Originally Posted by Teus Tuinenburg

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

Dear sir

Tonnage of thanks

1) My specific quiry in previous thread was you said

, now is this (95 Degc) the dTempdiff ? ■

Dear kj,

In the form of a formula:

dTempdiff = TempMix1 - TempAmbient – TempMix2 + TempAmbient = TempMix1 – TempMix2

Yes, that is correct.

have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

In the form of a formula:

dTempdiff = TempMix1 - TempAmbient – TempMix2 + TempAmbient = TempMix1 – TempMix2

Yes, that is correct.

have a nice day

Teus

If so , then what is the value of below variable (please refer your corresponding result of dTempdiff = 95 Degc) in quantifying dTempdiff ?

a) HeatRestFactPipeWall :- You already said it is 0.1032 Kcal/hr/ Deg C/m2 . Please correct me?

b) TempDifference :- 155-35= 120 Deg c . Please correct me?

c) dPipeMass :- Please specify?

d) dSedimentMass :- Please specify? ■

Dear kj,

Kcal/sec/ Deg C/m2

155-35=120. correct.

dPipeMass is the amount of moving material in a pipe section with a length of dL

dSedimentMass is the amount of sedimented material in a pipe section with a length of dL

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Kcal/sec/ Deg C/m2

155-35=120. correct.

dPipeMass is the amount of moving material in a pipe section with a length of dL

dSedimentMass is the amount of sedimented material in a pipe section with a length of dL

Have a nice day

Teus

Sir as you said

HeatRestFactPipeWall = = 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

U= 1/ ( 1/hi+1/k pipe+ 1/ho)

Where

a) hi= Internal heat transfer coefficient

b) ho= External heat transfer coefficient

Now to calculate hi , I use formula:-

Nu (nusselt no for turbulent ) = hi d/k

Hence hi = Nu*k/d

Nu = (0.025)Re^0.79)(Pr^0.42)* ((u/uw)^0.11)

Now my quiry is

a) Is this the right approach for our pneumatic conveying system ? . If not please explain ?

c) What do I do to calculate External heat transfer coefficient

d) Eventually would you please enlighten me simple equation , as how did you arrive

HeatRestFactPipeWall = 0.1032 Kcal/sec/ Deg C/m2 ■

Dear kj,

Probably, this approach is too scientific for practical use.

In pneumatic conveying installations you will find a variety of pipelines (steel painted pipes as in cement conveying installations or insulated pipes or underground pipes, etc.)

Each pipe has a different heat transfer factor and calculating this factor would take a lot of calculating power.

I eliminated this problem by calculating the heat transfer coefficient back from field measurements.

This was sufficient for the majority of applications, which I was dealing with.

I think that applying scientific theories in engineering always uses a “fudge factor” to link the theory to the practice.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Probably, this approach is too scientific for practical use.

In pneumatic conveying installations you will find a variety of pipelines (steel painted pipes as in cement conveying installations or insulated pipes or underground pipes, etc.)

Each pipe has a different heat transfer factor and calculating this factor would take a lot of calculating power.

I eliminated this problem by calculating the heat transfer coefficient back from field measurements.

This was sufficient for the majority of applications, which I was dealing with.

I think that applying scientific theories in engineering always uses a “fudge factor” to link the theory to the practice.

Have a nice day

Teus

Thanks

Sir in our previous therad i raised a quiry of using the LMTD instead of tempDifference

Would you please justify , why dont we use LMTD instead of tempdiff Iin above formula ?

LMTD:- log mean temperature difference (also known by its acronym LMTD) is used to determine the temperature driving force for heat transfer in flow systems, ■

Dear kj,

LMTD = (TempDiffBegin – TempDiffEnd) / ln(TempDiffBegin/ TempDiffEnd)

In this case TempDiffBegin is known and TempDiffEnd is to be calculated.

The value of LMTD is now unknown and therefore, this formula is useless in this case.

Also have a look at:

http://en.wikipedia.org/wiki/Logmea...uredifference

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Attached the requested calculations.

Success

Teus

Sir

I recall this therad once again in effort to get your attention few of my quiries

1) Please refer attached sheet (Fly ash conveying 1200 mtr) . In this sheet there is possible condensation at part 4,6, & 7 as the RH is almost 100% . Now if there is possible condensation , then what is the extra precaution you have considered to adress the same in design stage ?

2) Secondly to determine the suspension velocity ie 1.29 m/s , it seems you have considered air density at atmospheric condition , however to arrive at the pickup velocity the density of air has to be corrected for local condition , which implies the density of air shall be apprx 2kg/m3 (cosnidering 2.75 bar pressure at pickup) . Please comment ? ■

Originally Posted by Teus Tuinenburg

href="showthread.php?p=68014#post68014" rel="nofollow">

Dear kj,

Attached the requested calculations.

