[QUOTE=Teus Tuinenburg;68628]Dear kj,

Sir plea to respond me against the below clarification

Mu represents the variable SLR

What is kgas and what shall be value

a)What is = (Abs(Suspensiondp) and how do I deter the value of same?

b)I presume , Productdp is partial pressure drop due to material ?

c)What is dPipeMass & dSedimentMass and how do I determine the value of same?

d)Is this (dTempAirHeatMat) is the consolidated temperature change accounted for all three heat transfer mode

a)What is HeatRestFactPipeWall

b)What is Diameter(t)

c)In above formulas , for heat transfer through convection , have you included the overall heat transfer coefficient and heat transfer coefficient in pipe?

d) Secondly will (TempDifference = TempMix – AmbientTemperatur) this be the final formula accounting all the heat transfer mode?. If so that means my underline objective shall be to determine the mixtemp at loacal condition and substracting the ambient tempearure will yiled the mixture temperature at local consition ■

Dear kj,

kgas is the specific heat ratio of a 2 atomic gas (Cp/Cv) and for air kgas=1.4

(Abs(Suspensiondp) is the pressure drop for keeping the product in suspension and is calculated from the local conditions in the pipe in relation to the local suspension velocity using the law of conservation of energy.

correct

dPipeMass & dSedimentMass are the moving mass of product and sediment mass of product in a pipe section dL, involved in the calculation of the heat content and the temperature changes.

Calculating these masses is based on capacity and product velocity.

dTempAirHeatMat is the temperature change of the product/gas mixture and added to the previous temperature gives the actual temperature. dTempAirHeatMat can be positive or negative.

HeatRestFactPipeWall is the combined heat resistance factor of the pipe wall in kCal/m2.degrC.sec

Diameter(t) is the diameter of the considered pipe section (t).

I have considered that all the calculated temperature differences are equalized over the product/gas mixture instantaneously.

(TempDifference = TempMix – AmbientTemperatur) is the temperature difference between the product/gas mixture, which determines the amount of heat that is dissipated through the pipe wall to the surroundings, resulting in a temperature decrease (or increase, when the environment is hotter than the product/gas mixture) of the product/gas mixture.

All the above is straight forward physics, applied to the technology of pneumatic conveying.

Although you need to be familiar with this science, you do not need to be a professor for it.

Common sense, persistence, experience, observative and a flexible and inventive mind are enough to understand and of course, a certain amount of relevant education and training.

The given formulas are (a very small) part of the computer program.

Success

Teus ■

L F Pumping (Europe) Ltd have both conveyed and reduced the moisture of flyash. With the patented LamiFlo Technology the maintenance of conveying and drying is extremely low, along with low cost and high capacity.

Flyash is a material that we have tested in a 50 tonne an hour LamiFlo Dryer unit. Due to its revolutionary design the same system will dry and convey at the same time.

As the LamiFlo Dryer uses air rather than heat from a direct heat source, we substantially reduce the carbon emissions compared with any traditional drying methods.

http://www.lfpumping.com ■

Sir kindly finf the below clarification'

[QUOTE=Teus Tuinenburg;68662]Dear kj,

A. Here is the absolute pressure is 1+Pressure drop for particular pipe section?

B. You said this formula is based on pV^k/T=constant . Where is volume appearing this formula ?

C. I presume the dpLength is the total pressure drop for particular pipe section

OK. Sir , what shall be the formula involved to determine the Abs(Suspensiondp)

OK. Sir , what shall be the formula involved to determine the Productdp

Moving mass of product and sediment mass of product , seems to be a interesting topic . Sir would you please elaborate the formula involved to determine the same ?. As taught by you I know to calculate the product velocity , but what shall be the equation with involved variables of capacity and velocity to determine the moving mass and sediment mass . Secondly in what manner pertinently the same (moving mass and sediment mass) effect the tempearure

Where exactly I shall add the same

A.Which temp is Tair . Is it the temp drop due to Air cools down with expansion. As calculated earlier

Sir what shall be the value of same . And how should I determine the same

Diameter(t) is the diameter of the considered pipe section (t).

Would you explain the same once again as. I could not able to interpret the same statement .

Sir I have flowcharted the steps involved with the algorithm . Please correct me with my calculation steps

Description Steps

Calculate Tair due to A. Air cools down with expansion1

Calculate dHeatMat 2

Calculate dTempAirHeatMat equilizing with dHeatMat 3

Calculate SpecHeatContMix 4

Calculate TempMix using SpecHeatContMix & i

ncorporating Tair as calculated in step 1 5

TempDifference = TempMix – AmbientTemperature6

Calculate dTempdiff 7This shall be the final temperature at end of node. ■

Dear sir

A.Here is the absolute pressure is 1+Pressure drop for particular pipe section?

B.You said this formula is based on pV^k/T=constant . Where is volume appearing this formula ?

C.I presume the dpLength is the total pressure drop for particular pipe section

OK. Sir , what shall be the formula involved to determine the Abs(Suspensiondp)

OK. Sir , what shall be the formula involved to determine the Productdp

Moving mass of product and sediment mass of product , seems to be a interesting topic . Sir would you please elaborate the formula involved to determine the same ?. As taught by you I know to calculate the product velocity , but what shall be the equation with involved variables of capacity and velocity to determine the moving mass and sediment mass . Secondly in what manner pertinently the same (moving mass and sediment mass) effect the tempearure

Where exactly I shall add the same

A.Which temp is Tair . Is it the temp drop due to Air cools down with expansion. As calculated earlier

Sir what shall be the value of same . And how should I determine the same

I have considered that all the calculated temperature differences are equalized over the product/gas mixture instantaneously.

Would you explain the same once again as. I could not able to interpret the same statement .

Sir I have flowcharted the steps involved with the algorithm . Please correct me with my calculation steps

Description Steps

Calculate Tair due to A.

