:\MathCement_demo_pdf\MC_V1_B303_Lab_25BIU.mcd
Page 1 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Input data
Two raw materials are given with the following composition, (Raw material No. 1 is Limestone, Raw material No. 2 is Marl)
Limestone Clay
SiO2 S1 1.42%:= S2 62.95%:=
Al2O3 A1 0.48%:= A2 18.98%:=
Fe2O3 F1 0.38%:= F2 7.37%:=
CaO C1 52.60%:= C2 1.40%:=
MgO M1 1.11%:= M2 0.98%:=
SO3 So1 0.85%:= So2 0.85%:=
LOI LOI1 43.16%:= LOI2 7.47%:=
Total1 S1 A1+ F1+ C1+ M1+ So1+ LOI1+:=
Total1 100 %=
Total2 S2 A2+ F2+ C2+ M2+ So2+ LOI2+:=
Total2 100 %=
Lime saturation factor desired in raw mix LSF 0.92:=
List of parameters used
Two raw materials are given with the following composition, (Raw material No. 1 is Limestone, Raw material No. 2 is Marl)
Constituent Matrl. 1 Matrl2
SiO2 S1 S2
Al2O3 A1 A2
Fe2O3 F1 F2
CaO C1 C2
MgO M1 M2
SO3 So1 So2
LOI LOI1 LOI2
Rest Balance R1 R2
Lime saturation factor desired in raw mix LSF
D:\MathCement_demo_pdf\MC_V1_B303_Lab_25BIU.mcd
Page 2 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
com2 0.2=com21
x 1+:=
com1 0.8=com1x
x 1+:=
com2parts of material no.2 in raw mix
com1parts of material no.1in raw mix
Where x represents parts of limestone or material no.1 apportioned to one part of clay or material no. 2
x 4.05=
x2.8 LSF⋅ S2⋅ 1.65 A2⋅+ 0.35 F2⋅+( ) C2−
C1 2.8 LSF⋅ S1⋅ 1.65 A1⋅+ 0.35 F1⋅+( )−:=
LSF 0.92=Set value of lime saturation factor
With this formula we calculate how many parts of limestone or material no.1 in the raw mix are apportioned to one part of clay or the material no. 2. Accordingly we get:
x2.8 LSF⋅ S2⋅ 1.65 A2⋅+ 0.35 F2⋅+( ) C2−
C1 2.8 LSF⋅ S1⋅ 1.65 A1⋅+ 0.35 F1⋅+( )−=
and solving for x:
LSF
xC1 C2+
x 1+⋅ 1.65
x A1⋅ A2+
x 1+⋅ 0.35
x F1⋅ F2+
x 1+⋅+
−
2.8 xS1 S2+
x 1+⋅
⋅
=
Inserting into Kind's formula the calculation symbols used previously, we get
Calculation algorithm
D:\MathCement_demo_pdf\MC_V1_B303_Lab_25BIU.mcd
Page 3 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
We now determine analysis of material no. 1and 2 in the raw mix
Limestone x com.1 Clay x com.2
S1c com1 S1⋅:= S1c 0.01= S2c com2 S2⋅:= S2c 0.12=
A1c com1 A1⋅:= A1c 0= A2c com2 A2⋅:= A2c 0.04=
F1c com1 F1⋅:= F1c 0= F2c com2 F2⋅:= F2c 0.01=
C1c com1 C1⋅:= C1c 0.42= C2c com2 C2⋅:= C2c 0=
M1c com1 M1⋅:= M1c 0.01= M2c com2 M2⋅:= M2c 0=
So1c com1 So1⋅:= So1c 0.01= So2c com2 So2⋅:=
D:\MathCement_demo_pdf\MC_V1_B303_Lab_25BIU.mcd
Page 4 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Fri Sep 23 4:05:55 PM 2005
References
Observation
com2 19.79 %=
com2 0.2=marl or component 2
com1 80.21 %=limestone or component 1 com1 0.8=
Results
LSF 0.92=
LSFC 1.65 A⋅ 0.35 F⋅+( )−
2.8 S⋅:=
The resulting Kind's lime saturation factor is
Total 1=Total S A+ F+ C+ M+ So+ LOI+:=
Now calculating the clinker composition based on raw mix, on loss free basis
f1
1 LOIm−:= f 1.56=
Clinker
S f Sm⋅:= S 0.21=
A f Am⋅:= A 0.06=
F f Fm⋅:= F 0.03=
C f Cm⋅:= C 0.66=
M f Mm⋅:= M 0.02=
So f Som⋅:= So 0.01=
LOI 0:=
D:\MathCement_demo_pdf\MC_V1_B303_Lab_25BIU.mcd
Page 5 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Vol.percentage of Hydrogen in flue gas
H2[ ] 0.0%:=
CO[ ] 0.02%:=Vol.percentage of Carbon monoxide in flue gas
CO2[ ] 20.5%:=Vol.percentage of Carbon dioxide in flue gas
O2[ ] 3.5%:=Vol.percentage of Oxygen in flue gas
Input data
TKTemp. of gases in deg. K
TTemp. of gases
List of parameters used
Nm3 Nm3
=kJ 103J:=
Nm3 m3
:=degC 1:=
User defined units
Let us consider a typical flue gas having the following volumetric composition as shown in input data.
It is known that molal volume of any gas at NTP is 22.4 m3. That means, say for example, oxygen has a molecular weight of 32 .So, 32 Kg. of oxygen will occupy a volume of 22.4m3
at NTP
Density of a mixture of dry gases at NTP can be calculated by knowing the volumetric percentages of individual gases present in the mixture.
