The product energy losses are calculated as a product pressure drop.
Therefore, it is important to know which phenomena causes these losses.
1)The energy losses along the pipe wall
2)The energy losses, caused by the number of particle collisions
3)The energy losses in the particle collisions.
4)The energy losses caused by the particle velocity because of the chosen gas velocity. Acceleration and Reynold number)
5)The energy losses caused by the material properties s.a. friction and breakage.
The particle energy losses must be calculated in a formula which accounts for the above 5 energy losses.
The energy losses under 5) appear to be influenced by the particle size distribution.
Smaller particle sizes have a lower collision energy loss than bigger particle sizes.
This can be proven by a theoretical of a 2-particle size mathematical exercise.
In a 2-particle particle distribution theoretical derivation, one can imagine the following.
-If the particles both have diameter of zero, the Solid Loss factor must be 0. (There are no particles, hence no energy losses)
-If the particles both have the diameter, the Solid Loss factor must be 0. (There particles have the same velocity, hence no energy losses)
-Between those conditions, the particles obtain different velocities because of the different particle sizes.
It is possible to calculate the impact forces and particle deformations between the small particle fraction and the bigger particle fraction of a product with a widespread real particle size distribution.
This must be done for a product with a small average particle size and a bigger particle size.
These calculation results then can be combined with field measurements of a material with different average particle sizes.
The field measurements supported the theory.
Therefore, it is important to measure or know the Solid Loss Factor of the particle size distribution of the specific product to be conveyed.
The product energy losses must then be converted in a gas pressure drop.
Which indicates that a “product” cannot have a pressure drop. ■
Product pressure drop in pneumatic conveying.
The product energy losses are calculated as a product pressure drop.
Therefore, it is important to know which phenomena causes these losses.
1)The energy losses along the pipe wall
2)The energy losses, caused by the number of particle collisions
3)The energy losses in the particle collisions.
4)The energy losses caused by the particle velocity because of the chosen gas velocity. Acceleration and Reynold number)
5)The energy losses caused by the material properties s.a. friction and breakage.
The particle energy losses must be calculated in a formula which accounts for the above 5 energy losses.
The energy losses under 5) appear to be influenced by the particle size distribution.
Smaller particle sizes have a lower collision energy loss than bigger particle sizes.
This can be proven by a theoretical of a 2-particle size mathematical exercise.
In a 2-particle particle distribution theoretical derivation, one can imagine the following.
-If the particles both have diameter of zero, the Solid Loss factor must be 0. (There are no particles, hence no energy losses)
-If the particles both have the diameter, the Solid Loss factor must be 0. (There particles have the same velocity, hence no energy losses)
-Between those conditions, the particles obtain different velocities because of the different particle sizes.
It is possible to calculate the impact forces and particle deformations between the small particle fraction and the bigger particle fraction of a product with a widespread real particle size distribution.
This must be done for a product with a small average particle size and a bigger particle size.
These calculation results then can be combined with field measurements of a material with different average particle sizes.
The field measurements supported the theory.
Therefore, it is important to measure or know the Solid Loss Factor of the particle size distribution of the specific product to be conveyed.
The product energy losses must then be converted in a gas pressure drop.
Which indicates that a “product” cannot have a pressure drop. ■
Teus