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GATE 2026 CH – Question 48

Mass Transfer · Absorption · 2 marks · Multiple choice

A gas stream with 2.01 mol % ammonia is to be scrubbed in a counter-current isothermal packed bed absorber using pure water to reduce its concentration to 0.01 mol %. Assume that dilute conditions apply, operating line is linear and the mass transfer coefficients are constant throughout the column. The liquid and the gas flows inside the absorber (in kmol m$^{-2}$ h$^{-1}$) are 1000 and 200, respectively. The equilibrium relationship is $y = 0.9x$, where, $y$ is the mole-fraction of ammonia in the gas phase and $x$ is that in the liquid. Under these conditions, the height of overall gas transfer unit ($H_{tOG}$) is 0.8 m and the number of overall gas transfer units ($N_{tOG}$) is given by the following integral.

$N_{tOG} = \int_{y_2}^{y_1} \frac{dy}{y - y^*}$

The integration is performed between the two ends of the column, and $y^*$ is the equilibrium mole fraction in the gas phase, corresponding to the composition of the liquid at corresponding location. Which one of the following is the minimum length of the packing (in m) necessary to achieve the desired scrubbing?

  1. 4.2
  2. 5.0
  3. 6.2
  4. 7.7

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Correct answer: (B) 5.0

Explanation

With pure water, the operating line is $x = \frac{G}{L}(y - y_2) = 0.2(y - 0.0001)$, so $y^* = 0.9x = 0.18(y - 0.0001)$ and $y - y^* = 0.82y + 0.000018$. Then $N_{tOG} = \int_{y_2}^{y_1}\frac{dy}{0.82y + 0.000018} = \frac{1}{0.82}\ln\frac{0.82 \times 0.0201 + 0.000018}{0.82 \times 0.0001 + 0.000018} = \frac{1}{0.82}\ln\frac{0.016500}{0.000100} = 6.23$. The packing height is $Z = H_{tOG} N_{tOG} = 0.8 \times 6.23 = 4.98 \approx 5.0$ m.