DEAR ALL ,

AS BEING MENTIONED IN THE ABOVE MENTIONED ARTICLES /SOLUTIONS THAT THE FUNDAMENTAL THING IS

PV/T =CONST , BUT THIS STATES THAT THE VOLUME EXPANTION DOESNOT DEPEND ON THE PIPE LENGTH IT JUST DEPEND UPON THE PRESURE DROP WHICH WILL BECOME FIXED WITH ANY LENGTH OF PIPE.

REGARDS

SACHIN ■

The PV/T relationship is true for each and every point in the pipeline - the pressure drop for each increment of pipe length changes, therefore the volume changes and thus the velocity changes

If you could manufacture a pipe line which had an internal tapered bore and the slope of the bore was the same relationship as the expansion of air, then you would have a pipeline with constant velocity. That's the rationale on which the concept of line stepping is predicated.

Regards ■

dear Hilbert

what you have told is true but if i use a blower at one end of a smaller length of a pipe line with the other end at atmospheric pressure and some other bigger pipe with with the same blower at one end and atmospheric pressure at the other end then in both the cases we take theortically the final velocity same in both the cases as the pressure drop in both the cases is same that is (blower pressure) - (atmospheric pressure)

please clarify

thanks and regards

sachin ■

Dear Sachin,

If you operate the same blower at 2 pipelines of which the diameter is the same and the pressure is the same, but the Length1 < Length2, then you are right.

v-beginL1 = v-beginL2

v-endL1 = v-endL2

But in this situation, the Loading Ratio LR1>LR2

In other words: capacity L1 > capacity L2

success ■

Sachin

If your material capacity is the same in both examples you gave, then the terminal velocity is the same but the pick-up velocities are different due to the higher pressure drop in the longer length line.

Regards ■

dear all ,

i believe that for the smaller pipe length the pressure drop per unit equivalent length is more as the the total pressure drop across the pipe ends shall get uniformaly devided and hence the pickup velocities and as well as the conveying velocities shall be more in the smaller pipelength

regards

sachin ■

Dear Sachin,

The pressure drop per unit of equivalent length , defined as dp/L, is for a smaller length L1 higher than for a longer length L2, providing dp is the same for both lengths.

The velocity in the pipe is calculated as (minor influences neglegted);

v = blowervolume/pipearea * pabs/1 * 273/(273+t)

Note that the length is not part of the velocity calculation.

As long as the entrance pressure is the same, the velocity is also the same.

At the end of the pipeline, the pressure ia atmospheric for both pipelines, so the end velocity is also the same.

Do not forget that, when you change the pipeline length, either the pressure drop is kept constant and then the capacity changes or the capacity is kept constant and then the pressure changes.

When the pressure changes, the velocities change (see formula)

Intermediate solutions are also possible.

have a nice day ■

take a simple example

100 mm ID pipe

75 m long

true density of product = 1000 Kg/m3

capacity = 10 mtph

With a blower of 15 m3/min and a pressure at the pick-up point of 0.5 barg, the pick up velocity is roughly 1270 m/min and the terminal velocity is 1886 m/min.

If I extend the line to 125 m long, and change nothing else, the pressure at the pick-up point increases to 0.88 barg, the pick-up velocity drops to 1015 m/min and the terminal velocity remains the same.

Regards ■

Dear Teus,

firstly i think the pressure drop in the specified case is same ie at both the ends of the pipelengths the the pressure differential is of gauge pressure of blower discharge .

please clarify if the actual pressure drop in both the pipelengths are different i.e one pipelength has greater no of rotory feeder or some other pressure drop points that weather that parameter comes into account or not , if yes then how .

thanks and regards

sachin ■

Dear Hilbert ,

thanks for the clarification , but I am not getting the pickup velocity i.e by flow rate/ cross-section area .

please clarify howcome you are changing the pressure at the pickup point when pressure differential across the pipe length is same for botth the cases .

thanks and regards

sachin ■

Sachin

My example assumed that capacity was the same in both case, that's why the longer length line has a higher pressure drop and lower pick up velocity

If you asssume the pressure drop to be constant, then you have accepted a lower capacity in the longer length line but have the same pressure drop, so the pick up velocities in that case could be the same.

