Research Article Mathematical Model of a Flash Drying...

Research ArticleMathematical Model of a Flash Drying Process

Erik W Aslaksen

Gumbooya Pty Ltd 3 Gumbooya Street Allambie Heights NSW 2100 Australia

Correspondence should be addressed to Erik W Aslaksen erikgumbooyacom

Received 16 September 2013 Revised 13 December 2013 Accepted 23 December 2013 Published 24 February 2014

Academic Editor Santiago Laın

Copyright copy 2014 Erik W Aslaksen This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The paper presents a basic model of the flash drying process as it is applied in a number of industrial applications and illustratesthis by means of a particular application the drying of subbituminous coal Besides its economic importance that application isrepresentative of those where the product is combustible so that the drying needs to be conducted in an inert atmosphere whichis achieved by recycling some of the drying gas A novel feature of the model is that it takes explicit account of the transport of heatand moisture within the coal particles The model provides the basis for the development of a tool to support the design of a flashdrying plant

1 Introduction

11 Background Flash drying is the process of drying partic-ulate matter by exposing it briefly (typically a few seconds)to a high temperature gas stream resulting in a rapid rateof evaporation without excessive heating of the productThe process is used in various industries such as the foodand wood processing industries and a considerable liter-ature exists on the design and modelling of the process[1] However a literature survey carried out in 2009 whenthis work was undertaken did not find any comprehensivemodels of the process suitable as a basis for plant design[2] In particular while individual research papers treat bothparticle size and shape (see Section 36) and particle-gas heattransfer (see Section 41) these results do not appear to havebeen integrated into a design modelThe present monographfocuses on the application of the process to the drying ofcoal but the basics of the process remain the same for itsapplication in other industries and themodel developed herehas a correspondingly general validity

There are at least three reasons for the focus on coaland in particular on the drying of subbituminous coal asan example for the purpose of developing a detailed modelsuitable for numerical calculationsThe first is that pulverisedcoal is a difficult material to handle and of the many dryersthat have been constructed over the last several decades onlyvery few remain in operation the rest having succumbed

to fire andor explosions or simply been abandoned due torelated operational problems Similar problems arise in someother industries for example in the drying of wood fibres

The second reason is that the last 20 years have seenrenewed interest in drying subbituminous coal with largedeposits being found in the USMidwest in China Mongoliaand on Kalimantan Subbituminous coal lies between ligniteand bituminous coal in this development sequence andis characterised by a water content in the range 10ndash40carbon content in the range 35ndash45 and calorific values inthe range 10ndash20MJkg Compared with bituminous coal ithas the advantage of low sulphur content (typ 025) andlow ash content (typ 5) but its disadvantage is the highwater content and thereby lower calorific value which isdetrimental to two important utilisations of the coal Firstlywhen the coal is burned in a boiler the flame temperatureis lower due to the water vapour and so the boiler surfacesneed to be larger to transfer the required amount of heatConversely if subbituminous coal is used in a boiler designedfor bituminous coal the power rating is reduced So as a con-sequence there would be considerable value in convertingsubbituminous coal to bituminous coal by removing some orall of the water content

Secondly the current concerns regarding CO2emissions

are driving efforts to increase the efficiency of coal as a fuelfor electricity generation and to capture and store the CO

2

Prominent among the former is the gasification of coal in

Hindawi Publishing CorporationJournal of Industrial MathematicsVolume 2014 Article ID 460857 16 pageshttpdxdoiorg1011552014460857

2 Journal of Industrial Mathematics

order to use it as a fuel in combined cycle plants and in orderto gasify lower rank coals such as subbituminous coal andbrown coal (or lignite) they need to be dried as a first stageof the process

12 Structure of the Paper Following the listing of somematerial parameter values in the next subsection the math-ematical model is developed in the following four sectionsSection 2 defines the process and its parameters and looks atits basic thermodynamic properties and operating character-istics Section 3 discusses themechanical interaction betweenthe coal particles and the drying gas and the influence of thesize distribution of the coal particles Section 4 develops amodel of the thermal interaction between the coal particlesand the drying gas including a number of detailed aspectsof this interaction Section 5 is concerned with several issuesthat need to be considered when creating a computer appli-cation for the numerical evaluation of the model

13 Parameter Values Throughout this monograph we shallbe assuming the coal to be a typical subbituminous coal withthe following characteristicsProximate Analysis

total moisture 323 (range 31ndash35)ash 39volatiles 335fixed carbon 303

Ultimate Analysis (Coal Has Been Dried but the Ash IsIncluded)

carbon 683hydrogen 526nitrogen 093oxygen 197sulfur 015total 9434ash 566

The specific heat capacity of this coal is 12 Jg∘CWe shall also be using the following constants and values

for material properties(a) standard atmosphere

(i) temperature 0∘C(ii) pressure 1013 kPa(iii) density 1293 kgm3

(b) composition of dry air (by mass) 7552 N2 2313

O2 129 Ar 006 various

(c) mean molecular weight of air 2897(d) vapour pressure of water as shown in Table 1(e) the heat of vaporisation of water 2255 kJgUnless otherwise stated all equations and expressions

assume that the variables are measured in SI units

Recyclegas

Hot gas

Dried coal

Combustiongas

Dryingcolumn

Fluegas

Crushedcoal

Figure 1 The drying circuit illustrating how a fraction of the gasexiting from the drying process is recycled and combined with thecombustion gas in order for the oxygen content of the hot gas thatenters the drying column to be below the explosion limit that isinert

2 Basic Properties and Relations

21 Drying Circuit and Associated Parameters As was indi-cated in the Introduction an important aspect of this partic-ular model is that the material to be dried that is coal dustis highly combustible This then requires the drying to takeplace in an inert atmosphere (see Section 37) and the inertatmosphere is created by recycling some of the gas exiting thedrying process andmixing itwith the combustion gas tomakeup the hot gas injected into the drying process

The drying circuit consists of a drying column in whichcrushed coal is injected into an upward flow of hot gas inthe lower part of the column At the top of the column amixture of gas and dried coal exits the column and the coal isseparated from the gas A proportion of the gas is exhaustedto atmosphere and the rest the recycle gas is mixed withthe combustion gas to produce the hot gas This is illustratedschematically in Figure 1

211 In All That Follows the Rate of Dried Coal ProducedWillBe Normalised to 1 kgs The parameters of this process are asfollows

moisture content of the crushed coal 119909moisture content of the dried coal 119910ambient temperature 119879

119886(∘C)

temperature of the hot gas 119879119909(∘C)

fraction of exit gas recycled 120578flow of combustion gas 119892 (kgs)

Journal of Industrial Mathematics 3

Table 1∘C 100 101 102 103 104 105 106 107 108 109 110kPa 1013 1050 1087 1126 1166 1208 1250 1293 1339 1385 1432

coal analysis (dry basis)

Carbon 1198881

Hydrogen 1198882

Oxygen 1198883

Ash 1198884

fraction of water in gas 120596fraction of nitrogen in dry gas 120582

The assumptions made about this process which arejustified by practical plant design considerations are asfollows

(i) The fuel used to generate the combustion gas is driedcoal

(ii) The burner operates with 20 excess air(iii) The nitrogen and sulphur content of the coal is

negligible (as far as the drying process is concerned)(iv) The gas temperature at the drying column exit is

140∘C(v) The average temperature of the dried coal is 110∘C(vi) Heat losses are 5 and occur mainly in the furnace

and hot gas part of the circuit(vii) The separation of dried coal from the gas is 100

efficient(viii) The distribution of crushed coal and gas is uniform

over the cross-section of the drying column at everylevel

(ix) Crushed coal and combustion air are both at ambienttemperature

Some of these assumptions will be reexamined in laterchapters and it is not difficult to see what the consequencesof a change to any one of these assumptions would beThe purpose of making these assumptions is to simplify thepresentation of the basic relations and to allow numericalanalysis to be carried out with a reasonable amount of effort

22 Basic Relationships Theamount of crushed coal enteringthe drying column is given by the expression

1 minus 119910

1 minus 119909

(1)

Consequently the amount of water to be extracted fromthe coal evaporated and exhausted to atmosphere per secondis given by the expression

119909 minus 119910

1 minus 119909

(2)

The total exhaust flow is this amount plus the combustiongas flow 119892 and so it follows that the recycle gas flow equals

(119892 +

119909 minus 119910

1 minus 119909

)

120578

1 minus 120578

(3)

and the hot gas flow is just this plus the combustion gasor

119892 + 120578 ((119909 minus 119910) (1 minus 119909))

1 minus 120578

(4)

23 Energy Balance The power required to produce 1 kgs ofdried coal has the following components

heating the coal from 119879119886to 11∘C (1 minus 119910) sdot 12 sdot (110 minus

119879119886)(kW)

heating the water from 119879119886to 100∘C 119909(1 minus 119910)(1 minus 119909) sdot

4184 sdot (100 minus 119879119886) (kW)

vaporising part of the water (119909 minus 119910)(1 minus 119909) sdot

2255 (kW)heating the vapour from 100∘C to 140∘C (119909 minus 119910)(1 minus

119909) sdot 205 sdot 40 (kW)

where the specific heat of coal has been taken as 12 Jg∘Cthat of superheated steam averaged over the temperaturerange of 100∘C to 140∘C as 205 Jg∘C and the heat ofvaporisation of water as 2255 kJg Adding these componentstogether the power required 119864 in kW is

119864 = 2337

119909 minus 119910

1 minus 119909

+ (1 minus 119910)

times [

119909

1 minus 119909

(4184 minus 001119879119886) + 132 minus 12119879

119886]

(5)

This energy has to be delivered by the hot gas as itstemperature declines from 119879

119909to 140∘C that is the difference

in enthalpy per kg of the gas at 119879119909and 140∘C multiplied by

the hot gas flow The specific enthalpy difference is given bythe expression

int

119879119909

140

119888119901(119879) 119889119879 (6)

and so it is necessary to determine 119888119901(119879) the heat capacity at

constant pressure as a function of temperatureUnless stated otherwise the hot gas shall be defined as

having the following composition (fractions by weight)

nitrogen 120582(1 minus 120596ℎ)

carbon dioxide (1 minus 120582)(1 minus 120596ℎ)

water 120596ℎ


where 120596ℎis the water content in the hot gas and the small

amount of oxygen present in the hot gas is counted inwith thenitrogen for the purpose of determining the specific heatThespecific heat of the hot gas is the weighted sum of the specificheats of these three componentsThe individual specific heatsare determined in Section 25 and the value of 120582 will bedetermined in the next subsection it remains to determinethe value of 120596

ℎ

Let the water content of the combustion gas be denotedby 120596119888(see next subsection) then the amount of water in the

hot gas is given by the expression

119892120596119888+

120578 (119909 minus 119910)

(1 minus 120578) (1 minus 119909)

(7)

