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Design of Packed Tower with Berl Saddle Packing

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7000 kg/hr of air, at a pressure of 7 atm abs and a temperature of 127^{o}C 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 10^{5}Z V_{s}^{1.82}ρ^{1.85})/D_{p}^{1.4 }

where Δp is the pressure drop in kgf/cm^{2}, Z is the bed height in meter, ρ is the density in g/cc, D_{p }is nominal diameter of Berl saddles in cm, V_{s }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

D_{p} = 2.5 cm

760 mm Hg = 1 kgf/cm^{2} = 1 atm

Δp = 500 mm Hg = (500/760) x 1 kgf/cm^{2} = 0.65 kgf/cm^{2}

Formula:

Ideal gas law:

PV = nRT

Formula given,

Δp = (1.65 x 10^{5}Z V_{s}^{1.82} ρ^{1.85})/D_{p}^{1.4}

**Calculations:**

**
**ρ= Mn/V = MP/(RT) = 29 x 7 x 1.01325 x 10^{5} / (8314 x (273 + 127) ) = 6.185 kg/m^{3} = 6.185 x 10^{-3} g/cc

Δp = (1.65 x 10^{5}Z V_{s}^{1.82}ρ^{1.85})/D_{p}^{1.4}

0.65 = (1.65 x 10^{5} x 6 x V_{s}^{1.82} x (6.185 x 10^{-3} )^{1.85} ) / 2.5^{1.4}

V_{s}^{1.82} = 4.603 x 10^{-5}

V_{s} = 0.029 m/sec.

Volumetric flow rate = mass flow rate/density = 1.944/6.185 = 0.3144 m^{3}/sec

Cross sectional area = Volumetric flow rate / V_{s} = 0.3144 / 0.029 = 10.84 m^{2}

Required Minimum Diameter (D) = (A x 4/π)^{0.5} = (10.84 x 4/π)^{0.5} = **3.715 m**.

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Last Modified on: 01-May-2024

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