Packed tower design: HTU and NTU

Why it matters

A packed absorber or stripper has no discrete stages: composition changes continuously with height. Its height is set by how difficult the separation is (number of transfer units, NTU) times how effective the packing is (height of a transfer unit, HTU). This split lets you change packing or flow rates and see at once which term moves, and it is the standard design method in exams and in industry.

Key ideas

Differential balance. Over a slice dZ of packing with interfacial area a per unit packed volume, the solute lost by the gas equals the solute transferred: G·dy = K_y·a·(y − y*)·dZ (dilute system, G as molar flux per unit cross-section). Integrating from the top (y₂) to the bottom (y₁) gives the packed height.

Splitting the integral.

• H_OG = G/(K_y·a) collects the flow rate and the packing's mass transfer performance. It has units of length and is fairly constant along a dilute column (typically 0.3–1.5 m).
• N_OG = ∫ dy/(y − y*) measures how hard the separation is: the change in composition divided by the average driving force. A large N_OG means a small driving force relative to the change required.
• Z = H_OG × N_OG. Equivalent pairs exist on each side: Z = H_G·N_G = H_L·N_L = H_OL·N_OL. Only matching pairs may be multiplied.

Evaluating N_OG.

• Linear equilibrium, dilute: the integral reduces to the change in y divided by the log-mean driving force, or to the Colburn equation in terms of the absorption factor A = L/(mG).
• Curved equilibrium: integrate numerically – plot 1/(y − y*) against y and find the area, or use the graphical (Baker) method.
• Concentrated gas: include (1 − y) terms; the integrand becomes (1 − y)_LM/[(1 − y)(y − y*)].

Combining film HTUs. Because 1/K_y = 1/k_y + m/k_x, the overall HTU is the gas HTU plus the liquid HTU weighted by the stripping factor mG/L. Gas-film control gives H_OG ≈ H_G.

HTU versus HETP. HETP is the packed height equivalent to one ideal stage. For straight operating and equilibrium lines, HETP and H_OG are linked through the stripping factor; they are equal only when mG/L = 1.

Effect of variables.

• Raising G (fixed L): H_OG rises roughly as G^0.2–0.3 (K_y·a rises less than G), and N_OG rises because A falls – the column gets taller.
• Raising L (fixed G): A rises, N_OG falls; H_L falls with better wetting. But L is limited by flooding and solvent cost.
• Better packing (larger effective a): lower H_OG.

Formulas

Packed height (dilute): Z = H_OG × N_OG H_OG = G / (K_y·a); H_G = G/(k_y·a); H_L = L/(k_x·a) N_OG = ∫ from y₂ to y₁ of dy/(y − y*)

• Z – packed height, m; G, L – molar fluxes, kmol/(m²·s); K_y·a, k_y·a, k_x·a – volumetric coefficients, kmol/(m³·s); y* – gas mole fraction in equilibrium with the local liquid.

Linear equilibrium (y* = m·x + b), log-mean form: N_OG = (y₁ − y₂)/Δy_LM, Δy_LM = (Δy₁ − Δy₂)/ln(Δy₁/Δy₂), Δy₁ = y₁ − y₁*, Δy₂ = y₂ − y₂*

Colburn equation (pure or lean solvent, linear equilibrium y = m·x): N_OG = ln[ ((y₁ − m·x₂)/(y₂ − m·x₂))·(1 − 1/A) + 1/A ] / (1 − 1/A), A = L/(m·G) If A = 1: N_OG = (y₁ − y₂)/(y₂ − m·x₂)

Combining HTUs: H_OG = H_G + (m·G/L)·H_L

HETP and H_OG (straight lines, S = m·G/L): HETP = H_OG · ln S / (S − 1) (HETP = H_OG when S = 1)

Worked examples

Example 1 – height of a dilute absorber (standard). Given: y₁ = 0.02, y₂ = 0.001, solute-free solvent (x₂ = 0), y* = 1.2x, L/G = 1.71 (molar), G = 0.02 kmol/(m²·s), K_y·a = 0.05 kmol/(m³·s).

1. Outlet liquid: x₁ = (y₁ − y₂)/(L/G) = 0.019/1.71 = 0.01111.
2. Driving forces: Δy₁ = 0.02 − 1.2 × 0.01111 = 0.00667; Δy₂ = 0.001 − 0 = 0.001.
3. Δy_LM = (0.00667 − 0.001)/ln(6.667) = 0.00567/1.897 = 0.002987.
4. N_OG = 0.019/0.002987 = 6.36 (Colburn with A = 1.425 gives the same 6.36).
5. H_OG = G/(K_y·a) = 0.02/0.05 = 0.40 m.
6. Z = 0.40 × 6.36 = 2.54 m.

N_OG = 6.36, H_OG = 0.40 m, Z ≈ 2.54 m of packing.

Example 2 – combining film HTUs and tightening the spec (GATE level). Given: same system, H_G = 0.30 m, H_L = 0.40 m, m = 1.2, L/G = 1.71. Find Z for 95 % removal (y₂ = 0.001) and for 99 % removal (y₂ = 0.0002).

1. H_OG = H_G + (mG/L)·H_L = 0.30 + (1.2/1.71) × 0.40 = 0.30 + 0.281 = 0.581 m.
2. A = 1.71/1.2 = 1.425; 1 − 1/A = 0.2982.
3. 95 %: N_OG = 6.36 (Example 1) → Z = 6.36 × 0.581 = 3.69 m.
4. 99 %: N_OG = ln[(0.02/0.0002)(0.2982) + 0.7018]/0.2982 = ln(30.52)/0.2982 = 3.418/0.2982 = 11.46.
5. Z = 11.46 × 0.581 = 6.66 m.

H_OG = 0.581 m; Z = 3.69 m for 95 % and 6.66 m for 99 % removal – the last 4 % of removal costs about 80 % more packing.

Common mistakes

• Multiplying H_OG by N_G (mismatched pair).
• Using arithmetic-mean driving force instead of the log mean.
• Putting the equilibrium value for the wrong end: y₁* is in equilibrium with x₁ (the liquid at the bottom), not with x₂.
• Using the stage formula (Kremser, ln A in the denominator) when N_OG (with 1 − 1/A in the denominator) is needed.
• Treating HETP and H_OG as equal when mG/L ≠ 1.
• Using mass fluxes with molar-based K_y·a.

For GATE CH

Typical questions: Z from H_OG and N_OG; N_OG from the log-mean driving force or Colburn's equation; H_OG from H_G and H_L; height change when removal or flow rate changes; HETP from H_OG. Questions often give the equilibrium line and ask for the outlet liquid concentration first. Practise the A = 1 special case.

Quick check

1. Write H_OG in terms of G and K_y·a.
2. What does a large N_OG indicate?
3. If mG/L = 1, how are HETP and H_OG related?
4. For A = 1, y₁ = 0.02, y₂ = 0.002 and pure solvent, what is N_OG?

Answers: 1. H_OG = G/(K_y·a); 2. a difficult separation – small driving force relative to the composition change; 3. they are equal; 4. (0.02 − 0.002)/0.002 = 9.