120596ℎ=

119892120596119888(1 minus 119909) (1 minus 120578) + 120578 (119909 minus 119910)

119892 (1 minus 119909) + 120578 (119909 minus 119910)

(8)

24 The Combustion Gas The quantity of combustion gasrequired to produce the energy 105E (taking into accountthe 5 heat loss) is determined by the calorific value of thedried coal which is again determined by the composition ofthe coal Using Dulongrsquos formula the heating value HV isgiven by

HV = 33950 sdot 1198881+ 144200 (119888

2minus

1198883

8

) (kJkg) (9)

where 119888119894are the component fractions by weight as defined in

Section 211For coal with a moisture content y there are then two

heating values a gross heating value HHV which assumesthat all combustion components are brought back to ambienttemperature and the net heating value LHV which assumesthat the water vapour in the combustion gas is not condensedand which is the one appropriate for the drying processConsider

HHV = HV (1 minus 119910) (10)

LHV = HHV minus 2400 (119910 + 9 sdot 1198882) (kJkg) (11)

Thenext step is to determine the amount of air required tocombust 1 kgs of dry coal The mass flow of dry air requiredfor the stoichiometric combustion of 1 kgs of coal is given by

1198821015840

119886= 435 sdot (267 sdot 119888

1+ 8 sdot 119888

2minus 1198883) (kgs) (12)

The combustion air will contain some moisture If therelative humidity is denoted by 120576 then a good approximationfor the water content 119903 in gm3 at 1013 kPa and ambienttemperature 119879

119886 is given by the expression

119903 = 120576 sdot (0021 sdot 1198792

119886+ 021 sdot 119879

119886+ 49) (13)

obtained by fitting a second-order function to data pointsprovided in the Handbook of Chemistry and Physics [4] Thestoichiometric quantity ofmoist combustion air119882

119886 can then

be closely approximated by

119882119886= 1198821015840

119886(1 +

119903

1200

) (14)

Table 2 Mass of combustion gas components resulting from thecombustion of 1 kg of dry coal with components 119888

119894and using an

excess air fraction 120581

Component Mass (kg)CO2 c1sdot367H2O 119888

2sdot 9 + 000083(1 + 120581)W

119886sdot 119903

N2 W119886(1+120581)sdot(077 minus 000083sdot119903)

O2 120590sdotW119886sdot023

Ash 1198884

Total (1+120581)sdotW119886+1

This value needs to be multiplied by a factor (1 + 120581)where 120581 is the excess air fraction as there needs to be someexcess oxygen in the combustion gas in order to ensurecomplete combustion and a correspondingly low level of COIn accordance with assumption (ii) of Section 21 120581 = 02

The composition of the combustion gas can now becalculated For 1 kg of dry coal combusted the mass of thevarious components is shown in Table 2

If the dried coal has a moisture content of 119910 then thevalues in Table 2 need to be multiplied by (1 minus 119910) in orderto apply to the combustion of 1 kg of dried coal (rather thandry coal)

The third step is to apply a small correctionAs the heatingvalues are defined assuming that the combustion gases arereturned to ambient temperature we would have to correctthis LHV by taking into account the energy lost through thecombustion gas exhausted to atmosphere at 140∘C Insteadwe add that energy lost per second and kg of dried coal to 119864If we take the specific heat capacity of the combustion gas tobe constant and equal to 1 Jg∘C this correction Δ

119890 equals

Δ119890

= 119892 sdot (140 minus 119879119886) (kJkg sdot s) (15)

The amount of fuel required to produce 1 kgs of driedcoal is therefore given by the expression

105 sdot 119864 + Δ119890

LHV(pu) (16)

and combining this with the earlier equations(2591112131415) determines the value of 119892 for anygiven coal

119892 =

105119864 (119882119886(1 + 120581) (1 minus 119910) + 1)

LHV minus (140 minus 119879119886) (119882119886(1 + 120581) (1 minus 119910) + 1)

(17)

The amount of nitrogen (plus oxygen and ash) producedby combusting 1 kg of dried (not dry) coal is determinedby the corresponding expressions in Table 2 or 119882

119886(097 minus

minus0000996 sdot 119902 + 0046 + 1198884)(1 minus 119910) and the expression for 120582 is

120582 = (119882119886(1016 minus 0000996 sdot 119903 + 119888

4) (1 minus 119910)

times (105119864 + 119892 (140 minus 119879119909)))

times (119892 sdot LHV)minus1

(18)


The water content of the combustion gas 120596119888 is also given

by the corresponding expression in Table 2 plus the watercontent in the dried coal

120596119888= ( (119910 + (9119888

2+ 000083 (1 + 120581)119882

119886sdot 119903) (1 minus 119910))

times (105119864 + 119892 (140 minus 119879119909)) )


(19)

Regarding the composition of the hot gas the fraction ofwater 120596

ℎ is now a function of 120578 only in accordance with (8)

However to determine the value of 120578 as a function of the hotgas temperature 119879

119909 we need to know the enthalpy of the hot

gas

25 Heat Capacities and Enthalpy As discussed inSection 23 we require the specific heat capacities asfunctions of temperature for the three components of thehot gas nitrogen carbon dioxide and superheated steamUsing the values given in Gas Tables by Keenan Chao andKaye [20] we find the following very good approximationsfor the temperature range of interest

nitrogen 119888119901

119899(119879) = 100951 + 0000213 sdot 119879 (Jg∘C)

carbon dioxide 119888119901

119888(119879) = 09293

+ 0000413 sdot 119879 (Jg∘C)

steam 119888119901

119904(119879) = 1664 + 00008 sdot 119879 + 324119879 (Jg∘C)

(20)

In addition there is the ash component The specificheat capacity of fly ash is according to published values(eg httpwwwscotashcom) about 075 Jg∘C and ispractically constant with temperature in the range of interesthere However as the ash content is less than 10 the fuelcoal makes up less than 10 of the combustion gas and thecombustion gas makes up less than half of the hot gas it is avery good approximation to just lump the ash in with the drygas (as was already done in (10))

The specific enthalpy of the dry gas component is 120582119888119901

119899 +(1 minus 120582)119888

119901

119888 and the specific enthalpy difference as defined by(6) of this component is given by the expression

ℎ119892= (09293 + 008021 sdot 120582) (119879

119909minus 140)

+ (00002065 minus 00001 sdot 120582) (119879119909minus 140)

2

(21)

The specific enthalpy difference of the superheated steamis given by the expression

ℎ119904= 1664 (119879

119909minus 140) + 00004(119879

119909minus 140)

2

+ 324 sdot ln (119879119909minus 140)

(22)

26 Operating Point The drying energy which is providedby the hot gas depends on the temperature of the gas and the

10

13

15

18

20

23

25

500 550 600 650 700 750 800 850 900

Hot

gas

flow

(kg

s)

Hot gas temperature (∘C)

Figure 2 Rate of flow 119865 in kgs (vertical axis) versus temperature119879119909 in ∘C (horizontal axis) of hot gas for the production of 1 kgs of

dried coal reducing the moisture from 33 to 6

rate of flow of gas That is the same amount of energy can beprovided by a lower temperature and a higher flow rate or ahigher temperature and a lower flow rate For given values of119909 and 119910 the choice of the hot gas temperature119879

119909 determines

the operating point of the dryerThe hot gas flow rate is given by (4) this multiplied by the

specific enthalpy difference must equal the required dryingpower 119864 or

(119892 + 120578

119909 minus 119910

1 minus 119909

) [(1 minus 120596ℎ) ℎ119892+ 120596ℎℎ119904] = (1 minus 120578) 119864 (23)

and inserting (8) for 120596ℎ we find that this relationship

becomes a quadratic equation determining the remainingvariable the recycle ratio 120578

1198861205782+ 119887120578 + 119888 = 0 (24)

where the coefficients are given by

119886 = 119892120596119888(1 minus 119909) (119909 minus 119910) (ℎ

119904+ ℎ119892) + (2ℎ

119892minus ℎℎ) (119909 minus 119910)

2

119887 = 1198922120596119888(1 minus 119909)

2(ℎ119892minus ℎ119904)

+ 119892 (1 minus 119909) (119909 minus 119910) (2ℎ119904+ ℎ119892(3 minus 120596

119888)) + 119864 (1 minus 119909)

119888 = 1198922(120596119888(1 minus 119909)

2(ℎ119904minus ℎ119892) + ℎ119892(1 minus 119909)

2) minus 119864 (1 minus 119909)

(25)

Once the recycle ratio 120578 has been determined the watercontent of the hot gas120596

ℎ is given by (8) and the relationship

between temperature 119879119909 and the rate of flow 119865 of the hot

gas is fixed an example is shown in Figure 2

3 Gas-Particle Dynamics

31 Particle Size Distribution

311 Introduction Material produced by crushing or millingis commonly characterised by a particle size distributiondetermined by sieving that is by passing a sample of thematerial through a succession of sieves 119894 = 1 119899 withdecreasing mesh openings 119886

119894 with 119886

119894gt 119886119894+1

For simplicity


Table 3 Isometric particle characteristics

IsometricParticle Max dimension Volume Surface Sphericity

Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806

Spherewithdiameter 119904

119904 = 119886 05236 1198863 31416 119886

2 10

Tetrahedronwith edges 119904 119904 = 119886cos (15∘) 013054 119886

3 18564 1198862 067

the mesh openings are assumed to be square Let the fractionof the sample contained in sieve 119894 be denoted by 119902

119894 in the limit

of 119886119894minus 119886119894minus1

rarr 0 we can introduce the function 119902(119886) with119902(119886)119889119886 being the fraction of material (by weight) between 119886

and 119886 + 119889119886The particles making up the sample in one sieve will

show a range of shapes and volumes and depending onthe application the sieving results have to be interpretedand applied appropriately In a number of applications it iscommon to treat the particles as if they were spherical witha diameter 119904 determined by the sieve opening that is 119904 = 119886and then correct for the fact that the particles are not reallyspherical by introducing a sphericity factor 120595 defined as theratio of the surface area of a sphere to that of the particle whenboth have the same volume The purpose of Section 31 is toexamine the effect of this approach in the case of flash dryingof particulatematter which involves both the dynamics of theparticles in a gas stream and their thermal interaction withthe gas

312 Isometric Particles Our investigation focuses on thecase where the particles are assumed to be isometric (iehave similar dimensions in three orthogonal directions)The choice is based both on previous published work inpneumatic conveying [5ndash7] and on a visual examination of asample of subbituminous coal that has passed through a cagemill that was adjusted to give nominally a minus3mm productand it is acknowledged that this limits the generality of thisinvestigation as the particle shape depends on both themate-rial and the manner of its processing with milling usuallygiving particle shapes closer to spherical than crushing [8]

Consider the largest (in dimension) isometric particlesthat will pass through a sieve with opening 119886 as shown inTable 3

