Gas absorption: minimum solvent rate and number of stages

Material balances and operating lines for counter-current absorbers on mole-fraction and solute-free bases, the minimum solvent rate and pinch, and ideal stages by graphical stepping and the Kremser equation.

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Why it matters

Gas absorption removes CO₂ and H₂S from natural gas, recovers ammonia and solvents, and cleans flue gases. The two first design decisions are how much solvent to circulate and how many equilibrium stages are needed. Too little solvent makes the separation impossible; too much wastes pumping and regeneration energy. Both follow from a material balance (operating line) and the equilibrium curve.

Key ideas

Counter-current absorber. Gas enters at the bottom (rich, y₁ or Y₁) and leaves at the top (lean, y₂ or Y₂); solvent enters at the top (x₂ or X₂, often solute-free) and leaves at the bottom (x₁ or X₁). Counter-current flow keeps a driving force along the whole column and allows the leaving liquid to approach equilibrium with the entering gas.

Solute-free basis. When the solute is not dilute, the total gas and liquid flows change down the column but the carrier gas G_s and solute-free solvent L_s do not. Using mole ratios Y = y/(1 − y) and X = x/(1 − x) makes the operating line straight: slope L_s/G_s. For dilute systems (y < about 0.05) you may use mole fractions and total flows directly.

Operating line. A solute balance over the top of the column gives a line through (X₂, Y₂) and (X₁, Y₁). For absorption it lies above the equilibrium curve (gas richer than equilibrium with the liquid); for stripping it lies below.

Minimum solvent rate. As L_s is reduced, the operating line swings towards the equilibrium curve. L_s,min is reached when the line first touches the equilibrium curve – usually at the rich end (X₁* in equilibrium with Y₁), but at a tangent point if the equilibrium curve is concave upward. At L_s,min an infinite number of stages would be needed (a pinch). Below it the specified separation is impossible however many stages are used. Practical designs use 1.2–2.0 × L_s,min (commonly 1.5).

Number of ideal stages.

  • Graphically: step off stages between the operating line and the equilibrium curve on X–Y axes, starting at either end.
  • Analytically for dilute systems with linear equilibrium y = m·x: the Kremser equation, using the absorption factor A = L/(m·G). A > 1 makes high recovery possible with a few stages; for A < 1 the maximum fractional absorption with infinite stages is A.
  • Actual trays = ideal stages / overall efficiency.

Effect of variables. Higher L (larger A) → fewer stages but more solvent. Higher G at fixed L → more stages. Lower temperature or higher pressure lowers m (more soluble gas) → fewer stages. Any lean-solvent solute content (x₂ > 0) raises the lowest achievable y₂ (= m·x₂).

Formulas

Mole ratios: Y = y/(1 − y), X = x/(1 − x), G_s = G(1 − y), L_s = L(1 − x)

Overall solute balance (operating line): G_s (Y₁ − Y₂) = L_s (X₁ − X₂); Y = Y₂ + (L_s/G_s)(X − X₂)

Minimum solvent (pinch at the rich end): (L_s/G_s)_min = (Y₁ − Y₂) / (X₁* − X₂), with X₁* in equilibrium with Y₁

Absorption factor and Kremser equation (dilute, y = m·x, constant L and G): A = L/(m·G) N = ln[ ((y₁ − m·x₂)/(y₂ − m·x₂))·(1 − 1/A) + 1/A ] / ln A (A ≠ 1) N = (y₁ − y₂)/(y₂ − m·x₂) (A = 1) Fraction absorbed: (y₁ − y₂)/(y₁ − m·x₂) = (A^(N+1) − A)/(A^(N+1) − 1)

  • G, L – molar flows, mol/s or kmol/h; y, x – mole fractions; m – equilibrium slope, dimensionless; N – ideal stages. Here subscript 1 is the rich (bottom) end and 2 the lean (top) end.

The same equations work in mole ratios if the equilibrium is linear in X and Y.