Statement of problem
This worksheet should be used to calculate density of of a mixture of dry and wet gases
About this topic
Topic: Calculation of density of a mixture of dry and wet gases
MC_V1_B304_Fuel_11BIU
Fuels & Combustion
MathCement 2000
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 1 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
MCO[ ] 28:=
MSO2[ ] 64:=Sulphur dioxide
Nitrogen MN2[ ] 28:=
Water vapour MH2 O⋅[ ] 18:=
Air (dry) Mair[ ] 28.965:=
MV 22.4m
3
kg:=Molal volume
Calculation algorithm
Standard density (at 0 degC and 760 mm of Hg)
Density of Oxygen γO2MO2[ ]
MV:=
γO2 1.43kg
m3
=
Density of Carbon dioxide
γCO2MCO2[ ]
MV:=
Vol.percentage of Sulphur dioxide in flue gas
SO2[ ] 5.3%:=
Vol.percentage of Nitrogen in flue gas
N2[ ] 70.08%:=
Type of gas : Typ = 1 for flue gas, Typ=2 for air
Typ 1:=
Dew point temperature of the sample gas tdew1 59degC:=
Caution : please read values of moisture in gas corresponding to dew point in from the graph in the calculation area.
Constants
Gas Molecular weights
Oxygen MO2[ ] 32:=
Hydrogen MH2[ ] 2:=
Carbon dioxide MCO2[ ] 44:=
Carbon monoxide
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 2 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
γair.dry 1.29kg
m3
=
γair.dryMair[ ]
MV:=
Density of air (dry)
γH2O 0.8kg
m3
=
γH2OMH2 O⋅[ ]
MV:=
Density of water vapour
γH2 0.09kg
m3
=
γH2MH2[ ]
MV:=
Density of Hydrogen
γN2 1.25kg
m3
=
γN2MN2[ ]
MV:=
Density of Nitrogen
γSO2 2.86kg
m3
=
γSO2MSO2[ ]
MV:=
Density of Sulphur dioxide
γCO 1.25kg
m3
=
γCOMCO[ ]
MV:=
Density of Carbon monoxide
γCO2 1.96kg
m3
=
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 3 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
tdew data 0⟨ ⟩:= tdew Dew point temperature in deg. C
Vapair data 1⟨ ⟩:= Vapair Kg of water vapour per Kg of dry air
Vapgas data 2⟨ ⟩:= Vapgas Kg of water vapour per Kg of dry gas
with30% Co2
20 30 40 50 60 70 80 900.01
0.1
1
Dew pt. temp. - deg.C
Kg
wat
er v
apou
r per
Kg
dry
air /
dry
gas
Vapair
Vapgas
tdew
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 4 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
m3 /kg dry gasVvap 0.16 m3
=
VvapVap
γH2O:=
Volume of vapour Vvap
Vgas1 0.68 m3
=
Vgas11kg
γgas.dry:=
Volume of 1 kg. dry gas Vgas1
Vap 0.13 kg=
Vap if Typ 1= Vapgas, Vapair,( ):=
Vapour in air or gas =Vap
read from the graph (approx. value)kg /kg dry gas Vapgas 0.126kg:=
read from the graph (approx. value)kg /kg dry air Vapair 0.151kg:=
tdew1 59 degC=
Dew point temperature of the sample gas
Calculation of density of moist gas when dew point temperature is known.
Vapair 0.077737=
Vapgas 0.066117=tdew 46.954=
Vapair 0.038542=
Vapgas 0.032781=tdew 34.885=
Practice : try reading thse values from the graph:
Use above graph to read out data for water vapour in Kg per Kg of dry air or gases corresponding to adew point temperature. (hint - Rt. click the mouse on the graph and select trace)
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 5 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
Density of moist gas γgas.wet
γgas.wet1kg Vap+
Vvap Vgas1+:= All volumes in NTP
NTP - Normal temp. of o deg.C and normal preassure of 760 mm Hgγgas.wet 1.35
kg
m3
=
Results All volumes in NTP
Density of the dry gas = γgas.dry γgas.dry 1.48kg
m3
=
Density of moist gas γgas.wet γgas.wet 1.35kg
m3
=
Observation
References
Fri Sep 23 4:13:03 PM 2005
D:\MathCement_demo_pdf\MC_V1_B304_Fuel_11BIU.mcd
Page 6 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
ρbBulk density of material
VBelt speed
bBelt width
αAngle of inclination for vertical lift
HVertical lift of the conveyor
LHorizontal run of the conveyor
List of parameters used
User defined units
A rubber belt conveyor has been designed convey crushed bulk material material .Depending upon the layout of the plant the conveyor may run only horizontally or in combination of horizal travel and vertical lift. We know the design parameters of the belt and also the material characteristics. Find 1) the total resistance to the movement of the belt2) effective tension in the belt3) carrying side and return side tension4) sag tension 5) power required to drive the belt
Statement of problem
This is worksheet can be used to find the basic parameters related to calculation of power required to drive a rubber belt conveyor.As belt conveyors are extensively used in a cement plant to carry crushed raw materials and crushed coal over fairly long distances, it will be useful to check these basic parameters.
About this topic
Topic: Power and tension calculations of rubber belt conveyor -
Page 6 of 8 Copyright 2000-2005 Softideas Pvt. Ltd.
Fri Sep 23 4:20:12 PM 2005
1)Chapter on Belt Conveyor -Bulk Solids Handling by C.R. Woodcock and J.S.Mason: Published by Leonard hill2)Chapter on "Handling of Solids" - Unit Operation of Chemical Engineering3)Hallmark Technologies, Pune
References
Observation
Pm 6108.17 W=Motor power rating
Po 5191.95 W=Operating power required at driving drum
Page 8 of 8 Copyright 2000-2005 Softideas Pvt. Ltd.
mmWg1
13.6mmHg:=
bar 760mmHg:=
mmHg mm:=mmHg 133.33Pa:=
kN 1000N:=
User defined units
p v1 a1
Throat Q
dt
d
Ventury Meter ht v2 a2
A horizontal pipeline is carrying water and it is desired to measure the flowrate. The pipeline is equipped with an in-line ventury meter.Diameter of the pipe is given and the static pressure before ventury is measured.Throat diameter of ventury is also known and the pressure head at the throat is also measured. The pressure loss due to friction in the ventury between inlet and throat is assumed as percentgae of pressure difference.