So if we again consider two sieves the top one withopening 119886 + 119889119886 and the bottom one with opening 119886 theparticles on the bottom sieve will have volumes ranging from02493 to 191 times that of a sphere with volume 120587119886

36 Or

by expressing the volumes in terms of the diameter of sphereswith the same volumes the diameters range from 06294119886 to1241119886

Thedistribution of particle sizes on the sievewith opening119886 is generally not known but in the absence of any furtherinformation the simplest assumption by far is that theprobability density function of particle volumes 119901

119886(119881) as

measured by the diameter of the sphere of equivalent volume

x

s a

1634a

pa(x)

Figure 3 The function 119901119886(119909) representing the contribution of the

fraction of the sample on the sieve with opening 119886 to the density ofparticles with a volume of 12058711990436

0 0241

10

08060670

minus03706

xa

120595(x)

Figure 4 The assumed linear dependence of the sphericity factor120595 on the deviation of diameters 119909 from the sieve opening 119886

is uniform over the range 02493 to 191 times that of a spherewith volume 119881

0= 12058711988636 That is the fraction of the sample

with diameters between 119904 and 119904+119889119904 on the sieve with opening119886 is given by 119901

119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886

Consider now the situation illustrated in Figure 3 inwhich particles contained in the sieve with opening 119886 = 119904 minus 119909

contribute to the density of particles with volume 12058711988636

which will be denoted by 119891(119904)The probability density function 119891(119904) is then given by the

expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)

and 119891(119904)119889119904 is the fraction of particles (by weight) withvolumes between 120587119904

36 and 120587(119904 + 119889119904)

36 The boundaries of

119901119904minus119909

(119909) are given by 119904 = 06294 (119904 minus 119909) and 119904 = 1241 (119904 minus 119909)or minus059119904 and 0194119904

The distribution 119891(119904) is of interest in many applicationsin particular because if 120588 is the density of the particlesubstance (eg 1300 kgm3 for coal) then the specific numberof particles (ie per unit mass) with diameter between 119904 and119904 + 119889119904 119899(119904)119889119904 is given by

119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)

The value of the sphericity is indicated in Table 3 and ifwe assume a linear relationship between sphericity and thevariable 119909 as shown in Figure 4 we find that the averagesphericity 120595 has the value 0818

The average sphericity of the particles is a somewhatcontentious issue and in light of the great differences in


Table 4 Sieving results on a sample of sub-bituminous coal milled to minus3mm together with the fractions calculated by fitting a Weibulldistribution with 120582 = 1065 and 119896 = 1115

Sieve number 1 2 3 4 5 6 7 8Mesh opening (mm) 2000 1000 0500 0250 0125 0063 0031 0016Fraction retained 0136 0268 0231 0124 0103 0079 0037 0022Calculated fractions 0155 0255 0243 0163 0091 0047 0023 0011

sphericity between particle shapes one might question ifusing an average is appropriate at all That will depend on theapplication (eg abrasion conveying drying combustionetc) but if an average value is used how should it bedetermined The shape distribution investigated above givesa value of 0825 for isometric particles However the valuesreported in the literature are 078 [9] and 073 [10] and in[11] it is suggested that if no more accurate information isavailable one should use 120595 = 07 The difference betweenthe two values 0825 and 07 is probably to be found inirregularities in real particle surfaces as compared with thesmooth surfaces of the geometric shapes

313 The Probability Density Function 119902(119886) In order todetermine the pdf 119891(119904) through integration as indicated by(27) we require an analytic expression for the pdf 119902(119886) andit is common practice to characterise particulate matter byapproximating the result of a sieving analysis by the two-parameter Weibull pdf (also called a Rosin-Rammler pdf)

119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)

where 120582 is the shape factor and 119896 is the scale factor Thefraction retained on the 119894th sieve is given by

119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)

fitting these values of 119876119894to the measured values then

determines 120582 and 119896To see what this means in practice consider the following

results obtained on subbituminous coal milled to minus3mmsize as shown in Table 4

The rms error of the fit shown in Table 4 is 21 but itis obvious that individual values differ considerably more asevidenced for example by the values for sieve number 8A pdf that gives a better fit is the following four-parameterfunction

119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)

and a comparison of the two functions with the sieving resultis shown in Figure 5 where the parameter values are 119888

1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083

314 The Diameter pdf Returning to our original quest thatof determining the function 119891(119904) defined in (27) we can nowuse the analytical expressions for 119902(119886) Using the Weibull pdf

0000

0050

0100

0150

0200

0250

0300

1 2 3 4 5 6 7 8Sieve number

Frac

tion

reta

ined

Figure 5 Comparison of the Weibull approximation (dark centralbars) and the four-parameter function (bars on the right) with themeasured values (bars on the left) showing that the four-parameterfunction provides a significantlymore accurate representation of themeasured values as compared with the Weibull distribution

0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s

Equivalent spherical particle diameter s (mm)

Figure 6 The full curve is the best fit of a Weibull pdf to theexperimental sieving results the dotted curve is the resulting pdfof spherical diameters

with the values of 120582 and 119896 given in the caption of Table 4numerical integration gives the result shown in Figure 6

The result is as one would expect a shift of the pdftowards smaller diameters and performing the same calcula-tion using the four-parameter function gives the result shownin Figure 7

In these two figures the shift looks almost insignificantbut if we form the ratio of the shifted values to the originalfor each value of 119904 as shown in Figure 8 we see that thecorrection is in the range plusmn15 Expressed in terms of theaverage diameter there is a shift from 0316mm to 0279mmor 117 in the case of the Weibull pdf and from 0273mmto 0242mm or 114 in the case of the four-parameterfunction

We shall return to this issue of particle shape and sizedistributions and their influence (if any) on the drying


0001002003004005006007

001 01 1 10

pdf v

alue

s


Figure 7The full curve is the best fit of a four-parameter pdf to theexperimental sieving results the dotted curve is the resulting pdf ofspherical diameters

08

09

1

11

12

13

001 01 1 10Sieve opening or particle diameter (mm)

Cor

rect

ion

fact

or

Figure 8 The correction factor that must be applied to the sievingresults in order to obtain the distribution of the equivalent sphericalparticle diameter 119904 assuming a uniform distribution of isometricparticle shapes The full curve applies to a Weibull distribution ofparticle sizes and the dotted curve to a four-parameter distribution

process as the development of our understanding of theprocess progresses first in Section 36 and then in Section 5

32 The Drag Force The force exerted by a gas stream withvelocity V

119903relative to the particle and density 120590 on a particle

with cross-section 119860 is given by the expression

119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)

where 119862119889 the drag coefficient is a factor that depends on the

shape and diameter of the particle as well as on V119903 Assuming

for the moment a spherical shape the dependence of 119862119889on

diameter and velocity is usually expressed in terms of theReynolds number Re

Re =

119904V119903

] (33)

where 119904 is the particle diameter and ] is the kinematic vis-cosity 120583120590 A commonly used expression for the dependenceof 119862119889on Re in the range of Re of interest (1 to 1000) is the

following [12]

119862119889= 04 +

26

(Re)08 (34)

We shall return to the issue of nonsphericity inSection 34 but to determine 119865V as a function of gasvelocity we first need the gas density and viscosity

Table 5 Densities of drying gas components

Component Density at 100∘C and 101 kPa (kgm3)CO2 1447N2 0915H2O 0596

33 Drying Gas Characteristics The viscosity of the threecomponents of the gas in the drying column (nitrogencarbon dioxide and steam) can be found in Appendix A of[3] and by fitting linear relations to these values in the rangeof temperature of interest 140ndash900∘C we find the followingdependence on temperature (note that for brevity the valuesof 120583 have been multiplied by 107)

nitrogen 120583 = 6485 + 0365 sdot 119879

carbon dioxide 120583 = 3673 + 0380 sdot 119879

steam 120583 = minus982 + 03606 sdot 119879

(35)

The viscosity of the gas is approximately equal to theweighted average of these viscosities or

120583 = 119886119879 + 119887 (36)

with

119886 = 0365 sdot 120582 (1 minus 120596) + 038 (1 minus 120582) (1 minus 120596) + 03606 sdot 120596

119887 = 6485 sdot 120582 (1 minus 120596) + 3673 (1 minus 120582) (1 minus 120596) minus 982 sdot 120596

(37)

For simplicity and as a very good approximation it willbe assumed that the pressure within the drying column isconstant and denoted by 119901

0 Denoting the densities of the

gas components at 100∘C and 101 kPa by 120590119894 and the partial

pressures by 119901119894 the density of the gas in the drying column

120590 at temperature 119879 is then finally given by

120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)

Values for the densities 120590119894are given in Table 5

Letsum = 12059618 + 120582(1 minus 120596)28 + (1 minus 120582)(1 minus 120596)44 then thepartial pressures are given by the expressions

nitrogen 00357 sdot 1199010sdot 120582(1 minus 120596)sum

carbon dioxide 00238 sdot 1199010(1 minus 120582)(1 minus 120596)sum

steam 00556 sdot 1199010sdot 120596sum

34 Equations of Motion With the characteristics of thedrying gas now determined the drag force acting on a singleparticle of diameter 119904 due to the relative velocity V

119903is also

determined and can be written in the form

119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)


where 119897 is the vertical coordinate within the drying columnmeasured upwards from the point at which the crushed coalis injected into the drying column and all the parameters onthe right-hand side are dependent on 119897

In addition to the drag force there is the force of gravity119865119892 acting on the particle it is given by

119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)

where 120588 is the density of the particle The equation of motionof the particle is then

119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)

where V is the upward velocity of the particle However thesubsequent numerical calculations require V as a function of 119897rather than of 119905 or119889V119889119897 = 119889V119889119905sdot119889119905119889119897 So consider a particlewith velocity V(119897) at position 119897 andmoving to 119897+119889119897 in the time119889119905 at which point its velocity is V(119897 + 119889119897) = V(119897) + 119889119897 then

119889V119889119897

=

119889V119889119905

sdot

1

V (42)

and the equation of motion is now

119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)

As it is usually assumed that the particles are injectedhorizontally into the drying column the initial condition isV(119897 = 0) = 0 in which case (43) implies that 119889V119889119897 = infinwhich is clearly not useful and the way around this apparentproblem is discussed in Section 52 However it is of interestto investigate the time and distance scales we are confrontedwith in this initial part of the drying process and to that endwe can define a characteristic distance 120575 as the distance aparticle would travel until its velocity reaches the gas velocityif the acceleration remained constant as given at 119905 = 0

For convenience in this investigation we shall take thedependence of 120583 and 120590 on the temperature 119879 [∘C] to beapproximated by the two expressions 120590 = 310(119879 + 273)

and 120583 = (150 + 0355119879) sdot 10minus7 That is we are neglecting

the influence of the changing gas composition but this is nottoo bad of an approximation as can be verified using valuespublished in Appendix A of [3] and is also demonstrated bythe actual calculations in the model (see Section 42) Withthese approximations the expression for 119862

