How to calculate pressure drop in t 1

[patrickraj]

How to calculate pressure drop in two phase separators.Relevent references?

Thanks in advance

 

[Guidoo]

Assumption is that you write about vapor/liquid separator

Reference: Carl Branan, "Rules of Thumb for Chemical Engineers", 2nd edition, Gulf Publishing,1998.page 131

Pressure drop equals pressure drop over inlet nozzle plus pressure drop over vapor outlet nozzle plus pressure drop over mist eliminator (if provided).

Pressure drop over inlet nozzle (assuming no special devices fitted) is appr. 0.5 x density x velocity[sup]2[/sup]
Pressure drop over vapor outlet nozzle is appr. 0.25 x density x velocity[sup]2[/sup]

Where: pressure drop is in Pascal (=0.00001 bar)
density is in kg/m[sup]3[/sup]
velocity is in m/s
For inlet nozzle you should use the mixed phase density and velocity, for outlet nozzle you should use the vapour density and velocity.

Pressure drop over mist eliminator is appr. 1 inch of water. This is about 250 Pa, so 0.0025 bar

 

[patrickraj]

HI Guidoo (Chemical),
Thanks for the very useful info.

 

[25362]

Just for the records 1 Pa = 1 kg/(m.s[sup]2[/sup])

 

[patrickraj]

HI Guidoo (Chemical),

Does it mean that there is no press drop in the vessel, other than mesh and i/l & o/l nozzles.

Thanks in advance.

 

[Guidoo]

Well, there must be some pressure drop, but it will be much lower than pressure drop over nozzles. You can look at the vessel as a largely oversized pipe.
Of course, it is based on the assumption that there are no internals in the vessel, apart from the mesh.

 

[patrickraj]

HI Guidoo (Chemical),

Encore thanks for your reply.