mc_v0_b000_toc_v7_0_1_pdf

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

Volume-1 : Cement plant process calculations

MC_V1_TOC Expanded Table of Contents -Vol.1

Book 200 Process calculation of plant sections

Book 201 MC_V1_B201_Quarry_TOC Quarry Section

BIU MC_V1_B201_Quarry_1BIU Quarry Deposit Estimation

MC_V1_B201_Quarry_1 Quarry Deposit Estimation

BIU MC_V1_B201_Quarry_2BIU Quarry - Excavator

MC_V1_B201_Quarry_2 Quarry - Excavator

Chapter-4: Back to Basics

MC_V0_B101_Units_TOC Units

BIU MC_V0_B101_units_01BIU About Built-in Units

BIU MC_V0_B101_units_02BIU User Defined Units

MC_V0_B102_Review_TOC Process Review

MC_V0_B102_Review_1TXT NOx chemistry

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_V1_B201_Quarry_3BIU_r3 Quarry - Dumper Calculations

MC_V1_B201_Quarry_3_r3 Quarry Dumper Calculations

Book 202 MC_V1_B202_Crushing_TOC Crushing Section

BIU MC_V1_B202_Crushing_1BIU_r5 Crusher and Auxiliaries Calculations

MC_V1_B202_Crushing_1_r5 Crusher and Auxiliaries Calculations

BIU MC_V1_B202_Crushing_2BIU Single Rotor Hammer or Impact Crusher Calculations

MC_V1_B202_Crushing_2 Single Rotor Hammer or Impact crusher calculations

BIU

MC_V1_B203_Preblending_4BIU Stockpile capacities-Kidney shaped windrows -triangular section

MC_V1_B203_Preblending_4 Stockpile capacities-Kidney shaped windrows -triangular section

Book 204 MC_V1_B204_Rawmill_TOC Raw Mill Section

MC_V1_B204_RawMill_1_r4 Bond Work Index of the various material based on bond test mill result dry basis

Calculation of Power at Ball Mill Shaft for Raw Material,based on Bond's Work Index

BIU MC_V1_B204_RawMill_2BIU

MC_V1_B204_RawMill_2 Calculation of power at Ball Mill shaft for Raw material, based on Bond's work index

Raw Mill and Auxiliary Equipment Capacity Calculations


Book 203 MC_V1_B203_Pre-blending_TOC Pre-blending Section

BIU MC_V1_B203_Preblending_1BIU Preblending and Stockpile Equipment - Calculations

MC_V1_B203_Preblending_1 Preblending and Stockpile Equipment Calculations

BIU MC_V1_B203_Preblending_2BIU Stockpile capacities-Longitudinal with trapizoidal cross section

MC_V1_B203_Preblending_2 Stockpile capacities-Longitudinal with trapizoidal cross section

BIU MC_V1_B203_Preblending_3BIU Stockpile capacities-Longitudinal with triangular cross section

MC_V1_B203_Preblending_3 Stockpile capacities-Longitudinal with triangular cross section

BIU

MC_V1_B204_RawMill_7BIU Calculation of Grinding Ball Size

MC_V1_B204_RawMill_7 Calculation of Grinding Ball Size

BIU MC_V1_B204_RawMill_8BIU Specific Heat of Raw Material as a Function of Temperature

MC_V1_B204_RawMill_8 Specific Heat of Raw Material as a Function of Temperature

Estimation of specific power consumption, at mill shaft, based on the operating parameters of close circuit ball mill


MC_V1_B204_RawMill_9 Estimation of specific power consumption, at mill shaft, based on the operating parameters of close circuit ball mill

Calculation of mill power demand for desired output at specified feed and product sizes, based on Bond Index.

BIU MC_V1_B204_RawMill_10BIU_r3

MC_V1_B204_RawMill_3 Raw Mill and Auxiliary Equipment Capacity Calculations

Raw Material Drying - Estimation of total moisture to be evaporated from feed


MC_V1_B204_RawMill_4_r3 Raw Material Drying - Estimation of total Moisture to be Evaporated from Feed

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


MC_V1_B204_RawMill_5 Raw Mill Heat Balance - Evaluation of Hot Gas Requirement for Drying

Calculation of Ball Mill Percentage Filling as per Measurement


MC_V1_B204_Rawmill_6 Calculation of Ball Mill Percentage Filling asper Measurement

BIU

MC_V1_B204_RawMill_13 Ball Mill Critical Speed and Ball Charge Calculations

BIU MC_V1_B204_RawMill_14BIU_r4 Vertical Roller Mill -- Calculation of Power

MC_V1_B204_RawMill_14_r4 Vertical Roller Mill -- Calculation of Power

BIU MC_V1_B204_RawMill_15BIU_r3 Radiation Loss in Raw Mill Heat Balance

MC_V1_B204_RawMill_15_r3 Radiation Loss in Raw Mill Heat Balance

Calculation of false air coming into grinding mill system


MC_V1_B204_RawMill_16 Calculation of false air coming into grinding mill system

Calculation of water pressure at nozzle over material bed of Vertical Roller Mill with supply from overhead tank


MC_V1_B204_RawMill_10_r4 Calculation of mill power demand for desired output at specified feed and product sizes, based on Bond Index.

BIU MC_V1_B204_RawMill_10_1BIU Calculation of mill output for different feed and product sizes

MC_V1_B204_RawMill_10_1 Calculation of mill output for different feed and product sizes

BIU MC_V1_B204_RawMill_11BIU Calculation of Efficiency of Dynamic Air Separator

MC_V1_B204_RawMill_11 Calculation of Efficiency of Dynamic Air Separator

BIU MC_V1_B204_RawMill_12BIU Average Piece Weight of Grinding Media

MC_V1_B204_RawMill_12 Average Piece Weight of Grinding Media

BIU MC_V1_B204_RawMill_13BIU Ball Mill Critical Speed and Ball Charge Calculations

Calculation of auxiliary drive rating of Ball Mill for raw grinding

MC_V1_B204_RawMill_21_r1 Calculation of auxiliary drive rating of Ball Mill for raw grinding

BIU MC_V1_B204_RawMill_22BIU_r2 Calculation of makeup charge or ball mills

MC_V1_B204_RawMill_22_r2 Calculation of makeup charge or ball mills

BIU MC_V1_B204_RawMill_23BIU_r1 Calculation of wear rate of grinding balls in ball mills

MC_V1_B204_RawMill_23_r1 Calculation of wear rate of grinding balls in ball mills

BIU MC_V1_B204_RawMill_24BIU Calculation of Recirculation Air in Raw Grinding Vertical Roller Mill

MC_V1_B204_RawMill_24 Calculation of Recirculation Air in Raw Grinding Vertical Roller Mill

MC_V1_B204_RawMill_17 Calculation of water pressure at nozzle over material bed of Vertical Roller Mill with supply from overhead tank

BIU MC_V1_B204_RawMill_18BIU Calculation of retention time of material in a two compartment ball mill

MC_V1_B204_RawMill_18 Calculation of retention time of material in a two compartment ball mill

BIU MC_V1_B204_RawMill_19BIU_r1 Rosin-Rammler equation -calculation of predicted residue on sieve and variaton calculations

MC_V1_B204_RawMill_19_r1 Rosin-Rammler equation -calculation of predicted residue on sieve and variaton calculations

BIU MC_V1_B204_RawMill_20BIU Calculation of mean diameter of fine ground particles

MC_V1_B204_RawMill_20 Calculation of mean diameter of fine ground particles


MC_V1_B205_Blending_4_r2 Calculation of blending factor by methods of standard deviation

Book 206 MC_V1_B206_Kilnfeed_TOC Kiln-feed Section

Locating feeding point of material into Preheater Gas ducts

BIU MC_V1_B206_Kilnfeed_1BIU_r3

MC_V1_B206_Kilnfeed_1_r3 Locating feeding point of material intoPreheater Gas ducts

Book 207 MC_V1_B207_Clinker_TOC Clinkerisation (kiln,pre-heater) Section

BIU MC_V1_B207_Clinker_1BIU_r3 Calculation of Kiln and Auxiliary Equipment Capacities.

MC_V1_B207_Clinker_1_r3 Calculation of Kiln and Auxiliary Equipment Capacities

Book 205 MC_V1_B205_Blending_TOC Blending and Storage Section

BIU MC_V1_B205_Blending_1BIU Calculation of Blending & Storage Silos and Auxiliary Equipment Capacity for Raw-meal preparation

MC_V1_B205_Blending_1 Calculation of Blending & Storage Silos and Auxiliary EquipmentCapacity for Raw-meal Preparation

BIU MC_V1_B205_Blending_2BIU_r3 Calculation of Kiln Feed Raw-meal Systems

MC_V1_B205_Blending_2_r3 Calculation of Kiln Feed Raw-meal Systems

BIU MC_V1_B205_Blending_3BIU Calculation of Kiln Feed Raw-meal Quantity

MC_V1_B205_Blending_3 Calculation of Kiln Feed Raw-meal Quantity

BIU MC_V1_B205_Blending_4BIU_r2 Calculation of blending factor by methods of standard deviation

BIU MC_V1_B207_Clinker_7BIU Kiln Tyre / Shell Ovality

MC_V1_B207_Clinker_7 Kiln Tyre / Shell Ovality

BIU MC_V1_B207_Clinker_8BIU_r4 Kiln Capacity check

MC_V1_B207_Clinker_8_r4 Kiln Capacity check

BIU MC_V1_B207_Clinker_9BIU Degree of Decarbonation of Raw Meal -Definition and calculation

MC_V1_B207_Clinker_9 Degree of Decarbonation of Raw Meal -Definition calculation

BIU MC_V1_B207_Clinker_10BIU Degree of Decarbonation of Raw Meal Calculation on Operating Data

MC_V1_B207_Clinker_10 Degree of Decarbonation of Raw Meal Calculation on Operating Data

BIU MC_V1_B207_Clinker_11BIU Reaction Enthalpy -Decarbonation and Clinkerisation

MC_V1_B207_Clinker_11 Reaction Enthalpy -Decarbonation and Clinkerisation

BIU MC_V1_B207_Clinker_2BIU_r3 Leakages through Kiln Air Seals

MC_V1_B207_Clinker_2_r3 Leakages Through Kiln Air Seal

BIU MC_V1_B207_Clinker_3BIU Calculation of Specific Heats of Clinker as a Function of Clinker Temperature

MC_V1_B207_Clinker_3 Calculation of Specific Heats of Clinker as a Function of Clinker Temperature

BIU MC_V1_B207_Clinker_4BIU Kiln Retention Time and Related Parameters

MC_V1_B207_Clinker_4 Kiln Retention Time and Related Parameters

BIU MC_V1_B207_Clinker_5BIU_r3 Kiln sinter zone cooling fans

MC_V1_B207_Clinker_5_r3 Kiln Sinter Zone Cooling Fans

BIU MC_V1_B207_Clinker_6BIU_r3 Calculation of nose ring cooling air for kiln inlet

MC_V1_B207_Clinker_6_r3 Calculation of nose ring cooling air for kiln inlet

Kiln Hydraulic Thruster - A Note

BIU MC_V1_B207_Clinker_17BIU Kiln Drive Specification

MC_V1_B207_Clinker_17 Kiln Drive Specification

BIU MC_V1_B207_Clinker_18BIU Calculation of expansion of Kiln in hot condition

MC_V1_B207_Clinker_18 Calculation of expansion of Kiln in hot condition

BIU MC_V1_B207_Clinker_19BIU Calculation of Temerature Profile of Kiln Shell in hot condition

