Dear kieuminhvn,

P=Ai*Vv/

Ai=255553*pa*log(pp/pa )

P=255553*pa*Vv/*log(pp/pa )

P=255553*1*4.15/0.7*log(5.3/1)=1097324 J⁄sec=1097 kW

Based on:

horizontal length = 220 m

vertical length = 42 m

number of bends = 8

pipe line diameter = 18” (438 mm)

2-vessel system

tank volume = 50 m3

raw meal volume = 45 m3

Oil filled screw compressor 4.15 m3/sec

Calculation results:

system capacity = 424 tph at 4.3 barg (With a chance of choking)

stable system capacity = 392 tph at 3.2 barg

have a nice day ■

Can you send me the document

hashanisuhejl@gmail.com ■

Dear Mr. Kieu Trang Minh

The formula are obviously for compressors, compressing at constant temperature (Isothermal compression) and for different unit systems.

In reality, the normally used compressors compress adiabatically or isochorically or a combination of the two.

At least the compression can be polytrophic.

The power calculation in pneumatic conveying is the last figure to determine, when the compressor type is known and the conveying pressure is known.

The SLR is just the ratio between Solid mass flow and Gas mass flow.

This figure is part of all pressure drop formulas in the pneumatic conveying calculation.

The value of the SLF is depending on the installation geometry, the conveying pressure and the material to be conveyed.

-Longer pipelines lower SLR

-higher pressures higher SLR

-high material friction losses lower SLR

The SLF value is not a basis for the pneumatic conveying calculation, but follows from the selected pneumatic conveying components and the conveyed material.

All the best wishes for the coming year for everybody. ■

There is a rule of thumb for estimating max. possible SLR (Solids to Air Ratio).

Physical background is roughly:

a) Air expands along the conveying pipe

b) Max. conveying pressure at beginning of conveying pipe defines max. amount of air (voidage)

Therefore approx.:

SLR = K x 1 / SQR (L_red)

with

K= apprx. 250-300 for coarser particles [Geldart A/B (fine sand, fine limestone)]

K= apprx. up to 400-550 for fine particles [Geldart A (very fine Cement, very fine Flyash etc.)]

and

L_red= L_P + L_K+ L_V

where

L_P = Total length of pipe (m)

L_K = Additional "length for bends" (no. of Bends x 5 - 8 ) (m)

L_V = Additional "length for vertical pipe section" (0,2 x vertical length) (m)

Example:

- Cement

- 160 m total pipe length

- thereof 40 m vertical

- 6 bends R = 5 => 6 x 7 m

L_P = 160 (m)

L_K = 6 x 7 = 43 (m)

L_V = 0,25 x 40 = 10 (m)

K = 500 (Cement)

=> SLR = 500 / SQR (160 + 43 + 10) = 34,2

Caution:

Use this formula only as a first shoot

Never base your calculation on this formula without a deep knowledge of all material characteristica

Greetings from Cologne, Germany

Klaus Schneider ■

Dear Mr Klaus Schneider,

How you derive the formula from:

a)Air expands along the conveying pipe

b)Max. conveying pressure at beginning of conveying pipe defines max. amount of air (voidage)

to

SLR = K x 1 / SQR (Lred)

is not clear to me, however, I also found a similar relationship

between the SLR and the conveying length, independent of the pipe diameter(s).

The formula, I found is:

SLR = K x 1 / (Lred)^a

whereby K and a are determined by:

-Material

-Conveying pressure

-gas velocity along the pipeline (single diameter pipe line or stepped pipeline)

I also use this relationship as the first capacity assumption in my pneumatic conveying calculation program.

Have a nice day ■

hi guys using an excel sheet i found the pressure drop and gas flow rate , Can you guys help me to find the power required the pressure drop i got was 0.17 bar and gas flow rate as 0.10920.

What all parameters are needed to calculate the power absorbed and with the pressure drop how can we check the vacum blower can handle it (pressure drop only gives the difference in pressure right ) if anyone out there please help me.

With regards

kiran varghese ■

.....gas flow rate as 0.10920........

Please mention the units. ■

Originally Posted by johngateley

.....gas flow rate as 0.10920........

Please mention the units.

Dear John

sorry for the late reply

units are in bar and flow rate in m3/s lets assume the value be 0.55 bar and 0.02713 m3/s ■

Dear kiranvarghese,

I understand that the mentioned pressure drop is for a vacuum system. (see your previous reply).

Vacuum = 0.55 bar

Then we must know how the mentioned flow rate is defined.

Is it the material flow rate or the airflow rate?

If that is the air flow rate, then:

Is the airflow given at atmospheric conditions?

(If so, what are the atmospheric conditions?)

Is the airflow given as the vacuum pump displacement?

In your case 0.02713 m3/sec # 1.6278 m3/min at a vacuum of 0.55 bar. (0.45 bar abs)

If it is a vacuum system, a PD blower would consume approx. 2 kW at a vacuum of 0.55 bar, with an outlet temperature of approx. 82 degrC

Whether you calculated your system correctly, cannot be answered.

Take care ■

Originally Posted by kiranvarghese

hi guys using an excel sheet i found the pressure drop and gas flow rate , Can you guys help me to find the power required the pressure drop i got was 0.17 bar and gas flow rate as 0.10920.

What all parameters are needed to calculate the power absorbed and with the pressure drop how can we check the vacum blower can handle it (pressure drop only gives the difference in pressure right ) if anyone out there please help me.

With regards

kiran varghese

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ++

Dear Kiran,

For calculating this power you can refer to any text book that gives this equation/method. Power is work done per unit of time. This work is being done by compressing the gas flow. Using adiabatic compression equation, you can calculate work done/per minute and then convert it into horse power.

I will be glad to send a copy of my article that gives this calculation method if you need one.

Regards,

Amrit Agarwal

Pneumatic Conveying Consulting

Email: polypcc@aol.com ■

Originally Posted by Amrit Agarwal

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ++

Dear Kiran,

For calculating this power you can refer to any text book that gives this equation/method. Power is work done per unit of time. This work is being done by compressing the gas flow. Using adiabatic compression equation, you can calculate work done/per minute and then convert it into horse power.

I will be glad to send a copy of my article that gives this calculation method if you need one.

Regards,

Amrit Agarwal

Pneumatic Conveying Consulting

Email: polypcc@aol.com

Dear Mr. Agarwal,

Can I have your article as you mentioned ?

I will not share with any others.

Thank you in advance.

Regards,

Danu

Email: danu7286@gmail.com ■