Mass transfer and absorption

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
  • #1
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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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  • #2
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)?
 
  • #3
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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  • #4
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?
 
  • #5
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
 
  • #6
Do you know how to work with these equations to solve your problem?
 
  • #7
Rahulx084 said:
Yes sir
Chestermiller said:
Do you know how to work with these equations to solve your problem?
Yes I guess
 
  • #8
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?
 
  • #9
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.
 
  • #10
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!
 
  • #11
Thanks for your efforts sir to make my concepts clear
 

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