Regards ■

DEAR ALL ,

I WAS GOING THROUGH THE ABOVE MENTIONED CALCULATIONS AND FOUND SOMETHING WHICH I AM ANABLE TO DIGEST I.E

84.95 nm3/min at 0.3885 bar vacuum or 0.6115 bar(a)

the density at the blower intake is then:

1.293 * 273/(273+20) * 0.6115 kg/m3

( I DONOT THINK IT SHOULD BE 1.293*273/(273+20) ONLY

The blower displacement is :

84.95/0.6115 = 138.92 m3/min equals 4908 CFM

UNABLE TO UNDERSTAND THIS FORMULEA CAN YOU EXPLAIN IT IN BRIEF .

CAN ANYBODY GIVE ME THE COPY OF THE VACUUM PRESSURE DROP CURVE YOU PEOPLE ARE TALKIN OF

REGARDS

SACHIN ARORA ■

Dear Sachin,

The conditions at the intake of the blower in vacuum mode are:

vacuum at inlet 0.3885 bar

p-inlet(absolute) at inlet is 1-0.3885 = 0.6115 bar(a)

Assuming the temperature at the inlet is 20 degrees Celsius.

The density is calculated as:

density = 1.293 * p(abs)/p(atm) * 273/(273+t)

density = 1.293 * 0.6115/1 * 273 / 293 = 0.7367 kg/m3

(Which can be expected, because thin air is lighter)

1.293 kg/m3 is the air density at 0 degrees Centgrade and atmospheric pressure.

The blower is a positive displacement pump and therefore displaces a constant volume of air under intake conditions per revolution.

best regards ■

Rob

going back to your original (3) questions

1. yes -- blowers are usually sized based on the intake conditions

The volume is expressed as ACFM -meaning the ACTUAL CFM at the inlet conditions of the blower.

2. Conversion of volumes from SCFM to ACFM at any point in the system is done by applying the relationship P1xV1/T1 = P2xV2/T2

Remember- the terms have to be in absolute units so V is CFM, P is PSIA (psig +14.7) and T is in Rankine or (F + 460). In the case of vacuum, absolute pressure = (29.92- vacuum gage)

3. Rather then use a blast gate to control volume by raising the pressure differential - you're much better off to change sheaves to have the blower run at the correct speed for the volume you need.

If we take your situation of 3000 SCFM and a blower inlet of 11.3 " Hg, (disregarding the effect of temperature) you would have (3000 x 29.92) / (29.92-11.3) = V2 = 4820 ACFM

The fact that the ACFM is greater then the SCFM is the reason the terminal conveying velocity in a pneumatic system is greater then the pick up velocity as long as you use the same diameter pipe.

Regards

Regards ■

Dear All ,

i am still anable to understand that howcome you have taken pressure at pickup of 0. 8 barg in the longer pipeline and of 0.5 barg in the shorter one .

secondly what my personal view is that the pressure drop per unit length in the Shorter pipeline will be more hence expansion of gases per unit pipelength is more in the shorter one but finally at the terminal point the the terminal velocity shall be same in both the cases.

secondly can anybody help in calculating the actual pressure drop which a designer shall consider i.e the blower/compressor capacity a designer have to consider if he knows the material to air ratio to be considered & the the pick up velocity to be considered and the conveying distance,i.e Equivalent length to be considered & the characterstics of the material to be conveyed .

thanks and regards

sachin ■

Dear Sachin,

If you have 2 different pipe lengths of the same diameter and with the same blower, then there can be 2 situations.

1)Capacity is the same

Then, the pressure is different (0.5 barg and 0.8 barg)and the begin velocity is different.

The end velocity is the same

Pressure drop per meter is the same, causing a higher pressure drop at a longer length.

2)Pressure is the same

Then, the capacity is different and the begin velocity is the same.

The end velocity is the same

Pressure drop per meter is different, caused by a lower Solid loading Ratio at a longer length.

lower pressure drop per meter * longer length = same pressure drop

Your second question comprises the complete knowledge of pneumatic conveying, which is too complex to explain in this forum in short.