Worked examples

Example 1 – dilute absorber (standard). Given: G = 100 kmol/h, y₁ = 0.02; 95 % of the solute is to be absorbed in pure water (x₂ = 0); equilibrium y = 1.2x; L = 1.5 L_min.

  1. y₂ = 0.02 × 0.05 = 0.001. x₁* = y₁/m = 0.02/1.2 = 0.01667.
  2. (L/G)_min = (0.02 − 0.001)/(0.01667 − 0) = 1.14 → L_min = 114 kmol/h.
  3. L = 1.5 × 114 = 171 kmol/h; x₁ = G(y₁ − y₂)/L = 100 × 0.019/171 = 0.0111.
  4. A = L/(m·G) = 171/(1.2 × 100) = 1.425.
  5. N = ln[(0.02/0.001)(1 − 1/1.425) + 1/1.425]/ln 1.425 = ln(5.965 + 0.702)/0.354 = ln 6.667/0.354 = 5.36.

L_min = 114 kmol/h; L = 171 kmol/h; N = 5.36, i.e. 6 ideal stages.

Example 2 – concentrated gas on a solute-free basis (GATE level). Given: 100 kmol/h of gas with 10 mol % solute; 90 % recovery; solute-free solvent; equilibrium in mole ratios Y* = 1.5X; L_s = 1.3 L_s,min.

  1. G_s = 100 × 0.9 = 90 kmol/h. Y₁ = 0.1/0.9 = 0.1111; Y₂ = 0.1 × 0.1111 = 0.01111.
  2. X₁* = Y₁/1.5 = 0.07407.
  3. L_s,min = G_s (Y₁ − Y₂)/X₁* = 90 × 0.1000/0.07407 = 121.5 kmol/h.
  4. L_s = 1.3 × 121.5 = 158.0 kmol/h; X₁ = 90 × 0.1000/158.0 = 0.0570.
  5. Equilibrium is linear in ratios, so Kremser applies with A = L_s/(m·G_s) = 158.0/(1.5 × 90) = 1.17.
  6. N = ln[(0.1111/0.01111)(1 − 1/1.17) + 1/1.17]/ln 1.17 = ln(1.453 + 0.855)/0.157 = 5.33.

L_s,min = 121.5 kmol/h; L_s = 158 kmol/h; N ≈ 5.3, i.e. 6 ideal stages.

Common mistakes

  • Using total flows and mole fractions for a concentrated gas – the operating line is then curved.
  • Taking the pinch at the rich end when the equilibrium curve is concave upward; the tangent pinch gives a larger L_min.
  • Writing L_min from the actual outlet liquid composition instead of the equilibrium value X₁*.
  • Defining A = mG/L (that is the stripping factor S = 1/A).
  • Rounding N down, or forgetting to divide by tray efficiency.
  • Forgetting that y₂ can never be lower than m·x₂ when the solvent is not pure.

For GATE CH

Expect: L_min or (L/G)_min from an equilibrium relation; outlet liquid composition from a balance; number of ideal stages with the Kremser equation; whether a recovery is feasible for a given A; and graphical-reasoning questions about pinches and the position of the operating line relative to equilibrium. Practise both mole-fraction (dilute) and mole-ratio (concentrated) versions.

Quick check

  1. Where does the operating line lie for absorption: above or below the equilibrium curve?
  2. How many ideal stages are needed at the minimum solvent rate?
  3. Define the absorption factor.
  4. With A = 0.8, what is the largest fraction of solute that can be absorbed with infinite stages (pure solvent)?

Answers: 1. above; 2. infinite; 3. A = L/(m·G); 4. 0.8 (80 %).

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is gas absorption in the context of mass transfer?Concept

    Gas absorption is a mass transfer process where a gas mixture is contacted with a liquid solvent to selectively dissolve one or more components of the gas. This process is used to separate specific gases from a mixture, often for purification or environmental control purposes.