MC_V1_B207_Clinker_19 Calculation of Temperature Profile of Kiln Shell in hot condition

BIU MC_V1_B207_Clinker_20BIU Calcination Function

MC_V1_B207_Clinker_20 Calcination Function

BIU MC_V1_B207_Clinker_12BIU Theoretical Heat of Clinker Formation - Calculation

MC_V1_B207_Clinker_12 Theoretical Heat of Clinker Formation - Calculation

Calculation of kiln mass under operating conditions

BIU MC_V1_B207_Clinker_13BIU

MC_V1_B207_Clinker_13 Calculation of Kiln mass under Operating Conditions

BIU MC_V1_B207_Clinker_14BIU Calculation of Kiln Torque under Operating Conditions

MC_V1_B207_Clinker_14 Calculation of Kiln Torque under Operating Conditions

BIU MC_V1_B207_Clinker_15BIU_r3 Calculation for Kiln Hydraulic Thruster

MC_V1_B207_Clinker_15_r3 Calculation for Kiln Hydraulic Thruster

Text MC_V1_B207_Clinker_16TXT

no files hereMC_V1_B208_ClCooling_1BIUBIU

Clinker Cooling SectionMC_V1_B208_ClCooling_TOCBook 208

Calculation of fresh cooling air and water to reduce the temperature of kiln (alkali) bypass gases before release to atmosphere

MC_V1_B207_Clinker_25

Calculation of fresh cooling air and water to reduce the temperature of kiln (alkali) bypass gases before release to atmosphere

MC_V1_B207_Clinker_25BIUBIU

Alkali bypass systems

Correlation Between Central Chord Angle or Charge Angle and Degree of Filling of Kiln

MC_V1_B207_Clinker_24

Correlation Between Central Chord Angle or Charge Angle and Degree of Filling of Kiln

BIU MC_V1_B207_Clinker_21BIU Calculation of false air infiltration into kiln preheater system

MC_V1_B207_Clinker_21 Calculation of false air infiltration into kiln preheater system

BIU MC_V1_B207_Clinker_22BIU Calculation of Alkali Bypass System

MC_V1_B207_Clinker_22 Calculation of Alkali Bypass System

BIU MC_V1_B207_Clinker_23BIU Calculation of Range of Bypass Quantity of Gases from Kiln to Limit Alkali concentration in Clinker

MC_V1_B207_Clinker_23 Calculation of Range of Bypass Quantity of Gases from Kiln to Limit Alkali concentration in Clinker

BIU MC_V1_B207_Clinker_24BIU

MC_V1_B209_Clstore_TOC Clinker Storage Section

BIU MC_V1_B209_Clstore_1BIU Calculation of Clinker Transport Equipment Capacity and Clinker Stockpile

MC_V1_B209_Clstore_1 Calculation of Clinker Transport Equipment Capacity and Clinker Stockpile

Book 210 MC_V1_B210_Cement_TOC Cement Grinding Section

BIU MC_V1_B210_Cement_1BIU Calculation of Cement Mill and Auxiliary Equipment

MC_V1_B210_Cement_1 Calculation of Cement Mill and Auxiliary Equipment

BIU MC_V1_B210_Cement_2BIU Calculation of Cement Grindability based on Ziesel Value

MC_V1_B210_Cement_2 Calculation of Cement Grindability based on Ziesel Value

BIU MC_V1_B208_ClCooling_2BIU Calculation of Grate Cooler Drive Power

MC_V1_B208_ClCooling_2 Calculation of GrateCooler Drive Power

BIU MC_V1_B208_ClCooling_3BIU Calculation of Grate Cooler Recuperation Efficiency -T.A. Duct Tapping from Cooler

MC_V1_B208_ClCooling_3 Calculation of Grate Cooler RecuperationEfficiency - T.A.Duct Tapping from Cooler

BIU MC_V1_B208_ClCooling_4BIU Calculation of Grate Cooler Recuperation Efficiency -T.A. Duct Tapping from Kiln Hood

MC_V1_B208_ClCooling_4 Calculation of Grate Cooler Recuperation Efficiency - T.A.Duct Tapping from Kiln Hood

BIU MC_V1_B208_ClCooling_5BIU Calculation of Grate Cooler Offset from Kiln axis

MC_V1_B208_ClCooling_5 Calculation of Grate Cooler Offset from Kiln axis

Book 209

MC_V1_B210_Cement_7 Sizing of Close Circuit Cement Mill

BIU MC_V1_B210_Cement_8BIU Cement Mill Output at Different Finenesses

MC_V1_B210_Cement_8 Cement Mill Output at Different Finenesses

BIU MC_V1_B210_Cement_9BIU Cement Mill Cooling Air

MC_V1_B210_Cement_9 Cement Mill Cooling Air

BIU MC_V1_B210_Cement_10BIU Mill Radiation Loss

MC_V1_B210_Cement_10 Mill Radiation Loss

BIU MC_V1_B210_Cement_11BIU Cement Mill Grinding Performance as a Function of Surface Generated

MC_V1_B210_Cement_11 Cement Mill Grinding Performance as a Function of Surface Generated

BIU MC_V1_B210_Cement_3BIU Cement Mill Heat Balance and Calculation of Hot Gas for Drying

MC_V1_B210_Cement_3 Cement Mill Heat Balance and Calculation of Hot Gas for Drying

BIU MC_V1_B210_Cement_4BIU Calculation of Cement Mill Output as a function of Grinding Media Load

MC_V1_B210_Cement_4 Calculation of Cement Mill Output as a function of Grinding Media Load

BIU MC_V1_B210_Cement_5BIU Estimation of grindability of Clinker based on the operating parameters

MC_V1_B210_Cement_5 Estimation of grindability of Clinker based onthe operating parameters

BIU MC_V1_B210_Cement_6BIU Calculation of Water Spray in the Mill for Cooling

MC_V1_B210_Cement_6 Calculation of Water Spray in the Mill for Cooling

BIU MC_V1_B210_Cement_7BIU Sizing of Close Circuit Cement Mill

The Barometric pressure at site based on the altitude above mean sea level

MC_V1_B301_Site_1BIUBIU

Plant Site MC_V1_B301_Site_TOCBook 301

Process design calculations - support systemsBook 300

Evaluation of effect of weighing accuracy of packing m/cMC_V1_B211_Cmtstore_2

Evaluation of effect of weighing accuracy of packing m/c

MC_V1_B211_Cmtstore_2BIUBIU

Calculation of Cement Transport and Silo and Auxiliary Equipment

MC_V1_B211_Cmtstore_1

BIU MC_V1_B210_Cement_12BIU Expected Power Draw of Cement Mill as a Function of Speed

MC_V1_B210_Cement_12 Expected Power Draw of Cement Mill as a Function of Speed

BIU MC_V1_B210_Cement_13BIU_r2 Clinker grinding mill - calculation of capacity and power

MC_V1_B210_Cement_13_r2 Clinker grinding mill - calculation of capacity and power

Book 211 MC_V1_B211_Cmtstore_TOC Cement Storage and Dispatch Section

BIU MC_V1_B211_Cmtstore_1BIU Calculation of Cement Transport and Silo and Auxiliary Equipment

MC_V1_B302_Dedust_25 Sketches for K - factors -set5

BIU MC_V1_B302_Dedust_3BIU Calculation of Pressure Losses in Ducting conveying powdery material

MC_V1_B302_Dedust_3 Calculation of Pressure Losses in Ducting conveying powdery material

BIU MC_V1_B302_Dedust_31BIU Calculation of Friction Losses in Ductings from Tables /Graphs

MC_V1_B302_Dedust_31 Calculation of Friction Losses in Ductings from Tables /Graphs

BIU MC_V1_B302_Dedust_4BIU Design Parameters for Design of Duct Systems

MC_V1_B302_Dedust_4 Design Parameters for Design of Duct Systems

Text MC_V1_B302_Dedust_5TXT Duct balancing

MC_V1_B301_Site_1 The Barometric pressure at site based on the altitude above mean sea level

Book 302 MC_V1_B302_Dedust_TOC Dedusting Systems

Text MC_V1_B302_Dedust_1 Estimation of Vent Air Volume

Text MC_V1_B302_Dedust_2TXT K - Factors





MC_V1_B302_Dedust_10 Electrostatic Precipitator- Specific Collection Area

BIU MC_V1_B302_Dedust_11BIU Electrostatic Precipitator- Performance Evaluation

MC_V1_B302_Dedust_11 Electrostatic Precipitator- Performance Evaluation

BIU MC_V1_B302_Dedust_12BIU Electrostatic Precipitator- To calculate Migration Velocity

MC_V1_B302_Dedust_12 Electrostatic Precipitator- To calculate Migration Velocity

BIU MC_V1_B302_Dedust_13BIU_r3 Chimney design

MC_V1_B302_Dedust_13_r3 Chimney Design

BIU MC_V1_B302_Dedust_14BIU Calculation of flow through duct based onPitot tube measurement

MC_V1_B302_Dedust_14 Calculation of flow through duct based on Pitot tube measurement

BIU MC_V1_B302_Dedust_6BIU Glass Bag House -Calculation of Gas Volume as Function of temperature.

MC_V1_B302_Dedust_6 Glass Bag House -Calculation of Gas Volume as Function of temperature.

BIU MC_V1_B302_Dedust_7BIU GCT and ESP - Calculation of Gas Volume

MC_V1_B302_Dedust_7 GCT and ESP - Calculation of Gas Volume

BIU MC_V1_B302_Dedust_8BIU Sizing of Gas Conditioning Tower

MC_V1_B302_Dedust_8 Sizing of Gas Conditioning Tower

BIU MC_V1_B302_Dedust_9BIU Electrostatic Precipitator- Deutsch Efficiency Formula

MC_V1_B302_Dedust_9 Electrostatic Precipitator- Deutsch Efficiency Formula

BIU MC_V1_B302_Dedust_10BIU Electrostatic Precipitator- Specific Collection Area

BIU MC_V1_B302_Dedust_20BIU Calculation relating to conversion of industrial gases from standard volume to operating volume and vice-versa

Book 303 MC_V1_B303_Lab_TOC Laboratory Investigations and Raw Mix Designs

BIU MC_V1_B303_Lab_1BIU Loss on Ignition of Kiln Feed Raw Meal

MC_V1_B303_Lab_1 Loss on Ignition of Kiln Feed Raw Meal

BIU MC_V1_B303_Lab_2BIU Silica Ratio and Requirement of Components

MC_V1_B303_Lab_2 Silica Ratio and Requirement of Components

BIU MC_V1_B303_Lab_3BIU Alumina Ratio and Requirement of Components

MC_V1_B303_Lab_3 Alumina Ratio and Requirement of Components

BIU MC_V1_B302_Dedust_15BIU Calculation of pressure loss in duct due to friction and bends

MC_V1_B302_Dedust_15 Calculation of pressure loss in duct due to friction and bends

BIU MC_V1_B302_Dedust_16BIU Velocity selection for ducting design

MC_V1_B302_Dedust_16 Velocity selection for ducting design

BIU MC_V1_B302_Dedust_17BIU Calculation for air curtain

MC_V1_B302_Dedust_17 Calculation for air curtain

BIU MC_V1_B302_Dedust_18BIU Calculation of migration velocity of operating ESP

MC_V1_B302_Dedust_18 Calculation of migration velocity of operating ESP

BIU MC_V1_B302_Dedust_19BIU Calculation of dedusting efficiency of a dust separator in terms of degrees of dust separation

