Differential pressure measurement with PDT in cyclone reactor
Differential pressure measurement with PDT in cyclone reactor
(OP)
Dear Sir
A PDT take a measurement across tap1 (EL 1500mm) and tap 2
(EL 1000mm) so that the differential pressure reading is taken across a known distance 0.5m. And the density of particle is from 400kg/m3 to 1000 kg/m3 in this reactor fluidized bed.
I wonder which formula I can used to calculated the differential pressure drop between these two tap?
Any help and comment will be appreciated.
Thank you so much.
A PDT take a measurement across tap1 (EL 1500mm) and tap 2
(EL 1000mm) so that the differential pressure reading is taken across a known distance 0.5m. And the density of particle is from 400kg/m3 to 1000 kg/m3 in this reactor fluidized bed.
I wonder which formula I can used to calculated the differential pressure drop between these two tap?
Any help and comment will be appreciated.
Thank you so much.





RE: Differential pressure measurement with PDT in cyclone reactor
I am not sure this is a silly question.
I wonder which formula in the following can be applied in this case
1. delta P=delta H/density*g
or just use
2. delta P [kg/cm2]= delta H [m] / (average density[kg/m3] / 10000)
which one is correct ?
Could any one give me a help ?
Thank you very much.
RE: Differential pressure measurement with PDT in cyclone reactor
where specific gravity is the ratio of the density of the material to the density of water.
Water's density is 1,000 Kg/m3
The material at 400Kg/m3 has an SG of 0.4
The material at 1000Kg/m3 has an SG of 1.0
If the 500mm level were at the maximum with 100% SG 1.0 material,
the DP measurement output would be DP*1.0 = 500mm
If the 500mm level were at the maximum with 100% SG 0.4 material,
the DP measurement output would be DP*0.4 = 200mm, and need correction.
The correction for actual level is DP measurement/SG.
200mm indicated output/0.4 = 500mm actual level.
I'm not clear how you will determine the 'average' SG of the blend.