What Is the Concept Behind Finding Mole Fraction in Gas Absorption Problems?

[Rahulx084]

A gas stream containing 3% A is passed through a packed column to remove 99% of A by absorption in the water . The absorber operates at 25 degree Celsius and 1atm and the gas and liquid rates are to be $20\frac{mol}{hr ft^2}$ and $100\frac{mol}{hr ft^2}$. Find the $(NTU)_{OG}$ , $(HTU)_{OG}$.
Equilibrium relation: $y^* =3.1x$



$K_x a$= $60\frac{mol}{hr ft^3}$
$K_y a$= $15\frac{mol}{ hr ft^3}$

In the given picture there is a question, I'm having huge confusion in finding out the mole fraction, I have put a solution to find the mole fraction in the picture, I know its wrong. This is where I'm stucked , I don't know the concept behind this . Please someone help me to get through this , where I'm wrong and what concept is used to find mole fraction in these questions or other varieties like this .

 

[Chestermiller]

A gas stream containing 3% A is passed through a packed column to remove 99% of A by absorption in the water . The absorber operates at 25 degree Celsius and 1atm and the gas and liquid rates are to be $20\frac{mol}{hr ft^2}$ and $100\frac{mol}{hr ft^2}$. Find the $(NTU)_{OG}$ , $(HTU)_{OG}$.
Equilibrium relation: $y^* =3.1x$



$K_x a$= $60\frac{mol}{hr ft^3}$
$K_y a$= $15\frac{mol}{ hr ft^3}$

In the given picture there is a question, I'm having huge confusion in finding out the mole fraction, I have put a solution to find the mole fraction in the picture, I know its wrong. This is where I'm stucked , I don't know the concept behind this . Please someone help me to get through this , where I'm wrong and what concept is used to find mole fraction in these questions or other varieties like this .

Let V and L be the molar flow rates per unit area (of column) of liquid and vapor. Let x represent the mole fraction of A in the gas phase, and let y represent the mole fraction of A in the liquid phase. Let $phi(z)$ represent the molar flow rate of A from the gas phase to the liquid phase per unit area of column at location z. Consider the section of the absorber between axial locations z and $z+\Delta z$. What is the mass balance over this interval of A in the gas phase an of A in the liquid phase (in terms of the parameters identified so far)?

 

[Rahulx084]

Let V and L be the molar flow rates per unit area (of column) of liquid and vapor. Let x represent the mole fraction of A in the gas phase, and let y represent the mole fraction of A in the liquid phase. Let $phi(z)$ represent the molar flow rate of A from the gas phase to the liquid phase per unit area of column at location z. Consider the section of the absorber between axial locations z and $z+\Delta z$. What is the mass balance over this interval of A in the gas phase an of A in the liquid phase (in terms of the parameters identified so far)?

Mass balance in the elemental region $dz$
$d(Vx)=d(Ly)=phi(z)$
Where $V$=molar flow rate of Vapour phase
$L$= Liquid molar flow rate
$x$=Mole fraction of A in gas phase
$y$= Mole fraction of A in liquid phase

 

[Chestermiller]

Here's my take on this. Let x* and y* be the concentrations of A at the gas-liquid interface. The liquid is flowing downward at rate L, and the gas is flowing upward at rate V. Let z be the vertical coordinate through the column. Let $\phi(z)$ the molar flow rate of A per unit height of column and per unit cross sectional area of column from the gas phase to the liquid phase. The mass balances on A are as follows:
$$L[x(z)-x(z+\Delta z)]=\phi \Delta z$$
$$V[y(z+\Delta z)-y(z)]=-\phi \Delta z$$
Taking the limit of these as $\Delta z$ approaches 0, we have:$$L\frac{dx}{dz}=-\phi$$
$$V\frac{dy}{dz}=-\phi$$
The interphase molar flow rate of A is related to the mole fractions in the liquid and vapor by:
$$\phi=K_ya(y-y^*)=K_xa(x^*-x)$$
The phase equilibrium relationship is $$y^*=Hx^*$$

Are you comfortable with this so far?

 

[Rahulx084]

$d(Vx)=d(Ly)=phi(z)$
Where $V$=molar flow rate of Vapour phase
$L$= Liquid molar flow rate
$x$=Mole fraction of A in gas phase
$y$= Mole fraction of A in liquid phase

Here's my take on this. Let x* and y* be the concentrations of A at the gas-liquid interface. The liquid is flowing downward at rate L, and the gas is flowing upward at rate V. Let z be the vertical coordinate through the column. Let $\phi(z)$ the molar flow rate of A per unit height of column and per unit cross sectional area of column from the gas phase to the liquid phase. The mass balances on A are as follows:
$$L[x(z)-x(z+\Delta z)]=\phi \Delta z$$
$$V[y(z+\Delta z)-y(z)]=-\phi \Delta z$$
Taking the limit of these as $\Delta z$ approaches 0, we have:$$L\frac{dx}{dz}=-\phi$$
$$V\frac{dy}{dz}=-\phi$$
The interphase molar flow rate of A is related to the mole fractions in the liquid and vapor by:
$$\phi=K_ya(y-y^*)=K_xa(x^*-x)$$
The phase equilibrium relationship is $$y^*=Hx^*$$

Are you comfortable with this so far?

Yes sir

 

[Chestermiller]

Do you know how to work with these equations to solve your problem?

 

[Rahulx084]

Yes sir

Do you know how to work with these equations to solve your problem?

Yes I guess

 

[Chestermiller]

Yes I guess

If you'd like more help, I'll be glad to provide it. What is your first step in the solution to this problem?

 

[Rahulx084]

If you'd like more help, I'll be glad to provide it. What is your first step in the solution to this problem?

Thank you sir , I got this now .
First I take initial basis for both the flow rates , I took 100kmol for the gas phase and 500kmol for liquid phase observing their flow rates , and then I calculated the mole fraction of solute in both entering as well as leaving stream from the given information and solved further using equilibrium relation.

 

[Chestermiller]

Thank you sir , I got this now .
First I take initial basis for both the flow rates , I took 100kmol for the gas phase and 500kmol for liquid phase observing their flow rates , and then I calculated the mole fraction of solute in both entering as well as leaving stream from the given information and solved further using equilibrium relation.

Excellent!

 

[Rahulx084]

Thanks for your efforts sir to make my concepts clear