BIU MC_V1_B303_Lab_8BIU Percentage Liquid and Burnability Index Cement Clinker - Calculation

MC_V1_B303_Lab_8 Percentage Liquid and Burnability Index Cement Clinker - Calculation

BIU MC_V1_B303_Lab_9BIU Total Carbonate Content in Kiln Feed Raw Meal- Calculation

MC_V1_B303_Lab_9 Total Carbonate Content in Kiln Feed Raw Meal- Calculation

BIU MC_V1_B303_Lab_10BIU To Calculate Quantity of CaO Required to Attain Specific Value of Total Carbonate Content in Kiln Feed Raw Meal

MC_V1_B303_Lab_10 To Calculate Quantity of CaO Required toAttain Specific Value of Total CarbonateContent in Kiln Feed Raw Meal

BIU MC_V1_B303_Lab_11BIU Kiln Dust Loss in Terms of Kiln Feed Raw Meal-and Apparant Degree of Calcination

MC_V1_B303_Lab_11 Kiln Dust Loss in Terms of Kiln Feed RawMeal-and Apparent Degree of Calcination

BIU MC_V1_B303_Lab_4BIU Lime Saturation Factor - Calculation

MC_V1_B303_Lab_4 Lime Saturation Factor Calculation

BIU MC_V1_B303_Lab_5BIU Hydraulic Ratio of Kiln Feed Raw Meal - Calculation

MC_V1_B303_Lab_5 Hydraulic Ratio of Kiln Feed Raw Meal - Calculation

BIU MC_V1_B303_Lab_6BIU Burnability Factor of Kiln Feed Raw Meal - Calculation

MC_V1_B303_Lab_6 Burnability Factor of Kiln Feed Raw Meal - Calculation

BIU MC_V1_B303_Lab_7BIU Complex Compounds of Cement Clinker - Calculation by Bogue's Formulae

MC_V1_B303_Lab_7 Complex Compounds of Cement Clinker - Calculation by Bogue's Formulae

Text MC_V1_B303_Lab_21TXT Separation of ESP and Filter Dust

BIU MC_V1_B303_Lab_22BIU Volatility of alkalies

MC_V1_B303_Lab_22 Volatility of alkalies

Raw-mix Design

BIU MC_V1_B303_Lab_23BIU Raw-mix Design for Two Components

MC_V1_B303_Lab_23 Raw-mix Design for Two Components

BIU MC_V1_B303_Lab_24BIU Raw-mix Design based on Hydraulic Module

MC_V1_B303_Lab_24 Raw-mix Design based on Hydraulic Module

BIU MC_V1_B303_Lab_25BIU Raw-mix Design based on Lime Saturation Factor

MC_V1_B303_Lab_25 Raw-mix Design based on Lime SaturationFactor

BIU MC_V1_B303_Lab_26BIU Three Component Raw-mix Design based on Lime Saturation Factor and Silica Ratio

Text MC_V1_B303_Lab_12TXT Quantity of Raw Material Samples for Lab. Investigation

Text MC_V1_B303_Lab_13TXT Coal Samples -A Point of View

Text MC_V1_B303_Lab_14TXT Purposes of Tests

Text MC_V1_B303_Lab_15TXT Tests Conducted at Physical Laboratory

Text MC_V1_B303_Lab_16TXT Chemical Tests at Chemical Lab.

Text MC_V1_B303_Lab_17TXT Mineralogical Investigations

Text MC_V1_B303_Lab_18TXT Fuel Investgation

Text MC_V1_B303_Lab_19TXT Raw Mix Investgation

Text MC_V1_B303_Lab_20TXT Burnability Investigation

BIU MC_V1_B303_Lab_31BIU Calculation of Effect of Coal Ash Absorption on Clinker Analysis

MC_V1_B303_Lab_31 Calculation of Effect of Coal Ash Absorption on Clinker Analysis

BIU MC_V1_B303_Lab_32BIU Calculation of Two Component Raw Mix Design Considering Effect of Coal Ash Absorption

MC_V1_B303_Lab_32 Calculation of Two Component Raw Mix Design Considering Effect of Coal Ash Absorption

BIU MC_V1_B303_Lab_33BIU Raw-mix design calculation with three components to match desired potential clinker compounds

MC_V1_B303_Lab_33 Raw-mix design calculation with three components to match desired potential clinker compounds

BIU MC_V1_B303_Lab_34BIU Evaluation of influence of coal ash on clinker composition

MC_V1_B303_Lab_26 Three Component Raw-mix Design based on Lime Saturation Factor and Silica Ratio

BIU MC_V1_B303_Lab_27BIU Calculation of Coal Ash Absorption

MC_V1_B303_Lab_27 Calculation of Coal Ash Absorption

BIU MC_V1_B303_Lab_28BIU No files here

Raw-mix design calculation with two components to match desired potential clinker compounds

BIU MC_V1_B303_Lab_29BIU

MC_V1_B303_Lab_29 Raw-mix design calculation with two components to match desired potentialclinker compounds

BIU MC_V1_B303_Lab_30BIU Calculation of Potential Clinker Composition

MC_V1_B303_Lab_30 Calculation of Potential Clinker Composition

Calculation of Coal Analysis and Heat Values Coal

MC_V1_B304_Fuel_2 Calculation of Coal Analysis and Heat Valuesof Coal

BIU MC_V1_B304_Fuel_3BIU_r3 Calculation of combustion air for burning of coal

MC_V1_B304_Fuel_3_r3 Calculation of combustion air for burning of coal

BIU MC_V1_B304_Fuel_4BIU Calculation of theoretical air for combustion of coal and products of combustion

MC_V1_B304_Fuel_4 Calculation of theoretical air for combustion of coal and products of combustion

BIU MC_V1_B304_Fuel_5BIU Calculation of Flame Temperature

MC_V1_B304_Fuel_5 Calculation of Flame Temperature

Text MC_V1_B304_Fuel_6TXT Reactivity of Coal as a Function of Fineness

MC_V1_B303_Lab_34 Evaluation of influence of coal ash on clinker composition

Material Properties

BIU MC_V1_B303_Lab_35BIU Properties of Bulk Solids - Bulk Density Determination

MC_V1_B303_Lab_35 Properties of Bulk Solids - Bulk Density Determination

Properties of Bulk Solids - Characteristics Influencing Behaviour Pattern

Text MC_V1_B303_Lab_36TXT

BIU MC_V1_B303_Lab_37BIU Properties of Bulk Materials - Methods of particle size analysis

Book 304 MC_V1_B304_Fuel_TOC Fuels and Combustion

BIU MC_V1_B304_Fuel_1BIU_r3 Calculation of Capacities in Coal Preparation Section

MC_V1_B304_Fuel_1_r3 Calculation of Capacities in CoalPreparation Section

BIU MC_V1_B304_Fuel_2BIU

MC_V1_B304_Fuel_12 Vertical Roller Mill for grinding coal- calculation of capacity and power

Hot air for drying moisture in Vertical Roller Mill for grinding coal


MC_V1_B304_Fuel_13 Hot air for drying moisture in Vertical Roller Mill for grinding coal

BIU MC_V1_B304_Fuel_14BIU Calculation of combustion air for burning of fuel oil

MC_V1_B304_Fuel_14 Calculation of combustion air for burning of fuel oil

BIU MC_V1_B304_Fuel_15BIU Calculation of combustion air for burning of gaseous fuels

MC_V1_B304_Fuel_15 Calculation of combustion air for burning of gaseous fuels

BIU MC_V1_B304_Fuel_16BIU Gaseous fuels - calculation of products of combustion

Gaseous fuels - calculation of products of combustion

MC_V1_B304_Fuel_16

BIU MC_V1_B304_Fuel_7BIU Specific heats of gases as function of temp. T

MC_V1_B304_Fuel_7 Specific heats of gases as function of temp. T

Text MC_V1_B304_Fuel_8TXT Coal Grinding-Process considerations

Text MC_V1_B304_Fuel_9TXT Coal Dosing and Firing-Process considerations

Text MC_V1_B304_Fuel_10TXT Coal Burner -Process considerations

BIU MC_V1_B304_Fuel_11BIU Calculation of density of a mixture of dry and wet flue gases

MC_V1_B304_Fuel_11 Calculation of density of a mixture of dry and wet flue gases

Vertical Roller Mill for grinding coal- calculation of capacity and power


BIU MC_V1_B304_Fuel_21BIU Calculation of total heat loss due to moisture in coal fired to kiln / furnace

MC_V1_B304_Fuel_21 Calculation of total heat loss due to moisture in coal fired to kiln / furnace

BIU MC_V1_B304_Fuel_22BIU Calculations to Fix Locating Co-ordinates of Kiln Burner

MC_V1_B304_Fuel_22 Calculations to Fix Locating Co-ordinates of Kiln Burner

BIU MC_V1_B304_Fuel_23BIU_r1 Determination of Fuel Oil Stock in a Cylindrical Horizontally Mounted Tank.

MC_V1_B304_Fuel_23_r1 Determination of Fuel Oil Stock in a Cylindrical Horizontally Mounted Tank.

MC_V1_B304_Fuel_24TXT Basic constituents of Coal

MC_V1_B304_Fuel_25TXT Heat values of fuels

BIU MC_V1_B304_Fuel_17BIU Combustion of fuel - calculation of chemical reactions

Combustion of fuel - calculation of chemical reactions

MC_V1_B304_Fuel_17

BIU MC_V1_B304_Fuel_18BIU Orsat analysis and calculation of air / fuel ratio as function of carbon burnout

MC_V1_B304_Fuel_18 Orsat analysis and calculation of air / fuel ratio as function of carbon burnout

BIU MC_V1_B304_Fuel_19BIU Combustion calculations for burning of fuel oil

MC_V1_B304_Fuel_19 Combustion calculations for burning of fuel oil

BIU MC_V1_B304_Fuel_20BIU Gaseous fuels - combustion calculations

MC_V1_B304_Fuel_20 Gaseous fuels - combustion calculations

Belt Conveyor -Length of Conveyor Belt in a Roll

BIU MC_V1_B305_Convey_5BIU Belt Conveyor --Length of Belt and Speed

MC_V1_B305_Convey_5 Belt Conveyor --Length of Belt and Speed

BIU MC_V1_B305_Convey_6BIU_r3 Rubber belt conveyor -calculation of belt width and capacity

MC_V1_B305_Convey_6_r3 Rubber belt conveyor -calculation of belt width and capacity

BIU MC_V1_B305_Convey_7BIU Design of Fluidised Gravity Conveyor for raw meal or cement

MC_V1_B305_Convey_7 Design of Fluidised Gravity Conveyor for raw meal or cement

Design of Fluidised Gravity Conveyor for raw mill return grits with fines

BIU MC_V1_B305_Convey_8BIU

Design of Fluidised Gravity Conveyor for raw mill return grits with fines

MC_V1_B305_Convey_8

Book 305 MC_V1_B305_Convey_TOC Conveying Systems

BIU MC_V1_B305_Convey_1BIU_r4 Vertical bucket elevator (chain type) calculation of capacity and power

MC_V1_B305_Convey_1_r4 Vertical bucket elevator (chain type)calculation of capacity and power

BIU MC_V1_B305_Convey_2BIU Deep Bucket or Deep Pan type Conveyors -Power and Filling

