Lecture 9 Flashcards
(11 cards)
Break moment
Okay in this lecture we will basically talk about stripping and absorption: What are their operating lines, and how can we determine their number of equilibrium stages etc etc
What is a general description of the absorption and stripping towers we will be working with?
For absorption in the countercurrent-flow, trayed tower shown in digital notes (a), stages are conventionally numbered from
the top, where the absorbent enters, to the bottom.
For the stripper in digital notes (b), stages are numbered from the bottom, where the stripping agent enters, to the top. P
Phase equilibrium is assumed between the vapour and liquid leaving each tray.
Assume for an absorber and a stripper that only one solute transfers. Let:
L′ = molar flow rate of solute-free absorbent, V′ = molar flow rate of solute-free gas (carrier gas) X = mole ratio of solute to solute-free absorbent in the
liquid, Y = mole ratio of solute to solute-free gas in the vapour
With these definitions, the values of L′ and V′ remain constant throughout the tower because only solute undergoes mass
transfer between phases.
What do the equilibrium curves of the solute look like?
It is shown in digital notes.
For the solute at any stage n, the K-value is
Kn = yn/xn = Yn∕(1 + Yn) / Xn∕(1 + Xn)
where Y = y∕(1 − y) and X = x∕(1 − x). *
The figure in the notes shows representative equilibrium X–Y plots. In general, the equilibrium curve will not be a straight line, but it will pass through the origin. If the solute undergoes an irreversible liquid-phase chemical reaction with the solvent to make a nonvolatile product, the equilibrium curve will be a straight line of zero slope, passing through the origin.
*We already defined X and Y to be the mole ratio of solute to solute-free absorbent in the liquid, and the mole ratio of solute to solute-free gas in the vapour. Sooo make sense when we say X =x/(1-x) since x is the mole fraction of the solute in the liquid and 1-x is the solute free solvent mole fraction
Break moment..
Since the derivation of the operating lines, min and max flow rates, and mostly everything for adsorption and stripper is practically the same we will do all derivation in parallel
What are the operating lines for stripping and absorbance?
The entering and leaving solute compositions and solute-free flow rates are paired. For the absorber, the pairs at the top are (X0, L′ ) and (Y1, V′) and the pairs at the bottom are (YN+1, V′) and (XN, L′). For the stripper, (XN+1, L′) and
(YN, V′) are at the top, and (Y0, V′) and (X1, L′) are at the bottom.
These terminal pairs relate to intermediate pairs passing streams between stages by solute material balancesSolute material balances are written around one end of the tower and an arbitrary intermediate equilibrium stage, n as shown in digital notes
the rest of the derivation is on written notes
How do the operating lines look like
They are shown in digital notes
The terminal points represent conditions at the top and bottom of the tower. For absorbers, the operating line is above the equilibrium line because, for a given solute
concentration, X, in the liquid, the solute concentration, Y, in the gas is always greater than the equilibrium value, thus
providing a mass-transfer driving force for absorption. For strippers, operating lines lie below equilibrium lines, thus
enabling desorption. Please take a close look at where is the top and where is the bottom, since as we will see later we always need to start our equilibrium stages from the bottom.
Another important point is the the operating line would never cross the equilibrium line
.
What are the maximum and minimum flow rates?
Operating lines for four different solute-free absorbent flow rates, L′, are shown in digital notes for a fixed solute-free gas
feed rate, V′. In each case, the solute concentration in the exiting gas, Y1, is the same. Therefore, each operating line passes through the terminal point, (Y1, X0), at the top of the column.
To achieve the desired value of Y1 for given YN+1, X0, and V′, the solute-free absorbent flow rate L′ must be between an ∞ absorbent flow with L′∕ V′= ∞, as represented by operating line 1 (THIS IS THE MAX FLOWRATE Xn =0), and a minimum absorbent rate (corresponding
to ∞ stages), L′ min, as represented by operating line 4, with the equilibrium curve and operating line intersecting at YN+1. This point of intersection is known as the pinch point (Derivation for Lmin in the written notes).
Intermediate operating lines 2 and 3, correspond to 2 and 1.5 times L′ min, respectively. The solute concentration in the outlet liquid, XN, depends on L′.
Note that the operating line can terminate at the equilibrium line as in operating line 4 of the figure, but cannot cross it because that would be a violation of the second law of thermodynamics.
What does the maximum nd minimum flow rate tell?
As the operating-line slope L′ ∕V′ is
increased, fewer equilibrium stages are required until zero is needed and the maximum flow rate. As L′/V′ decreases, more stages are required until L′ min ∕ V′ is reached, at which the operating line and the equilibrium curve intersect at a pinch point located on the horizontal line for YN+1. At L′ min, an infinite number of stages is required
how can we determine the number of equilibrium stages (step 1)?
First, we have to determine where to start drawing our stages from top or bottom. Usually, we always start at the bottom especially when we have an efficiency of less than 100%. Now in the book for the absorber, they make the statement that the two passing streams are determined by the operating line, and the vapour and liquid streams leaving a stage are determined by the equilibrium line (look in digital notes) and because of this we should start at the top of column and work down but Micheal said we always start from bottom stage and I will ask and confirm.
how can we determine the number of equilibrium stages (step 2)?
Diagrams in digital notes
Now we can construct a staircase, Moving up the staircase, steps off the number of stages required for a given solute-free absorbent flow rate corresponding to the slope of the operating line, which in digital notes is (L′∕ V′) = 1.5(L′min∕ V′).
Starting at the point (Y1, X0) on the operating line, a horizontal move is made to the right to (Y1, X1) on the equilibrium curve. From there, a vertical move
is made to (Y2, X1) on the operating line. The staircase is climbed until the terminal point (YN+1, XN) on the operating line is reached. This point corresponds to the passing streams at the bottom end of the column. As shown in Figure 6.12a, the stages are counted at the points on the equilibrium curve. In this case, N = 3.
The stages required for stripping a solute are determined analogously to absorption. An illustration is shown in Figure 6.12b. note that V′ min is determined from
the slope of the operating line that passes through the points (Y0, X1) and (YN, XN+1)
One more leftover slide
digital notes