Air cools down with expansion 1

Calculate dHeatMat 2

Calculate dTempAirHeatMat

equilizing with dHeatMat 3

Calculate SpecHeatContMix 4

Calculate TempMix using SpecHeatContMix &

incorporating Tair as calculated in step 1 5

TempDifference = TempMix – AmbientTemperature6

Calculate dTempdiff 7This shall be the final fi Temperature at end of node.

Success

Teus[/QUOTE] ■

Dear kj,

A.absolute pressure = ambient pressure + compressor pressure – pressure drop + 1

B.The V in the formula is the volume.

C.Correct.

Suspensiondp and Productdp are calculated by equalizing the change in internal energy in the gas with the energy change of the particles.

This equality is derived from the equilibrium of forces.

Checking the pipe wall velocity in relation to the local suspension velocity and considering that, when the wall velocity divided by the local suspension velocity becomes too low, sedimentation starts.

Sedimentation stops when the gas velocity in the remaining pipe section increases sufficiently to carry the particles. The solid loss factor increases thereby, due to the rough surface of the sediment compared to a smooth pipe wall.

It is not equation, it is a check, requiring an iterative algorithm.

The mass of material is storing heat and thereby influencing the temperature.

In the already given formula:

TempMix = (cpGas * Tair + MatSpecHeatCont * 4.1875 * Mu * (TempMix + dTempAirHeatMat)) / (cpGas + MatSpecHeatCont * 4.1875 * Mu)

Tair is the air temperature. (Not a temperature drop)

The heat resistance factor can be estimated, using thermodynamic laws and material heat related property data. Field measurements can also generate the required data by reverse calculating.

The same procedure as is used for heat exchangers.

The value depends on the pipe structure and f.i. insulation.

Heat equalization requires time and I assumed that the time needed for heat equalization is zero.

A simplification of reality.

Your calculation schedule is not completely clear to me.

There must be a clear distinction between heat, temperature and temperature change.

I admire your efforts to pursuit a working pneumatic conveying calculation algorithm.

By experience, I understand the complexity of such a project. It took me at least 25 years of theory building, testing, programming, refining, testing again, discovering phenomena, incorporating more and more physical influences, testing again, adapting to installation types, solving condensation problems, verifying new tests, convincing other people of new ideas about lower velocities and pressures, research on internal product flow in pressure tanks, bringing vacuum conveying and pressure conveying into one calculation algorithm with the same parameters, studying thermodynamics, molecular gas theory, interactions between particles, energy flows, generating a Zenz diagram by calculation, etc.

Success

Teus ■

Dear sir

Where is the V in this formula appearing?

But sir , would you please help me in determining the same? . I mean what is the derivation involved to calculate the suspension and product dp

But how do I use the same concept in determining the suspension mass of product. I can evaluate the possibility of suspension as taught by you ie sediment start when the Vwall is less than Vsuspension . But how do I relate the same interpretation in determining the suspension mass & moving mass values .

Would you please help me to checking using iterative method , to determine sediment

Eventually my quiries centres down to determine the quantity of sediment mass and moving mass which can be used in your stated formula. Please help me ?

Ok got our point , but there is a dual stage involved One is sediment & another is moving mass. I any particular section both or either of the case may occur , but contextually and fundamentally what is the individual influence on temperature by both the situation(sediment & moving mass)

In this formula there is another TempMix to determine the TempMix itself . What which temp is this and how do we determine

Tair is the air temperature. (Not a temperature drop)

Would you please elaborate the derivation or formula involved in this

I still didn’t understand . Would you please explain ?

Your calculation schedule is not completely clear to me.

There must be a clear distinction between heat, temperature and temperature change.

Thanks very much for you kind effort and putting helping me about core fundamental involved in pneumatic conveying system . As I said , one tied cannot lift all the boat but trying every effort to get the things is very important and I learn from you during lot of my thread interaction with you . Contextually and directionally I do not want to miss any of you thread and posting as the same thrust my and inspires to to taking the things forward . I know , iam doing may not be the appropriate way to go with , but I do not care of appropriateness .Every time I learn something new . There is too many articles and journals related to pneumatic conveying , but truly speaking you thread did put me to appropriate level

Sir would you please suggest any specific book I can purchase and is really appropriate in this filed

Thanks ■

Dear kj,

The formula is Temperature*pressure^(kgas/(kgas-1)) = Constant (adiabatic compression)

This can also be read in Mr Agarwal’s article.

The moving mass is related to the Solid Loading ratio.

The sedimented mass is calculated with the bulk density and the lost cross section area, leaving enough cross section area for conveying at sufficient high velocities.

If the wall velocity is too low, then reduce the conveying cross section by 1%.

Then check the wall velocity again.

If the wall velocity is still too low, then reduce the conveying cross section by another 1%.

a.s.o.

If the wall velocity is Ok, then the conveying cross section is known and thereby the occupied cross section also.

I consider no difference between the 2

TempMix is the equalized temperature of the material gas mixture.

In the program, this Tempmix2 is recalculated by the increased material temperature (TempMix1 + dTempAirHeatMat)

Energy loss = k * d(temperature) * Area * d(t)

k is the heat resistance factor

Normally, heat takes time to equalize between heat sources of different temperatures.

In the calculation, I assume this equalizing time is zero.

Searching the web will give you some (well-known) books.

Whether they are good or not is not for me to judge.

Success

Teus ■

Dear sir

That means , moving mass = SLR (Solid loading ratio). But what shall be the folrmula to calculate the sediment mass?. Sir iam in vague , please help me in calculating these two variable parameter at local condition

Sir is formula for HeatRestFactPipeWall

TempMix is the equalized temperature of the material gas mixture.

In the program, this Tempmix2 is recalculated by the increased material temperature (TempMix1 + dTempAirHeatMat)

Regards ■

Dear kj,

Moving mass = SLR * moving air mass. (by definition)

The sedimented mass is calculated with the bulk density and the lost cross section area, leaving enough cross section area for conveying at sufficient high velocities.

If the wall velocity is too low, then reduce the conveying cross section by 1%.

Then check the wall velocity again.