MC_V1_B305_Convey_2 Deep Bucket or Deep Pan type Conveyors -Power and Filling

BIU MC_V1_B305_Convey_3BIU Screw Conveyor

MC_V1_B305_Convey_3 Screw Conveyor

BIU MC_V1_B305_Convey_4BIU Belt Conveyor -Length of Conveyor Belt in a Roll

MC_V1_B305_Convey_4

BIU MC_V1_B305_Convey_13BIU Various selection charts for power and tension calculations of rubber belt conveyor -

MC_V1_B305_Convey_13 Various selection charts for power and tension calculations of rubber belt conveyor -

Book 306 MC_V1_B306_Fans_TOC Air and Gas Handling (Fans, Blowers & Compressors

BIU MC_V1_B306_Fans_1BIU Calculation of Fan Motor Power

MC_V1_B306_Fans_1 Calculation of Fan Motor Power

BIU MC_V1_B306_Fans_2BIU Calculation of Fan Operating Volume

MC_V1_B306_Fans_2 Calculation of Fan Operating Volume

Design of Fluidised Gravity Conveyor for cement mill return grits with fines


Design of Fluidised Gravity Conveyor for cement mill return grits with fines

MC_V1_B305_Convey_9

Design of Fluidised Gravity Conveyor for raw mill return grits without fines


Design of Fluidised Gravity Conveyor for raw mill return grits without fines

MC_V1_B305_Convey_10

BIU MC_V1_B305_Convey_11BIU Design of Fluidised Gravity Conveyor for cement mill grits without fines

MC_V1_B305_Convey_11 Design of Fluidised Gravity Conveyor for cement mill grits without fines

BIU MC_V1_B305_Convey_12BIU Power and tension calculations of rubber belt conveyor -

MC_V1_B305_Convey_12 Power and tension calculations of rubber belt conveyor -

MC_V1_B306_Fans_7BIU Calculation of velocity pressure and pressure head development due to flow of air in a duct

MC_V1_B306_Fans_7 Calculation of velocity pressure and pressure head development due to flow of air in a duct

BIU MC_V1_B306_Fans_8BIU Calculation of system resistance in a ducting system and selecting an appropriate fan based on their characteristic curves

MC_V1_B306_Fans_8 Calculation of system resistance in a ducting system and selecting an appropriate fan based on their characteristic curves

BIU MC_V1_B306_Fans_9BIU Calculation and plotting of system resistance curves by introduction of change in system design in a ducting system

MC_V1_B306_Fans_9 Calculation and plotting of system resistance curves by introduction of change in system design in a ducting system

BIU MC_V1_B306_Fans_10BIU Calculation of fan pressure development as a function of gas temperature.

MC_V1_B306_Fans_10 Calculation of fan pressure development as a

Text MC_V1_B306_Fans_3TXT Design of Large Fans -Requirement of Motor Data

BIU MC_V1_B306_Fans_4BIU Calculation of linear expansion of shafts -during hot running conditions

MC_V1_B306_Fans_4 Calculation of Expansion of Fan shaft -During Hot Condition

BIU MC_V1_B306_Fans_5BIU Calculation of Fan Pressure and Power Rating

MC_V1_B306_Fans_5 Calculation of Fan Pressure and Power Rating

BIU MC_V1_B306_Fans_6BIU Calculation of pressure development by fans and blowers

MC_V1_B306_Fans_6 Calculation of pressure development by fans and blowers

BIU

BIU MC_V1_B307_Matrls_1BIU Ventilation of Ball mills and cooling air calculation

MC_V1_B307_Matrls_1 Ventilation of Ball mills and cooling air calculation

BIU MC_V1_B307_Matrls_2BIU Solving with LMTD -heat transfer in cooler or dryer.

MC_V1_B307_Matrls_2 Solving with LMTD -heat transfer in cooler or dryer.

BIU MC_V1_B307_Matrls_3BIU Soil Properties Calculations

MC_V1_B307_Matrls_3 Soil Properties Calculations

BIU MC_V1_B307_Matrls_4BIU Evaluation of hot gas utilisation factor in a drying system

MC_V1_B307_Matrls_4 Evaluation of hot gas utilisation factor in a drying system

MC_V1_B306_Fans_10 Calculation of fan pressure development as a function of gas temperature.

BIU MC_V1_B306_Fans_11BIU Calculation of fan pressure development at a new location - considering change in barometric pressure

MC_V1_B306_Fans_11 Calculation of fan pressure development at a new location - considering change in barometric pressure

BIU MC_V1_B306_Fans_12BIU Calculation of fan output data under test conditions.

MC_V1_B306_Fans_12 Calculation of fan output data under test conditions.

BIU MC_V1_B306_Fans_13BIU Calculation of fan output as a function of change in speed

MC_V1_B306_Fans_13 Calculation of fan output as a function of change in speed

Book 307 MC_V1_B307_Matrls_TOC Materials and Processing -General Folder

MC_V1_B308_Hydraulics_3BIU

MC_V1_B308_Hydraulics_3 Calculation of static pressure in a pipeline based on measurement by U-Tube Manometer

BIU MC_V1_B308_Hydraulics_4BIU Water jet calculation in cement grinding mill

MC_V1_B308_Hydraulics_4 Water jet calculation in cement grinding mill

BIU MC_V1_B308_Hydraulics_5BIU External water jet calculation in cement grinding mill

MC_V1_B308_Hydraulics_5 External water jet calculation in cement grinding mill

BIU MC_V1_B308_Hydraulics_6BIU Efficiency calculation of a centrifugal pump in operation

MC_V1_B308_Hydraulics_6 Efficiency calculation of a centrifugal pump in operation

Pressure and power calculation of a centrifugal pump to pump water

BIU MC_V1_B308_Hydraulics_7BIU

BIU MC_V1_B307_Matrls_5BIU Estimation of cement requirement for building brick wall

MC_V1_B307_Matrls_5 Estimation of cement requirement for building brick wall

Book 308 MC_V1_B308_Hydraulics_TOC Plant Hydraulics

BIU MC_V1_B308_Hydraulics_1BIU Quality of Plant Cooling Water

MC_V1_B308_Hydraulics_1 Quality of Plant Cooling Water

BIU MC_V1_B308_Hydraulics_2BIU Hydraulic jack -calculation of power requirement

MC_V1_B308_Hydraulics_2 Hydraulic jack -calculation of power requirement

Calculation of static pressure in a pipeline based on measurement by U-Tube Manometer

BIU

Fluid flow - calculation of flow through pipeline with ventury meter

MC_V1_B308_Hydraulics_11 Fluid flow - calculation of flow through pipeline with ventury meter

BIU MC_V1_B308_Hydraulics_12BIU Selection of pipeline in pumping system design

MC_V1_B308_Hydraulics_12 Selection of pipeline in pumping system design

BIU MC_V1_B308_Hydraulics_13BIU Energy in liquids -an approach to Barnaulli's equation

MC_V1_B308_Hydraulics_13 Energy in liquids -an approach to Barnaulli's equation

BIU MC_V1_B308_Hydraulics_14BIU Calculation of pressure, downstream in a pipeline

MC_V1_B308_Hydraulics_14 Calculation of pressure, downstream in a pipeline

Pressure and power calculation of a centrifugal pump to pump water

MC_V1_B308_Hydraulics_7

BIU MC_V1_B308_Hydraulics_8BIU Fluid flow - calculation of Raynold's number and friction factor

MC_V1_B308_Hydraulics_8 Fluid flow - calculation of Raynold's number and friction factor

BIU MC_V1_B308_Hydraulics_9BIU Fluid flow - calculation of flow between three interconnected tanks in star formation

MC_V1_B308_Hydraulics_9 Fluid flow - calculation of flow between three interconnected tanks in star formation

Fluid flow - calculation of work done by a pump to empty a sump and lift the water to over head tank


Fluid flow - calculation of work done by a pump to empty a sump and lift the water to over head tank

MC_V1_B308_Hydraulics_10


Pump systems

Net Positive Suction Head NPSH

BIU MC_V1_B308_Hydraulics_20BIU Calculation of Net Positive Suction Head (NPSH) for pumplocated above reservoir level.

MC_V1_B308_Hydraulics_20 Calculation of Net Positive Suction Head (NPSH) for pumplocated above reservoir level.

BIU MC_V1_B308_Hydraulics_20BIUa Calculation of Net Positive Suction Head (NPSH) for pumplocated above reservoir level.

Using global variable

BIU MC_V1_B308_Hydraulics_21BIU Calculation of Net Positive Suction Head (NPSH) for pump located below reservoir level.

MC_V1_B308_Hydraulics_21 Calculation of Net Positive Suction Head (NPSH) for pump located below reservoir level.

BIU MC_V1_B308_Hydraulics_15BIU Pressure loss in a pipeline due to fluid flow

MC_V1_B308_Hydraulics_15 Pressure loss in a pipeline due to fluid flow

BIU MC_V1_B308_Hydraulics_16BIU Calculation of power required for a pump to deliver water through a pipeline system

MC_V1_B308_Hydraulics_16 Calculation of power required for a pump to deliver water through a pipeline system

Text MC_V1_B308_Hydraulics_17TXT Pumping terminologies

BIU MC_V1_B308_Hydraulics_18BIU Pumping laws

MC_V1_B308_Hydraulics_18 Pumping laws

MC_V4_B201_KilnPH_1BIU Preheater exhaust gases- calculation

MC_V4_B201_KilnPH_1a Preheater Exhaust Gases -with oil firing

BIU MC_V4_B201_KilnPH_1aBIU Preheater Exhaust Gases -with oil firing

MC_V4_B201_KilnPH_2 Pressure loss in kiln and preheater

BIU MC_V4_B201_KilnPH_2BIU Pressure loss in kiln and preheater

MC_V4_B201_KilnPH_3 Preheater fan rating

BIU MC_V4_B201_KilnPH_3BIU Preheater fan rating

MC_V4_B201_KilnPH_4 Tertiary air and ducting to precalciner

BIU MC_V4_B201_KilnPH_4BIU Tertiary air and ducting to precalciner

Book 309 MC_V1_B309_Insulation_TOC Insulation

BIU MC_V1_B309_Insulation_1BIU General Specifications for Insulation in Cement Plant

MC_V1_B309_Insulation_1 General Specifications for Insulation in Cement Plant

Volume-4 : Simulation model for pyro-processing

MC_V4_TOC Expanded Table of Contents - Vol.4

Book 200 Process calculation of plant sections

Book 201 MC_V4_B201_KilnPH_TOC Kiln Preheater Section

MC_V4_B201_KilnPH_1 Preheater exhaust gases- calculation

BIU

MC_V4_B201_KilnPH_9 Input data calculation for file B201_KilnPH_10

BIU MC_V4_B201_KilnPH_9BIU Input data calculation for file B201_KilnPH_10BIU

BIU MC_V4_B201_KilnPH_10BIU Heat gas and mass balance in 5 stage preheater with precalciner

MC_V4_B201_KilnPH_11 Radiation losses from kiln

BIU MC_V4_B201_KilnPH_11BIU Radiation losses from kiln

BIU MC_V4_B201_KilnPH_12BIU Calculation of false air into the system

BIU MC_V4_B201_KilnPH_13aBIU Kiln heat balance

MC_V4_B201_KilnPH_5 Primary air fan for coal firing in kiln

BIU MC_V4_B201_KilnPH_5BIU Primary air fan for coal firing in kiln

MC_V4_B201_KilnPH_6 Mass balance in a single cyclone preheater stage

BIU MC_V4_B201_KilnPH_6BIU Mass balance in a single cyclone preheater stage

MC_V4_B201_KilnPH_7 Mass balance in multiple preheater stages

BIU MC_V4_B201_KilnPH_7BIU Mass balance in multiple preheater stages

MC_V4_B201_KilnPH_8 Gas balance in preheater

BIU MC_V4_B201_KilnPH_8BIU Gas balance in preheater

Over all heat balance, gas balance and mass balance in the pyro section-6 stage preheater -74p