Success

Teus

Sorry i forget to attch the file , please refer the enclosed file

Attachments

flyash conveying 1200m (PDF)

■

Dear kj,

Cooling and drying the convey air.

At 2.75 bar, the air density is approx. (2.75+1) * 1.2 = 4.5 kg/m3.

Indeed, for the pickup velocity at the beginning of the pipeline where the pressure is 2.75 bar, the pickup velocity has to be corrected for the local density.

Success

Teus ■

Sir

Buit what happens to process in such case (condensation) part 4,6,7 . I mean would there be any effect on process (could be extra drag force ) ? If so , what shall be measure to address the same

This implies the revised pickup velocity is

Suspension velocity = SQRT ((4/3 * (3)^-5*2270/(0.04*4.5)) = 0.544 m/s

d particle= 30 micron

particle dens= 2270

density air @ local cond = 4.5

Hence pick velocity = 6*0.29 = 3.2 m/sec

Please correct me ? ■

Dear kj,

In the example file there is 120 kg/hr condensed water at 142000 kg/hr fly ash.

That means approx. 0.084% in weight.

Unless the condensed water stays at the same location, binding the fly ash, (which is not likely to be the case) there will be no problem. The material flow will remove potentially wet fly ash.

A more hygroscopic material is probably less tolerant to moisture.

The calculated pickup velocity at 2.75 bar is correct and a good start for the first design calculations.

Have a nice day

Teus ■

Dear sir

Iam posting below matter in this thread as my earlier post doesnt appear in relevant therad

YOu said

Great . This implies the pressure drop at the bend is only gas pressure drop which will be more than gas deltap in linear pipe , because of increase gas velocity on account of sediment formed which lead to increase block area and deducted from flowing area . Please correct me with my presumption?

.

I re-initiate the matter articulating that angular position in bend depends on the material velocity & material concentration . Now as you said concentration follows from the flow rate and the velocity . Now I may not be correct with my previous perception , If so

a) Than in what way the concentration influence the angular position of particle in bend?

b) You said concentration follows from the flow rate and the velocity. If I define the concentration correctly , it is the amount of material present in total mixture . If so , would you please mathematically express once again the relation of material velocity and concentration (specially at the bend) ?

1) Ok , once the material lost the momentum in bend , this implies we require extra momentum force to move the lost momentum particles which is being taken care after the bend . But sir , how we are making sure that the lost momentum particle is still moving in the bend ?

2) What is restored interaction after bend . ?

3) Now if I understand correctly the re-acceleration pressure drop after bend is to re-acclerate the material (having less velocity due to momentum loss ) to its steady state (acceleration in in linear pipe) . Please correct me ?

4) For a given pipe area of 0.017 m2 for 150mm pipe & solid flow of 8.3 (kg/sec) & vproduct at bend:- 2.9m/s & BD=750kg/m3

Filled area = 8.3/(2.9*750)= 0.0038m2

Hence area for air :- 0.017(pipe area)-0.0038(filled area) = 0.0132 m2

Please correct me with the derived value of filled area

Ok. But how did you determine conveying energy (253.9) ? . Sir please emphasize on mathematical expression.

a) Sir , as per you experience what is value of alpha & beta?

b) Once I determine material loss coefficient , where shall I use the same ?

Sir my quiry is still persist as above

Sir my quiry is still persist as above

For an = -[ffriction * (vrel^2/Radius-g*sin(a))-g*cos(a)]

Now if V product1(at inlet of bend) at inlet of bend :- 3 m/s

vrel = (8-3) = 5 m/s

Radius of bend :- 1mtr

Then :- Angular position:- 0.003 = 0.17 Deg

an = -0.5 * ((5)^2/1 -9.8*sin0.17 – 9.8*cos0.17)) = -7.6m/s2

Hence Vproduct2 (at bend) = 3-7.6*0.001=2.99m/s

Sir please correct me with the derived vproduct2 at bend ?

Sir you said Particle surface water is not considered in the calculations . But sir In general increasing the moisture content of a powder decreases its ability to flow smoothly. Increased surface moisture caused increased surface tension, which caused cohesion between particles . Indeed the fly ash as you taught is inherent in nature ,but having this perception in mind , do you see any apparent implication of surface moisture in fly ash in flowibility ?. If so , where shall we account the same

thansk in anticipation ■

Originally Posted by Teus Tuinenburg

Dear kj,

In the example file there is 120 kg/hr condensed water at 142000 kg/hr fly ash.

That means approx. 0.084% in weight.

Unless the condensed water stays at the same location, binding the fly ash, (which is not likely to be the case) there will be no problem. The material flow will remove potentially wet fly ash.

A more hygroscopic material is probably less tolerant to moisture.

The calculated pickup velocity at 2.75 bar is correct and a good start for the first design calculations.