If the wall velocity is still too low, then reduce the conveying cross section by another 1%.

a.s.o.

If the wall velocity is Ok, then the conveying cross section is known and thereby the occupied cross section also.

Calculate the reduction cross-section as filled with sedimented material against bulk density

The heat HeatRestFactPipeWall k can be estimated, using thermodynamic laws and material heat related property data.

Adding up the respective heat resistances s.a heat transfer coefficient from air to wall + 1/heat conduction coefficient through pipe material + heat transfer coefficient between pipe material and paint + 1/heat conduction coefficient through paint + heat transfer coefficient from paint to outside air + radiation coefficient.

Field measurements can also generate the required data by reverse calculating.

The same procedure as is used for heat exchangers.

The value depends on the pipe structure and f.i. insulation.

TempMix2 = TempMix1 + dTempAirHeatMat

TempMix1 is the mixture temperature at the begin of the considered length dL

TempMix2 is the mixture temperature at the end of the considered length dL

TempMix1 of the next considered length dL is TempMix2 of the next dL

These are standard calculations, which are widely used in other disciplines, mainly heat exchangers.

A standard book of thermo dynamics as used at every or any professional education institute will suffice.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

I calculated the temperature changes as described in the replies #43 and #45 of this thread.

The mentioned general formulas in #45 are applied on the pneumatic conveying calculations for each time increment dt.

That is all.

In the example, the temperature of 155 degrC is calculated from the flyash of 150 degrC and the compressed aur temperature of 265 degrC.

Because of the high mixture temperature and the low ambient temperature, the cooling is significant and in the intake section, the temperature drops down to 66 degrC.

Success

Teus

Sir , let me take this oppurtunity to de-structure quiries into part and get your response on that . Trust me this will help me out to reach some constructive conculsion .

I list out the following heat reaction involved and request you to please mention formula. Also please correct me with any additional heat transfer involved

a) Air cools down with expansion.

Please mention the formula?

b) Air heats up because of friction resistance

Please mention the formula?

c) Air heats up because of pressure drop caused by keeping the product in suspension

Please mention the formula?

d) Product: heats up because of collision- and friction losses.

Please mention the formula?

e) The mixture cools down because of heat convection and radiation to the environment.

Please mention the formula?

f) Mixture is exchanging heat with surrounding

Please mention the formula?

g) heat transfer from the conveying air to the pipe wall

Please mention the formula?

Sir please -2 patronage me in following the correct wind

2) In formula of heat transfer by collision ie

* m * (vel2^2 – vel1^2) = h(specific heat air) * mair * dT + h(specific heat material) * mmaterial * dT ,

you taught there is a mathematical trick to determine the vel2 (velocity after collision) vel1 (velocity before collision). Would you explain the mathematical trick involve to determine the same ? ■

Dear kj,

pV^k/T=constant

Energy= dp(air) * Volume

Energy= dp(suspension) * Volume

Energy= dp(product loss) * Volume

Energy loss = k * d(temperature) * Area * d(t)

Energy loss = k * d(temperature) * Area * d(t)

Energy loss = k * d(temperature) * Area * d(t)

In general, all the formulas are based on thermodynamics and follow the rule of “conservation of energy”.

Calculating the velocity slip is one of the things that I preserve until I decide to publish. (with the risk that somebody else finds the solution also)

Best regards

Teus ■

Sir please find the enclosed quiry . Please help me in determining my ultimate objective of potential condensation of mixture in long conveying system

Originally Posted by Teus Tuinenburg

Dear kj,

pV^k/T=constant

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy= dp(air) * Volume

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy= dp(suspension) * Volume

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy= dp(product loss) * Volume

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy loss = k * d(temperature) * Area * d(t)

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy loss = k * d(temperature) * Area * d(t)

Would you please specify the detail formula involve in this to determine the temperature chamge

Energy loss = k * d(temperature) * Area * d(t)

Would you please specify the detail formula involve in this to determine the temperature chamge

In general, all the formulas are based on thermodynamics and follow the rule of “conservation of energy”.

Calculating the velocity slip is one of the things that I preserve until I decide to publish. (with the risk that somebody else finds the solution also)

In all above formula you have not considered the formula of heat transfer by collision ie

* m * (vel2^2 – vel1^2) = h(specific heat air) * mair * dT + h(specific heat material) * mmaterial * dT .

Secondly sir please help me out in identifying the material velocity after collision

Best regards

Teus

■

Dear kj,

The relevant formulas are:

SpecHeatContMix = (cpGas + MatSpecHeatCont * 4.1875 * MuLocal) / (1 + MuLocal)

dHeatMat = (Abs(Suspensiondp) + Productdp + Gasdp) * GasVolume * 10 * dt

dTempAirHeatMat = dHeatMat / (MatSpecHeatCont * 4.1875 * (dPipeMass - dSedimentMass) * 1000)

dTair = (Tair + 273) / ((AbsolutePressure / (AbsolutePressure - dpLength / 10000)) ^ ((kGas - 1) / kGas)) - (Tair + 273)

Tair = Tair + dTair

TempMix = (cpGas * Tair + MatSpecHeatCont * 4.1875 * Mu * (TempMix + dTempAirHeatMat)) / (cpGas + MatSpecHeatCont * 4.1875 * Mu)

Tair = TempMix

TempDifference = TempMix – AmbientTemperature

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

TempMix = TempMix + dTempdiff

Tair = TempMix

These formulas have to be embedded in the iteration algorithm over each length dL calculated with the local particle velocity and d(t).

Success

Teus ■

[QUOTE=Teus Tuinenburg;68762]Dear kj,

Moving mass = SLR * moving air mass. (by definition)

Does that means , moving mass = SLR (Solid loading ratio). But what shall be the folrmula to calculate the sediment mass?. Sir please help me in calculating sedimant and moving mass

Shall i take it in a way that summation of all these resistance will yield the overall heat transfet coefficient? ■

Dear kj,

By definition:

SLR = material mass flow/air mass flow in (kg/sec)/ (kg/sec)

hence:

material mass flow = SLR * air mass flow in kg/sec

The cross section area of the pipe that is used for conveying is checked for the condition that the wall air velocity is well above the local material suspension velocity.

