Re: Power Consumption In Pneumatic Conveying
Dear Shibo,
1)
The formula, you are referring to, is the power, needed to compress the flow rate from pinlet to poutlet isothermally.
The right notation is: "power=165*gas flowrate*ln(Pcompressor/Pinlet)"
The relation is derived in thermodynamics and can be applied to a compressor that compresses the gas isothermally (cooled while compressed).
This is not directly related to pneumatic conveying.
2)
The diagram, you are referring to is the Zenz diagram.
The region left to the lowest pressure drop is called dense phase conveying and the region to the right is the is called dilute- or lean phase conveying.
However, the energy consumption is not proportional to the pressure but is a function of the pressure and the airflow.
A higher pressure with a lower airflow can require a lower power.
Energy consumption is normally expressed as Power/(tons/hr) = kW/(tons/hr) = kWh/ton.
If you calculate the Zenz diagram and the energy consumption at various airflows, then the lowest energy consumption per ton is in the dense phase region.
This is caused by the fact that the conveying pressure is a function of the sum of partial pressure drops of which some of them are a function of the velocity^2 and SLR^(alpha) and traveling time in the pipe.
If the airflow decreases, then the pressure drops related to velocity also decrease.
If the airflow decreases, then the pressure drops related to SLR and traveling time in the pipe increase.
The sum of the 2 can decrease or increase.
In the region, where the pressure drop does not change much, the influence of the airflow on the power consumption is the strongest.
This causes always the lowest energy consumption to be in the dense phase region, provided that the material can be conveyed in the dense phase mode.
Attached a calculated Zenz diagram is reference.
Have a nice day
Teus
Attachments
zenzdiagram400mportlandcement_1 (PDF)
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Teus
Re: Power Consumption In Pneumatic Conveying
Dear shibo
as the formula "power=165*gas flowrate*ln(Pcompressor/Pinlet", we can understand the power is related with flowrate and the pressure,
so if the vendor said their dense phase pneumatic conveying system is saving power, it can not always be right. you should compare the dense phase and dilute phase power consumption using above formula, then, you can get the right result.
any question contact with me: jcgjcg168@163.com
BR
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Re: Power Consumption In Pneumatic Conveying
Thanks for your clarification. It is greatly helpful.
I have an another question. Now it is clear that power is a function of pressure and air fowrate. However, "power=165*gas flowrate*ln(Pcompressor/Pinlet)" seems to have to treat air as compressible. It also requires the pressure of the compressor to calculate the power required. Provided that I have in hand a zenz diagram. It may be obtained from some bend or straight pipe section, involving only fully-developed flow. How can I calculate the corresponding power consumptions according to the zenz diagram? Here, I only know particle properties, gas velocity, and pressure gradient across a fully-developed flow pipe section.
I just read a paper. It gives some relations for energy consumption, which are very different from the above one. It says for the developed section, E=dp/dl * air velocity. For the developing section, E= p * air velocity. Your comments on these two relations would be greatly appreciated. ■
Re: Power Consumption In Pneumatic Conveying
Dear shibo,
"power=165*gas flowrate*ln(Pcompressor/Pinlet)" is a power calculation of an isothermal compressor.
Other type of compressors (screw compressors, blowers or piston compressors) require other formulas.
Air is compressible.
E=dp/dl * air velocity = N/m^2 /m* m/sec = N/m^2sec ??? (Energy is in Nm and power is in Nm/sec)
E= p * air velocity= N/m^2 * m/sec = N/msec ??? (Energy is in Nm and power is in Nm/sec)
To calculate power from the Zenz diagram use the formula:
dp*Volume/sec = Internal energy/sec = N/m2 * m3/sec = Nm/sec.
You will find that the smallest rectangle between the origin of coordinates and a point on the Zenz diagram coincides with the lowest energy consumption.