  2. 2.Explain the concept of minimum solvent rate in gas absorption.Concept

    For a fixed gas rate and specified inlet and outlet gas compositions, reducing the solvent rate swings the operating line towards the equilibrium curve. The minimum solvent rate is reached when the operating line first touches the equilibrium curve – usually at the rich end, where the leaving liquid would be in equilibrium with the entering gas, or at a tangent point if the curve is concave upward. At that pinch the driving force is zero and infinite stages or height would be needed, so real columns use about 1.2 to 2 times the minimum.

  3. 3.How is the number of stages in a gas absorption process determined?Concept

    First fix the operating line from a solute balance, using mole ratios on a solute-free basis if the gas is concentrated. Then step off ideal stages between the operating line and the equilibrium curve on an X–Y diagram, starting from either end. For dilute systems with linear equilibrium y = m·x the Kremser equation gives the number directly from the absorption factor A = L/(mG) and the inlet and outlet compositions. Actual trays are the ideal stages divided by the overall tray efficiency.

  4. 4.Why is it important to determine the minimum solvent rate in gas absorption?Application

    The minimum solvent rate is the feasibility limit: below it the specified recovery cannot be achieved with any number of stages. It is also the reference for choosing the actual rate, typically 1.2–2 times the minimum. Close to the minimum the column needs very many stages (high capital cost); far above it, solvent pumping and regeneration costs rise and the leaving solvent is more dilute, so the optimum lies in between.

  5. 5.What happens if the solvent rate is below the minimum solvent rate in a gas absorption process?Application

    The operating line would cross the equilibrium curve, which means that somewhere in the column the gas would have to be leaner than equilibrium with the liquid – impossible for absorption. The specified outlet gas composition therefore cannot be reached with any number of stages or any packed height. The column will still absorb some solute, but the outlet gas will be richer than specified and the liquid will approach equilibrium with the inlet gas at the bottom.

  6. 6.Explain why counter-current flow is often used in gas absorption columns.Application

    Counter-current flow is used in gas absorption columns because it maximizes the driving force for mass transfer. In this configuration, the gas and liquid phases flow in opposite directions, allowing the most concentrated solvent to contact the most concentrated gas, enhancing the efficiency of the absorption process.

  7. 7.What is the impact of increasing the number of stages in a gas absorption process?Application

    Increasing the number of stages in a gas absorption process generally improves the separation efficiency, allowing for a higher degree of absorption. However, it also increases the complexity and cost of the system, so a balance must be found between efficiency and cost.

  8. 8.Calculate the minimum liquid-to-gas ratio for absorbing ammonia from a gas containing 10 % ammonia by volume down to 1 %, using solute-free water, if the equilibrium is approximated as y = 0.5x (treat as dilute).Numerical

    At the minimum, the leaving liquid is in equilibrium with the entering gas: x₁* = y₁/m = 0.10/0.5 = 0.20. A solute balance gives (L/G)_min = (y₁ − y₂)/(x₁* − x₂) = (0.10 − 0.01)/(0.20 − 0) = 0.45. So for every mol/s of gas, at least 0.45 mol/s of water is needed. At 10 % solute a rigorous design would use mole ratios on a solute-free basis, which changes the number slightly.

  9. 9.A gas absorption column operates with a solvent flow rate of 20 mol/s. If the minimum solvent rate is 15 mol/s, is this a reasonable operating point?Application

    The ratio L/L_min = 20/15 = 1.33, inside the usual design range of about 1.2–2, so the separation is feasible with a finite number of stages and the choice is reasonable. Whether it is optimal depends on the trade-off between column height (more stages close to the minimum) and solvent circulation and regeneration cost. At 1.33 times the minimum the number of stages is fairly large, so a check of the stage count and cost would be prudent.

  10. 10.Describe the role of the equilibrium line in determining the number of stages in a gas absorption process.Concept

    The equilibrium line represents the relationship between the concentration of the gas in the liquid phase and the gas phase at equilibrium. It is used in conjunction with the operating line to graphically determine the number of theoretical stages required for the desired separation in methods like the McCabe-Thiele method.

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