If the wall velocity is too low, then the flowing cross section is reduced until this condition is met.

The non flowing cross section is assumed to be filled with material at bulk density.

This is a matter of checking the condition and changing the flowing diameter until the condition is met. Not a formula, but a matter of try and error.



I also take this thread as an opportunity to wish you, all the members and visitors of this forum and Mr Reinhard Wohlbier all the best for the coming year 2011.



Merry Christmas

Teus ■

Dear kj,

By definition:

SLR = material mass flow/air mass flow in (kg/sec)/ (kg/sec)

hence:

material mass flow = SLR * air mass flow in kg/sec

The cross section area of the pipe that is used for conveying is checked for the condition that the wall air velocity is well above the local material suspension velocity.

If the wall velocity is too low, then the flowing cross section is reduced until this condition is met.

The non flowing cross section is assumed to be filled with material at bulk density.

This is a matter of checking the condition and changing the flowing diameter until the condition is met. Not a formula, but a matter of try and error.

The summation of all the heat conductance resistances will yield the overall heat transfer coefficient



I also take this thread as an opportunity to wish you, all the members and visitors of this forum and Mr Reinhard Wohlbier all the best for the coming year 2011.



Merry Christmas

Teus ■

If so , then the sediment mass will alwyas be zero as there wouldnt be any sedimentation. Iam little bit vague may be iam wrong in my interpretation , would you articulate on teh same matter . As in effort to keep th ewall velocity above suspension velocity , we have to maintain the cross section , doing such practice will ensure that , the matreial velocity is above suspension velocity and there would not be any sedimentation , henec sedimentation mass is always zero

The summation of all the heat conductance resistances will yield the overall heat transfer coefficient



I also take this thread as an opportunity to wish you, all the members and visitors of this forum and Mr Reinhard Wohlbier all the best for the coming year 2011.



Merry Christmas

Teus[/QUOTE] ■

That means sediment mass will always be zero , since this execise will ensure that the wall velcoity is always high than product suspension velocity . Sir would you articuate on my this interpretation?

Will there be any case where the sediment mass occurs?



I also take this thread as an opportunity to wish you, all the members and visitors of this forum and Mr Reinhard Wohlbier all the best for the coming year 2011.



Merry Christmas

Teus[/QUOTE] ■

Dear kj,

Now I understand the misunderstanding.

The sediment will reduce the flowing cross section in the pipe.

The remaining diameter is considered the “new” wall of the pipe.

This “new diameter” is then used to calculate the “new wall velocity”, which will be higher.

When the “new wall velocity” is high enough to prevent sedimentation, the pneumatic conveying calculation continues with the “new diameter”.

As the “new wall” now consists of sediment instead of a smooth steel pipe wall, the solid loss factor is also higher.

A single diameter pressure system starts sedimentation in the beginning of the pipe line, where the high pressure causes low velocities. Towards the end of the pipeline, the pressure drops, the air volume increases, the air velocity increases and the wall velocity also increases, preventing sedimentation.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Now I understand the misunderstanding.

The sediment will reduce the flowing cross section in the pipe.

The remaining diameter is considered the “new” wall of the pipe.

This “new diameter” is then used to calculate the “new wall velocity”, which will be higher.

When the “new wall velocity” is high enough to prevent sedimentation, the pneumatic conveying calculation continues with the “new diameter”.

As the “new wall” now consists of sediment instead of a smooth steel pipe wall, the solid loss factor is also higher.

A single diameter pressure system starts sedimentation in the beginning of the pipe line, where the high pressure causes low velocities. Towards the end of the pipeline, the pressure drops, the air volume increases, the air velocity increases and the wall velocity also increases, preventing sedimentation.

Have a nice day

Teus

Sir

As you said , the pickup velocity is factor times the suspension velocity , henec at the begening also the pipe wall velcoity is higher than suspension velocity , hence no sedimenmtation. For instance i referred you any of the calculation for pneumatic conveying , in your calculation sheet you have maintain the Vwall higher than suspension velocity which prevents sedimentation . Now my concern is , in such situation sedimen mass becomes un-influence and becomes zero , is int it? . If iam wrong , please help in determining the sediment mass ■

Dear kj,

If the wall air velocity is sufficiently above the local particle suspension velocity then there is no sedimentation.

If not, then there will form e sediment layer until the remaining cross section generates enough air velocity to keep the particles in suspension.

The sediment layer stays un-influenced.

When this happens in the beginning of the pipeline at high pressures, it is possible that, when the pressure drops at the end of the cycle, that the air velocity increases to such an extent that the particles are picked up again and that the pipe is cleaned out.

The pump characteristics (airflow=function(pressure)) are very important in such situations.

A centrifugal fan would block right away, because the airflow drops significantly when the pressure increases due to f.i. sedimentation.

Have a nice day

Teus ■

Sir

In formula above , is the dt is time incerment of 0.0001sec? . If so how should i detremine the same ? ■

Dear kj,

Yes, dt is meant as the time increment.

dt is a chosen value. A very small increment causes many calculations and thereby a long calculation time, though very accurate.

A too large increment causes the calculations to run out of realistic values (s.a. a negative absolute pressure) and the calculations gives an error.

A value of dt = 0.001 sec is OK for normal calculations, dt = 0.0001 sec for regions with high accelerations and dt = 0.00005 sec for materials with a very low suspension velocity (5-10 micron powders)

Good luck

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Yes, dt is meant as the time increment.

dt is a chosen value. A very small increment causes many calculations and thereby a long calculation time, though very accurate.

A too large increment causes the calculations to run out of realistic values (s.a. a negative absolute pressure) and the calculations gives an error.

