Question

Make a preliminary design for a horizontal separator to separate a mixture of air and water; flow-rates: air 11500 kg/h, water 8000 kg/h; operating pressure 28 bar at temperature 110 C . And prepare a data sheet for the equipment
Air density 29 kg / cubic meter

Answer 1

Solution
From steam tables, at 28 bar: saturation temperature 230.057 0C
liquid density is 933.828 kg/m3
,and
Air density is 29 kg/m3

• Settling velocity is given by following formula

u​​​​​​t​​​= 0.07[(L -v)/v]0.5
Where ut is settling velocity

L is liquid density

v is air density

Put the respective values in the above equation.

u​​​​​​t = 0.07[(933.828-29)/29]0.5

u​​​​​​t = 0.391m/s

A demister pad is not specified as the separation of water from air is not critical.

So

u​​​​​t =0.15× 0.3672 m/s

ut =0.05865m/s

• Volumetric flow rate of air

V = 11500/ 29 m³/h = 396.55 m³/ h = 396.55/3600 m³/s

V = 0.110153256 m³/s

• The minimum allowable diameter of the separator is given by

D= [(4V)/π•ut]0.5

So let's calculate minimum allowable diameter

D = [ (4×0.110153256)/π•0.05865]0.5 m

D = 1.5464 m

• Water volumetric flow rate L

L= 8000/933.828 m³/h = 8.5669 m³/h = 8.5669/3600 m³/s

L= 0.0023797 m³/s

If we allow a minimum of 10 minutes hold up for water

• Volume of vessel required= L ×10×60 = 0.0023797× 60 m³

Volume = 0.14278 m³

  

• Liquid depth required ,h =

h = 0.14278 m³/[ 0.25π (1.5464)²]

h= 0.07603 m

Data sheet:-

Parameter	value
Settling velocity	0.05865m/s
Volumetric flow rate of air	0.110153256 m³/s
Diameter of separator	1.5464m
Volumetric flow rate of water	0.0023797 m³/s
Volume of vessel for 10 min hold up	0.14278 m³
Liquid depth required	0.07603 m