Have a nice day

Teus

Dear sir

I happen to determine the conveying condition (temp mix) IN WINTER SEASON (10DEG C) at after 50mtr travel and below is observation

1) Condition A :- Inlet ambient air conditions:

Inlet ambient air conditions:

Ambient pressure = 1 bar(a)

Ambient temperature = 10 degrC

RH = 100%

Calculated water vapor content = 0.007625 kg/kg dry air

2) Condition B :- Compressed air conditions:

Compressed air conditions:

Pressure = 3.5 bar(a)

Temperature = 50 degrC

RH-34%

Calculate water vapor content = 0.007625 kg/kg dry air

Condensed water=0

3)Condition C :- Then the air mixes with the fly ash (150 degrC).

The mixture temperature becomes approx. 125 degrC The convey air conditions become then:

Pressure = 3.5 bar(a)

Temperature = 125 degrC

Again RH-0%

4) Air conditions after 50mtr in pipeline :

Pressure = 3.5 bar(a)

Temperature = 25 degrC

Calculate water vapor content = 0.005604 kg/kg dry air

Condensate mass :- 0.007625-0.005604 = 0.002062 kg/kg dry air

Now atmospheric condition being in winter season will yield a condensate mass of 0.002062 kg/kg dry air which will potentially create the problem in conveying pipeline . Now to address the issue I have two option

Except using air dryer what shall be extra measure has to be incorporated in design state ?

THANKS ■

Dear kj,

Condensation is influenced by:

- temperature

- pressure

- water content

Preventing condensation is done by controlling these three parameters.

- Increase temperature (is not always possible)

- Decrease pressure (Lower pressure design)

- Decrease water content (Drying compressed air)

Whether the condensation is a problem, determine the amount of condensed water in relation to the amount of conveyed material. Normally this ratio is so low, that it is acceptable.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

Condensation is influenced by:

- temperature

- pressure

- water content

Preventing condensation is done by controlling these three parameters.

- Increase temperature (is not always possible)

- Decrease pressure (Lower pressure design)

- Decrease water content (Drying compressed air)

Whether the condensation is a problem, determine the amount of condensed water in relation to the amount of conveyed material. Normally this ratio is so low, that it is acceptable.

Have a nice day

Teus

Dear sir

I require your consideration on my attached calculated temperature drop in pneumatic conveying

a) Ambient temp/Inlet air temp to comp = 50Degc

b) All other parameters are stipulated in attached sheet

Now my quiry is

In last second node of node length 100mtr the dTempdiff is 45 degc which yield the Tempmix2= Tempmix1- dTempdiff= 35.02 Deg c . Now here the 35.02 seems to be lower than atmospheric temperature which should not be . Iam sure iam wrong with my calculated procedure , but would you pelase help m in correcting the same ?

Attachments

temperature drop (PDF)

■

Dear kj,

If a calculated mixture temperature is lower than the ambient temperature, that can be caused by a too long considered node length.

Remind that the applied method (numerical integration) is meant for small steps.

In your attached sheet, I noticed that the “flow at start node kg/hr” is not the same as “flow at end node kg/hr”.

The law of continuity for a closed system forbids this, unless there is mass added to or removed from the considered system.

Success

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

If a calculated mixture temperature is lower than the ambient temperature, that can be caused by a too long considered node length.

Remind that the applied method (numerical integration) is meant for small steps.

In your attached sheet, I noticed that the “flow at start node kg/hr” is not the same as “flow at end node kg/hr”.

The law of continuity for a closed system forbids this, unless there is mass added to or removed from the considered system.

Success

Teus

Tonnage of thanks for your generosity

I take pleasure to once again pickup the matter iam referring to

Please refer the attached pdf sheet and shall be read in conjunction with pdf sheet attached in previous thread. Here as suggested by you I have determined stepwise calculation with notified result . As apprehended length is not too long , rather it is 100mtr . Adhering with all the process parameter ( as attached in my previous thread) Tempmix2 at end of node2 = 31.42 Degc ( less than ambient temp )

1) Imay be wrong in stepwise calculation , but please correct me with correct result

2) Also would you please forward the anticipated calc sheet keeping all the process parameter as same ? This will help me in comparing my result with likely result and then can discover the error

Attachments

temp drop for node2 (PDF)

■

Dear kj,

At the beginning of the length dL, there is one mixture temperature, where the air temperature equals the material temperature.

In your example, the air temperature is 45.86 degrC and the material temperature is 77.34 degrC.

This should be approx. 70 degrC, depending on the loading ratio.

In step 4, you are still using a length of 100 m, which is far too long, leading to a too big temperature drop.

I do not have a calculation sheet for the calculation as you have done. The temperature calculation in my program is embedded in the complete calculation algorithm.

Success

Teus ■