This problem involves calculating the final ethanol concentration in the spent water from a counter-current absorption column. We will use the concepts of minimum liquid flow rate and the operating line.
Absorption Basics and Given Data
We are given:
- Inlet gas mole fraction of ethanol: $Y_1 = 0.05$
- Desired outlet gas mole fraction of ethanol: $Y_2 = 0.005$
- Inlet liquid composition (pure water): $X_0 = 0$
- Equilibrium relationship: $Y^* = 2X$ (where $Y^*$ and $X$ are mole fractions)
- $CO_2$ is insoluble, and the column is isothermal.
- The actual water flow rate ($L$) is twice the minimum water flow rate ($L_{min}$).
We need to find the mole fraction of ethanol in the spent water, $X_L$. We'll work with mole fractions ($Y$ for gas, $X$ for liquid) and assume $G$ and $L$ represent the molar flow rates of the inert gas ($CO_2$) and liquid respectively, allowing us to use the ratio $L/G$.
Minimum Liquid Flow Rate Calculation
The minimum liquid flow rate ($L_{min}$) occurs when the operating line is tangent to the equilibrium curve at the gas inlet condition. This defines a "pinch point".
- Find Equilibrium Liquid Composition at Gas Inlet ($X_1^*$): Using the equilibrium relation $Y^* = 2X$ at $Y_1 = 0.05$: $0.05 = 2X_1^* \implies X_1^* = \frac{0.05}{2} = 0.025$.
- Calculate Minimum Liquid-to-Gas Flow Rate Ratio ($L_{min}/G$): This ratio is the slope of the line connecting the inlet gas condition ($Y_1$, $X_1^*$) and the outlet gas condition ($Y_2$, $X_0$). $\frac{L_{min}}{G} = \frac{Y_1 - Y_2}{X_1^* - X_0} = \frac{0.05 - 0.005}{0.025 - 0} = \frac{0.045}{0.025} = 1.8$.
Actual Flow Rate and Operating Line
The problem states that the actual water flow rate is twice the minimum ($L = 2 L_{min}$). Therefore, the actual liquid-to-gas flow rate ratio is:
$ \frac{L}{G} = 2 \times \frac{L_{min}}{G} = 2 \times 1.8 = 3.6 $.
The operating line represents the relationship between the gas and liquid compositions at different points in the column. Its slope is $L/G$. The operating line passes through the points $(X_0, Y_2)$ and $(X_L, Y_1)$.
Spent Water Composition Determination
We use the slope of the operating line connecting the overall gas inlet and liquid outlet compositions:
$ \frac{L}{G} = \frac{Y_1 - Y_2}{X_L - X_0} $.
Substitute the known values:
$ 3.6 = \frac{0.05 - 0.005}{X_L - 0} $.
$ 3.6 = \frac{0.045}{X_L} $.
Solve for $X_L$:
$ X_L = \frac{0.045}{3.6} = 0.0125 $.
Final Result
The calculated mole fraction of ethanol in the spent water is $0.0125$. This matches closely with Option B.
Important Questions from Absorption
- A gas stream with $2.01 \text{ mol }\%$ ammonia is to be scrubbed in a counter-current isothermal packed bed absorber using pure water to reduce its concentration to $0.01 \text{ mol }\%$. 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 $\text{kmol m}^{-2}\text{ h}^{-1}$) are 1000 and 200, respectively. The equilibrium relationship is $y = 0.9 x$, 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 $0.8 \text{ m}$ and the number of overall gas transfer units ($N_{tOG}$) is given by the following integral.
$$ N_{tOG} = \int_{y_1}^{y_2} \frac{dy}{y - y^*} $$
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? Benzene is removed from air by absorbing it in a non-volatile wash-oil at 100 kPa in a counter-current gas absorber. Gas flow rate is 100 mol/min, which includes 2 mol/min of benzene. The flow rate of wash-oil is 50 mol/min. Vapor pressure of benzene at the column conditions is 50 kPa. Benzene forms an ideal solution with the wash-oil and the column is operating at steady state. Gas phase can be assumed to follow ideal gas law. Neglect the change in molar flow rates of liquid and gas phases inside the column.
For this process, the value of the absorption factor (up to two decimal places) is ________________- A liquid $L$ containing a dissolved gas $S$ is stripped in a countercurrent operation using a pure carrier gas $V$. The liquid phase inlet and outlet mole fractions of $S$ are $0.1$ and $0.01$, respectively. The equilibrium distribution of $S$ between $V$ and $L$ is governed by $y_e = x_e$, where $y_e$ and $x_e$ are the mole fractions of $S$ in $V$ and $L$, respectively. The molar feed rate of the carrier gas stream is twice as that of the liquid stream. Under dilute solution conditions, the minimum number of ideal stages required is ________ (in integer).
- Pure gas P is being absorbed into a liquid. The dissolved P undergoes an irreversible reaction in the liquid film. The reaction is first order with respect to P. Which one of the following represents the concentration profile of P in the liquid film at steady state?
A countercurrent absorption tower is designed to remove 95% of component A from an incoming binary gas mixture using pure solvent B. The mole ratio of A in the inlet gas is 0.02. The carrier gas flow rate is 50 kmol/h. The equilibrium relation is given by $Y = 2X$, where Y and X are the mole ratios of A in the gas and liquid phases, respectively. If the tower is operated at twice the minimum solvent flow rate, the mole ratio of A in the exit liquid stream is ________ (rounded off to three decimal places).