Also visit the BulkBog at the following links:
Pneumatic conveying, Performance and Calculations:
https://news.bulk-online.com/?p=65
Dense phase- or dilute phase pneumatic conveying:
https://news.bulk-online.com/?p=238
Pneumatic conveying, turbo- or positive displacement air mover:
https://news.bulk-online.com/?p=309
Energy consumption per ton of a pneumatic conveying system:
https://news.bulk-online.com/?p=331
Pneumatic conveying, an unexpected relationship.
https://news.bulk-online.com/?p=445
Pneumatic unloaders: Problems to avoid
https://news.bulk-online.com/?p=74
Influence of electro static charge on pneumatic conveying.
https://news.bulk-online.com/?author=15
Bacterial heating of cereals and meals.
https://news.bulk-online.com/?p=216
Have a nice day
Teus ■
Teus
Re: Power Consumption In Pneumatic Conveying
Dear Teus,
Thanks for your helps.
Would you please direct me to some papers and books which detail the calculation of power or energy under different conditions? ■
Re: Power Consumption In Pneumatic Conveying
Dear shibo,
The power calculation can be made when the following data are available (or calculated):
1)Airflow
2)Conveying pressure
3)Type of compressor
Airflow and conveying pressure follow from the pneumatic conveying calculation.
(Ask Mr Agarwal for his article "Theory and Design of Dilute Phase Pneumatic Conveying Systems" and spreadsheet calculation)
Also see thread:
https://forum.bulk-online.com/showthread.php?t=19268
The type of compressor is a choice, whereby the conveying pressure is a ruling parameter.
Normally for low pressure (below 1 bar) a positive displacement blower is used and for pressures between 1 and 3.5 bar a screw compressor with internal compression is chosen.
They have the advantage of being simple in design and reliable in operation.
However, each compressor type has it own calculation formulas.
A blower operates under isochoric conditions.
A screw compressor with internal compression, compresses internally under adiabatic conditions and then expands or compresses under isochoric conditions to the outlet pressure.
Other type of compressors s.a. vane compressors, turbines, piston compressors, water ring pumps, etc are also possible.
Have a nice day
Teus ■
Teus
Re: Power Consumption In Pneumatic Conveying
Dear Teus,
It seems to you that type of compressor is prerequisite to energy calculation. Is my understanding correct? if this case, no offense, this sounds strange as we may have difficulty to separately assess energy consumptions in different parts of pipelines? For example, there is a 100m pipeline consisting of two 50-m horizontal pipes and one bend. The bend is installed between the two horizontal pipes. I only want to know the energy consumption or power of the bend. How can I calculate it?
Best regards,
Shibo ■
Re: Power Consumption In Pneumatic Conveying
Dear Shibo,
The calculated energy from the Zenz diagram (dp * Volume) represents the internal energy of the conveying air that is transferred to the material particles and losses.
The compressor creates that internal energy by its compression principle.
There 2 energy consumption issues here.
The compressor energy consumption and the transferred internal energy to the conveyed product.
If we are only considering the internal energy, then you are right, the compressor plays no role.
Calculating the energy consumption of a bend:
The material is entering the bend with a material velocity vin
The material is changed in direction against the outer wall of the bend, where friction decelerates the material to the material velocity vout.
A slight pressure drop due to gas losses is created.
Due to the complete separation of material and gas, there is no interaction between the gas and the material and no internal energy is transferred to the material.
While there is no internal energy transfer, there is also no pressure drop.
After the bend, the material with a velocity (vout) is reaccelerated to the gas velocity minus the slip velocity.
This re-acceleration energy is considered the pressure drop that is caused by the bend.
To calculate this pressure drop, it is necessary to know vin and vout and then calculate the acceleration energy until the material velocity reaches the quasi stationary condition.
( In reality, the material is always accelerated, due to the expansion of the air in a constant diameter pipe)
Have a nice day
Teus ■
Teus
How To Calculate Power Required For Bulk Pneumatic Conveying
Dear Mr. Teus,
I have in hand an old manual in czech writen by prof. Drazan for calculation of power required for pneumatic conveying, in which we have a formula : P=Ai*Vv/ efficiency of the system, where
Ai =255553*pA*lg(pp/pA)
where : P is power required, Ai is specific work for isothermal compression J/m3; Vv is flow rate of air required for conveying m3/s; efficiency of the system is between 0.6 to 0.8; pA is atmospheric pressure = 0.1 MPa; pp is compressor pressure; By using this formula the power required for pneumatic conveying is very low. I am calculating the system to convey 500 M3 /h of raw meal for 262 m long including height and other pressure loss, the pressure required is 4.3 bar and air flow rate is about 4.15 m3/s. Applying this formula the power required is only 108 kW. Can you tell me what is wrong with this formula ?