A value of dt = 0.001 sec is OK for normal calculations, dt = 0.0001 sec for regions with high accelerations and dt = 0.00005 sec for materials with a very low suspension velocity (5-10 micron powders)

Good luck

Teus

Sir , wish you very happy new year. Hoping to persistently help the needers

Below is few quiries

Sir I tried to find the same from Mr Agarwal article which says DeltaPproduct= DeltaPair*k*R

But this implies there is a linear relation between Air and product deltaP .

Would you please explain the relation to calculate Productdp by equalizing the change in internal energy in the gas with the energy change of the particles

Here shall I take the compressor pressure a pressure drop across the node? Or is it the compressor pressure just after delivery of compressor?

a) In above formula why we have divided dpLength by 10000 ? If so , then why it is not appearing in formula of absolute pressure? ie absolute pressure = ambient pressure + compressor pressure – pressure drop + 1

b) In absolute pressure formula (ie absolute pressure = ambient pressure + compressor pressure – pressure drop + 1) why we have added the +1

In this formula why did we multiply mass with 1000

Is this the temperature drop due to radiation ?

a) In above formula , why the sign is minus (-) in prefix

b) Will this (dTempdiff) be a final temperature ? or do I have to substract this from Tempmixbegin ■

Dear kj,

I calculated a cement system for 3 different pressures and determined the ratio between product pressure drop and gas pressure drop.

Capacity------Product dp---SLR------air dp----------convey pressure-------k= dpProduct/dpAir.R

228.4----------14864----------38.3------366----------------25000------------------1.06

204.7----------11205---------33.7-------418----------------20000------------------0.795

176.9----------7510------- --28.5-------488----------------15000------------------0.54

Conclusion: the Friction multiplier for the solids conveyed (K) is not constant for each Solid Loading Ratio

Calculating the various energies of the material and the conveying gas is an application of nature laws and is very extensive and complex, however doable.

absolute pressure= absolute pressure at considered location in the pipeline

compressor pressure = compressor pressure in bar(g)

pressure drop = pressure drop between compressor pressure and considered location in the pipeline.

+1 was too much

dpLength is in mmWC and to convert the pressure to bar, divide by 10000

To convert the values to the right units.

This is the temperature difference between the moving material/gas mixture in the pipe and the environment outside the pipe.

a)the minus(-) indicates the direction of the difference.

b)dTempdiff is a temperature change and needs to be added (because of the minus sign) to the Tempmixbegin.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

I calculated a cement system for 3 different pressures and determined the ratio between product pressure drop and gas pressure drop.

Capacity------Product dp---SLR------air dp----------convey pressure-------k= dpProduct/dpAir.R

228.4----------14864----------38.3------366----------------25000------------------1.06

204.7----------11205---------33.7-------418----------------20000------------------0.795

176.9----------7510------- --28.5-------488----------------15000------------------0.54

Conclusion: the Friction multiplier for the solids conveyed (K) is not constant for each Solid Loading Ratio

Calculating the various energies of the material and the conveying gas is an application of nature laws and is very extensive and complex, however doable.

absolute pressure= absolute pressure at considered location in the pipeline

compressor pressure = compressor pressure in bar(g)

pressure drop = pressure drop between compressor pressure and considered location in the pipeline.

+1 was too much

dpLength is in mmWC and to convert the pressure to bar, divide by 10000

To convert the values to the right units.

This is the temperature difference between the moving material/gas mixture in the pipe and the environment outside the pipe.

a)the minus(-) indicates the direction of the difference.

b)dTempdiff is a temperature change and needs to be added (because of the minus sign) to the Tempmixbegin.

Success

Teus

In attempt to fine tune the calculation followimng is input and output result , plea to have your view on same .

INPUT IS

[LIST=1][*]Node length :- 50mtr[*]Pipe dia :- 304mm[*]Material mass flow or dPipeMass :- 36 kg/sec[*]dSedimentMass :- 0 kg/sec[*]Tair initial:- 50 Degc[*]T material initial :- 150 Deg c[*]Compressor pressure :- 3 bar[*]Compressed air pressure drop from machine to suction point of pipe :- 1 bar[*]Atmospheric pressure :- 1bar[*]Absolute pressure :- 3+1-1=2 Bar[*]Kgas :- 1.4[*]Suspensiondp :- 0 bar[*]Productdp :- 0.07 bar[*]Gasdp :- 0.0023 bar[*]Gas mass flow :- 1.3 kg/sec[*]Sp heat of ash and air :- 0.25 kcal/kg & 0.24 kcal/kg respectively [*]HeatRestFactPipeWall = 1.032 Kcal/hr/ Deg C

Now my out put taking all units in consideration :-

[LIST=1]

1) dTair = (Tair + 273) / ((AbsolutePressure / (AbsolutePressure - dpLength / 10000)) ^ ((kGas - 1 / kGas)) - (Tair + 273) = - 3 Deg c

2) Tair = Tair + dTair= 47 Deg C

3) dHeatMat = (Abs(Suspensiondp) + Productdp + Gasdp) * GasVolume * 10 * dt = 0.2 kj/sec

4) dTempAirHeatMat = dHeatMat / (MatSpecHeatCont * 4.1875 * (dPipeMass - dSedimentMass) * 1000) = 0.0034

5) TempMix1 starting of node :- 146 Deg C

6) TempMix2 end of node :- TempMix1 + dTempAirHeatMat = 146.59

7) TempDifference = TempMix – AmbientTemperature= 106.59

8) dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass)) = -138.05 Deg C

Now my quiry is

1) Would you please check the same calculation and make a comment ?

2) Secondly calculated dTempdiff seems to be wrong , would you please correct me ?

3) Now once I have dTempdiff , then what shall be my final temperature at node end ?

4) I presume iam wrong in calculating HeatRestFactPipeWall as 1.032 Kcal/hr/ Deg C . Would you please correct me fith correct figure , taking assumption of the any parameter if missing ? ■

Dear kj,

In the iterative calculations, node length = product velocity * d(time)

3+1-1 = 3 instead of 2

Dimension is 0.1032 Kcal/hr/ Deg C/m2

dTempdiff is a result of cooling and therefore the final temperature at the end of the node is dTempdiff lower in temperature than originally calculated.