Statement of problem
This worksheet can be used to find the flow rate through a pipeline with the introduction of ventury meter
About this topic
Topic: Fluid flow - calculation of flow through pipeline with ventury meter
Page 6 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
water column
hpVapour pressure of fluid at the given temperature
hfitCummulative head loss due to pipe fittings and friction in terms of height of water column
water column
Sp. weight of fluid flowing in the pipe γ
vFlow velocity in suction pump
z1Vertical height of centre line of pump suction line above the water level in the ground water tank
gGravitational constant
List of parameters used
MathCement 2000Used Gloval variable to display results along side input data
Plant Hydraulics
MC_V1_B308_Hydraulics_20BIUa
Topic: Calculation of Net Positive Suction Head (NPSH) for pump located above reservoir level.
About this topic
This worksheet can be used to calculate Net Positive Suction Head before a pump, which is used for pumping water from a reservoir and delivering to a discharge tank.Pump is located so that the centre line is above the water level in the reservoir
Statement of problem
A pump is required to pump water from a ground level reservoir to an elevated tank.The location of the pump above the reservoir and also the height of the tank are given. Pumping rate in the suction line is known. The details of head losses due to pipe fittings and friction in the suction line is also known.The atmospheric pressure, temperature of water under suction and corresponding specific weight and vapour pressure are given.Calculate NPSH for the pump.
Page 1 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
NPSH 7.12 m=Net Positive Suction Head
Result preview
p 101356Pa≡Static pressure on the fluid surfaceUsed Gloval variable to display results along side input data. Caution: Don't redefine global variable within the body of calculations to avoid confusion.
hp 0.654m≡Vapour pressure of fluid at the given temperature in terms of height of water column
water column
hfit 1.1m≡Cummulative head loss due to pipe fittings and friction in terms of height of water column
water column
Sp. weight of fluid flowing in the pipe γ 9830.11N
m3
≡
v 1.1m
s≡Flow velocity in suction pump
z1 1.5− m≡Vertical height of centre line of pump suction line above the water level in the ground water tank.Caution: Use (-) sign if water level below pump centreline
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
MathCement_PYRO
Kiln & Preheater Section
MC_V4_B201_KilnPH_7BIU
Topic: Mass Balance in multiple preheater stages
Caution - switch to manual calculation mode ( deselect math>automatic calculation) .Since the length of this calculationis very big -you maynot get consistant result under automatic mode of calculation.
About this topic
In this topic we have worked out detailed procedure for calculating mass balance of multiple stages of preheater .The example is based on the following:1) preheater with 5 stages ( count from bottom most cyclone)2) precalciner3) Tertiary air through separate duct.4) Tertiary air tapped from kiln hood5) Fuel type is coal,oil or gas
ConventionA suspension type cyclone preheater is divided into stages.In our calculation we will call the lowest stage as stage 1 and subsequent higher stages as 2, 3 ,4 etc.A stage consists of a cyclone,the gas duct leading to the cyclone and meal chute below the cyclone.
Feed / Meal
DustGas
Meal
Cyclone separator
Gas duct
Hot gas Dust
Typical cyclone stage
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 1 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
F 0.005kg
kg_rawmeal⋅:=
Reaction enthalpy( heat ) of clinker - RW RW 450kcal
kgcl⋅:=
Starting temperature of decarbonation - TA TA 1063K:=
End temperature for decarbonation - TE TE 1163K:=
Circulating dust load at kiln inlet -SOE SOE 0.3kg
kgcl:=
Bypass quantity of gases at kiln inlet - BY BYntp 0.00m3
:=
Percent fuel firing in precalciner -VC VC 55%:=
Cyclone efficiency -stage wise
Cyclone efficiency -stage 1- ηst1 ηst1 70%:=
Cyclone efficiency -stage 2- ηst2 ηst2 70%:=
Cyclone efficiency -stage 3- ηst3 ηst3 80%:=
Statement of problemUser defined units
B-IU Calculation using Mathcad's built-in units
This calculation use Mathcad's builtin units.So you will not find conventional unit conversion factors in the equations used.
Mio_tonne 106tonne:= kgcl kg:= kg_rawmeal kg:=
kg_coal kg:= kg_CO2[ ] kg:= degC 1:=
kg_fuel kg:= kg_ash kg:= cu_mntp m3
:=
List of parameters used
All temperatures in Kelvin =K K 273 degC+=
Input Data
Loss on ignition of raw meal - GV GV 0.35:=
Moisture in raw meal -F
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 2 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
Fpc 0.102kg_fuel
kgcl=
kg. fuel / kg.clFpc VC Freq⋅:=
Fuel fired to precalciner -Fpc
Freq 0.19kg_fuel
kgcl=
Freqηf
Hu:=
Fuel requirement -Freq
η f 797kcal
kgcl=
η f 797kcal
kgcl=
η f 797kcal
kgcl:=
Let's assume fuel efficiency ,i.e heat release by fuel per unit mass of clinker -η f
Hth 420kcal
kgcl=
Hth RW 30kcal
kgcl−:=
Theoretical heat of formation of clinker -Hth
Calculation
We start with certain assumed values of1) Fuel consumption2) Temperature of tertiary air Establish preheater bottom stage conditions.Then calculate mass balance of individual stages to find the unknowns.