119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)

where the relative velocity is now simply the hot gas velocityV and the function 119891(119879) is closely approximated by theexpression

119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)

The acceleration at 119905 = 0 is then given by the expression

[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)

Table 6 The characteristic length at the injection point of thecrushed coal in meters as function of the hot gas temperature 119879in ∘C and the particle diameter 119904 in mm for a hot gas velocity of25ms

ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718

and the characteristic distance is just V2 divided by twice thisacceleration As an example the characteristic length (m) as afunction of gas temperature (∘C) and particle diameter (mm)is shown in Table 6 for the case of a hot gas velocity of 25ms

As the values in Table 6 demonstrate for very smalldiameters (eg 119904 lt 01mm) the acceleration at the injectionpoint is so great that it is practically impossible to model theparticle trajectorieswith any useful accuracy and the simplestapproach is to say that the particles attain their ultimatevelocity (the gas velocity minus the entrainment velocity seenext section) in a linear manner within the first ldquoslicerdquo of thedrying column (see Section 52 for a definition of ldquoslicerdquo)

As the particle travels upward in the drying column itdries by ejecting water in the form of steam and the watercontent of the particle initially equal to 119909 decreases to 119908There are then two extreme possibilities either the particlevolume remains unchanged and the density decreases or thevolume decreases in proportion to the ejected water andthe density remains unchanged The actual situation may beanywhere between these two extremes and to account for thiswe introduce the shrinkage parameter 120576 with 0 le 120576 le 1such that a value of 1 means that the particle shrinks by allthe volume of ejected water and a value of 0 means that thereis no shrinkage

Consider a coal particle to consist of two componentswater (initially a fraction 119909) and the rest (about half-and-halffixed carbon and volatile matter with a small amount of ashand some air in the form of porosity) Initially the particlehas a diameter 119904

0 a volume 05236 sdot 119904

0

3 and a density 1205880

and the mass of the rest is 05236 (1 minus 119909)12058801199040

3 Assuming theabsence of any devolatilisation the mass of this componentremains constant as the water is evaporated and let 119908 bethe fraction of the water remaining Then for the case 120576 =0 the density is given by 120588 = 120588

0(1 minus 119909(1 minus 119908)) and the

diameter remains unchanged for the case 120576 = 1 the diameteris given by 119904 = 119904

0(1 minus 119909120588

0(1 minus 119908))

13 and the density by 120588 =

1205880(1 minus 119909(1 minus 119908))(1 minus 119909120588

1015840

0(1 minus 119908)) For the general case

120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840

0is the density relative to that of water (ie 1205881015840

0=

12058801000) This shrinking is discussed further in Section 54


Table 7 The entrainment velocity V0 in ms as a function of the

particle diameter (columns in mm) and the gas temperature (rowsin ∘C)

01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365

35 Entrainment Velocity The entrainment velocity V0 for

a coal particle in the drying column is that gas velocity forwhich the particle does notmove in the vertical direction thatis the velocity at which the drag force equals the gravitationalforce or

119865119889(V0) minus 981119898 = 0 (48)

where 119898 is the mass of the particle and the gravitationalacceleration equals 981ms2 Or using (32) for 119865

119889and if we

for the purpose of this subsection assume an average densityfor the coal particles in the drying columnof 1300 kgm3 thenwe obtain

V0= 1304radic

119904

120590119862119889

(49)

If we further for the present purpose set 120576 = 0 and use(44) for 119862

119889 (49) becomes a nonlinear equation for V

0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)

A set of results is shown in Table 7For the purpose of the numerical model of the drying

process (see Section 42) the results in Table 7 can be approx-imated by the following expression

V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)

The drying column diameter can then be chosen so as toensure that the drying gas velocity is a certain amount abovethe normalised entrainment velocity the multiplication

factor is the entrainment assurance factor with a valuetypically in the range 125ndash15

As discussed in the previous section small particles (iewith 119904 lt 01mm) are accelerated so rapidly at the injectionpoint that the numerical integration of the equation ofmotionbecomes impractical Instead we shall simply assume thatthey attain the entrainment velocity linearly in the firstintegration step (or ldquoslicerdquo) and set the entrainment velocityequal to 02ms

It is interesting to note some entrainment velocitiesreported in the literature In Wypych [13] we find foralumina pellets with a diameter of 01mm in air at 20∘C and101 kPa the two values 074ms and 079ms whereas forcrushed coal at 6mm diameter the values are 1565ms and1589ms

36 The Significance of Particle Shape As already discussedin Section 31 the fact that the crushed coal particles injectedinto the drying column are not spherical while most of thetheoretical and experimental information relates to sphericalparticles has led to a somewhat peculiar situation First of allif the particles are characterised by the diameter of sphericalparticles of the same volume (or mass) the distribution (bymass) of these ldquoequivalentrdquo particles differs slightly from thatobtained directly from the sieving results Secondly many ofthe results obtained for these ldquoequivalentrdquo spherical particleshave to be corrected for the lack of sphericity A review of thesubject matter of particulate flows as of the year 2000 can befound in [14 15]

Starting with the dynamics any shape departing fromthat of a sphere will have a drag coefficient greater thanthat of a sphere of the same volume and while the body ofexperimentally determined values of the drag coefficient isnot as extensive as for spherical particles there is a reasonableunderstanding of the influence of particle shape much of itbased on the seminal experiments of Pettyjohn and Chris-tiansen [16] In particular it is accepted that the influenceof nonsphericity on the drag coefficient can be describedby a single parameter the previously introduced sphericity120595 and a function proposed by Ganser [17] and Haider andLevenspiel [18] has been shown by Hartman et al [5] to givean excellent representation of the experimental data Basedon data presented in [14 15] the influence of sphericity onthe drag coefficient as a function of the Reynolds numbercan be determined and is shown in Figure 9

37 Oxygen Level Asmentioned in Section 11 in those caseswhere the material to be dried is combustible and potentiallyexplosive in pulverised form a central issue in the control ofthe drying process is maintaining an inert atmosphere or inother words maintaining a low oxygen content in the dryinggas The oxygen content is determined by a number of theprocess parameters but the general situation can be assessedin terms of the basic process diagram shown in Figure 10(refer also to Figure 1)

Then using the notation for the flows indicated inFigure 10 and using the notation introduced in Section 21


000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r

Figure 9 The ratio of drag coefficient for a nonspherical particleto that of a sphere as a function of the Reynolds number (Re) forsphericity values of (from bottom to top) 095 09 08 06 and 04based on data given in [3]

b OR

120578b OR

g OC

a

(1 minus 120578)b OR

OH

Figure 10 Gas circuit with flows (by weight) and oxygen contents

we obtain the following expressions relating to the variousquantities

119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)

The oxygen content of the combustion gas 119874119862 will

depend on the fuel used to generate the gas as expressed bythe quantity of air needed for stoichiometric combustion butto a first approximation it can be taken to be the followingfunction of the excess air 120581

119874119862= 023

120581

11 + 120581

(56)

which with assumption (ii) in Section 21 gives 119874119862

asymp 35The value of 119892 is determined by the fuel and given by (17)

Heat conduction zone

0Distance from centre

100∘C

30∘C

0∘Cs2

t1

t2

T

Water evaporation zone

Steam transport zone

Figure 11 Simplified representation of the thermal processes takingplace in a single particle of diameter 119904 at some time 119905

2after injection

into the drying column with the temperature shown as a functionof distance from the centre of the particle The temperature curvelabelled 119905

1indicates the situation just after the particle is injected

into the drying column

and so for any given case the oxygen level 119874119867 can be

determined for the range of operating points and provideassurance that it will remain below the explosion limit

4 Heat Transfer

41 Heat Transfer between the Gas and a Single ParticleA particular feature of the present model is that it takesexplicit account of the fact that the majority of the moistureis contained within the particles This is in contradistinctionto most commercial designs which assume that the moistureis mainly surface moisture [19]

Consider a single coal particle taken for the followingdevelopment to be spherical with diameter s It is initially atambient temperature say 30∘C but on introduction into thedrying column its surface temperature will go immediatelyto just over 100∘C and water on or near the surface startsto evaporate as shown in Figure 11 As time progresses theevaporation moves inward and the volume between theevaporation and the surface serves to transport the steamwhich becomes gradually more superheated to the surfaceFrom the evaporation zone heat is also conducted inwardheating the coal there and the water contained in it

The picture presented in Figure 11 contains the majorassumption that as long as there is water left in the particle toevaporate the surface temperature will not rise appreciablyabove 100∘C That is the effective thermal resistance of whatis labelled as the steam transport zone in Figure 11 is muchless than the thermal resistance equivalent to the heat transferbetween the gas and the surface of the particle

The energy (heat) transferred from the gas to a single par-ticle of diameter 119904 per second is reasonably well represented


by the following expression [3]

119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)

and ℎ1015840 = heat transfer coefficient (Wm2∘C) 119896 = thermal

conductivity of gas (Wm ∘C) 120572 = thermal diffusivity k120590c119901

(m2s) 120583 = viscosity (Nsm2) 120583119904= viscosity at the tem-

perature of the particle surface (110∘C) and ] = kinematicviscosity 120583120590 (Nsdotssdotmkg)

The value of Pr can be taken as constant and equal to 07(see eg Table 2 of [20]) An expression for 120583 was given in(37) and the thermal conductivity 119896 as given for example inthe Table cited above can be approximated by the expression(271 + 0052 sdot 119879) sdot 10minus3 The Reynolds number Re wasintroduced in (33)

With this the energy transferred per second to a particlewith diameter 119904 and located at a position in the drying columnwhere the gas is characterised by 120582 120596 and 119879 and where therelative velocity is V

119903can be written as

ℎ (119904 120582 120596 119879 V119903) = (851 + 0163 sdot 119879) (119879 minus 110) sdot 119904 sdot 10

minus3

times [2 + 0867 (04 sdot Re12 + 006 sdot Re23)

times (

120583

120583119904

)

14

]

(59)

We can now check if the assumption implicit in Figure 11is justified By evaluating (59) we find the energy transferredper second ℎ(119904) (W) to a particle of diameter 119889 at a gas tem-perature of 618∘C (an arbitrary but typical gas temperature atthe beginning of the drying process) is as shown in the secondcolumn of Table 8 The third column displays the equivalentthermal resistance 119877

119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be

the thickness of a surface layer of the particle expressed asa fraction of 119904 then the thermal resistance of this layer 119877

119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)

where we have taken the thermal conductivity of the subbi-tuminous coal to be 033 Wm∘C [21] If this is evaluated for120575 = 0103 which corresponds to the layer having a volumeequal to half that of the particle (and which can be takenas an average distance the heat has to penetrate in orderto evaporate water) the results are displayed in the fourthcolumn

