4. A soluble gas is absorbed in water using a packed tower. The equilibrium relationship may be taken as ye=0.06xe. Terminal conditions are as follows: Top Bottom x 0 0.08 y 0.001 0.009 If Hx=0.24 m and Hy=0.36 m. What is the height of the packed section? (15 Points) 5. A gas stream containing 3.0 percent A is passed through a packed column to remove 99 percent of the A by absorption in water. The absorber will operate at 25°C and 1 atm, and the gas and liquid rates to be 20 mol/h-ft² and 100 mol/h-ft², respectively. Mass-transfer coefficients and equilibrium data are given below: y* = 3.1 x at 25°C kxa = 60 mol/h-ft².unit mol fraction kya = 15 mol/h-ft².unit mol fraction Find Noy, Hoy, and ZT, assuming isothermal operation and neglecting changes in gas and liquid flow rates. What percent of the total resistance is in the gas phase? (15 Points)
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First, we need to calculate the number of transfer units (Noy) and the height of a transfer unit (Hoy). Noy is given by the equation: Noy = (y1 - y2) / (y* - y2) where y1 is the initial mole fraction of A in the gas, y2 is the final mole fraction of A in the Show more…
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Question 3. A counter-current packed-bed absorption column is to be designed to remove sulphur dioxide (SO2) from an air stream using pure water. The column operates at atmospheric pressure and room temperature, 25°C. The inlet gas at a flow rate of 120 kmol/h contains 6 mol% of SO2. The column is designed to reduce SO2 concentration in exit gas to 0.5%. The overall mass transfer coefficient based on gas phase (K'y) is 3x10^-5 mol/(m^2 s mole fraction of A / mole inert). The absorption column has a dimension of 1.5 m^2 cross-sectional area and 3.5 m packed height and operates at 50% efficiency. The equilibrium relationships of SO2 in the gas and liquid phase can be represented in the table as follows where; Table 1. Equilibrium data of Water-SO2 system. x, mole fraction of A in liquid phase | y, mol fraction of A in gas phase 0 | 0 0.0002 | 0.00668 0.0004 | 0.01336 0.0006 | 0.02004 0.0008 | 0.02672 0.0010 | 0.03340 0.0012 | 0.04008 0.0014 | 0.04676 0.0016 | 0.05344 0.0018 | 0.06012 0.0020 | 0.06680 (i) Write the operating line equation of this process. [2 MARKS] (ii) Calculate the minimum and actual amount of water required for this absorption. (Rule of thumb: the solvent rate used is 1.5 of the minimum solvent rate). [8 MARKS] (iii) Estimate the number of theoretical and actual stages of absorption required. [4 MARKS] (iv) Determine the flux of mass transfer at the top of the column. [3 MARKS] (v) Calculate the number of transfer unit required and height of transfer unit for this absorption column. [5 MARKS] (vi) Discuss the effect of solvent flowrate in sizing and operating the absorption column. [3 MARKS]
Sri K.
A partial pressure difference of CO2 over water will promote transfer across the liquid-gas interface and the dissolution of CO2 into the liquid. Some of that CO2 will hydrate to form carbonic acid (H2CO3). As aqueous CO2 is far more concentrated, it is convenient to express [H2CO3] and CO2 (aq) collectively as CO2 (aq). The equilibrium concentration of CO2 in the water is governed by Henry's Law and where fCO2 is the fugacity of CO2 and fCO2 = pCO2 for practical purposes. [CO2 (aq)] = kCO2 • fCO2 P = kH.C or C = kP(g) pCO2 = kCO2(T) • CCO2 = 1/SCO2 • CCO2 Table 1: Carbon Dioxide Solubility in Water C (mol/L) S (mol kg⁻¹ /atm⁻¹) P (bar)/(atm) k (mol/L•atm) T (°C) 3.70 x 10⁻² 1.00 20.0 6.80 x 10⁻² 20.0 1.00 3.40x10⁻² 25.0 C = solubility of a gas at a fixed temperature in a particular solvent (M or mL gas/L); S = solubility coefficient, the inverse functions of Henry's Law; k = Henry's Law constant (often in units of M/atm) and P = partial pressure of the gas (often in units of atm) (a) Current pCO2 (35 Pa) and 22°C today compared to during the last glacial maximum (~18 ka BP) when the climate was cooler and pCO2 was about 20 Pa, what was the approximate average temperature?
Gas from a petroleum distillation column, has a concentration of H2S reduced from 0.03kmol H2S/kmol of inert hydrocarbon gas to 1 % of this value by scrubbing with a triethanolamine-water solvent in a counter-current tower, operating at 300° K and atmospheric pressure. The equilibrium relation for the solution may be assumed to be a straight line and taken as ye =2x. The solvent enters the tower free of H2S and leaves containing 0.013 kmol of H2S/ kmol of solvent. If the flow of inert gas is 0.015 kmol/s.m2 of tower cross section, calculate: a) the height of the absorber necessary, and b) The number of transfer units Noy required The overall coefficient for absorption Ky = 0.04 kmol/sm3 of tower volume Hints yb= 0.03kmole /kmol inert gas= 0.03kmole/(1+0.03) kmole total gas stream Va /S≈ Vb /S = 0.015 kmol/s.m2 (since the fraction of gas removed is very little) Answers: Noy = 20.8 and Hoy = 0.375 m, and Zoy= 7.8 m
Adi S.
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