How to interpret “HETP”: Height Equivalent of a Theoretical Plate

There are two different contacting methods used for separation processes that involve either gas/vapor contacting a liquid (e.g., gas absorption, distillation, stripping, etc.) or two immiscible liquids( e.g. liquid extraction).

One is a “continuous contact” operation, such as the packed towers in the lab and pilot plant, where the two phases are always in contact (and the device is designed to make this as efficient as possible).  The other is an “equilibrium stage” operation where the phases are flowing counter-currently and are alternatively mixed to allow equilibrium to be achieved, and then separated.  Since they are flowing in opposite directions, the gas from a lower stage is then mixed with a liquid from a higher stage while the liquid is contacted with the from a lower stage.   The photo below, from Treybal:  Mass Transfer Operations (1980) shows how this can work with the gas/vapor flowing upward through the trays.  At each tray the composition of the gas changes because of the contact with a liquid of a different composition than the previous tray.

For a tray column, a graphical construction of the process could be shown as “steps” on a y-x (i.e., mole fractions) plot.  Each step,  is a line between the “operating line” (the component mass balance) and the equilibrium line.  This graphical construction was “invented” by Warren McCabe  and Ernest Thiele (a former ND faculty member) while they were graduate students at MIT (much to the chagrin of E.V.Murphree who was the leading researcher into distillation of this time!)  

The next figure makes it clear that the compositions of the streams passing each other between trays (stages), e.g., 2, y3, are located on the operating line and the compositions of the streams in equilibrium on a tray, e.g., x3, y3,  are located on the equilibrium line.  

The number of stages for a required separation will depend on the inlet and outlet compositions, the ratio of gas to liquid flow (L/G -- which is the slope of the operating line) and the slope and location of the equilibrium line (or curve).

The procedure for determining the number of stages is to locate the lines and starting/ending compositions and then draw the steps.

Tray towers are are used quite often for distillation where they can be > 5m in diameter if needed.  

Trays can also be used for absorption or stripping, but this limited to very large flow situations, perhaps for natural gas “sweetening” (i.e., H2S, SO2 removal).

Issue at hand

There is some practical  interest in answering the question: for a given separation, would a tray tower or a packed tower be taller? Thus we might wish to know both the number of trays needed for a given separation (e.g., CO2 from N2) as well as the total height of packing for this same separation.  However, for the MEA -- CO2 system, the equilibrium line is essentially horizontal and hence a McCabe - Thiele construction cannot be done.  Further, we may have no experimental information about using trays.  

So we would like to answer the question and compare packed and tray columns with only information that we could get from a packed tower. 

Following Kremser, the number of equilibrium stages for a given separation for the case of linear operating and equilibrium “lines” can be derived as:

ntrays=Log[(y1-y1s)/(y2-y2s)]/Log[A]

Where A = L/( m G), m is the slope of the equilibrium line and L and G are the molar flow rates of the liquid and gas.  The bottom of the tower is “1” and the top “2”.  Thus y1 is the inlet gas and y1s is the value of y in equilibrium with the liquid at the bottom.  Note that if the solvent feed to the column is completely cleared of solute, y2s =0.

We can pick some numbers.  Suppose that A = 1.2, y1 =.1 and y2 = .001, we would calculate 16 trays.  (You can flip the numerator if you wish to get rid of the - sign.  Also note that I used the mass balance to determine y1s:  L (x1-x2) = G ( y1-y2), and that y1s = m x1.

Log[ ( .001)/( .1 - 10/12 ( .1-.001))]/Log[1.2]

As expected as we make the separation easier, the number of transfer trays would drop.

ntrays1/.{ y1->.1, y2->.01, A->1.2}

From McCabe et al.

The substitutions give:

ntrays1 = ntrays/.{y2s->0, y1s-> 1/A ( y1-y2)}

You might be worried about the value of A going to infinity for the MEA system.  Don’t worry.  We just need the ratio  of this formula and the standard “NTU” formula for a packed column -- which has a similar A--> ∞ problem!

HETP comparison

The quantity HETP is thus the height of the packing, divided by the number of equilibrium stages that would result according to the equilibrium stage calculation for a packed column of this height.  This is equation (9) in the Imperial writeup for the gas absorber.  

Note however that in the Imperial gas absorber writeup, the comparison between n and  NOG, equation (13) is incorrect.  The n and NOG should be swapped.   
We will get the correct result next.

NTU or NOG

The number of transfer units,  is (∫_(Subscript[y, 1])^(Subscript[y, 2]))(1/((y - y*)))dy.  For the case of a constant slope m, the mass balance and y* = m x can be used to get y* all along the column to allow integration as:

ntu = Log[ y1/y2 ( 1- 1/A) + (1/A)]/( 1- 1/A)

Take ratio of trays to transfer units

nratio = ntrays1/ntu

Some simple numbers.  Our previous example:

nratio/.{ y1->.1, y2->.001, A-> 1.2}

So the number of trays is less than the number of transfer units which would be

ntu/.{ y1->.1, y2->.001, A-> 1.2}

As expected as we make the separation easier, the number of transfer units would drop.

ntu/.{ y1->.1, y2->.01, A-> 1.2}

We can show the ratio over a range of A

Plot[nratio/.{ y1->.1, y2->.001},{A, .1,20},PlotRange->{0,.5}]

Limit[nratio,A-> ∞]

So if A<1, then NTU < number of trays.  Further, we see that as A--> ∞, the ratio is well behaved at a value of about 0.5.

This will allow an easy comparison of HETP data for the packing (from Koch literature) and the “k” that you calculate from the NTU for the pilot plant.  

That is:

NTU HTU = H
n HETP = H
thus if n/NTU = .5, then

HTU = .5 HETP

From the chart it looks like HETP for the “350 X” packing is about 14 inches.    Thus the HTU that you expect is

htucalc  = .5 14 in  30.5 cm/(12 in) m/100/cm

From HW example problem we are getting higher values...

For the entire column:

htuexamp = 5 1.37 m/(Log[.063/.00049])

For the bottom section.

htuexamp = 1 1.37 m/(Log[.063/.031])

7 For a middle section

htuexamp = 1 1.37 m/(Log[.031/.0136])

Conclusion

The packing is apparently under performing in this run.  

??  Are the flow rates in the optimal range for the packing?
Does MEA give a higher HETP than NaOH solution or whatever the data are for?