The data presented in Table 8 shows that our assumption119877119904119877119888

≫ 1 is not justified except for very small particles

Table 8 Heat transferred per second to a particle of diameter 119904h(s) in a gas of temperature 618∘C and the corresponding surfacethermal resistance 119877

119904 and internal coal resistance 119877

119888 as well as the

ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788

It appears that the surface temperature is not always slightlyabove 100∘C but somewhere between 100∘C and 100 + (618 minus

100)119877119888(119877119904+119877119888)∘C and that wemay not be able to neglect the

superheating of the steam inside the particle To investigatethe latter issue let the heat flow that reaches the evaporationzone be denoted by ℎ

119890(Js) then the amount of steam

generated per second in the particle equals ℎ1198902255000 (kgs)

The energy per second used to superheat this steam to thesurface temperature 119879

119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where

the specific heat capacity of steam has been taken to be equalto 2 kJkg ∘C Consequently the influence of the superheatingon the drying of a particle is a reduction by a factor givenapproximately by the expression

ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)

This reduction is not negligible but our approach is nowto at first neglect it in the determination of ℎ and then applyit as a correction factor once a value of ℎ has been found bythe following successive approximation method

Consider a particle of diameter 119904 Assuming that themoisture is distributed uniformly throughout its volumethen if the remaining fraction of the moisture is 119908 thedistance of the evaporation zone from the surface is equal to119904(1minus119908

13)2 Using (60) the corresponding effective thermal

resistance 119877119888 is then given by

119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)

The surface temperature 119879119904 is just 100 + ℎ sdot 119877

119888and

inserting this into (59) we can in principle solve for thecorrected value of ℎ However in that equation 119879

119904not only

is substituted for our assumed value of 110∘C but also entersinto the expression for 120583

119904 so that we cannot get an explicit

expression for ℎ Instead we assume a value ℎlowast calculate ℎand then reduce the difference between ℎ and ℎ

lowast by successiveapproximation Once the value of ℎ has been determinedthe correction factor in (61) is applied before calculating theamount of water evaporated

The steam leaving the particle is now already superheatedto the surface temperature119879

119904 and so needs only to be further


superheated to the gas temperature This is calculated using(22) but with 119879 minus 119879

119904substituted for 119879

119909minus 140

42 Macroscopic Heat Transfer Turning now from a singleparticle to the two-component fluid (gascoal) in the dryingcolumn consider a columnwith internal cross-section119860 andlength 119871 (measured from the coal injection point) At anylevel 119897 in the column let the density of coal consisting ofparticles with 119904 in the range 119904 to 119904 + 119889119904 be denoted by 120585(119904 119897)119889119904

(measured in kgm3) then

120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)

is the total density of coal particles at the level 119897 Also as theparticles move with a velocity V(119904 119897) the total flow of coalthrough a cross-section of the column at level 119897 is given bythe expression

119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)

and at the injection point 119897 = 0 this must equal the rate ofcrushed coal injection to be denoted by 119872

0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)

However as the vertical particle velocity at that point iszero it would seem to imply that the density is infinite Toovercome this situation which arises only as a result of ouridealised treatment of the injection process our numericalcalculation will consider ldquoslicesrdquo of the drying column withthickness 119889119897 and use the average velocity within that slice (seeSection 53)

Themass of a particle with diameter 119904 is 05236120588sdot1199043 so the

density (in particles per m3) of particles with 119904 in the range119904 + 119889119904 is 120585(119904 119897)119889119904 divided by this quantity Then for a ldquoslicerdquoof the drying column of thickness 119889119897 the energy transferredbetween gas and the coal particles with diameters in the range119904 to 119904 + 119889119904 per second119867(119904 119897)119889119897 sdot 119889119904 is given by

119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)

In (66) V119903is determined by using (14) and 120588(119904 119897) is

determined by (15) Of course both quantities depend onthe drying gas velocity V

0(119897) which itself depends on 119867(119897)

so because of this coupling as well as the highly nonlinearnature of the expressions for the various quantities involvedin (32) there is no analytic expression for 119867(119897) Therefore inorder to convert the model of the flash drying process devel-oped in the foregoing into a practical tool for plant designone needs to create a corresponding computer applicationThat is in itself not a major or difficult task but the nextsection raises a number of issues that need to be taken intoconsideration when developing the program

Table 9

The index (119868) means

The value in the middleof the 119868-th slice for

The value at theend

of the 119868-th slice forAcceleration Temperature

Velocity

5 Numerical Evaluation Issues

51 The Independent Variables The basic function of anynumerical representation of the model is to carry out thenumerical integration of (66) which in effect means thecoupled integration of the dependent variables involvedin that equation such as temperatures velocities and gascompositions These dependent variables are functions ofthe two independent variables 119897 position along the dryingcolumnmeasured from the injection point and 119904 the particlediameter These two continuous variables are representedas discrete variables by two indices (119868) for position alongthe column and (119869) for the particle diameter The positionintegration steps take into account that gas temperature andparticle velocities vary much more rapidly at the bottom ofthe column than towards the top the ldquoslicesrdquo of the columnhave centre values 119883(119868) and step sizes DL(119868) (both asfractions of the column length 119871) and for a choice of 24ldquoslicesrdquo a suitable division is shown in Table 10

Note that when a variable say 119886(119868) is a function ofposition along the column it can mean either its value in themiddle of the 119868th slice or at the end of the slice as shown inTable 9

The particle diameter 119904 is represented by discrete classescorresponding to the sieving results An example of anapplication with eight classes (ie 119869 = 1 to 8 with 1corresponding to the largest particles and 8 to the smallestas used already in Table 8) is shown in Table 11 if the sievingresults are initially available in some other form they musteither be converted to the chosen number of classes or theapplication must be designed to handle a variable number ofclasses

52 Integration Algorithm In developing a program animportant observation is that of the many parameters thatvary with position within the drying column only two arerapidly varying the gas temperature 119879(119868) and the particlevelocities V(119869 119868) Consequently the changes in these twovariables from the start of a ldquoslicerdquo to the end of the ldquoslicerdquoare determined using what is in effect an improved Eulerintegration algorithm first based on the values of all the otherparameters at the beginning of the ldquoslicerdquo then at the end ofthe ldquoslicerdquo and then using the average to compute the values ofthese two variables at the end of the ldquoslicerdquoThe values of all theother parameters are determined at the end of this algorithmA step in the integration therefore takes the form shown inFigure 12

The differential equation for the velocity function is givenby (41) however for the numerical integration we need to


V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean

Calculateheat transfer T(I)

Calculate allother parameters

All parameter

CalculateCalculate

CalculateTmean



I minus 1 IProcess-step I

T(I)

T(I)

Calculateheat transfer

Tmean

values I minus 1

Fm (J I minus 1) (J I)

(J I)

mean(J)

mean(J)

Fm(J I)

Figure 12 Calculation process for a single step in the numerical integration of (66)

proceed as follows Consider themovement of a particle from119897 to 119897+119889119897 in the time 119889119905 with the velocity changing from V(119897) toV(119897+119889119897) by 119889V under the acceleration119865119898Then 119889V = 119889119905sdot119865119898

and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)

The solution to this equation is

V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)

53 Initial Values In order to start the recursive integrationprocess it is necessary to generate the starting values that isfor 119868 = 1 As wasmentioned following (43) it is generally truethat V(119868 = 0) = 0 which is not an admissible value as far as theintegration algorithm is concerned so a different approach isrequired for the first step It consists of first calculating thegas density at 119909 = 0 and then the resultant gas velocity (asthe mass flow is known from the input data) The particleaccelerations at 119909 = 0 then follow from (46) and this valueis used to determine the particle velocities and densities at119909 = 119909(1) The heat transfer in the short distance from 119909 = 0

to119909(1) is calculated using themean of these particle velocitiesand densities

54 Unequal Drying Particles with smaller diameters willdry faster than the larger ones and below some diameter theparticles will be completely dry prior to reaching the end ofthe drying column This is taken into account by first calcu-lating the initial (ie at injection) amount of water containedin each particle class and then keeping track of the cumulativeamount of water evaporated from each class At the pointwhere the cumulated amount reaches the initial amount theheat required to elevate the coal temperature from its surfacetemperature to the gas temperature is subtracted from the gasand from then on the coal in that class gives up heat as itprogresses upward in the drying column in equilibrium withthe gas Heating the coal in a class within a single ldquoslicerdquo isof course an approximation that leads to a local distortionin the temperature profile but the error in the overall dryingprocess is small

55 Particle Shrinkage The simple picture of coal particlesprogressing upward with only their water content and tem-perature changing is complicated by the fact that as water isevaporated the coal particles shrink and so a fraction of theparticles in a class move into the next lower class (ie highervalue of 119869)

Consider the ldquoslicerdquo of the column labelled by 119868 anddenote the mass throughput of coal in the size class 119869 inthe bottom of the ldquoslicerdquo by 119872(119869 119868 minus 1) (refer to (64)) andthe amount of water evaporated from the size class 119869 in that


Table 10 The subdivision of the length of the drying columninto 24 discrete ldquoslicesrdquo for the purpose of performing a numericalintegration of (66)

119868 119883 Step1 0001 0000ndash00022 0003 0002ndash00043 0010 0004ndash00164 0022 0016ndash00285 0034 0028ndash00406 0050 004ndash0067 0070 006ndash0088 0090 008ndash0109 0110 010ndash01210 0130 012ndash01411 0150 014ndash01612 0170 016ndash01813 0190 018ndash02014 0225 020ndash02515 0275 025ndash03016 0325 030ndash03517 0375 035ndash04018 0425 040ndash04519 0475 045ndash05020 0550 050ndash06021 0650 060ndash07022 0750 070ndash08023 0850 080ndash09024 0950 090ndash100

Table 11 For the purpose of the numerical calculations the particlesize 119904 is divided into discrete classes These have been chosen soas to correspond to the number and mesh sizes of sieves commonlyused to determine the particle size distribution of the crushed coal

119869 119904 Δ119904

1 3 2ndash52 15 1-23 075 05ndash14 0375 025ndash055 01875 0125ndash0256 009375 00625ndash01257 0046875 003125ndash006258 0023438 001ndash003125

ldquoslicerdquo by DW(119869 119868) As a result of the evaporation the massof the coal is reduced by DW(119869 119868) that is119872(119869 119868) = 119872(119869 119868 minus

1) minus DW(119869 119868) and the volume of the coal particles will alsobe reduced However experience shows that the reductionis somewhat less than the volume of evaporated water asindicated by the shrinkage factor 120576 introduced in Section 34the volume of coal is reduced by 120576 sdot DW1000 where the

density of water is 1000 kgm3 and the particle density at thetop of the ldquoslicerdquo 120588(119869 119868) is given by the following expression

120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)