MC_V4_B203_Pyro_3BIUBIU



Over all heat balance, gas balance and mass balance in the pyro section -5 stage preheater-69p

MC_V4_B203_Pyro_2



Pyro Processing ( Kiln, Preheater & Cooler ) Section

MC_V4_B203_Pyro_TOCBook 203

Calculation of grate area and bed height for clinker cooler

MC_V4_B202_Cooler_3

Clinker cooler gas balance and heat balance MC_V4_B202_Cooler_2BIUBIU

Clinker cooler gas balance and heat balance MC_V4_B202_Cooler_2

Clinker cooler -cooling air calculation MC_V4_B202_Cooler_1

Clinker Cooler SectionMC_V4_B202_Cooler_TOCBook 202

MC_V4_B301_Heat_3BIU Calculation of total heat required to evapoate moisture

Volume-5 : Fluid power


Book 200 Air and Gas Handling

Book 201 MC_V5_B201_Fans_TOC Pneumatics and compressed air systems

Compressed air systems

Pneumatics basics

BIU MC_V5_B201_Fans_1BIU Calculation of mass of gases stored in a vessel

MC_V5_B201_Fans_1 Calculation of mass of gases stored in a vessel

Book 300 Process design calculations - support systems

Book 301 MC_V1_B301_Site_TOC Plant Site

BIU MC_V1_B301_Site_1BIU The Barometric pressure at site based on the altitude above mean sea level

The Barometric pressure at site based on the altitude above mean sea level

MC_V1_B301_Site_1

Book 302 MC_V4_B302_Heat_TOC Heat & Combustion

BIU MC_V4_B301_Heat_1BIU Combustion calculations for burning of fuel oil

BIU MC_V4_B301_Heat_2BIU Gaseous fuels- combustion calculations

BIU

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


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


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


Calculation of Future Value of a Fixed Deposit or Investment and Recurring Deposit Separately or Together


Calculation of Future Value of a Fixed Deposit or Investment and Recurring Deposit Separately or Together


Calculation of Rates of Interest on an Investment when Final Balance After the End of Deposit Period is Known


Calculation of Rates of Interest on an Investment when Final Balance After the End of Deposit Period is Known


Calculation of Nominal Rate of Interest on an Investment when Effective Rate of Interest is Known


Calculation of Nominal Rate of Interest on an Investment when Effective Rate of Interest is Known


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


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.


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.


MC_V23_B204_Fuels_1BIU No files here

BIU MC_V23_B204_Fuels_2BIU Burner calculation - 3- channel coal firing burner

MC_V23_B204_Fuels_2 Burner calculation - 3- channel coal firing burner

Book 205 MC_V23_B205_Ducts_TOC Chutes & Ducts

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 V0_A001_TablesCharts_9BIU Dimension, Units, Variables & Constants, Degree of Freedom etc

BIU V0_A001_TablesCharts_10BIU Pressure Related Basic Calculations

BIU V0_A001_TablesCharts_11BIU Calculations to Convert Column of Fluid to Pressure

Text V0_A001_TablesCharts_12TXT Introduction toPsychrometry

BIU V0_A001_TablesCharts_13BIU Bulk Densities of Materials

BIU V0_A001_TablesCharts_14BIU Distributed Loads on Platforms for Civil Design

BIU V0_A001_TablesCharts_15BIU Foundation Pockets Size as a Function of Bolt Size

Appendix 1 : Tables and charts

MC_V0_A001_TablesCharts_TOC Tables and charts

BIU V0_A001_TablesCharts_1BIU Separation Factor of ESP / Filter Dust

BIU V0_A001_TablesCharts_2BIU Humidity Chart - Finding Dew Point and Humidity

BIU V0_A001_TablesCharts_3BIU Steam Tables - Finding Dew Point and Humidity

V0_A001_TablesCharts_4 Reactivity of coal

BIU V0_A001_TablesCharts_5BIU Charts and Tables-Conversion Tables and Factors

BIU V0_A001_TablesCharts_6BIU Calculation of Hard Grove Index from Laboratory Result and conversion .

BIU V0_A001_TablesCharts_7BIU Charts and Tables-Wagner,Sieve.Blaine Relations

MC_V0_A002_Sciencedata_9TXT Properties of Metals - Poission's Ratio

Text MC_V0_A002_Sciencedata_10TXT Properties of Metals - Modulus of Elasticity

Text MC_V0_A002_Sciencedata_11TXT Properties of Metals - Melting Point in Degrees Kelvin at Atmospheric Pressure

Text MC_V0_A002_Sciencedata_12TXT Properties of Common Solids - Density

Text MC_V0_A002_Sciencedata_13TXT Properties of Common Solids - Specific Gravity

Text MC_V0_A002_Sciencedata_14TXT Properties of Common Solids - Specific Heat

Text MC_V0_A002_Sciencedata_15TXT Properties of Common Solids - Thermal Conductivity

Text MC_V0_A002_Sciencedata_16TXT Properties of Gases - Specific Gravity

MC_V0_A002_Sciencedata_TOC Scientific data

Text MC_V0_A002_Sciencedata_1TXT Basic Sciences Fundamental Physical Constants

Text MC_V0_A002_Sciencedata_2TXT Basic Sciences Temperature Scales

Text MC_V0_A002_Sciencedata_3TXT Basic Sciences Periodic Table of Elements

Text MC_V0_A002_Sciencedata_4TXT Calculus Reference Formulae Table of Integral Formulas

Text MC_V0_A002_Sciencedata_5TXT Properties of Metals - Thermal Conductivity at Atmospheric Pressure ,250C

Text MC_V0_A002_Sciencedata_6TXT Properties of Metals - Specific Gravity

Text MC_V0_A002_Sciencedata_7TXT Properties of Metals - Co-efficient of Linear Expansion per Degree C

Text MC_V0_A002_Sciencedata_8TXT Properties of Metals - Electrical Resistivity

Text

Circle

BIU MC_V0_A003_Mensuration_AP_6BIU Inscribed Circle

BIU MC_V0_A003_Mensuration_AP_7BIU Circumscribed Circle

BIU MC_V0_A003_Mensuration_AP_8BIU Sector of a Circle

BIU MC_V0_A003_Mensuration_AP_9BIU Circular Segment

BIU MC_V0_A003_Mensuration_AP_10BIU Regular Polygon

BIU MC_V0_A003_Mensuration_AP_11BIU Inscribed Polygon

BIU MC_V0_A003_Mensuration_AP_12BIU Circumscribed Polygon

BIU MC_V0_A003_Mensuration_AP_13BIU Parabolic Segment

BIU MC_V0_A003_Mensuration_AP_14BIU Ellipse

Text MC_V0_A002_Sciencedata_17TXT Properties of Gases - Specific Heat at 1.0 Atm Pressure, 20ºC, Except as Noted

Text MC_V0_A002_Sciencedata_18TXT Properties of Gases - Molecular Weightat 1.0 Atm Pressure, 20ºC, Except as Noted

Text MC_V0_A002_Sciencedata_19TXT Properties of Common Liquids - Specific Heat

MC_V0_A003_Mensuration_TOC Mensuration

MC_V0_A003_Mensuration_AP_TOC Areas and perimeters

BIU MC_V0_A003_Mensuration_AP_1BIU Triangle

BIU MC_V0_A003_Mensuration_AP_2BIU Rectangle

BIU MC_V0_A003_Mensuration_AP_3BIU Parallelogram

BIU MC_V0_A003_Mensuration_AP_4BIU Trapezoid

BIU MC_V0_A003_Mensuration_AP_5BIU

MC_V0_A003_Mensuration_VS_6BIU Frustum of Right Circular Cone

BIU MC_V0_A003_Mensuration_VS_7BIU Sphere

BIU MC_V0_A003_Mensuration_VS_8BIU Spherical Cap

BIU MC_V0_A003_Mensuration_VS_9BIU Spherical Triangle

Did you know?(DYK001)Click here to find out





Did you know?(DYK006)

MC_V0_A003_Mensuration_VS_TOC Volumes and surfaces

BIU MC_V0_A003_Mensuration_VS_1BIU Rectangular Prism

BIU MC_V0_A003_Mensuration_VS_2BIU Parallelepiped

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

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.

D:\MathCement_demo_pdf\MC_V0_B000_Foreword_01TXT.mcd

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".

Dr.A.K.ChatterjeeWhole-time DirectorThe Associated Cement Cos.Ltd.'Cement House', 121, M.K.RoadMumbai 400 020, India

Mumbai , July 27, 2001

Fri Sep 23 12:58:29 PM 2005

D:\MathCement_demo_pdf\MC_V0_B000_Foreword_01TXT.mcd


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


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.



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



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.

References

Fri Sep 23 10:44:39 AM 2005

D:\MathCement_demo_pdf\MC_V0_B000_Models_03TXT.mcd


MathCement

Fri Sep 23 10:44:39 AM 2005



MathCement

References



MathCement

Picture is inserted instead of paste to reduce file size

Page set up- A-4, Margins- Left 1.2in,top & bottom 1in, right 0.75in.

Region check done 10/5/05

Rev 4Checked for assignment in symbol area 10/5/05

Rev 3Spell check done 10/5/05

4p/1h Rev -210/5/05Rev 2MC_V0_B000_Models_01TXT

SP 3-4/12/0211 /12/02Rev 1

4p/2h Rev -0BIU 3-4/12/02Rev -0V1_B0_CH1_Modelling_1



MathCement



VdDumper volumetric capacity required

BcsExcavator Shovel Bucket Capacity (volumetric) based on standard availability

BnNumber of shovel buckets to fill a Dumper (normal- 3 to 8 )

HtHopper to hold material equivalent to crusher feed in terms of time

BDLSBulk density of uncrushed stone (limestone) 1.5 -1.6

Crusher capacity Qcr

To find Dumper volumetric capacity

List of parameters used

User defined units Top

For transporting quarried stone to crushing plant by dumpers, calculate number of dumpers, vol. capacities, hopper capacity for crusher.

Statement of problem

Results

Calculation algorithm

Input data


Important bookmarks (Double click on linked regions below to go to sections •directly)

Use this worksheet to estimate capacity requirement of dumpers in the quarry

About this topic

Topic: Quarry - Dumper Calculations

MC_V1_B201_quarry_3BIU_r3

Quarry Section

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B201_quarry_3BIU_r3.mcd


Crusher capacity

Bulk density of uncrushed stone (limestone) 1.5 -1.6

BDLS 1.5tonne

m3

:=

Hopper to hold material equivalent to crusher feed in terms of time

Ht 15min:=

Number of shovel buckets to fill a Dumper (normal- 3 to 8 )

Bn 5:=

Excavator Shovel Bucket Capacity (volumetric) based on standard availability

Bcs 4.5m3

:=

Standard dumper volumetric capacity Vds 25m3

:=

Dumper turn around time Tdtr 20min:=

Top


Calculate required volumetric capacity of •Dumper

Vd

Vd Bcs Bn⋅:=

Dumper capacity Wds Wds

Capacity of dumping required Cdh

Minimum number of Dumpers required to feed crusher

Nd

Hopper capacity for crusher -( volumetric) based on desired holding time Ht

Hv1

Hopper capacity -( volumetric) based on minimum turnaround time for dumper - Hv2

Selected hopper capacity - the higher value between Hv1 and Hv2

Hv

Maximum possible capacity of each dumper. Wmax

Input data

Estimation of Number of Dumpers and Capacity and hopper capacity.

