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Exercise:
Packed Column
Air containing 1.6 vol% SO2 is scrubbed at 1 atm with pure water in a
packed column of 1.5-m2 cross-sectional area and 3.5m height, packed
with No. 2 plastic Super Intalox saddles. Total gas flow rate is 0.062
kmol/s, liquid flow rate is 2.2 kmol/s, and outlet gas SO2 concentration
is y = 0.004. At the column temperature, the equilibrium relationship is
y* = 40x.
a) What is L/Lmin?
b) Calculate NOG and compare your answer to that for the number of
theoretical stages required.
c) Determine HOG and the HETP from the operating data.
d) Calculate KGa from the data, based on a partial pressure driving force
as in Item 2 of Table shown in the next slide.
LECTURE 6: SHORT CUT METHOD 1
LECTURE 6: SHORT CUT METHOD 2
Answer (a)
Compute material balance. SO2-free inlet air rate = 0.062(1-0.016) =
0.061 kmol/s
SO2 inlet rate in feed gas = 0.062(0.016) = 0.001 kmol/s = V'
SO2 outlet rate in gas = 0.061(0.004/0.996) = 0.00025 kmol/s
SO2 rate in outlet water = 0.001 - 0.00025 = 0.00075 kmol/s
Fraction absorbed = 0.00075/0.001 = 0.75 or 75%
From Eq. L'min = V'K(fraction absorbed) = 0.061(40)(0.75) = 1.83 kmol/s
Therefore, L/L'min = 2.2/1.83 = 1.20
LECTURE 6: SHORT CUT METHOD 3
Answer (b)
Take A = L/KV = 2.2/[(40)(0.062)] = 0.887,
From Eq.
Solving, Nt = 4. For NOG, use Eq. (6-89) with yin = 0.016, yout = 0.004, xin
=0.0,
LECTURE 6: SHORT CUT METHOD 4
Answer (c)
Given height of packing = 3.5 m = lT
From Eq., HETP = lT /Nt = 0.875 m
From Eq., HOG = lT /NOG = 0.926 m
LECTURE 6: SHORT CUT METHOD 5
Answer (d)
From Table shown in slide 2,
◦ = 0.045 kmol/s-m3-atm
LECTURE 6: SHORT CUT METHOD 6

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Excercise pc 20172018

  • 1. Exercise: Packed Column Air containing 1.6 vol% SO2 is scrubbed at 1 atm with pure water in a packed column of 1.5-m2 cross-sectional area and 3.5m height, packed with No. 2 plastic Super Intalox saddles. Total gas flow rate is 0.062 kmol/s, liquid flow rate is 2.2 kmol/s, and outlet gas SO2 concentration is y = 0.004. At the column temperature, the equilibrium relationship is y* = 40x. a) What is L/Lmin? b) Calculate NOG and compare your answer to that for the number of theoretical stages required. c) Determine HOG and the HETP from the operating data. d) Calculate KGa from the data, based on a partial pressure driving force as in Item 2 of Table shown in the next slide. LECTURE 6: SHORT CUT METHOD 1
  • 2. LECTURE 6: SHORT CUT METHOD 2
  • 3. Answer (a) Compute material balance. SO2-free inlet air rate = 0.062(1-0.016) = 0.061 kmol/s SO2 inlet rate in feed gas = 0.062(0.016) = 0.001 kmol/s = V' SO2 outlet rate in gas = 0.061(0.004/0.996) = 0.00025 kmol/s SO2 rate in outlet water = 0.001 - 0.00025 = 0.00075 kmol/s Fraction absorbed = 0.00075/0.001 = 0.75 or 75% From Eq. L'min = V'K(fraction absorbed) = 0.061(40)(0.75) = 1.83 kmol/s Therefore, L/L'min = 2.2/1.83 = 1.20 LECTURE 6: SHORT CUT METHOD 3
  • 4. Answer (b) Take A = L/KV = 2.2/[(40)(0.062)] = 0.887, From Eq. Solving, Nt = 4. For NOG, use Eq. (6-89) with yin = 0.016, yout = 0.004, xin =0.0, LECTURE 6: SHORT CUT METHOD 4
  • 5. Answer (c) Given height of packing = 3.5 m = lT From Eq., HETP = lT /Nt = 0.875 m From Eq., HOG = lT /NOG = 0.926 m LECTURE 6: SHORT CUT METHOD 5
  • 6. Answer (d) From Table shown in slide 2, ◦ = 0.045 kmol/s-m3-atm LECTURE 6: SHORT CUT METHOD 6