119872 (119869 119868 minus 1) 120588 (119869 119868 minus 1) minus 120576 sdot DW (119869 119868) 1000

(69)

The shrinkage in volumemeans a reduction in the particlediameters denoted by shift (119869 119868) and it is given by theexpression

Shift (119869 119868) = 119904 (119869) [1 minus [1 minus 120576

120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)

The result of this process is therefore twofold Firstly theparticles lose some of their mass so that the particle masstransport rate 119872(119869 119868) as introduced in (64) is reduced byDW(119869 119868) Secondly the effect of the shift in the boundariesis that a corresponding fraction of the mass in size class 119869 ismoved into class 119869+1 so that there is a gradual change in theparticle size distribution toward smaller diameters as the coalmoves upward in the column as expressed by the algorithm

119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)

The overall effect is to change the coal density accordingto the following generalised version of (63)

120585 (119869 119868) =

2 sdot 119872 (119869 119868 minus 1) sdot 119902 (119869 119868 minus 1)

119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion

The model of the flash drying process presented in thispaper was initially developed for the specific purpose ofdrying subbituminous coal and was successfully used forthat purpose in the case of a 60MW plant although theresults of that application remain restricted by confidentialityagreements However the present version is of sufficientgenerality to be easily applied to the flash drying of othersubstances

The model when represented in the form of a computerprogram is a valuable tool for the basic design of flash dryingplant in that for given material and throughput parametervalues it allows trade-offs between plant parameters suchas drying column length and cross-section gas temperatureand recycle ratio It also permits investigations regardingthe sensitivity of a given design to variations in the inputmaterial parameters and the development of suitable controlstrategies It is also very useful in performing a design trade-off between the cost of high-temperature materials and thecost of handling increased gas flows as indicated by Figure 2

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper


Acknowledgments

It is a pleasure to acknowledge the support of this workthrough reviews and comments by the following personsW G Kalb (of TraDet Inc) K Clark (of White Energy)and C Needham and B Smeaton (both of SKM) Thevaluable comments of the anonymous reviewer are alsoacknowledged

References

[1] S M El-Behery W A El-Askary K A Ibrahim and M HHamed ldquoPorous particles drying in a vertical upward pneu-matic conveying dryerrdquoWorld Academy of Science Engineeringand Technology vol 53 pp 1337ndash1351 2009

[2] Y Li -H and J L Skinner ldquoDevelopment and validation ofa process simulator for drying subbituminous coalrdquo ChemicalEngineering Communications vol 49 pp 99ndash118 1988

[3] F P Incopera and D P DeWitt Fundamentals of Heat TransferJohn Wiley amp Sons 3rd edition 1990

[4] R C Weast Handbook of Chemistry and Physics CRC Press54th edition 1973

[5] M Hartman O Trnka K Svoboda and V Vesely ldquoInfluenceof particle shape on the drag coefficient of isometric particlesrdquoCollection of Czechoslovak Chemical Communications vol 59pp 2583ndash2594 1994

[6] S Laın and M Sommerfeld ldquoA study of the pneumatic convey-ing of non-spherical particles in a turbulent horizontal channelflowrdquo Brazilian Journal of Chemical Engineering vol 24 no 4pp 535ndash546 2007

[7] S Martin and L Santiago ldquoTransport characteristics of isomet-ric non-spherical particles in turbulent flowrdquo El Hombre y laMaquina no 30 2008

[8] E Kaya R Hogg and S R Kumar ldquoParticle shape modifi-cation in comminutionrdquo KONA vol 20 pp 188ndash195 2002httpwwwkonaorjpsearch20 188pdf

[9] J P Matthews S Eser P G Hatcher and A W ScaronildquoThe shape of pulverized bituminous vitrinite coal particlesrdquoHosokawa Powder Technology Foundation KONA vol 25 pp145ndash152 2007

[10] J H Perry Ed Chemical Engineersrsquo Handbook McGraw-Hill6th edition 1984

[11] M L de Souza-Santos Solid Fuels Combustion and GasificationModeling Simulation and Equipment Marcel Dekker 2004

[12] S Keys and A J Chambers Scaling Pneumatic ConveyingCharacteristics for Pipeline Pressure The Centre for Bulk Solidsand Particulate Technologies University of Newcastle 1996

[13] P Wypych Prediction amp Scale-Up of ldquoSteady-Staterdquo OperatingConditions Lectures on Pneumatic Conveying University ofWollongong

[14] M Sommerfeld Theoretical and Experimental Modelling ofParticulate Flows A Lecture Series Von Karman Institute forFluid Dynamics 2000

[15] R D Marcus A R Reed and A J Chambers PneumaticConveying of Bulk Solids Course Notes TUNRA Bulk SolidsHandling Research Associates Department of MechanicalEngineering University of Newcastle Newcastle Australia1986

[16] E S Pettyjohn and E B Christiansen ldquoEffect of particleshape on free-settling rates of isometric particlesrdquo ChemicalEngineering Progress vol 44 no 2 pp 157ndash172 1948

[17] G H Ganser ldquoA rational approach to drag prediction ofspherical and nonspherical particlesrdquo Powder Technology vol77 no 2 pp 143ndash152 1993

[18] A Haider and O Levenspiel ldquoDrag coefficient and terminalvelocity of spherical and nonspherical particlesrdquo Powder Tech-nology vol 58 no 1 pp 63ndash70 1989

[19] ldquoGEA Process Engineering Incrdquo httpwwwniroinccomfood chemicalflash drying conceptsasp

[20] J H Keenan and J Kaye Gas Tables John Wiley amp Sons 1946[21] J M Herrin and D Deming ldquoThermal conductivity of US

coalsrdquo Journal of Geophysical Research B vol 101 no B11 pp25381ndash25386 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


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Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


order to use it as a fuel in combined cycle plants and in orderto gasify lower rank coals such as subbituminous coal andbrown coal (or lignite) they need to be dried as a first stageof the process

12 Structure of the Paper Following the listing of somematerial parameter values in the next subsection the math-ematical model is developed in the following four sectionsSection 2 defines the process and its parameters and looks atits basic thermodynamic properties and operating character-istics Section 3 discusses themechanical interaction betweenthe coal particles and the drying gas and the influence of thesize distribution of the coal particles Section 4 develops amodel of the thermal interaction between the coal particlesand the drying gas including a number of detailed aspectsof this interaction Section 5 is concerned with several issuesthat need to be considered when creating a computer appli-cation for the numerical evaluation of the model

13 Parameter Values Throughout this monograph we shallbe assuming the coal to be a typical subbituminous coal withthe following characteristicsProximate Analysis

total moisture 323 (range 31ndash35)ash 39volatiles 335fixed carbon 303

Ultimate Analysis (Coal Has Been Dried but the Ash IsIncluded)

carbon 683hydrogen 526nitrogen 093oxygen 197sulfur 015total 9434ash 566

The specific heat capacity of this coal is 12 Jg∘CWe shall also be using the following constants and values

for material properties(a) standard atmosphere

(i) temperature 0∘C(ii) pressure 1013 kPa(iii) density 1293 kgm3

(b) composition of dry air (by mass) 7552 N2 2313

O2 129 Ar 006 various

(c) mean molecular weight of air 2897(d) vapour pressure of water as shown in Table 1(e) the heat of vaporisation of water 2255 kJgUnless otherwise stated all equations and expressions

assume that the variables are measured in SI units

Recyclegas

Hot gas

Dried coal

Combustiongas

Dryingcolumn

Fluegas

Crushedcoal

Figure 1 The drying circuit illustrating how a fraction of the gasexiting from the drying process is recycled and combined with thecombustion gas in order for the oxygen content of the hot gas thatenters the drying column to be below the explosion limit that isinert

2 Basic Properties and Relations

21 Drying Circuit and Associated Parameters As was indi-cated in the Introduction an important aspect of this partic-ular model is that the material to be dried that is coal dustis highly combustible This then requires the drying to takeplace in an inert atmosphere (see Section 37) and the inertatmosphere is created by recycling some of the gas exiting thedrying process andmixing itwith the combustion gas tomakeup the hot gas injected into the drying process

The drying circuit consists of a drying column in whichcrushed coal is injected into an upward flow of hot gas inthe lower part of the column At the top of the column amixture of gas and dried coal exits the column and the coal isseparated from the gas A proportion of the gas is exhaustedto atmosphere and the rest the recycle gas is mixed withthe combustion gas to produce the hot gas This is illustratedschematically in Figure 1

211 In All That Follows the Rate of Dried Coal ProducedWillBe Normalised to 1 kgs The parameters of this process are asfollows

moisture content of the crushed coal 119909moisture content of the dried coal 119910ambient temperature 119879

119886(∘C)

temperature of the hot gas 119879119909(∘C)

fraction of exit gas recycled 120578flow of combustion gas 119892 (kgs)


Table 1∘C 100 101 102 103 104 105 106 107 108 109 110kPa 1013 1050 1087 1126 1166 1208 1250 1293 1339 1385 1432


Carbon 1198881

Hydrogen 1198882

Oxygen 1198883

Ash 1198884













1 minus 119910

1 minus 119909

(1)


119909 minus 119910

1 minus 119909

(2)


(119892 +

119909 minus 119910

1 minus 119909

)

120578

1 minus 120578

(3)


119892 + 120578 ((119909 minus 119910) (1 minus 119909))

1 minus 120578

(4)



119879119886)(kW)


4184 sdot (100 minus 119879119886) (kW)



119909) sdot 205 sdot 40 (kW)


119864 = 2337

119909 minus 119910

1 minus 119909

+ (1 minus 119910)

times [

119909

1 minus 119909

(4184 minus 001119879119886) + 132 minus 12119879

119886]

(5)





int

119879119909

140

119888119901(119879) 119889119879 (6)






water 120596ℎ




ℎ



119892120596119888+

120578 (119909 minus 119910)

(1 minus 120578) (1 minus 119909)

(7)

120596ℎ=


119892 (1 minus 119909) + 120578 (119909 minus 119910)

(8)


HV = 33950 sdot 1198881+ 144200 (119888

2minus

1198883

8

) (kJkg) (9)




HHV = HV (1 minus 119910) (10)



1198821015840

119886= 435 sdot (267 sdot 119888

1+ 8 sdot 119888

2minus 1198883) (kgs) (12)



119903 = 120576 sdot (0021 sdot 1198792

119886+ 021 sdot 119879

119886+ 49) (13)


119886 can then


119882119886= 1198821015840

119886(1 +

119903

1200

) (14)


119894and using an



2sdot 9 + 000083(1 + 120581)W

119886sdot 119903


O2 120590sdotW119886sdot023

Ash 1198884

Total (1+120581)sdotW119886+1





119890 equals

Δ119890



105 sdot 119864 + Δ119890

LHV(pu) (16)


119892 =

105119864 (119882119886(1 + 120581) (1 minus 119910) + 1)


(17)