Calculation algorithm
Fash 30%:=Ash in fuel -Fash
Hu 4300kcal
kgcl:=Heat value of fuel to precalciner and kiln - Hu
ηst5 92%:=Cyclone efficiency -stage 5- ηst5
ηst4 80%:=Cyclone efficiency -stage 4- ηst4
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 3 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
RMCO2 GVc RMF⋅:=
CO2 in kiln feed raw meal - RMCO2kg_CO2[ ] 1 kg=
Assume total loss on ignition is entirely due to release of CO2 from raw meal
RMF 1.48kg_rawmeal
kgcl=
RMF1 ASHtot−
1 GVc−:=
Raw meal requirement for producing clinker -RMF
GVc 0.36=
GVc 11
1
1 GV−1−
1 ASHtot−1+
−:=
Loss on ignition ,corrected for ash -GVc
ASHtot 0.06kg_ash
kgcl=
ASHtot Fash Freq⋅:=
Total ash absorbed in clinker - ASHtot
ASHkiln 0.03kg_ash
kgcl=
ASHkiln Fash Fkiln⋅:=
Ash going into kiln with fuel -ASHkiln
See sketch 1_1_kiln_7_drg2
ASHpc 0.03kg_fuel
kgcl=
ASHpc Fash Fpc⋅:=
Ash going into precalciner with fuel -ASHpc
Fkiln 0.08kg_fuel
kgcl=
Fkiln Freq Fpc−:=
Fuel fired to kiln -Fkiln
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 4 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
kg
MI1M1
ηst1:=
Quantity of material entering cyclone 1 - MI1
M1 1.35kg
kgcl=
M1 RMki SOE+:=
Material at discharge chute of cyclone -1 / preheater stage 1- M1
SOE 0.3kg
kgcl=
Quantity of circulating dust at kiln inlet -SOE
RMki 1.05kg
kgcl=
RMkiRMcal.kiln
1 LOIki−:=
Quantity of raw meal at kiln inlet - RMki
LOIki 0.07=
LOIkiTE tkf.rm−
TE TA−GVc⋅:=
Loss on ignition of raw meal at kiln inlet -LOIki
Loss on ignition as linear relation to temperature difference between start and end temperature of decarbonation
tkf.rm 1143 K=
tkf.rm 1143K:=
Assume temperature at kiln inlet for calcined raw meal - tkf.rm
RMcal.kiln 0.975kg
kgcl=
RMcal.kiln 1 ASHkiln−:=
Raw meal quantity fully calcined (loss free basis)in kiln -RMcal.kiln
RMCO2 0.54kg_CO2[ ]
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 5 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
M2 2.06kg
kgcl=
M2 MI1 CO2pc+( ) MI1ash SOE+( )−:=
Quantity of material discharged through meal chute from cyclone stage 2 - M2
CO2pc 0.46kg
kgcl=
CO2pc RMCO2 RMki.co2−:=
CO2 released in precalciner -CO2pc
RMki.co2 0.08kg
kgcl=
RMki.co2 RMki RMcal.kiln−:=
CO2 in raw meal entering kiln -RMki.co2
This is calculated by calculating the presence of CO2 in raw meal entering the kiln and then subtracting it from total amount of CO2 present in raw meal feed.
CO2 released in precalciner -
MI1CO2 1.02kg
kgcl=
MI1CO2 RMki MI1ash−:=
The quantity of raw meal that has lost CO2 in precalciner- MI1CO2
MI1ash 0.03kg
kgcl=
MI1ash ASHpc:=
Ash coming into raw meal from precalciner -MI1ash
S1 0.58kg
kgcl=
S1 MI1 1 ηst1−( )⋅:=
Dust at exhaust from cyclone stage 1- S1
MI1 1.93kg
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 6 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
SOE 0.3kg
kgcl=RMki 1.05
kg
kgcl=RMcal.kiln 0.97
kg
kgcl=
ASHpc 0.03kg
kgcl=ASHkiln 0.03
kg
kgcl=
RMki.co2 0.08kg
kgcl=
S2MI2
M3
S1 M2
MI1
ASHpc
SOE ASHkilnM1 RMki.co2
RMki RMcal.kiln
M1 1.35kg
kgcl=
MI2 2.95kg
kgcl=M2 2.06
kg
kgcl=
S2 0.88kg
kgcl=S1 0.58
kg
kgcl=
S2 0.88kg
kgcl=
S2 MI2 1 ηst2−( )⋅:=
Dust at exhaust from cyclone stage 2- S2
MI2 2.95kg
kgcl=
MI2M2
ηst2:=
Quantity of material entering cyclone 2 - MI2
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 7 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
S3 293.45kg
kgcl=
kg / kg.clS3 MI3 1ηst3
100−
⋅:=
Dust at exhaust from cyclone stage 3- S3
MI3 295.82kg
kgcl=
MI3M3
ηst3
100
:=
Quantity of material entering cyclone 3 - MI3
M3 2.37kg
kgcl=
M3 MI2 S1−:=
Quantity of material discharged through meal chute from cyclone stage 3 - M3
See sketch 1_1_kiln_7_drg3
S3 0.59kg
kgcl=
S3 MI3 1 ηst3−( )⋅:=
Dust at exhaust from cyclone stage 3- S3
MI3 2.96kg
kgcl=
MI3M3
ηst3:=
Quantity of material entering cyclone 3 - MI3
M3 2.37kg
kgcl=
M3 MI2 S1−:=
Quantity of material discharged through meal chute from cyclone stage 3 - M3
See sketch 1_1_kiln_7_drg3
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 8 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
S4 MI4 1 ηst4−( )⋅:=
Dust at exhaust from cyclone stage 4- S4
MI4 368.67kg
kgcl=
MI4M4
ηst4:=
Quantity of material entering cyclone 4 - MI4
M4 294.94kg
kgcl=
M4 MI3 S2−:=
Quantity of material discharged through meal chute from cyclone stage 4 - M4
See sketch 1_1_kiln_7_drg1
M2 2.06kg
kgcl=
S1 0.58kg
kgcl=
ηst2 70 %=
S2 0.88kg
kgcl=
MI2 2.95kg
kgcl=
M3 2.37kg
kgcl=
M2 2.06kg
kgcl=
ηst3 80 %=
MI3 295.82kg
kgcl=
S3 293.45kg
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 9 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
