# Mass transfer and absorption

• Rahulx084
In summary, a gas stream containing 3% A is passed through a packed column at 25 degrees Celsius and 1 atm to remove 99% of A by absorption in water. The gas and liquid rates are ##20\frac{mol}{hr ft^2}## and ##100\frac{mol}{hr ft^2}##, respectively. The ##(NTU)_{OG}## and ##(HTU)_{OG}## can be found using the equilibrium relation ##y^*=3.1x## and the values of ##K_x a## and ##K_y a##. The concept of mass balance is used to find the mole fraction of A in both the gas and liquid phases, and

#### Rahulx084

Thread moved from the technical forums, so no Homework Template is shown
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 .

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Rahulx084 said:
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)?

Chestermiller said:
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

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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 said:
##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 said:
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

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

Rahulx084 said:
Yes sir
Chestermiller said:
Do you know how to work with these equations to solve your problem?
Yes I guess

Rahulx084 said:
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?

Chestermiller said:
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.

Rahulx084 said:
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!

Thanks for your efforts sir to make my concepts clear