The advice would be to buy some books, dealing with this subject.

best regards ■

Dear Sir ,

Please suggest me some book which will cover maximum no of aspects related to the specified topic , i have already read a few books but i am still not clear with the subject and till date i am anable to bring the results into practice.

Secondly i need your help in calculating the power consuption per tonne of material transfered in a pneumatic conveying system

specifically i would like to to know how to calculate power consumtion during loading and unloading cycles in a pressure conveying system , does this thing is also applicable in case i use ablower for some batch type system .

Thanks and Regards

Sachin ■

dear Sachin,

The article of mr Agarwal is a good start.

If you search the internet for “pneumatic conveying calculations”, you will find a lot of books to consider.

Do not expect to become an expert over a week.

If you want to know the energy consumption per ton conveyed, you have to calculate first:

1)pressure drop

2)capacity

3)compressor type( blower, screw compressor, compression regime)

4)vessel size (pressurizing time)

5)purge time of pipe line

6)etc.

The energy consumption per conveyed ton is then :

average power over the cycle time / capacity in kW/(tons/hr) = kWh/ton

success ■

The books by Dr. David Mills and Dr. George Klinzing are very good sources. You can also contact Powder Bulk Engineering to obtain a complete set of Pneumatic Conveying Points to Ponder which address every aspect of pneumatic conveying system and component design.

We offer several different levels of in plant training programs for the theory, application, system design and auxiliary components associated with pneumatic conveying systems. Several forms of sizing software are also available. We use the theory as a background but focus on practical solutions with real experience.

If you are interested, send me an e-mail to the address below.

Regards

Jack ■

dear all,

please tell me regarding the free fall velocity of ash under gravity ,

the ash is only 8-10 hour old for an FBC boiler having moisture content of 1-2 % & at a temperature of 100 degree celcius

thanks and regards

sachin ■

1.Q1 given on the exhauster curve is at max. vacuum at

the intake of the exhauster. You are correct!

2.cfm to scfm Sorry we only do metric system!!

3.If you increase you vacuum your volumetric flow rate

will decrease for given rpm. Check the exhauster curve to

calculate the decrease. ■

dear Rob,

You make it too difficult for yourself.

1)

I assume a positive displacement blower.

Then the blower data give the displaced volume under the stated intake conditions.

2)

The displaced mass of air is the displaced volume times the blower intake density.

Converting this into SCFM gives you the air volume at the intake of the vacuum pipeline.

3)

The displacement of the blower decreases a little (due to increased internal leakage)

at higher vacuum.

The vacuum limit is set by the outlet temperature.

To help you specifically, you need to inform us more in depth (installation,blower)

I donot understand the function of the blastgate. By opening you should destroy the vacuum.

success

teus ■

dear Rob

To avoid misunderstandings, the definitions are:

CFM = cubic feet per minute under any conditions

(SI units m3/min)

CFM = cubic feet per minute under atmospheric conditions of 20 degr C and 1 bar absolute.

(SI units nm3/min, n stand for normal)

You say that the blower performs.

3000 SCFM at 11.3” Hg vacuum

The inlet conditions are then :

In SI units:

84.95 nm3/min at 0.3885 bar vacuum or 0.6115 bar(a)

the density at the blower intake is then:

1.293 * 273/(273+20) * 0.6115 kg/m3

The blower displacement is :

84.95/0.6115 = 138.92 m3/min equals 4908 CFM

The confusing factor is the S in SCFM

If you need 2585 SCFM for the pneumatic conveying system,

the 3000 SCFM is sufficient.

The first advice could be to recalculate the pneumatic conveying system with

3000 SCFM and find out if that works to.

Bleeding air through a blast gate is a waste of energy.

Moreover the blast gate bleeds different volumes of air at different vacuums.

best regards

teus ■

The simple answer is YES !!!!!!

Look at the blower curve and follow the line of

your required pressure drop to the required volumetric

flowrate note the rpm. Get a new pully to match the rpm.

Problem solved.

Chocking the exhauster inlet in not the way to do it. Why do you

want to waste KW's !!! ■