S4 MI4 1 ηst4−( )⋅:=
S4 73.73kg
kgcl=
Quantity of material discharged through meal chute from cyclone stage 5 - M5
M5 MI4 S3−:=
M5 75.22kg
kgcl=
Quantity of material entering cyclone 5 - MI5
MI5M5
ηst5:=
MI5 81.76kg
kgcl=
Dust at exhaust from cyclone stage 5- S5
S5 MI5 1 ηst5−( )⋅:=
S5 6.54kg
kgcl=
Quantity of material entering through meal chute to gas duct to cyclone stage 5 as fresh feed - Mfeed
Mfeed MI5 S4−:=
Mfeed 8.02kg
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 10 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
kg
M1 1.35kg
kgcl=Material at discharge chute of
cyclone -1 / preheater stage 1- M1
SOE 0.3kg
kgcl=Quantity of circulating dust at kiln
inlet -SOE
RMki 1.05kg
kgcl=Quantity of raw meal at kiln inlet - RMki
RMcal.kiln 0.975kg
kgcl=Raw meal quantity fully calcined
(loss free basis)in kiln -RMcal.kiln
RMF 1.48kg_rawmeal
kgcl=Raw meal requirement for producing
1kg. clinker -RMF
ASHtot 0.06kg_ash
kgcl=Total ash absorbed in clinker - ASHtot
ASHkiln 0.03kg_ash
kgcl=Ash going into kiln with fuel -ASHkiln
ASHpc 0.03kg_ash
kgcl=Ash going into precalciner with fuel
-ASHpc
Results
S3 293.45kg
kgcl=
M4 294.94kg
kgcl=
ηst4 80 %=
MI4 368.67kg
kgcl=
S4 73.73kg
kgcl=
M5 75.22kg
kgcl=
ηst5 92 %=
Mfeed 8.02kg
kgcl=
MI5 81.76kg
kgcl=S5
MI5η st5
M feed
Cyclone stage -5
M5 S4
MI4η st4
S3
Cyclone stage -4
M4
S5 6.54kg
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 11 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
Dust at exhaust from cyclone stage 3- S3
S3 293.45kg
kgcl=
Quantity of material discharged through meal chute from cyclone stage 4 - M4
M4 294.94kg
kgcl=
Quantity of material entering cyclone 4 - MI4
MI4 368.67kg
kgcl=
Dust at exhaust from cyclone stage 4- S4
S4 73.73kg
kgcl=
Quantity of material discharged through meal chute from cyclone stage 5 - M5
M5 75.22kg
kgcl=
Quantity of material entering cyclone 5 - MI5
MI5 81.76kg
kgcl=
Dust at exhaust from cyclone stage 5- S5
S5 6.54kg
kgcl=
Quantity of material entering through meal chute to gas duct to cyclone stage 5 as fresh feed - Mfeed
Mfeed 8.02kg
kgcl=
We can now establish the overall mass balance as shown in the sketch below:
Quantity of material entering cyclone 1 - MI1
MI1 1.93kg
kgcl=
Dust at exhaust from cyclone stage 1- S1 S1 0.58kg
kgcl=
Ash coming into raw meal from precalciner -MI1ash
MI1ash 0.03kg
kgcl=
The quantity of raw meal that has lost CO2 in precalciner- MI1CO2
MI1CO2 1.02kg
kgcl=
Quantity of material discharged through meal chute from cyclone stage 2 - M2
M2 2.06kg
kgcl=
Quantity of material entering cyclone 2 - MI2
MI2 2.95kg
kgcl=
Dust at exhaust from cyclone stage 2- S2
S2 0.88kg
kgcl=
Quantity of material discharged through meal chute from cyclone stage 3 - M3
M3 2.37kg
kgcl=
Quantity of material entering cyclone 3 - MI3
MI3 295.82kg
kgcl=
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 12 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
Observation
S1 0.58kg
kgcl=ASHpc 0.03
kg_fuel
kgcl=ASHkiln 0.03
kg_ash
kgcl=RMki 1.05
kg
kgcl=
M1 1.35kg
kgcl=RMcal.kiln 0.975
kg
kgcl=SOE 0.3
kg
kgcl=
MI1 1.93kg
kgcl=
S2 0.88kg
kgcl=
M2 2.06kg
kgcl=
MI2 2.95kg
kgcl=
S3 293.45kg
kgcl=
M3 2.37kg
kgcl=
MI3 295.82kg
kgcl=
M4 294.94kg
kgcl=
S4 73.73kg
kgcl=
MI4 368.67kg
kgcl=
M5 75.22kg
kgcl=
S5 6.54kg
kgcl=
MI5 81.76kg
kgcl=
Mfeed 8.02kg
kgcl=
1 kg. clinker
RMkiASHkiln
RMki.co2
ASHpcM1SOE
MI1M2S1
MI2M3S2
M4MI3S3
MI4M5
S4
MI5S5
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 13 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
Observation
References
Fri Sep 23 5:38:13 PM 2005
D:\MathCement_demo_pdf\MC_V4_B201_KilnPH_7BIU.mcd
Page 14 of 14 Copyright 2000-2005 Softideas Pvt. Ltd.
mbar 100Pa:=mmWg1
13.6mmHg:=
kmol 1000mol:=mbar 0.01lbf
in2
=
degC 1≡bar 760mmHg:=
mbar 100N
m2
⋅:=
mmHg 132.95 Pa=
mmHg1
25.4in_Hg⋅:=
Nm3 m3
:=in_Hg 3376.86 Pa=
User defined units Top
Consider a compressor to be delivering a certain volume compressed of air expressed in terms of FAD (free air delivery) at a certain pressure. The compression takes place polytropically and the polytropic index is given. Calculate theoretical power requirement for compressing the air to the desired pressure.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This worksheet can be used to calculate the power requirement for compressing air polytropically. It should be remembered that polytropic type of compression which falls
between isothermal and adiabatic or isentropic follows the rule: P Vn
⋅ constant= n represents polytropic index and normally lies between 11.0 and 1.4 covering all types of changes in volume .