119886(097 minus


120582 = (119882119886(1016 minus 0000996 sdot 119903 + 119888

4) (1 minus 119910)

times (105119864 + 119892 (140 minus 119879119909)))


(18)




120596119888= ( (119910 + (9119888

2+ 000083 (1 + 120581)119882

119886sdot 119903) (1 minus 119910))

times (105119864 + 119892 (140 minus 119879119909)) )


(19)





gas


nitrogen 119888119901

119899(119879) = 100951 + 0000213 sdot 119879 (Jg∘C)


119888(119879) = 09293

+ 0000413 sdot 119879 (Jg∘C)

steam 119888119901

119904(119879) = 1664 + 00008 sdot 119879 + 324119879 (Jg∘C)

(20)



119899 +(1 minus 120582)119888

119901


ℎ119892= (09293 + 008021 sdot 120582) (119879

119909minus 140)

+ (00002065 minus 00001 sdot 120582) (119879119909minus 140)

2

(21)


ℎ119904= 1664 (119879

119909minus 140) + 00004(119879

119909minus 140)

2

+ 324 sdot ln (119879119909minus 140)

(22)


10

13

15

18

20

23

25

500 550 600 650 700 750 800 850 900

Hot

gas

flow

(kg

s)





119909 determines



(119892 + 120578

119909 minus 119910

1 minus 119909

) [(1 minus 120596ℎ) ℎ119892+ 120596ℎℎ119904] = (1 minus 120578) 119864 (23)



1198861205782+ 119887120578 + 119888 = 0 (24)


119886 = 119892120596119888(1 minus 119909) (119909 minus 119910) (ℎ

119904+ ℎ119892) + (2ℎ

119892minus ℎℎ) (119909 minus 119910)

2

119887 = 1198922120596119888(1 minus 119909)

2(ℎ119892minus ℎ119904)


119888)) + 119864 (1 minus 119909)

119888 = 1198922(120596119888(1 minus 119909)

2(ℎ119904minus ℎ119892) + ℎ119892(1 minus 119909)

2) minus 119864 (1 minus 119909)

(25)








119894 with 119886

119894gt 119886119894+1

For simplicity




Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806


119904 = 119886 05236 1198863 31416 119886

2 10


3 18564 1198862 067


119894 in the limit

of 119886119894minus 119886119894minus1







36 Or



119886(119881) as


x

s a

1634a

pa(x)



0 0241

10

08060670

minus03706

xa

120595(x)





119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886




expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)


36 and 120587(119904 + 119889119904)


119901119904minus119909



119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)








119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Table 1∘C 100 101 102 103 104 105 106 107 108 109 110kPa 1013 1050 1087 1126 1166 1208 1250 1293 1339 1385 1432


Carbon 1198881

Hydrogen 1198882

Oxygen 1198883

Ash 1198884













1 minus 119910

1 minus 119909

(1)


119909 minus 119910

1 minus 119909

(2)


(119892 +

119909 minus 119910

1 minus 119909

)

120578

1 minus 120578

(3)


119892 + 120578 ((119909 minus 119910) (1 minus 119909))

1 minus 120578

(4)



119879119886)(kW)


4184 sdot (100 minus 119879119886) (kW)



119909) sdot 205 sdot 40 (kW)


119864 = 2337

119909 minus 119910

1 minus 119909

+ (1 minus 119910)

times [

119909

1 minus 119909

(4184 minus 001119879119886) + 132 minus 12119879

119886]

(5)





int

119879119909

140

119888119901(119879) 119889119879 (6)






water 120596ℎ




ℎ



119892120596119888+

120578 (119909 minus 119910)

(1 minus 120578) (1 minus 119909)

(7)

120596ℎ=


119892 (1 minus 119909) + 120578 (119909 minus 119910)

(8)


HV = 33950 sdot 1198881+ 144200 (119888

2minus

1198883

8

) (kJkg) (9)




HHV = HV (1 minus 119910) (10)



1198821015840

119886= 435 sdot (267 sdot 119888

1+ 8 sdot 119888

2minus 1198883) (kgs) (12)



119903 = 120576 sdot (0021 sdot 1198792

119886+ 021 sdot 119879

119886+ 49) (13)


119886 can then


119882119886= 1198821015840

119886(1 +

119903

1200

) (14)


119894and using an



2sdot 9 + 000083(1 + 120581)W

119886sdot 119903


O2 120590sdotW119886sdot023

Ash 1198884

Total (1+120581)sdotW119886+1





119890 equals

Δ119890



105 sdot 119864 + Δ119890

LHV(pu) (16)


119892 =

105119864 (119882119886(1 + 120581) (1 minus 119910) + 1)


(17)


119886(097 minus


120582 = (119882119886(1016 minus 0000996 sdot 119903 + 119888

4) (1 minus 119910)

times (105119864 + 119892 (140 minus 119879119909)))


(18)




120596119888= ( (119910 + (9119888

2+ 000083 (1 + 120581)119882

119886sdot 119903) (1 minus 119910))

times (105119864 + 119892 (140 minus 119879119909)) )


(19)





gas


nitrogen 119888119901

119899(119879) = 100951 + 0000213 sdot 119879 (Jg∘C)


119888(119879) = 09293

+ 0000413 sdot 119879 (Jg∘C)

steam 119888119901

119904(119879) = 1664 + 00008 sdot 119879 + 324119879 (Jg∘C)

(20)



119899 +(1 minus 120582)119888

119901


ℎ119892= (09293 + 008021 sdot 120582) (119879

119909minus 140)

+ (00002065 minus 00001 sdot 120582) (119879119909minus 140)

2

(21)


ℎ119904= 1664 (119879

119909minus 140) + 00004(119879

119909minus 140)

2

+ 324 sdot ln (119879119909minus 140)

(22)


10

13

15

18

20

23

25

500 550 600 650 700 750 800 850 900

Hot

gas

flow

(kg

s)





119909 determines



(119892 + 120578

119909 minus 119910

1 minus 119909

) [(1 minus 120596ℎ) ℎ119892+ 120596ℎℎ119904] = (1 minus 120578) 119864 (23)



1198861205782+ 119887120578 + 119888 = 0 (24)


119886 = 119892120596119888(1 minus 119909) (119909 minus 119910) (ℎ

119904+ ℎ119892) + (2ℎ

119892minus ℎℎ) (119909 minus 119910)

2

119887 = 1198922120596119888(1 minus 119909)

2(ℎ119892minus ℎ119904)


119888)) + 119864 (1 minus 119909)

119888 = 1198922(120596119888(1 minus 119909)

2(ℎ119904minus ℎ119892) + ℎ119892(1 minus 119909)

2) minus 119864 (1 minus 119909)

(25)








119894 with 119886

119894gt 119886119894+1

For simplicity




Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806


119904 = 119886 05236 1198863 31416 119886

2 10


3 18564 1198862 067


119894 in the limit

of 119886119894minus 119886119894minus1







36 Or



119886(119881) as


x

s a

1634a

pa(x)



0 0241

10

08060670

minus03706

xa

120595(x)





119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886




expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)


36 and 120587(119904 + 119889119904)


119901119904minus119909



119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)








119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












ℎ



119892120596119888+

120578 (119909 minus 119910)

(1 minus 120578) (1 minus 119909)

(7)

120596ℎ=


119892 (1 minus 119909) + 120578 (119909 minus 119910)

(8)


HV = 33950 sdot 1198881+ 144200 (119888

2minus

1198883

8

) (kJkg) (9)




HHV = HV (1 minus 119910) (10)



1198821015840

119886= 435 sdot (267 sdot 119888

1+ 8 sdot 119888

2minus 1198883) (kgs) (12)



119903 = 120576 sdot (0021 sdot 1198792

119886+ 021 sdot 119879

119886+ 49) (13)


119886 can then


119882119886= 1198821015840

119886(1 +

119903

1200

) (14)


119894and using an



2sdot 9 + 000083(1 + 120581)W

119886sdot 119903


O2 120590sdotW119886sdot023

Ash 1198884

Total (1+120581)sdotW119886+1





119890 equals

Δ119890



105 sdot 119864 + Δ119890

LHV(pu) (16)


119892 =

105119864 (119882119886(1 + 120581) (1 minus 119910) + 1)


(17)


119886(097 minus


120582 = (119882119886(1016 minus 0000996 sdot 119903 + 119888

4) (1 minus 119910)

times (105119864 + 119892 (140 minus 119879119909)))


(18)




120596119888= ( (119910 + (9119888

2+ 000083 (1 + 120581)119882

119886sdot 119903) (1 minus 119910))

times (105119864 + 119892 (140 minus 119879119909)) )


(19)





gas


nitrogen 119888119901

119899(119879) = 100951 + 0000213 sdot 119879 (Jg∘C)


119888(119879) = 09293

+ 0000413 sdot 119879 (Jg∘C)

steam 119888119901

119904(119879) = 1664 + 00008 sdot 119879 + 324119879 (Jg∘C)

(20)



119899 +(1 minus 120582)119888

119901


ℎ119892= (09293 + 008021 sdot 120582) (119879

119909minus 140)

+ (00002065 minus 00001 sdot 120582) (119879119909minus 140)

2

(21)


ℎ119904= 1664 (119879

119909minus 140) + 00004(119879

119909minus 140)

2

+ 324 sdot ln (119879119909minus 140)

(22)


10

13

15

18

20

23

25

500 550 600 650 700 750 800 850 900

Hot

gas

flow

(kg

s)





119909 determines



(119892 + 120578

119909 minus 119910

1 minus 119909

) [(1 minus 120596ℎ) ℎ119892+ 120596ℎℎ119904] = (1 minus 120578) 119864 (23)



1198861205782+ 119887120578 + 119888 = 0 (24)


119886 = 119892120596119888(1 minus 119909) (119909 minus 119910) (ℎ

119904+ ℎ119892) + (2ℎ

119892minus ℎℎ) (119909 minus 119910)

2

119887 = 1198922120596119888(1 minus 119909)

2(ℎ119892minus ℎ119904)


119888)) + 119864 (1 minus 119909)

119888 = 1198922(120596119888(1 minus 119909)

2(ℎ119904minus ℎ119892) + ℎ119892(1 minus 119909)

2) minus 119864 (1 minus 119909)

(25)








119894 with 119886

119894gt 119886119894+1

For simplicity




Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806


119904 = 119886 05236 1198863 31416 119886

2 10


3 18564 1198862 067


119894 in the limit

of 119886119894minus 119886119894minus1







36 Or



119886(119881) as


x

s a

1634a

pa(x)



0 0241

10

08060670

minus03706

xa

120595(x)





119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886




expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)


36 and 120587(119904 + 119889119904)


119901119904minus119909



119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)








119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












120596119888= ( (119910 + (9119888

2+ 000083 (1 + 120581)119882

119886sdot 119903) (1 minus 119910))

times (105119864 + 119892 (140 minus 119879119909)) )