(Superposition of calculations, allowed, because d(time) is taken very small.

Have a nice day

Teus ■

[QUOTE=Teus Tuinenburg;69131]Dear kj,

Sir iam trying hard to integrate the concept of time duration in visual basic , but is very difficult . I will raise the concern in the relevent therad , but for a time being i have splitted the complete length of conveying pipe in appropriate node length .

Would you please elaborate how

Ok That means i have to calculate Pabs,dT air,Tair,dheat mat,dTempAirHeatMat,Temp mixTemp diff,dTempdiff at both the end of node length (ie pickup & termincal) isnt it?

Secondly as you said the final temperature is (dTempdiff at pickup of pipe -dTempdiff end of pipe), in such case then the value of dTempdiff shall be positive instead of negative . Please correctme .

This is something iam not familier with and yet to exercise for the same . But do you have any specific concern

Right now in the calculation of dHeatMat = (Abs(Suspensiondp) + Productdp + Gasdp) * GasVolume * 10 * dt , i have not considered the dt & assumed the P*dv total for one node length = kcal/sec . Please correct me with my assumption , do i need take number 10 in account? as the same is appearing as a part of calculation

In relevant thread you ssid using the dryer will yield to loss in energy .

I interpretate like this

1) Using air dryer will create the additional pressure drop and this being addressed by increasing the compressed air pressure propertionally by 8%

2) Indeed there is loss in temperature of compressed air too , but is negligible

Now other than above points , do you have any other concern in regard of energy loss ■

Dear kj,

By experience, I know that it is not easy to build a complex pneumatic conveying program in Visual Basic.

Using Visual Basic will become a second nature after a while and you will recognize the intellectual beauty of it.

Splitting the complete length of the conveying pipe in appropriate node length, is hardly “splitting”.

Splitting the conveying length in small sections (dL), which I did in small time increments (dt) is to simulate integration (numerical integration).

The numerical integration becomes inaccurate when the time increment (or length increment) is taken too big. (Result values can run out of control)

This value represents the amount of heat passing through the pipe wall in 1 hr per degree temperature difference between the internal temperature and the environment temperature (Heat radiation neglected) per square meter of pipe circumference area.

This value is very depending on the condition of the pipe wall. (F.i. insulation or submerged in the ground)

I use this figure as a result of calculation and experience in cement conveying installations, where the pipe temperature drops significantly in the first 10m to 20m.

Indeed, all these temperatures and temperature drops and increases are required to calculate the temperature at the end of the pipe length dL.

If you give “cooling” a positive value, then you have to subtract.

If you give “cooling” a negative value, then you have to add.

The numerical integration becomes inaccurate when the time increment (or length increment) is taken too big. (Result values can run out of control)

Pressure drop for elevation does not contribute to temperature changes and should be left out from

P*dv total.

The figure “10” is required to get the dimensions and value right.

1)correct.

2)if you use a refrigerating cooler, the temperature drop is significant and temperature is energy. In the pneumatic conveying calculation, the temperature changes mostly closely to the material temperature. (Not always negligible)

3)Not at the moment.

Success

Teus ■

sir

ground)

Sir would you please detail teh formula to calculate the same

You mean the dTempdiff at pickup point of node length subracting with dTempdiff at end point of node lenth , will yeld final temperature at node end.

Didnt understand ■

This is the derivation of the heat transfer through a series of heat-flow resistances.

Build from formulas like Qheat=heatconductivity * delta(T)*Area*Time

This can refer to heat transfer between 2 surfaces or heat transfer through a material.

see: http://en.wikipedia.org/wiki/Heatconductivity

The dTempdiff at end of node length subtracting with dTempdiff from starting point of node length, together with all the other temperature influences , will yield final temperature at node end.

To convert mmWC to Pascal

Have a nice day.

Teus ■

To convert mmWC to Pascal

Sir is this is like this

Pressure (pascal)*volume (m3/sec)= Kj/hr

Please comment on unit if iam wriong

Have a nice day.

Teus[/QUOTE] ■

.

Sir

Sir is the derivation is like this

Pressure (pascal) * volume(m3/sec)= Kj/hr

Please accentuate on unit if wrong

Have a nice day.

Teus[/QUOTE] ■

Dear kj,

N/m2 * m3/sec = Nm/sec = J/sec = Watt

To make this kJ/hr the multiplication factor is:

3600/1000=3.6

1 J/sec = 3.6 kJ/hr

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

N/m2 * m3/sec = Nm/sec = J/sec = Watt

To make this kJ/hr the multiplication factor is:

3600/1000=3.6

1 J/sec = 3.6 kJ/hr

Success

Teus

1) As suggested by you The dTempdiff at end of node length subtracting with dTempdiff from starting point of node length, together with all the other temperature influences , will yield final temperature at node end. But I have

a) Now since the dTempdiff at both end seems to be in decimal , hence I have considered , the Temp diff at end - dTempdiff at end will yield the final temperature at node . Please correct me with my algorithm ? If wrong , then what shall be corrected measures adopted in attempt to come-out with final node temperature

It is my humble request if you may please consider the pressure drop and air as a base approximation value and emphasize on my temp drop calculation and ■

Originally Posted by guddu

href="showthread.php?p=69397#post69397" rel="nofollow">

1) As suggested by you The dTempdiff at end of node length subtracting with dTempdiff from starting point of node length, together with all the other temperature influences , will yield final temperature at node end. But I have

a) Now since the dTempdiff at both end seems to be in decimal , hence I have considered , the Temp diff at end - dTempdiff at end will yield the final temperature at node . Please correct me with my algorithm ? If wrong , then what shall be corrected measures adopted in attempt to come-out with final node temperature

It is my humble request if you may please consider the pressure drop and air as a base approximation value and emphasize on my temp drop calculation and

Sir please read the aboev statement in conjuction with attached drg , as ai missed out attaching the same in my previous thread

Attachments

heat loss along fly ash convey pipe (PDF)

■

Dear kj,

The mathematical approach of calculating the pressure drop, velocity change and temperature change is based on numerical calculation.