To find Dumper volumetric capacity

Qcr 1200tonne

hr:=



Nd 12=Minimum number of Dumpers required to feed crusher

Nd 12=

Function ceil returns the smallest integer greater than or equal to x. x must be real.

Nd ceilQcr 1.1⋅

Cdh

:=

NdMinimum number of Dumpers required to feed crusher

Dumpers should be able to feed the crusher continuously at a rate slightly higher than the crusher intake.

Next , calculate the number of dumpers required to feed the crusher•

Cdh 112.5tonne

hr=Dumping capacity

Cdh 112.5tonne

hr=

CdhWds

Tdtr:=

CdhCapacity of dumping

Tdtr 20 min=

Dumper turn around time- i.e the average time taken by a dumper to laod,travel to crusher ,unloadand return to loading point -in minutes

Calculate dumping capacity •

Wds 37.5 tonne=

Wds Vds BDLS⋅:=

Dumper capacity Wds

Vds 25 m3

=Standard dumper volumetric capacity

Dumper capacities are determined by manufacturing standard. So select the nearest standard capacity dumper.

Next , select standard capacity Dumper from •manufacturer catalogue

Vd 22.5 m3

=Required volumetric capacity of Dumper

Vd 22.5 m3

=



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•



Vd 22.5 m3

=

Dumper capacity required, based on turn around time Wds 37.5 tonne=

Capacity of dumping required Cdh 112.5tonne

hr=

Minimum number of Dumpers required to feed crusher based on dumping capacity

Nd 12=

Hopper capacity for crusher -( volumetric) based on desired holding time Hv1 200 m

3=

Hopper capacity -( volumetric) based on minimum turnaround time for dumper - Hv2 266.67 m

3=

Selected hopper capacity - the higher value between Hv1 and Hv2

Hv 266.67 m3

=

Maximum possible capacity of each dumper.Wmax Wmax 200 tonne=

Select dumper capacity to fall between Wmax and Wds

Top

Results

Given data

Qcr 1200tonne

hr=Crusher capacity

Bulk density of uncrushed stone (limestone) 1.5 -1.6

BDLS 1.5tonne

m3

=

Hopper to hold material equivalent to crusher feed in terms of time

Ht 15 min=

Number of shovel buckets to fill a Dumper (normal- 3 to 8 )

Bn 5=

Excavator Shovel Bucket Capacity (volumetric) based on standard availability

Bcs 4.5 m3

=

Standard dumper volumetric capacity Vds 25 m3

=

Dumper turn around time Tdtr 20 min=

Derived data

Dumper volumetric capacity required



Observation

References

Fri Sep 23 1:34:59 PM 2005



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.



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.


Results


Input data



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

MC_V1_B202_Crushing_3BIU

Crushing section

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B202_Crushing_3BIU.mcd


Top

Pn 8.52 kW=Power requirement of crusher at new output as a fresult of changes in product size

Pn 8.52 kW=

Pn Mfn KR⋅1

Dpbn

1

Dpa−

⋅:=

PnNow we like to find new power requirement•

KR 14.58m

3

s2

=

KR

P

Mf

1

Dpb

1

Dpa−

:=

or

P

MfKR

1

Dpb

1

Dpa−

⋅=

Let us consider Rittinger's equation:•


Top

Dpbn 3.8mm:=New product size of rock -volume-surface mean dia.

Mfn 10tonne

hr:=New feed rate to be tried

P 8.95kW:=Power required by crusher for current output

Mf 15tonne

hr:=Current mass feed rate to the crusher

Dpb 5mm:=Product size of rock -volume-surface mean dia.

Dpa 19mm:=Feed size of rock -volume-surface mean dia.

Input data



Fri Sep 23 1:41:29 PM 2005

1)Chapter on "Size Reduction" - Unit Operation of Chemical Engineering2)Hallmark Technologies, Pune

References

Observation

Top

Pn 8.52 kW=New power requirement

Derived data

Dpbn 3.8 mm=New product size of rock -volume-surface mean dia.

Mfn 10tonne

hr=New feed rate to be tried

P 8.95 kW=Power required by crusher for current output

Mf 15tonne

hr=Current mass feed rate to the crusher

Dpb 5 mm=Product size of rock -volume-surface mean dia.

Dpa 19 mm=Feed size of rock -volume-surface mean dia.

Given data

Results



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.


Results


Input data



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_5BIU

Crushing section

MathCement 2000



Fva

Half the distance between the roll surfaces d

Radius of maximum size of feed particle R

ωAngular speed of the rolls

Linear velocity of rolls v

Actual feed rate (mass flow) Mva

Input datar 500mm:=Radius of rolls of the crusherwr 0.4m:=Width of roll face

Distance between the roll surfaces S 12.5mm:=

Angle of nip θ 31deg:=

User defined units

Hz1

s:=


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)



As 5000 mm2

=Cross sectional area of material flow through the rolls

As 5000 mm2

=

As S wr⋅:=

AsNow calculate the cross sectional area of •material flow through the rolls

R 25.36 mm=Radius of maximum size of feed particle

R 25.36 mm=

Rr d+

cos α( ) r−:=

therefore :

cos α( ) r d+

r R+=

αθ

2:=

RRadius of maximum size of feed particle

d 6.25 mm=dS

2:=

dHalf the distance between the roll surfaces

RFind the radius of maximum size of feed •particle


Top

Feed material density ρ 2500kg

m3

:=

f 2Hz:=Roll running frequency

q 12%:=Actual throughput capacity as percent of theoretical throughput



Top

Mva 9.42kg

s=Actual feed rate (mass flow)

Mva 9.42kg

s=

Mva ρ Fva⋅:=

MvaActual feed rate (mass flow)•

Fva 3.77 103−

×m

3

s=Actual feed rate (volumetric)

Fva 3.77 103−

×m

3

s=

Fva q Fvth⋅:=

FvaActual feed rate (volumetric)•

Fvth 0.03m

3

s=Theoretical feed rate (volumetric)

Fvth 0.03m

3

s=

Fvth v As⋅:=

FvthTheoretical feed rate (volumetric)•

v 6.28m

s=

v r ω⋅:=

vLinear velocity of rolls

ω 12.57rad

s=

ω 2 π⋅ f⋅:=

Angular speed of the rolls ω



Fri Sep 23 1:44:26 PM 2005

1)Chapter on "Size Reduction" - Unit Operation of Chemical Engineering2)Hallmark Technologies, Pune

References

Observation

Top

Mva 9.42kg

s=Actual feed rate (mass flow)

v 6.28m

s=Linear velocity of rolls

Angular speed of the rolls ω 12.57rad

s=

R 25.36 mm=Radius of maximum size of feed particle

Derived data

Feed material density ρ 2500kg

m3

=

f 2 Hz=Roll running frequency

q 12 %=Actual throughput capacity as percent of theoretical throughput

θ 31 deg=Angle of nip

S 12.5 mm=Distance between the roll surfaces

Width of roll face wr 0.4 m=Radius of rolls of the crusher r 0.5 m=

Given data

Results



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.


Results


Input data



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


R

ωAngular speed of the rolls


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


q 12:= %

User defined units

Hz1

s:=


rRadius of rolls of the crusher

wrWidth of roll face

Distance between the roll surfaces S

Angle of nip θ


q

Roll running frequency f

ρFeed material density

Theoretical feed rate (volumetric) Fvth

Actual feed rate (volumetric) Fva


Radius of maximum size of feed particle



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


Find the radius of maximum size of feed •particle

R


dS

2:= d 6.25= mm

Radius of maximum size of feed particle R

αθ

2

π

180⋅:=

cos α( ) r d+

r R+=



Fva

Fvaq

100Fvth⋅:=

Fva 3.77 103−

×= m3/s

Actual feed rate (volumetric) Fva 3.77 103−

×= m3/s

Actual feed rate (mass flow)• Mva

Mva ρ Fva⋅:=

Mva 9.42= kg/s

Actual feed rate (mass flow) Mva 9.42= kg/s

Top

ω 2 π⋅ f⋅:=

ω 12.57= rad/s


v r 103−

⋅ ω⋅:=

v 6.28= m/s

Theoretical feed rate (volumetric)• Fvth

Fvth v As⋅ 106−

⋅:=

Fvth 0.03= m3/s

Theoretical feed rate (volumetric) Fvth 0.03= m3/s

Actual feed rate (volumetric)•



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


q 12= %

Roll running frequency f 2= Hz

ρ 2500= kg/m3



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.


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.


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_16BIU

Raw Mill Section

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B204_Rawmill_16BIU.mcd


CO2 17.2%:=Carbon dioxide measured at point

CO2st1 19.7%:=Carbon dioxide measured at point 1 =CO2st1

O2st2 9.5%:=Oxygen measured at point 2 =O2st2

O2st1 8%:=Oxygen measured at point 1 =O2st1

Important: Please enter values as "0" if measument is not done

n 1:=Atleast one set of values must be available

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

G1 35m

3

s:=Gas flow rate at point 1 =G1

Let us calculate leakage air between two points" 1 and 2 "in a system under suction.The gas flows from 1 to 2.

Input data

CO2airCarbon dioxide in air =CO2air

O2airOxygen in air =O2air

CO2st2Carbon dioxide measured at point 2 =CO2st2

CO2st1Carbon dioxide measured at point 1 =CO2st1

O2st2Oxygen measured at point 2 =O2st2

O2st1Oxygen measured at point 1 =O2st1

nNumber of measurement =n



Carbon dioxide measured at point 2 =CO2st2

CO2st2 17.2%:=

Oxygen in air =O2air O2air 21%:=

Carbon dioxide in air =CO2air CO2air 0%:=


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

FACO21_2CO2st1 CO2st2−

CO2st2 CO2air−:=

FACO21_2 14.53 %=



Average of measurement FA1_2

FA1_2FAO1_2 FACO21_2+

n:=

FA1_2 27.58%=

Gas flow rate at point 2 including false air G2

G2 G1 FA1_2 G1⋅+:=

G2 44.65m

3

s=

Quantity of false air entry between point 1 and 2 =FA

FA G2 G1−:=

FA 9.65m

3

s=

Results

Quantity of false air entry between point 1 and 2 =FA FA 9.65m

3

s=

Observation

References



nNumber of measurement =n

G1Gas flow rate at point 1 =G1



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.


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


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:= %


CO2st1 19.7:= %


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


CO2st1


CO2st2

Oxygen in air =O2air O2air

Carbon dioxide in air =CO2air CO2air

Input data


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:=



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).