(19)





gas


nitrogen 119888119901

119899(119879) = 100951 + 0000213 sdot 119879 (Jg∘C)


119888(119879) = 09293

+ 0000413 sdot 119879 (Jg∘C)

steam 119888119901

119904(119879) = 1664 + 00008 sdot 119879 + 324119879 (Jg∘C)

(20)



119899 +(1 minus 120582)119888

119901


ℎ119892= (09293 + 008021 sdot 120582) (119879

119909minus 140)

+ (00002065 minus 00001 sdot 120582) (119879119909minus 140)

2

(21)


ℎ119904= 1664 (119879

119909minus 140) + 00004(119879

119909minus 140)

2

+ 324 sdot ln (119879119909minus 140)

(22)


10

13

15

18

20

23

25

500 550 600 650 700 750 800 850 900

Hot

gas

flow

(kg

s)





119909 determines



(119892 + 120578

119909 minus 119910

1 minus 119909

) [(1 minus 120596ℎ) ℎ119892+ 120596ℎℎ119904] = (1 minus 120578) 119864 (23)



1198861205782+ 119887120578 + 119888 = 0 (24)


119886 = 119892120596119888(1 minus 119909) (119909 minus 119910) (ℎ

119904+ ℎ119892) + (2ℎ

119892minus ℎℎ) (119909 minus 119910)

2

119887 = 1198922120596119888(1 minus 119909)

2(ℎ119892minus ℎ119904)


119888)) + 119864 (1 minus 119909)

119888 = 1198922(120596119888(1 minus 119909)

2(ℎ119904minus ℎ119892) + ℎ119892(1 minus 119909)

2) minus 119864 (1 minus 119909)

(25)








119894 with 119886

119894gt 119886119894+1

For simplicity




Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806


119904 = 119886 05236 1198863 31416 119886

2 10


3 18564 1198862 067


119894 in the limit

of 119886119894minus 119886119894minus1







36 Or



119886(119881) as


x

s a

1634a

pa(x)



0 0241

10

08060670

minus03706

xa

120595(x)





119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886




expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)


36 and 120587(119904 + 119889119904)


119901119904minus119909



119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)








119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Cubewith edges 119904 119904 = 119886 119886

3 6 1198862 0806


119904 = 119886 05236 1198863 31416 119886

2 10


3 18564 1198862 067


119894 in the limit

of 119886119894minus 119886119894minus1







36 Or



119886(119881) as


x

s a

1634a

pa(x)



0 0241

10

08060670

minus03706

xa

120595(x)





119886(119881(119904)) sdot 119889119904 = 119862

119886sdot 119889119904 and as

int

1241119886

06294119886

119862119886sdot 119889119904 = 1 (26)

we find that 119862119886= 1634119886




expression

119891 (119904) = 1634int

0194119904

minus059119904

119902 (119904 minus 119909)

119904 minus 119909

119889119909 (27)


36 and 120587(119904 + 119889119904)


119901119904minus119909



119899 (119904) 119889119904 = 191

119891 (119904)

120588 sdot 1199043119889119904 (28)








119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














119902 (119886 119896 120582) =

120582

119896

(

119886

119896

)

120582minus1

119890minus(119886119896)

120582

(29)


119876119894= 119890minus(119886119894119896)120582

minus 1198901(119886119894minus1119896)120582

(30)





119902 (119886 119888) =

2

11988811198882+ 11988831198884

[1198881119890minus(1198861198882)(1 minus 119890

minus(1198861198882))

+1198883119890minus(1198861198884)(1 minus 119890

minus(1198861198884))]

(31)


1=

0675 1198882= 0829 119888

3= 1758 and 119888

4= 0083


0000

0050

0100

0150

0200

0250

0300


Frac

tion

reta

ined


0

001

002

003

004

005

006

001 01 1 10

pdf v

alue

s








0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0001002003004005006007

001 01 1 10

pdf v

alue

s



08

09

1

11

12

13


Cor

rect

ion

fact

or






119865119889= 119862119889119860

120590 sdot V119903

2

2

(32)





Re =

119904V119903

] (33)


following [12]

119862119889= 04 +

26

(Re)08 (34)





nitrogen 120583 = 6485 + 0365 sdot 119879


steam 120583 = minus982 + 03606 sdot 119879

(35)


120583 = 119886119879 + 119887 (36)

with



(37)






120590 =

1

101

sdot

373

119879 + 273

3

sum

119894=1

119901119894120590119894 (38)







119903is also


119865119889(119897) = [0157 + 102 sdot (

120583

119904V119903120590

)

08

] 1199042120590V2119903 (39)




119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119865119892(119897) = 981 sdot 119898 with 119898 = 05236120588 sdot 119904

3 (40)


119889V119889119905

=

1

119898

(119865119889minus 119865119892) (41)


119889V119889119897

=

119889V119889119905

sdot

1

V (42)


119889V119889119897

=

1

119898 sdot V(119865119889minus 119865119892) (43)





119889 (34) becomes

119862119889= 04 +

119891 (119879) sdot 10minus3

(119904V)08 (44)


119891 (119879) = 8956 sdot 10minus61198792+ 001747 sdot 119879 + 3 (45)


[04 +

119891 (119879) sdot 10minus3

(119904V)08]

0179 sdot V2

(119879 + 273) 119904

minus 981 (46)


ldquo119879 119904rdquo 01 02 05 10 20 50500 010 032 130 360 1033 80600 010 032 133 379 1133 116700 010 031 134 394 1222 177800 010 031 135 405 1300 303900 009 030 135 413 1366 718





0 a volume 05236 sdot 119904

0






0(1 minus 119909120588

0(1 minus 119908))



1015840


120588 = 1205880

1 minus 119909 (1 minus 119908)

1 minus 120576 sdot 119909 sdot 1205881015840

0(1 minus 119908)

119904 = 1199040(1 minus 120576 sdot 119909 sdot 120588

1015840

0(1 minus 119908))

13

(47)

where 1205881015840


0=





01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












01 02 05 10 20 50100 033 088 276 533 885 1527200 030 082 273 557 960 1698300 028 077 267 570 1020 1845400 026 073 260 577 1067 1974500 025 070 253 579 1104 2089600 024 067 246 578 1132 2191700 023 064 239 574 1154 2283800 022 062 232 568 1171 2365



119865119889(V0) minus 981119898 = 0 (48)


119889and if we


V0= 1304radic

119904

120590119862119889

(49)



0 of the

form

V20+ 119886V120

minus 119887 = 0 (50)

with

119886 = 00025 sdot 119904minus08

sdot 119891 (119879)

119887 = 137 sdot 119904 (119879 + 273)

(51)



V0= 1198861199042+ 119887119904 minus 03 (52)

with

119886 = 0000307 sdot 119879 minus 0537

119887 = 000086 sdot 119879 + 568

(53)










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










000

200

400

600

800

1000

1200

1 10 100 1000Re

Dra

g fa

ctor

mul

tiplie

r


b OR

120578b OR

g OC

a


OH



119874119867

=

119892 sdot 119874119862+ 120578 sdot 119887 sdot 119874

119877

119892 + 120578 sdot 119887

119887 = (119892 +

119909 minus 119910

1 minus 119909

)

1

1 minus 120578

(54)

119874119877=

119892 sdot 119874119862

119892 + (119909 minus 119910) (1 minus 119909)

(55)



119874119862= 023

120581

11 + 120581

(56)





100∘C

30∘C

0∘Cs2

t1

t2

T




2after injection






4 Heat Transfer







119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119873119906 = 2 + (04 sdot Re12 + 006 sdot Re23) Pr04(120583

120583119904

)

14

(57)

where

119873119906 =

ℎ1015840sdot 119904

119896

Pr =

]120572

(58)









minus3


times (

120583

120583119904

)

14

]

(59)


119904= (119879 minus 110)ℎ(119904) (∘CW) Let 120575 be


119888 is

given by

119877119888(120575) =

1

033 sdot 4120587

int

1199042

(1199042)(1minus2120575)

119889119909

1199092 (60)







ratio 119877119904119877119888

119904 (mm) h(s) (W) 119877119904(∘CW) 119877

119888(∘CW) 119877

119904119877119888

3 45913 85 83 10215 16102 241 167 145075 05534 703 333 2110375 01982 1962 667 29401875 00734 5295 1333 397009375 00274 14211 2667 5330046875 00107 36313 5333 6810023438 00046 84018 10666 788







119904 is 2ℎ

119890(119879119904minus 100)2255 (Js) where


ℎ119890

ℎ

=

ℎ119890

ℎ119890+ ℎ119904

=

1

1 + (119879119904minus 100) 1125

(61)





119877119888=

0965

119904

sdot

1 minus 11990813

11990813

(62)


119888and


119904not only










119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119909minus 140



120585 (119897) = int

1199040

0

120585 (s 119897) 119889119904 (63)


119872(119897) = 119860int

1199040

0

120585 (119904 119897) sdot V (119904 119897) 119889119904 (64)


0 and

120585 (119904 119897 = 0) 119889119904 = 1198720

119902 (119904) 119889119904

119860 sdot V (119904) (65)




119867(119904 119897) 119889119897119889119904 = 191 sdot ℎ (119904 120582 120596 119879 V119903) 120585 (119904 119897)

sdot 120588minus1

(119904 119897) sdot 119904minus3

sdot 119860 sdot 119889119897 sdot 119889119904

(66)





Table 9



The value at theend


Velocity








V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










V(JI )Calculate

T(I)

V(JI )

Calculate

Vmean(J)

TmeanTmean



All parameter

CalculateCalculate

CalculateTmean




T(I)

T(I)


Tmean

values I minus 1


(J I)

mean(J)

mean(J)

Fm(J I)



and

119889119897 = (V (119897) +119889V2

) 119889V sdot

119898

119865

(119889V)2 sdot119898

2

+ 119898 sdot V119889V minus 119865119889119897 = 0

(67)


V (119897 + 119889119897) = (V2 (119897) + 2119889119897 sdot

119865

119898

)

12

(68)










119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119869 119904 Δ119904





120588 (119869 119868) =

119872 (119869 119868 minus 1) minus DW (119869 119868)


(69)



120588(119869 119868 minus 1) sdot DW(119869)

1000 sdot 119872(119869 119868 minus 1)

]

13

]

(70)


119902 (119869 119868) = 119902 (119869 119868 minus 1) [1 minus

Shift (119869 119868)

Δ119904 (119869)

]

+ 119902 (119869 minus 1 119868 minus 1) [

Shift (119869 minus 1 119868)

Δ119904 (119869)

]

(71)


120585 (119869 119868) =


119860 sdot (V (119869 119868 minus 1) + V (119869 119868)) (72)

6 Conclusion






Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Acknowledgments


References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Mathematical Model of a Flash Drying...

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