Thanks for your help and looking forward to seeing your reply soon.
Originally Posted by Teus Tuinenburg
The calculated energy from the Zenz diagram (dp * Volume) represents the internal energy of the conveying air that is transferred to the material particles and losses.
The compressor creates that internal energy by its compression principle.
There 2 energy consumption issues here.
The compressor energy consumption and the transferred internal energy to the conveyed product.
If we are only considering the internal energy, then you are right, the compressor plays no role.
Calculating the energy consumption of a bend:
The material is entering the bend with a material velocity vin
The material is changed in direction against the outer wall of the bend, where friction decelerates the material to the material velocity vout.
A slight pressure drop due to gas losses is created.
Due to the complete separation of material and gas, there is no interaction between the gas and the material and no internal energy is transferred to the material.
While there is no internal energy transfer, there is also no pressure drop.
After the bend, the material with a velocity (vout) is reaccelerated to the gas velocity minus the slip velocity.
This re-acceleration energy is considered the pressure drop that is caused by the bend.
To calculate this pressure drop, it is necessary to know vin and vout and then calculate the acceleration energy until the material velocity reaches the quasi stationary condition.
( In reality, the material is always accelerated, due to the expansion of the air in a constant diameter pipe)
Have a nice day
Teus
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Re: Power Consumption In Pneumatic Conveying
Dear Shibo,
I calculate:
Ai = 255553 * 1 * 4.15 * log(5.3/1) = 768127 J/sec = 768 kW
(If the compressor is isothermal, which is not the case for most compressors)
I also calculated your raw meal installation, based on the following data:
220 m horizontal
42 m vertical
8 bends
raw meal : 88 micron
Double tank system: 42 m3 tank volume – 34 m3 raw meal volume.
Based on the compressor airflow and the conveying pressure of 4.3 barg, A 16 “ (387 mm) pipeline is selected.
The resulted raw meal conveying capacity is then calculated at:
System capacity approx. 375 tons/hr at 4.5 barg.
However, the capacity= function (pressure) curve is becoming rather flat around this pressure.
That increases the risk of choking.
The calculated safe operational system conveying rate is estimated at:
316 tons/hr at 3.1 barg
Have a nice day ■
Teus
Re: Power Consumption In Pneumatic Conveying
Originally Posted by Teus TuinenburgI calculate:
Ai = 255553 * 1 * 4.15 * log(5.3/1) = 768127 J/sec = 768 kW
(If the compressor is isothermal, which is not the case for most compressors)
I also calculated your raw meal installation, based on the following data:
220 m horizontal
42 m vertical
8 bends
raw meal : 88 micron
Double tank system: 42 m3 tank volume – 34 m3 raw meal volume.
Based on the compressor airflow and the conveying pressure of 4.3 barg, A 16 “ (387 mm) pipeline is selected.
The resulted raw meal conveying capacity is then calculated at:
System capacity approx. 375 tons/hr at 4.5 barg.
However, the capacity= function (pressure) curve is becoming rather flat around this pressure.
That increases the risk of choking.
The calculated safe operational system conveying rate is estimated at:
316 tons/hr at 3.1 barg
Have a nice day
Dear Teus
Am having an vacum for the pneumatic conveying system so what will be the equation i need to use for calculating power consumed.
with regards
kiran varghese ■
Power Consumption in Pneumatic Conveying
Dear All,
I am puzzled by some issues regarding the relation between pressure drop and power consumption in pneumatic conveying and would appreciate your hints and suggestions. Below is the detail of my questions.
(1) What is the relation between pressure drop and power?
In Mills' book, it writes "power=165*gas flowrate*ln(Pinlet/Pcompressor)"
How this equation is derived? what kind of conditions it can be applied?
(2) Dense-phase flow or slug flow is often referred to as a flow mode with lower power consumption. Let's have a look at pressure drop-gas velocity diagram. In such a diagram, for a constant solid flowrate, the pressure drop of slug flow (at low gas velocity) may be much larger than that of dispersed or suspension flow (at high gas velocity). Why we still say "slug flow has lower power consumption"?
I am looking forward to your helps.
Cheers,
Shibo ■