About this topic
Topic: Calculation of theoretical power to compress air polytropically
MC_V5_B201_Fans_3BIU
Pneumatics and Compressed Air Systems
MathCement_Fluid Power
D:\MathCement_demo_pdf\MC_V5_B201_Fans_3BIU.mcd
Page 1 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
Free air delivery Vfad 3m
3
min:=
Gauge pressure of compressed air Pg 7bar:=
Atmospheric pressure Pb 1bar:=
Top
Calculation algorithm
Polytropic equation to be followed for calculation
P Vn
⋅ constant=
It follows, then,
P1 V1n
⋅ P2 V2n
⋅=
Work done W1
W1n
n 1−P2 V2⋅ P1 V1⋅−( )=
where
P1 represents initial pressure of air (absolute)
V1 represents initial volume of air
List of parameters used
Polytropic index of compression n
Free air delivery Vfad
Gauge pressure of compressed air Pg
Atmospheric pressure Pb
Compressed volume of air flow V2
Theoretical requirement of power Wp
Work done in compressing air W1
Input dataPolytropic index of compression n 1.3:=
D:\MathCement_demo_pdf\MC_V5_B201_Fans_3BIU.mcd
Page 2 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
V2 0.61m
3
min=Compressed volume of air flow
Derived data
Pg 7 bar=Gauge pressure of compressed air
Vfad 3m
3
min=Free air delivery
n 1.3=Polytropic index of compression
Pb 1 bar=Atmospheric pressure
Given data
Results
Top
Wp 13.48kW=
Wp W1:=
WpSo, theoretical requirement of power
W1 13.48kW=
W1n
n 1−P2 V2⋅ P1 V1⋅−( ):=
Therefore , work done
V2 0.61m
3
min=
V2 V1P1
P2
1
n
⋅:=
P2 8 bar=
V1 3m
3
min=V1 Vfad:=P2 Pb Pg+:=
P1 1 bar=P1 Pb:=
In our example we can write
V2 represents final volume of air
P2 represents final pressure of air (absolute)
D:\MathCement_demo_pdf\MC_V5_B201_Fans_3BIU.mcd
Page 3 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
min
Theoretical requirement of power Wp 13.48kW=
Work done in compressing air W1 13.48kW=
Top
Observation
References
Fri Sep 23 5:26:51 PM 2005
D:\MathCement_demo_pdf\MC_V5_B201_Fans_3BIU.mcd
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
PgGauge pressure of compressed air
List of parameters used
mbar 100Pa:=mmWg1
13.6mmHg:=
kmol 1000mol:=mbar 0.01lbf
in2
=
degC 1≡bar 760mmHg:=
mbar 100N
m2
⋅:=
mmHg 132.95 Pa=
mmHg1
25.4in_Hg⋅:=
Nm3 m3
:=in_Hg 3376.86 Pa=
User defined units
Top
Consider a compressor delivering compressed air at a certain rate at a defined gauge pressure. The average demand of the system is less than the capacity. Allowable lower limit of gauge pressure of compressed air is known. The receiver capacity is also given. As the receiver pressure falls to lower limit the compressor comes on load and continues till the pressure reaches the upper limit. For the given set of data, find 1) how many times the compressor will come on load per hour. Assume temperature of air is kept constant
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This worksheet can be used to calculate how many times a compressor will come on load to meet a certain demand of compressed air.
About this topic
Topic: Calculation of frequency at which compressor would come on load as function of system demand.
MC_V5_B201_Fans_8BIU
Pneumatics and Compressed Air Systems
MathCement_Fluid Power
D:\MathCement_demo_pdf\MC_V5_B201_Fans_8BIU.mcd
Page 1 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Average system demand for compressed air (f.a.d.)
Qcon 7m
3
min:=
Minimum gauge pressure of compressed air
Pgmin 6bar:=
Receiver volume Vr 6m3
:=
Top
Calculation algorithm
Average maximum pressure in pipeline in absolute scale
P1abs
P1abs Pg Pb+:=
P1abs 8 bar=
Average minimum pressure in pipeline in absolute scale
P2abs
P2abs Pgmin Pb+:=
P2abs 7 bar=
Let's calculate the volumes of free air stored in the receiver of compressed air at the maximum and minimum gauge pressure.
Pg
Atmospheric (barometric) pressure Pb
Flow rate of compressed air in terms of free air delivery
Q
Average system demand for compressed air (f.a.d.)
Qcon
Minimum gauge pressure of compressed air
Pgmin
Input data
Gauge pressure of compressed air Pg 7bar:=
Pb 1bar:=Atmospheric (barometric) pressure
Flow rate of compressed air in terms of free air delivery
Q 10m
3
min:=
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Page 2 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
tCharging time
Qchg 3m
3
min=
Qchg Q Qcon−:=
So the difference between delivery and consumption would equal volume charged by the compressor in unit time
Qcon 7m
3
min=
The circuit consumes per unit time
Q 10m
3
min=
The compressor can deliver per unit time
The compressor comes on load when the pressure falls to lower limit value and goes off load when the pressure reaches the upper limit value.
Vdiff 6 m3
=
Vdiff Vr1 Vr2−:=
VdiffThus the difference in free air volume stored in the receiver at maximum and minimum pressure
Vr2 42 m3
=
Vr2P2abs Vr⋅
Pb:=
Vr2Free air volume at minimum gauge pressure
Vr1 48 m3
=
Vr1P1abs Vr⋅
Pb:=
Vr1Free air volume at maximum gauge pressure
D:\MathCement_demo_pdf\MC_V5_B201_Fans_8BIU.mcd
Page 3 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Top
Note : If the cycle time is too short leading to very frequent start stop operation for compressor, consider increasing the receiver capacity or extend range of maximum and minimum pressure.