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



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


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


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

MC_V1_B207_Clinker_9BIU

Clinkerisation Section

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B207_Clinker_9BIU.mcd


CDCO2RM CO2CM−

CO2RM:=

CD Degree of calcination

CO2CM 4 %=CO2CM CO2 content in Partially Calcined Meal

CO2RM 33.96 %=CO2RM CO2 content in Raw Meal

CO2RM 33.96 %=

Mol. wt of CO2 = 44

Mol.wt. of CaCO3 =100

Mol. Wt. of MgCO3 = 84

CO2RM CaCO3RM44

100⋅ MgCO3RM

44

84⋅+

:=

CO2RM Let us denote CO2 content in Raw Meal

TC 0.77=

TC CaCO3RM MgCO3RM+:=

TC Total carbonate


CO2CM 4%:=CO2CM CO2 content in Partially Calcined Meal

MgCO3RM 0.01kg

kgraw:=MgCO3RM Content of MgCO3 in 1 kg. Raw Meal

CaCO3RM 0.76kg

kgraw:=CaCO3RM Content of CaCO3 in 1 kg. Raw Meal

D 0.2kg

kgcl:=Recirculating dust = D kg/kg cl

Input data



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:

Recirculating dust = D kg/kg cl kgcl 1kg:=D 0.2

kg

kgcl=

nett or actual degree of calcination CDnett

CDnett

CD

1 CD−

1

1 CD−D+

:=



References

Observation

CDnett 86.19 %=Nett or actual degree of calcination

Degree of calcination ( refined) apparent CDR 91.9 %=

CD 88.22%=Degree of calcination -apparent

CO2RM 33.96 %=CO2 content in Raw Meal

TC 77 %=Total carbonate

CO2CM 4 %=CO2 content in Partially Calcined Meal

MgCO3RM 0.01kg

kgraw=Content of MgCO3 in Raw Meal

CaCO3RM 0.76kg

kgraw=Content of CaCO3 in Raw Meal

D 0.2kg

kgcl=Recirculating dust

Results

CDnett 86.19 %=

1 CD−



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


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



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



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


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


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:=


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



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




So2c 0=

LOI1c com1 LOI1⋅:= LOI1c 0.35= LOI2c com2 LOI2⋅:= LOI2c 0.01=

Total1c S1c A1c+ F1c+ C1c+ M1c+ So1c+ LOI1c+:= Total1c 0.8=

Total2c S2c A2c+ F2c+ C2c+ M2c+ So2c+ LOI2c+:= Total2c 0.2=

100% Raw mix

Sm S1c S2c+:= Sm 0.14=

Am A1c A2c+:= Am 0.04=

Fm F1c F2c+:= Fm 0.02=

Cm C1c C2c+:= Cm 0.42=

Mm M1c M2c+:= Mm 0.01=

Som So1c So2c+:= Som 0.01=

LOIm LOI1c LOI2c+:= LOIm 0.36=

Totalm Sm Am+ Fm+ Cm+ Mm+ Som+ LOIm+:= Totalm 1=

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⋅:=



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:=



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


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.


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


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


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



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

=



Density of the dry gas = γgas.dry

γgas.dry O2[ ] γO2⋅ CO2[ ] γCO2⋅+ CO[ ] γCO⋅+ SO2[ ] γSO2⋅+ N2[ ] γN2⋅+:=

γgas.dry 1.48kg

m3

=

Dew point temperature = tdew

data

20

30

40

50

60

70

80

85

0.015

0.028

0.05

0.09

0.155

0.29

0.57

0.95

0.0123

0.024

0.043

0.075

0.13

0.24

0.475

0.73

:=

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



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)



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



ρbBulk density of material

VBelt speed

bBelt width

αAngle of inclination for vertical lift

HVertical lift of the conveyor

LHorizontal run of the conveyor


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


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 -

MC_V1_B305_Convey_12BIU

Conveying Systems

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B305_Convey_12BIU.mcd


Mass of carrying side idlers per unit of length mic

Mass of return side idlers per unit of length mir

Secondary resistance co-efficient Ksr

Efficiency of conveying n

Drive factor Kd

Sag of snubbed drive S

KsSag factor

Fri Sep 23 4:21:48 PM 2005

Input data

Horizontal run of the conveyor L 80m:=

Vertical lift of the conveyor H 11m:=

Angle of inclination for vertical lift α 8deg:=

msConveyor carrying rate for design

Width of troughing idlers Wt_idl

Pitch of troughing idlers pt_idl

Width of return idlers Wr_idl

Pitch of return idlers pr_idl

Co- efficient of friction between belt and drum µ

Belt wrap angle over drum θ

Belt friction co-efficient µr1

Load friction co-efficient µr2

Transmission efficiency between motor and drive pulley η t

Input by selection

Mass of belt per unit of length mb



Load friction co-efficient µr2 0.025:=

Transmission efficiency between motor and drive pulley η t 85%:=

Input by selection - for selection of some of the following data refer to earlier file

Mass of belt per unit of length mb 20kg

m:=

Mass of carrying side idlers per unit of length mic 30kg

m:=

Mass of return side idlers per unit of length mir 2.5kg

m:=

Secondary resistance co-efficient Ksr 0.9:=

Efficiency of conveying n 90%:=

Drive factor Kd 1.9:=

Sag of snubbed drive S 3%:=

Sag factor Ks 4.2:=

Fri Sep 23 4:21:48 PM 2005

Belt width b 800mm:=

Belt speed V 1.5m

s:=

Bulk density of material ρb 1000kg

m3

:=

ms 8kg

s:=Conveyor carrying rate for design

Width of troughing idlers Wt_idl 152mm:=

Pitch of troughing idlers pt_idl 1.0m:=

Width of return idlers Wr_idl 102mm:=

Pitch of return idlers pr_idl 3.0m:=

Co- efficient of friction between belt and drum µ 0.1:=

Belt wrap angle over drum θ 240deg:=

Belt friction co-efficient µr1 0.025:=



T2 6576.46 N=

T2 Kd Te⋅:=

T2Therefore slack side tension

Te 3461.3 N=

Te Fr:=

TeEffective tension in the belt

Fr 3461.3 N=

Fr FH 1 Ksr+( )⋅ Fsl+:=

FrTotal resistance to motion

Ksr 0.9=Secondary resistance co-efficient

Fsl 575.32 N=

Fslms

Vg⋅ H⋅:=

Load slope resistance to motion of belt Fsl

FH 1518.93 N=

FH µr1 mic mir+ 2mb cos α( )⋅+( ) L⋅ g⋅ ⋅ µr2ms

V⋅ g⋅ L⋅+:=

FHMain resistance to belt motion

Me 5768.86 kg=

Me mic mir+ 2mb cos α( )⋅+( ) L⋅:=

MeTotal effective mass of all movine part are abtained by summation




V 1.5m

s=Belt speed

b 0.8 m=Belt width

α 8 deg=Angle of inclination for vertical lift

H 11 m=Vertical lift of the conveyor

L 80 m=Horizontal run of the conveyor

Results

Pm 6.11 kW=

PmTe V⋅

η t:=

PmMotor power rating

Transmission efficiency between motor and drive pulley

Tight side tension T1

T1 1 Kd+( ) T2⋅:=

T1 19071.75 N=

Sag tension on carrying side Tsag_c

Tsag_c Ks pt_idl⋅ mbms

V+

⋅ g⋅:=

Tsag_c 1043.43 N=

Sag tension on return side Tsag_r

Tsag_r Ks pr_idl⋅ mb⋅ g⋅:=

Tsag_r 2471.28 N=

Operating power required at driving drum Po

Po Te V⋅:=

Po 5.19 kW=

η t



η t 85 %=

Mass of belt per unit of length mb 20kg

m=

Mass of carrying side idlers per unit of length mic 30kg

m=

Mass of return side idlers per unit of length mir 2.5kg

m=

Secondary resistance co-efficient Ksr 0.9=

Efficiency of conveying n 0.9=

Drive factor Kd 1.9=

Sag of snubbed drive S 3 %=

Sag factor Ks 4.2=

Main resistance to belt motion FH 1518.93 N=

Load slope resistance to motion of belt Fsl 575.32 N=

Bulk density of material ρb 1000kg

m3

=

ms 8kg

s=Conveyor carrying rate for design

Width of troughing idlers Wt_idl 0.15 m=

Pitch of troughing idlers pt_idl 1 m=

Width of return idlers Wr_idl 0.1 m=

Pitch of return idlers pr_idl 3 m=

Co- efficient of friction between belt and drum µ 0.1=

Belt wrap angle over drum θ 240 deg=

Belt friction co-efficient µr1 0.03=

Load friction co-efficient µr2 0.03=

Transmission efficiency between motor and drive pulley



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

Tsag_r 2471.28 N=Sag tension on return side

Tsag_c 1043.43 N=Sag tension on carrying side

T1 19071.75 N=Tight side tension

T2 6576.46 N=Therefore slack side tension

Te 3461.3 N=Effective tension in the belt

Fr 3461.3 N=Total resistance to motion



Input data area locked for demo folder



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.


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

MC_V1_B308_Hydraulics_11BIU

Plant Hydraulics

MathCement 2000

D:\MathCement_demo_pdf\MC_V1_B308_Hydraulics_11BIU.mcd


Area of cross section at throat of ventury

v1Velocity of water at inlet to ventury

v2Velocity of water at the throat of ventury

Input data

Static pressure at inlet to ventury p 150kN

m2

:=

Static head at throat of ventury -in terms of column of mercury

ht 400− mm:=

Percentage of pressure loss in ventury Ploss 3%:=

Dia. of pipeline d 300mm:=

dt 100mm:=Dia. of throat

Density of water ρw 1000kg

m3

:=

Density of mercury ρHg 13600kg

m3

:=


Static pressure at inlet to ventury p

Static head at throat of ventury -in terms of column of mercury

ht

Percentage of pressure loss in ventury Ploss

Dia. of pipeline d

dtDia. of throat

Density of water ρw

Density of mercury ρHg

Flow rate Q

a1Area of cross section at inlet to ventury

a2



∆php1

ρ g⋅

p2

ρ g⋅−=

v22

2 g⋅

v12

2 g⋅−

p1

ρ g⋅

p2

ρ g⋅−

1 Ploss−( )⋅=

∆ph 20.74 m=

∆ph p1h p2h−:=

Difference of static pressure head ∆p

p2h 5.44− m=

p2h htρHg

ρw⋅:=

hptpressure head at throat in terms of water column

p1h 15.3 m=

p1hp

ρw g⋅:=

hppressure head at inlet in terms of water column

p1

ρ g⋅

p2

ρ g⋅−

1 Ploss−( )⋅v2

2

2 g⋅

v12

2 g⋅−=

or

p1

ρ g⋅

v12

2 g⋅+

p2

ρ g⋅

v22

2 g⋅+ Ploss

p1

ρ g⋅

p2

ρ g⋅−

⋅+=

substituting for "loss"

loss Plossp1

ρ g⋅

p2

ρ g⋅−

⋅=

p1

ρ g⋅

v12

2 g⋅+

p2

ρ g⋅

v22

2 g⋅+ loss+=

Since the pipe and ventury are horizontal there is no difference in geodatic heights.