This approach calculates the real pressures, velocities and temperatures in small steps.

When the step is taken too big (as you do with 50 m length), then the used approach is not valid anymore and integration formulas are to be used.

Building a numerical calculation algorithm that represents a physical phenomena, or a number of physical phenomena’s, requires a thorough set up of partial effects, which are superimposed at the end.

In the case of pneumatic conveying, I choose for a time increment dt.

The length increment is calculated as dL= vproduct * dt

A length increment dL is also a possible choice as a step increment.

I understand that you have all the temperature effects now lined up and that you are combining these in a calculating algorithm. That is just a mathematical job to do.

You can use your own mnemonics to indicate the variables.

Your program results, you can verify with real installations and if these do not match, then you have to search for the mistakes.

Here you are more or less on your own, because this work takes place in your brain and that is very difficult to read by other people.

Success

Teus ■

Dear sir

Would you explain the application approach of integration in our followed procedure and how do we use it ?

2)As enumerated by you TempMix2 = TempMix1 + dTempAirHeatMat . But in calculatng the Tempmix1 , we have considered Tair at start and not the Tair at end (due to temp drop attributed to air expansion) . This implies the temperature drop due to expansion is being neglected . Please correct me with my assumption ?

3) Please refer my previous calculation (attached pdf) , for the given input parameter , dTempdiff at end of node ie 0.03 seems to be very small for a given pipeline of 50mtr , value . To arrive this value I have followed your algorithm is

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

dTempdiff at end = (0.1032 Kcal/hr/ Deg C/m2/*4.1875*110 Degc*3.14*0.304 mtr*50Degc)/(1.04kj/kg*80000kg/hr) =0.03 Deg C

Now would you please check the same and correct me with actual figure ?. Indeed I have not considered the time increment , but this shall is apprx value for complete length of 50mtr

4) Also in formula :- dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass)) , TempDifference is being considered instead of LMTD (LOG MEAN TEMP DIFF – A driving force ) , would you please explain ?

5) Now to arrive at final temperature at node end , you said The dTempdiff at end of node length subtracting with dTempdiff from starting point of node length, together with all the other temperature influences , will yield final temperature at node end. But my observation is this dTempdiff is temperature drop due to all the temperature influence and to arrive at final node temperature this dTempdiff at end shall be subtracted from TempMix1 will yield the final temperature at end node. Please articulate ?

Your program results, you can verify with real installations and if these do not match, then you have to search for the mistakes.

Iam using this algorithm to check apprx value of one of the ongoing project ( as a academic interest) and in attempt to determine the possible condensation , hence i dont have real installation to check the same as a result need your technical support

kj ■

Dear kj,

If the calculated “dTempAirHeatMat” does not include the temperature drop, due to air expansion, then you are correct.

should be:

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t)^2 * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

You have to use a figure for the time increment dt.

“(dPipeMass - dSedimentMass)” is the material, present in the considered pipe section in kg.

You used probably the conveying rate in kg/hr.

Check your dimesions again.

The method is:

Start: absolute temperature isTempMix1

Calculation: partial pressure changes (expansion, friction, cooling, etc.). Can be + or -

End: end temperature is TempMix2 = TempMix1 + partial pressure changes

Best regard

Teus ■

Sir as taught by you

dTempAirHeatMat = dHeatMat / (MatSpecHeatCont * 4.1875 * (dPipeMass - dSedimentMass) * 1000)

This formula do no include the temperature drop due to expansion . Please articulate ?

should be:

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t)^2 * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

You have to use a figure for the time increment dt.

Indeed I have used Kg/hr instead of kg (mass in pipe) . Iam not structurally poised and familiar with the time increment concept as the same approach requires time and appropriate operational method , which I yet to familiarize , but to get the quick approximation I have considered the total mass travel per hour . Considering my non familiar issue , if you have any further observation on my approach , please articulate ?

Here I have stressed out to determine the final temperature at end of node , but to get

At one stage you said dTempdiff at end of node length subtracting with dTempdiff from starting point of node length, together with all the other temperature influences , will yield final temperature at node end and now TempMix2 = TempMix1 + partial pressure changes

Please tell me which aproach I shall follow in attempt to achieve my underline objective ie final temperature at end of node

Sir still awaiting for my quiry as why we have not considered LMTD in calculating dTempAirHeatMat

Best regard

Teus[/QUOTE] ■

Dear kj,

I have to correct a mistake in my previous reply. (Sorry)

The method is:

Start: absolute temperature is TempMix1

Calculation: partial temperature changes (expansion, friction, cooling, etc.). Can be + or -

End: end temperature is TempMix2 = TempMix1 + partial temperature changes

What do you mean with “LMTD” ?

For the method of numerical calculations, which is used to calculate a pneumatic conveying system, look at:

http://en.wikipedia.org/wiki/Numericalanalysis

have a nice day

Teus ■

1) Sir is the partial temperature changes will be (dT air +dTempAirHeatMat deg c +dTempdiff ) ? If yes then following quiries arre associated :-

a) In dTempdiff formula , whether the Tempdiff shall be taken for start or end of node ?

b) If Tempdiff to be considered at end of node than which Temp mix ( Temp mix at end or Temp mix at begin) shall I consider to calculate Tempdiff ?

2) I am still awaiting for response against my earlier query

3)

LMTD:- Log mean temperature difference .

4) Sir would you please teach me stepwise procedure to use all the lined up formula as taught by you

Thanks in anticipation ■

Dear kj,

1)

In a formula, the known values of variables are positioned on the right of the equal sign (=).

These variable values are calculated in the previous calculation and represent the conditions at the begin of the considered d(Length).

The, to be calculated, values at the end of the considered d(Length) are on the left side of the equal sign.

2)

Correct, the temperature drop due to expansion has to be accounted for separately.