Nst 21=
Nst1hr
tcyc:=
NstSo the number of times the compressor would come on load per hour
tcyc 2.86 min=
tcyc t tdisch+:=
Thus total time between the compressor going on load or cycle time
tcyc
tdisch 0.86 min=
tdischVdisch
Qcon:=
tdischDischarge time from receiver
QconSystem consumption rate for compressed air
Vdisch 6 m3
=
Vdisch Vdiff:=
VdischVolume available for discharging
On the other hand the receiver discharges compressed air thus reducing its pressure
t 2 min=
tVdiff
Qchg:=
D:\MathCement_demo_pdf\MC_V5_B201_Fans_8BIU.mcd
Page 4 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
Fri Sep 23 5:28:41 PM 2005
Power Pneumatics by Michalel J. Pinches and Brian J. CallearReferences
Observation
Top
Nst 21=So the number of times the compressor would come on load per hour
tcyc 2.86 min=Thus total time between the compressor going on load or cycle time
tdisch 0.86 min=Discharge time
t 2 min=Charging time
Derived data
Vr 6 m3
=Receiver volume
Pgmin 6 bar=Minimum gauge pressure of compressed air
Qcon 7m
3
min=
Average system demand for compressed air (f.a.d.)
Q 10m
3
min=Flow rate of compressed air in terms
of free air delivery
Atmospheric (barometric) pressure Pb 1 bar=
Pg 7 bar=Gauge pressure of compressed air
Given data
Results
D:\MathCement_demo_pdf\MC_V5_B201_Fans_8BIU.mcd
Page 5 of 5 Copyright 2000-2005 Softideas Pvt. Ltd.
mbar 100Pa:=mmWg1
13.6mmHg:=
kmol 1000mol:=mbar 0.01lbf
in2
=
degC 1≡bar 760mmHg:=
mbar 100N
m2
⋅:=
mmHg 132.95 Pa=
mmHg1
25.4in_Hg⋅:=
Nm3 m3
:=in_Hg 3376.86 Pa=
User defined units
Top
A cylindrical actuator can receive compressed air under varying pressure within a range. The minimum sustained gauge pressure is given. Find the minimum bore for the cylinder required to develop a force to be applied on a mechanical work piece.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
This worksheet can be used to calculate the required bore of a cylinder in a compressed air system to activate the piston so that a predefined force can be developed to clamp a workpiece .Minimum cylinder pressure is known and also the atmospheric pressure is known and cylinder efficiency is known. The compressed air entering the cylinder at a certain pressure transmit the same pressure on to the piston. More and more compressed air come into the cylinder, as the piston moves, to maintain the pressure. The system is used to transmit the force through the piston rod. Cylinder bore being fixed ,the force can reduce in case of falling pressure. Calculation ,therefore, should be carried out for minimum sustained pressure..
About this topic
Topic: Calculating compressed air cylinder bore to be able to develop a desired clamping force
MC_V5_B201_Fans_16BIU
Pneumatics and Compressed Air Systems
MathCement_Fluid Power
D:\MathCement_demo_pdf\MC_V5_B201_Fans_16BIU.mcd
Page 1 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
d4 Acy⋅
π:=
rewriting
Acyπ d
2⋅
4=
dLet diameter of cylinder
Acy 61.86 cm2
=
AcyF
Pgmin Eff⋅:=
AcyWe first determine area of cross section of cylinder required to be acted on
Calculation algorithm
Top
F 3000N:=Force required to be transmitted by piston rod
Pgmin 5bar:=Minimum sustained pressure in cylinder
Eff 96%:=Cylinder transmission efficiency
Input data
dBore of the cylinder
FForce required to be transmitted by piston rod
PgminMinimum sustained pressure in cylinder
EffCylinder transmission efficiency
List of parameters used
13.6
D:\MathCement_demo_pdf\MC_V5_B201_Fans_16BIU.mcd
Page 2 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
Fri Sep 23 5:31:28 PM 2005
References
Observation
Top
Acy 61.86 cm2
=Area of cross section of cylinder
d 88.75 mm=Bore of the cylinder
Derived data
F 3000 N=Force required to be transmitted by piston rod
Pgmin 5 bar=Minimum sustained pressure in cylinder
Eff 96 %=Cylinder transmission efficiency
Given data
Results
Top
Select nearest standard cylinder bore
d 88.75 mm=
π
D:\MathCement_demo_pdf\MC_V5_B201_Fans_16BIU.mcd
Page 3 of 3 Copyright 2000-2005 Softideas Pvt. Ltd.
monthyr
12:=
USD 48 Rs=
USDRs
Rs_to_USD:=
Rate of exchange USD to Rupees = Rs_to_USD Rs_to_USD1
48:=
Rs 1≡
User defined units
Top
Surplus fund is invested in a fixed deposit with a bank. Bank pays a nominal yearly rate of interest. The principal is compounded at monthly or predefined frequency.Given the final balance amount at the end of the deposit period, find nominal rate of yearly interest and also effective rate of interest.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
When you save money in a bank or invest in finance scheme a certain fixed sum, you earn interest. Interest rates are normally indicated on yearly basis and is called nominal interest. The principal amount invested, however, increase every month as interest amount is credited. It is also normal that new monthly balance including interest earned in the previous month earns interest. This is known as compounding of interest (earning interest on interest) which results into higher effective earning. In this worksheet, you workout effective rate of interest and nominal rate of interest when the final balance at the end of the deposit period is known.
About this topic
Topic: Calculation of Rates of Interest on an Investment when Final Balance After the End of Deposit Period is Known
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
monthyr
12:=
USD 48 Rs=
USDRs
Rs_to_USD:=
Rate of exchange USD to Rupees = Rs_to_USD Rs_to_USD1
48:=
Rs 1≡
User defined units
Top
Determine the current value of a machine that you can afford to purchase.
You approach a bank and find out the rate of interest they would be charging annually.
Your budget permits a certain monthly payments that can be made over a certain period of time. You want to buy a machine now, with the help of a bank loan.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
You can then determine the current value of a machine that you can afford to purchase.
You can approach a bank and find out the rate of interest they would be charging annually.
