Sizing an absorption column

Gas absorption · A, NTU and packed height

A packed scrubber has to take a dilute solute from 2 % down to 0.1 % in the gas. The design flows are 45 kmol/h of lean solvent against 30 kmol/h of gas, the equilibrium line over this dilute range runs at a slope of 1.2, and the vendor's data for this packing and system gives a height of a transfer unit of 0.45 m. Find the absorption factor the flows establish, the number of transfer units the 20:1 cleanup requires, and the packed height to order.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

L = 45 kmol/h — Liquid molar flow
V = 30 kmol/h — Gas molar flow
m = 1.2 — — Equilibrium line slope
y₁ = 0.02 — — Inlet gas mole fraction
y₂ = 0.001 — — Outlet gas mole fraction
H_OG = 0.45 m — Height of a transfer unit

Determine

(a)the absorption factor
(b)the transfer units the separation needs
(c)the packed height

Step 1 of 3(a) · solve for Absorption factor

A = L/mV is the ratio of what the liquid can carry to what equilibrium demands it carry — the column's economics in one number. These flows give A = 1.25, in the classic design band: above 1 so the separation can go as deep as you like, below 2 so the solvent bill stays sane.

Rearranged for A

A = \frac{L}{m V}

L

m

V

Your values, in your units

A = \frac{\left( 45\ \text{mol/s} \right)}{\left( 1.2 \right) \cdot \left( 30\ \text{mol/s} \right)}

Answer

A = 1.25

in

Carried onward at full precision, not this rounded figure.

Open the Absorption Factor solver →

Step 2 of 3(b) · solve for Number of transfer units

The Colburn relation prices the cleanup: a 20:1 reduction at A = 1.25 costs 7.84 transfer units. The A-dependence is brutal near 1 — the same separation at A = 1.05 would need nearly twice the units, which is why a little extra solvent buys a lot less column.

Rearranged for N_OG

N_{OG} = \frac{\ln \left[ \dfrac{y_1}{y_2} \left( 1 - \dfrac{1}{A} \right) + \dfrac{1}{A} \right]}{1 - \dfrac{1}{A}}

A1.25 —carried from step 1

y_1

y_2

Your values, in your units

N_{OG} = \frac{\ln \left[ \dfrac{\left( 0.02 \right)}{\left( 0.001 \right)} \left( 1 - \dfrac{1}{\left( 1.25 \right)} \right) + \dfrac{1}{\left( 1.25 \right)} \right]}{1 - \dfrac{1}{\left( 1.25 \right)}}

Answer

N_{OG} = 7.8431

in

Carried onward at full precision, not this rounded figure.

Open the Transfer Units for Dilute Absorption solver →

Step 3 of 3(c) · solve for Packed height

HTU is where all the messy reality lives — packing geometry, wetting, diffusivities — measured by the vendor so the designer can multiply: 7.84 units at 0.45 m each is 3.53 m of packing. Order it as 4 m and bank the margin against the vendor's optimism.

Rearranged for Z

Z = H_{OG} \, N_{OG}

H_OG

N_OG7.8431 carried from step 2

Your values, in your units

Z = \left( 0.45\ \text{m} \right) \cdot \left( 7.84308 \right)

Answer

Z = 3.5294\ \text{m}

in

Carried onward at full precision, not this rounded figure.

Open the Packed Column Height from HTU and NTU solver →

Why this order

The transfer-unit method splits column design into the two questions that have different owners. How HARD is the separation — N_OG — belongs to thermodynamics: flows, equilibrium slope, and the concentration ratio, nothing else. How FAST does this hardware separate — H_OG — belongs to the packing vendor and the lab. Their product is height. Keeping them apart is what lets one measured HTU serve every duty, and one computed NTU survive a packing substitution.

The number to respect is the absorption factor. At A below 1 the column hits a wall: equilibrium pinches, and NO height of packing reaches deep cleanup — the equation says so by blowing up. Near A = 1 the units climb steeply, and the practical wisdom of designing at A ≈ 1.2–2 is just this curve read as economics: solvent is an operating cost, packed metres are capital, and A is the dial that trades one for the other.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.