7000 kg/hr of air, at a pressure of 7 atm abs and a temperature of 127oC is to be passed through a cylindrical tower packed with 2.5 cm Berl saddles. The height of the bed is 6 m. What minimum tower diameter is required, if the pressure drop through the bed is not to exceed 500 mm of mercury?
For Berl saddles, Δp = (1.65 x 105Z Vs1.82ρ1.85)/Dp1.4
where Δp is the pressure drop in kgf/cm2, Z is the bed height in meter, ρ is the density in g/cc, Dp is nominal diameter of Berl saddles in cm, Vs is the superficial linear velocity in m/sec.
Data:
Mass flow rate = 7000 kg/hr = 1.944 kg/sec
Height of bed (Z) = 6 m
Dp = 2.5 cm
760 mm Hg = 1 kgf/cm2 = 1 atm
Δp = 500 mm Hg = (500/760) x 1 kgf/cm2 = 0.65 kgf/cm2
Formula:
Ideal gas law:
PV = nRT
Formula given,
Δp = (1.65 x 105Z Vs1.82 ρ1.85)/Dp1.4
Calculations:
ρ= Mn/V = MP/(RT) = 29 x 7 x 1.01325 x 105 / (8314 x (273 + 127) ) = 6.185 kg/m3 = 6.185 x 10-3 g/cc
Δp = (1.65 x 105Z Vs1.82ρ1.85)/Dp1.4
0.65 = (1.65 x 105 x 6 x Vs1.82 x (6.185 x 10-3 )1.85 ) / 2.51.4
Vs1.82 = 4.603 x 10-5
Vs = 0.029 m/sec.
Volumetric flow rate = mass flow rate/density = 1.944/6.185 = 0.3144 m3/sec
Cross sectional area = Volumetric flow rate / Vs = 0.3144 / 0.029 = 10.84 m2
Required Minimum Diameter (D) = (A x 4/π)0.5 = (10.84 x 4/π)0.5 = 3.715 m.
Last Modified on: 01-May-2024
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