We can write the Barnoulli's equation between inlet and throat of ventury:




Solving for v2 using Solve Block of Mathcad

v2 ∆ph( ) 1 Ploss−( )⋅

a2

a1v2⋅

2

2 g⋅+

2⋅ g⋅=

Substituting in eqn 1 and rewriting:

Substituting for v1 and rewriting the equation 1, we get expression for v2 :

v1a2

a1v2⋅=

a2 0.01 m2

=

a2 πdt

2

4⋅:=

a1 0.07 m2

=

a1 πd

2

4⋅:=

v1a2

a1v2⋅=

v2velocity of water at the throat of ventury

v1velocity of water at inlet to ventury

a2area of cross section at throat of ventury

area of cross section at inlet to ventury a1

where:

a1 v1⋅ a2 v2⋅=

from equation of continuity

∆ph( ) 1 Ploss−( )⋅ 20.11 m=

----------------------------eqn 1v2

2

2 g⋅

v12

2 g⋅− ∆ph( ) 1 Ploss−( )⋅=

or



ρHg 13600kg

m3

=Density of mercury

ρw 1000kg

m3

=Density of water

dt 100 mm=Dia. of throat

d 300 mm=Dia. of pipeline

Ploss 3 %=Percentage of pressure loss in ventury

ht 400− mm=Static head at throat of ventury -in terms of column of mercury

p 150kN

m2

=Static pressure at inlet to ventury

Results

Q 0.16m

3

s=

Q v2 a2⋅:=

QFlow rate

v1 2.22m

s=

v1a2

a1v2⋅:=

v2 19.99m

s=

v2 Find v2( ):=

use function Find

v2 ∆ph( ) 1 Ploss−( )⋅

a2

a1v2⋅

2

2 g⋅+

2⋅ g⋅=

Given

v2 6m

s:=guess value:



Flow rate Q 0.16m

3

s=

a1 0.07 m2

=area of cross section at inlet to ventury

area of cross section at throat of ventury a2 0.01 m2

=

v1 2.22m

s=velocity of water at inlet to ventury

v2 19.99m

s=velocity of water at the throat of ventury

Observation

References Civil Engineering Hydraulics :page 81 -fluid flow concepts

Fri Sep 23 5:01:54 PM 2005



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


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


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.

User defined units

kN 1000N:= kPa 1000 Pa⋅:=

mmHg 133.33Pa:=

bar 760mmHg:=

mmWg1

13.6mmHg:=

D:\MathCement_demo_pdf\MC_V1_B308_Hydraulics_20BIUa.mcd


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

g 9.81m

sec2

≡Gravitational constant

Input data

NPSHNet Positive Suction Head

pStatic pressure on the fluid surface



hfit 1.1 m=Cummulative head loss due to pipe fittings and friction in terms of height of 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

g 9.81m

s2

=Gravitational constant

Results

NPSH can be thus interpreted as the total suction head above the vapour pressure of the fluid

NPSH 7.12 m=

note : z1 is subtracted if the water

level is below the pump centre lineNPSH

p

γ

v2

2g+

z1+ hfit− hp−≡

NPSH is given by the following equation


Discharge tank

Suction line Ref. line

Pump z1

Ground tank



Vapour pressure of fluid at the given temperature

hp 0.65 m=

Static pressure on the fluid surface p 101356 Pa=

Net Positive Suction Head NPSH 7.12 m=

Observation

References Fluid Mechanics for Engineering Technology by Irving Granet pp 290

Fri Sep 23 5:04:35 PM 2005



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


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%:=



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

:=


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



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.


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



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



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=



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=



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−( )⋅:=


MI2 2.95kg

kgcl=

MI2M2

ηst2:=




S3 293.45kg

kgcl=

kg / kg.clS3 MI3 1ηst3

100−

⋅:=


MI3 295.82kg

kgcl=

MI3M3

ηst3

100

:=


M3 2.37kg

kgcl=

M3 MI2 S1−:=



S3 0.59kg

kgcl=

S3 MI3 1 ηst3−( )⋅:=


MI3 2.96kg

kgcl=

MI3M3

ηst3:=


M3 2.37kg

kgcl=

M3 MI2 S1−:=





S4 MI4 1 ηst4−( )⋅:=


MI4 368.67kg

kgcl=

MI4M4

ηst4:=


M4 294.94kg

kgcl=

M4 MI3 S2−:=



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=



S4 MI4 1 ηst4−( )⋅:=

S4 73.73kg

kgcl=


M5 MI4 S3−:=

M5 75.22kg

kgcl=


MI5M5

ηst5:=

MI5 81.76kg

kgcl=


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=



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=




S3 293.45kg

kgcl=


M4 294.94kg

kgcl=


MI4 368.67kg

kgcl=


S4 73.73kg

kgcl=


M5 75.22kg

kgcl=


MI5 81.76kg

kgcl=


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:


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=


M2 2.06kg

kgcl=


MI2 2.95kg

kgcl=


S2 0.88kg

kgcl=


M3 2.37kg

kgcl=


MI3 295.82kg

kgcl=



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



Observation

References

Fri Sep 23 5:38:13 PM 2005



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=


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.


Results


Input data



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


Free air delivery Vfad 3m

3

min:=

Gauge pressure of compressed air Pg 7bar:=

Atmospheric pressure Pb 1bar:=

Top


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


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:=



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)



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



PgGauge pressure of compressed air


mbar 100Pa:=mmWg1

13.6mmHg:=


in2

=


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


Results


Input data



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.






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


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


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:=



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



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:=



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=


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



mbar 100Pa:=mmWg1

13.6mmHg:=


in2

=


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.


Results


Input data



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






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


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


13.6



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



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.


Results


Input data



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

MC_V22_B201_Interest_3BIU

Interest

MathCement_Business

D:\MathCement_demo_pdf\MC_V22_B201_Interest_3BIU.mcd


Pr 151049Rs:=

Effective rate of interest

Top

Calculation algorithmTotal duration of deposit dt

dtn

ny:= n 18=

ny 121

yr=

dt 1.5 yr=

YINominal rate of yearly interest

Balance at the end of the period of deposit Pr

Pr P 1YI

ny+

n⋅=

YIPr

P

1

n

1−

ny⋅:=

YI 15.4393%

yr=

List of parameters usedPPrincipal amount invested

rYearly nominal rate of interest

Number of compounding periods n

Duration of each compounding period of deposit d

Input data

P 120000 Rs⋅:=Principal amount invested

Number of compounding periods n 18:=

Number of compounding periods in a year ny12

yr:=

Final balance at the end of deposit period



n 18=Number of compounding periods

Principal amount invested P 120000 Rs=

Given data

Results

Top

APY 16.58%=

APR 16.58 %=

APY R:=APR R:=

Annual percentage of return (APR) or annual percentage yield (APY) provided n=12

n 18=

Number of compounding done during the period of deposit =n

r 0.011

month=

rYI

12month

yr

:=

rMonthly rate of nominal interest

YI 15.44%

yr=

R 16.5801 %=

RYI

ny1+

ny yr⋅

1−:=

Rewriting

YI R 1+( )

1

ny yr⋅1−

ny⋅=

REffective rate of interest

yr



Fri Sep 23 5:11:56 PM 2005

The Mathcad treasury: Interest rates, Mathsoft Engineering and education Inc.

References

Observation

Top

r 1.29%

month=Monthly rate of nominal interest

Pr 1.51 105

× Rs=Balance at the end of deposit period

R 16.58 %=Effective rate of interest

YI 15.44%

yr=Nominal yearly interest

Derived data

Pr 151049 Rs=Final balance at the end of deposit period

ny 121

yr=Number of compounding periods in a year



monthyr

12:=

USD 48 Rs=

USDRs

Rs_to_USD:=


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.


Results


Input data



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

MC_V22_B202_Annuities_4BIU

Annuities

MathCement_Business

D:\MathCement_demo_pdf\MC_V22_B202_Annuities_4BIU.mcd


P 4yr:=Period of repayment of borrowed fund

Monthly savings to be deposited to bank m 200Rs:=

Top

Calculation algorithmWe shall base our calculation on the formula to calculate future value of an ordinary annuity.

Monthly rate of interest

YI 12 %=r

YI

12:=

r 1 %=

Monthly deposit to be made to bank = m

Number of deposits to bank =n

n P12

yr⋅:=

n 48=

V m1 r+( )

n1−

r⋅:=

V 12244.52 Rs=

month12

:=


Final sum of money paid i.e total outgo V

YIYearly rate of interest

Period of repayment of borrowed fund P

Monthly rate of interest r

Monthly savings to be deposited to bank m

nNumber of deposits to bank

Input data

YI 12%:=Yearly rate of interest



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.

V 12244.52 Rs=

So the future value of the annuity is =V



Derived data

Final sum of money paid to fund V 12245 Rs=

Present value of the annuity, or the amount of loan fundto purchase the machine

Vp 7595 Rs=

Top

Observation

ReferencesMathematics all around, by Thomas L. Pirnot, Chapter 6.3 -Annuities

Fri Sep 23 5:19:37 PM 2005



lifeTotal expected useful life

costTotal cost of procurement of machine / asset


monthyr

12:=

USD 48 Rs=

USDRs

Rs_to_USD:=


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.


Results


Input data



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

MC_V22_B203_Depreciation_3BIU

Depreciation

MathCement_Business

D:\MathCement_demo_pdf\MC_V22_B203_Depreciation_3BIU.mcd


Book value = cost - accumulated depreciation

i 0life

yr..:=

iLet us represent number of completed years by variable

Depreciation

Dp 7.69 %=

using 1 in place of 100 due to Mathcad built in units Dp1

life

yr

:=

DpDepreciation percent by straight-line method

Depreciation_percent100

Expected_useful_life=

Depreciation percent by straight-line method

Assuming that depreciation will be calculated only at the end of the year

Dtot 900 Rs=

Dtot cost Vsalvage−:=

Total amount to be depreciated over the lifetime Dtot


Top

ty 3yr:=Define the period after which book value should be found

Vsalvage 100Rs:=Expected salvage value at the end of useful life

life 13yr:=Total expected useful life

cost 1000Rs:=Total cost of procurement of machine / asset

Input data

VsalvageExpected salvage value at the end of useful life



Dty cost BVn−:=

DtyDepreciation during the defined period

BVn 786.53 Rs=

dividing by "yr" to delete unitnty

yr:=

BVnTo find the book value after a particular period replace the value of n vis-a-vis ty in the input data block

indexed variableBV1 923.08 Rs=

indexed variableBV0 1000 Rs=

BVi

1000

923.08

852.07

786.53

726.02

670.18

618.62

571.04

527.11

486.57

449.14

414.59

382.7

353.26

Rs

=

BVi 1+ BVi Di 1+−:=

Di 1+ BVi Dp⋅:=

Di

Book value at the end of 1st. year is calculated by subtracting the depreciation from the cost at the beginning of the year or the procurement cost.

BV1

BV1 cost cost Dp⋅−:= subscripted variable

BV1 923.08 Rs= subscripted variable

Book value after 2nd year is calculated by subtracting depreciation from the depreciated value at the end of the previous year

BV2

BV2 BV1 BV1 Dp⋅−:= subscripted variable

BV2 852.07 Rs= subscripted variable

We can, therefore, make the following generalised equation

BV0 cost:= indexed variable

BVi 1+ BVi BVi Dp⋅−:=

Depreciation at end of each year



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

BVi 1+ BVi Di 1+−:=

BV1 923.08=BVi

1000

923.08

852.07

533.27

472.77

416.92

365.37

317.78

273.85

233.31

195.88

161.33

129.44

100

Rs

=



References

Observation

Top

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

=Depreciation values after each year of operation

BVn 533.27 Rs=Book value after the desired period

Total amount actually depreciated over the lifetimei

Di∑ 900 Rs=

Dtot 900 Rs=Total amount to be depreciated over the lifetime

Derived data

ty 3yr:=Define the period after which book value should be found

Vsalvage 100Rs:=Expected salvage value at the end of useful life

life 13yr:=Total expected useful life

cost 1000Rs:=Total cost of procurement of machine / asset

Given data

Results

Top



ReferencesBusiness mathematics by Charles D. Miller, Stanley A. Salzman, Gary Clendeman

Fri Sep 23 5:21:45 PM 2005



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?(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

material changes during the process). Source:

Equipment classes

mc_v0_b000_toc_v7_0_1_pdf

Documents

kidney shaped windrows

lime saturation factor

fluidised gravity conveyor

carrying side idlers

return side idlers

problemfalse air entry

recurring deposit separately

equated monthly instalment