4)

Lining up all the formulas is a job that took me the last 2 years busy and redoing that will take me again a few months to repeat 40 MB of programming.

have a nice day

Teus ■

1) This implies that temperature at end of node is

Tempmix2=Tempmix1+ (dT air+ dTempAirHeatMat + dTempdiff )

In my calculation

a) Tempmix1 = 144 Deg C

b) dT air = - 2 Degc

c) dTempAirHeatMat= 0.3

d) dTempdiff = -0.03

Hence Tempmix2 = 144-2-0.03+0.3=141.9 Deg C .Please correct me with my basic equation to quantify tempearure at end of node

2) Secondly please refer attached calculation as furnished by you quite some time back , in this case for first route length of 146 mtr there is a huge temperature drop from 155 Deg C to 66 Deg c , however in second node of same route length 146 mtr there is marginal temperature drop from 66 Deg C to 43 Deg c . Would you please justify the same

3) Also as reference to attached file , would you please put the relevant resultant figure due to part calculation to determine Tempmix 2 = 66 Deg C?

I mean please put the value on right side of below equation which lead to node end temp 66 Deg c as mentioned in calculation sheet

Tempmix 2 = Tempmix1……. ?+ (dT air……….?+ dTempAirHeatMat…..? + dTempdiff……….? )

Attachments

flyash conveying 1200m (PDF)

■

Dear kj,

The calculated temperature drop is related to the mixture temperature.

-The temperature increase of the material, due to friction has to be equalized with the conveying air.

-The temperature decrease of the air, due to expansion has to be equalized with the conveyed material.

-The temperature decrease, due to pipe cooling has to be equalized with the conveyed material and air.

For the example in your attached file it comes down to:

-The temperature increase of the material, due to friction is approx. 0.4 degrC.

-The temperature decrease of the air, due to expansion is approx. 7 degrC, which results in a mixture temperature decrease of approx. 0.5 degrC.

-The temperature decrease, due to pipe cooling is approx. 95 degrC. over the considered pipe length of 146 m.

The next pipe length has a much lower temperature difference with the environment and cools down less degrees.

When the temperature of the mixture approaches the temperature of the environment, the temperature will stay approx. constant.

Remember, that these values are calculated in a numerical logarithm and are not calculated in one step over a length of 146 m.

Have a nice day

Teus ■

1) Ok. That means while quantifying the mixture temperature drop as you explained below

, iam ensuring that the effect of both the temp drop due to expansion and temp increase due to friction is equalized , but what is equation for equalization of temperature decrease, due to pipe cooling with the conveyed material and air ? . Is this the dTempdiff ?

1) Sir I presume the 95 Deg c is dTempdiff . If yes the same seems to be very high , may be iam wrong , but to make the things simplicity would you please put the related value of below in below vformula , which lead to pipe cooling of 95 Deg C

dTempdiff = -(HeatRestFactPipeWall * 4.1875 * TempDifference * 3.141596 * Diameter(t) * dLength * dt) / (SpecHeatContMix * (dPipeMass - dSedimentMass))

2) May be if iam wrong in interpretation of above point no 2 ,but in such case please-please , tell me and arrange the resultant value of all the formula as explained earlier (dTair , dTempAirHeatMat, dTempdiff) in one equation in attempt to quantify temperature at end of node length of 146 mtr ie 66 Degc as referred to previous attached file

Thanks in antcicipation ■

Dear kj,

The temperature drop over a length of 146 from 155 degrC to 65 degrC does not seem to be unrealistic to me. Consider the pipe as a tube cooler of 146 m of length.

Filling in the formula with the relevant values does not work.

The temperature drop is calculated in small lengths (dLength) to approximate the cooling curve along the pipe.

If you use 146 m for dLength, the approximation is far from valid.

Because an approximation, numerical method is used, one formula is not possible and a set of formulas has to be applied (see reply #54 of this thread)

In reply #54 of this thread, the relevant formulas (part of the computer algorithm) are given.

I suggest that you evaluate the involved physics and rename the mnemonics of the variable according to your own understanding. That would be much more understandable for you. (and less to others )

Success

Teus ■

The temperature drop over a length of 146 from 155 degrC to 65 degrC does not seem to be unrealistic to me. Consider the pipe as a tube cooler of 146 m of length.

Sir in previous therad you said

. Is this dTempdiff (due to convection,conduction & radiation) ?

I understand and fully agree with you in regard of approach of time increment , to determine the temperature drop

Now , if I comply the same approach , then though I have all formula lineup , but how do I sequence it is something iam struggling with .

You said

Tempmix2= Tempmix1 + Temperature changes due to ( Expansion, friction, cooling )

Now to quantify the Temperature changes due to ( Expansion, friction, cooling ) , do I add all the temperature changes (dTair, dTempAirHeatMat, dTempdiff) ?

I esteem with the great kindness for your kind consideration ■

Dear kj,

The heat balance in pneumatic conveying is made by the expansion energy of the air, which covers the energy used for losses.

The expansion energy of the air is the equal to the energy used for losses.

The expansion of the air cools down the air and the lost energy in conveying, heats up the mixture.

The overall result is a slight change in temperature of the mixture.

At high mixture temperatures, the heat loss and thereby the temperature drop are high.

At lower temperatures, the heat loss and thereby the temperature drop are low.

When the mixture temperature approaches the environmental temperature, the mixture temperature stays fairly constant.

dTempdiff is the change in temperature difference between the mixture and the environment over a length dL.

Yes, you superimpose all the changes in the temperature of the mixture.

Better to say: Add all the involved energies that cause temperature changes of the mixture and then divide the resulting energy between the air mass and the material mass.

(see reply 79 of this thread)

The temperature issue in pneumatic conveying is just a fraction of all the involved physics and mathematics. This knowledge has to be sorted out by the respective engineers, which will build their skill and value for companies.

This makes the knowledge a commercial asset and you can only give away knowledge just once.

Transferring knowledge is done by universities.

This is why companies keep their calculations sealed from the customers.

Success

Teus ■