When your budget permits a certain monthly payments that can be made over a certain period of time, you may attempt to buy a machine now, with the help of a bank loan.
About this topic
Topic: Calculation to find, present value of an annuity
Page 2 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
m 200Rs=Monthly savings to be deposited to bank
Period of repayment of borrowed fund P 4yr=
Yearly rate of interest YI 12%=
Given data
Results
Top
Vp 7594.79 Rs=
Vp Find Vp( ):=
Vp 1 r+( )n
⋅ m1 r+( )
n1−
r
⋅=
Given
Vp 567:=
guess value for Vp
Solve for Vp
Vp 1 r+( )n
⋅ m1 r+( )
n1−
r
⋅=
or
V Vp 1 r+( )n
⋅=
We can now write the following equation
So we can define the present value of an annuity as the amount that we can borrow ,knowing the amount of monthly payments, the rate of interest and the number of payments or installments.
Let the present value be Vp
But this is the total outgo of your fund over the repayment period and does not mean that you can buy a machine of this current value. To find out the amount of money the bank will advance as a loan, which will be in tern the current value of the machine you can purchase , you have to find the current value of this annuity.
Page 4 of 4 Copyright 2000-2005 Softideas Pvt. Ltd.
lifeTotal expected useful life
costTotal cost of procurement of machine / asset
List of parameters used
monthyr
12:=
USD 48 Rs=
USDRs
Rs_to_USD:=
Rate of exchange USD to Rupees = Rs_to_USD Rs_to_USD1
48:=
Rs 1≡
User defined units
Top
Also find the book value after a certain defined period
A cement company decides to buy a hammer crusher to crush coal. It is decided to use reducing value depreciation method for depreciation calculation. For the given data find the depreciation value.
Statement of problem
Results
Calculation algorithm
Input data
Statement of problem
Important bookmarks (Double click on linked regions below to go to sections •directly)
Use this worksheet to calculate depreciation of machinery or other assets by reducing balance method
About this topic
Topic: Depreciation Calculation of Assets by Reducing Balance Method
Page 3 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
Dty 213.47 Rs=
i
Di∑ 646.74= Dtot 900=
Following condition is applied to balance total depreciation by considering the salvage value to be true and the calculated depreciation on declining balance
Dn if
i
Di∑ Dtot− 0≥ Dni
Di∑ Dtot−
+, Dni
Di∑ Dtot−
−,
:=
Di
0
76.92
71.01
318.8
60.5
55.85
51.55
47.59
43.93
40.55
37.43
34.55
31.89
29.44
Rs
=
D0 0=
D1 76.92 Rs=
D3 318.8 Rs=
i
Di∑ 900= Dtot 900=Dn 318.8=
Recalculate book values due to adjustment in depreciation
Page 6 of 6 Copyright 2000-2005 Softideas Pvt. Ltd.
Did you know?(DYK001)Recent research shows that many industrial wastescan be successfully used as replacement of cement
in preparation of concrete.This can reduce current (2005)demand for cement by 25%.This will serve to minimise
disposal problem associated with such industrial wastes and reduce global energy demand.
Source: World Cement: January 2005
Cement substitute
Did you know?(DYK002)Working with cement and concrete consumes 40% of
the global energy demand.And a reduction of cement and concrete usage by up to 25% by alternative material
could save 10% of total energy requirements.This also means a reduction of 400 million MT of CO2 release per
annum.Source: World Cement: January 2005
Energy and environment
Did you know?(DYK003)Fans used in cement plants are mostly of centrifugal
type.The flow characteristics, through a fan, are governed by three basic rules:
1) Flow rate is proportional to speed.2) Pressure developed is a square function of speed.
3) Power required varies as cube of speed. Source:
Fan laws
Did you know?(DYK005)Atomic weight of any element, when expressed in
grams, contains 6.02X1023 numbers of atoms of that element.Thus, 16 gm of Oxygen or 14 gm of Nitrogen will
have equal number of their respective atoms and this number will be equal to 6.02X1023, also called Avagadro
Number.Source:
Atomic weight and Avagadro Number.
Did you know?(DYK005)Atomic weight of any element, when expressed in
grams, contains 6.02X1023 numbers of atoms of that element.Thus, 16 gm of Oxygen or 14 gm of Nitrogen will
have equal number of their respective atoms and this number will be equal to 6.02X1023, also called Avagadro
Number.Source:
Atomic weight and Avagadro Number.
Did you know?(DYK006)Average pressure exerted by Earth's atmosphere, at sea level, is equal to the pressure exerted by a 760mm tall column of Mercury at its base, .This pressure is called
"Normal atmospheric pressure" or "Standard atmospheric pressure" and has a value of 101325 Pa
(Pascals).It's also expressed as 1 atm or 1kgf/cm2
Source:
Atmospheric pressure
Did you know?(DYK007)Natural gas is a mixture of hydrocarbon gases mainly of
Methane.But it may also contain Ethane, Propane, Butane and Pentane.It is called natural gas as it occurs
naturally in nature, under the Earths' surface. Natural gas is a fossil fuel like coal and is formed from the remains of
plants, animals and micro-organisms under intense pressure and heat over millions of years.
Source:
Gaseous fuels -Natural gas
Did you know?(DYK008)Natural gas, which erupts through fissures in the earth's crust, from below the surface of the Earth can ignite due to natural phenomenon like lightning.Such resultant fires awed the early civilization.Early Greeks assigned divinity to such natural spring of fire and created a temple around one such eruption. This is known as Oracle of Delphi.
Source:
"D:\Mathcement_July05\Images\oracle at delphi"
Gaseous fuels -Natural gas
Did you know?(DYK009)There are only three classes of equipment in a cement
plant. These are 1) Conveying equipment, 2) Storage equipment and 3) Process equipment.
Again process equipment can be classified in two two types i.e. 1)Physical process,(when composition of
incoming material does not change during the process) 2) Chemical process(when composition of incoming