Any good column design starts with MATH. That's what engineers do. It's not hit & miss. If you can handle

simple math, follow along with this...

Any smallish separation device (your column) can be worked out on paper mathematically, providing you know the vapor and packing parameters.

Mike Nixon and Tony Ackland (engineers & pioneer founders of this hobby) gave us these parameters many years ago, after much research & testing. So let's use their findings and not reinvent the wheel. I have put together the following essay for the benefit of would-be designers...

Vapor parameters:

Vapor generation rate (@ 1kW power input) = 45.77 cubic inch/second. (source: M.Nixon M.Eng.)

Ideal Column speed = 12in (min) <-> 18in (max). (source: M.Nixon M.Eng.)

Packing parameters:

Typical HETPs for common packings are :

Packing HETP

S/ Steel Wool Scrubbers = 0.13 m

Marbles (10mm diameter) = 0.33 m

6mm Ceramic Raschig Rings = 0.24 m

13mm Ceramic Raschig Rings = 0.38 m

(source: Tony Ackland Ch.Eng.)

The volume of a column L inches long, with cross sectional area A square inches, is V = L x A cubic inches.

As 1 kW produces 45.77 cubic inches vapor each second, W kilowatts produces 45.77 x W cubic inches each second.

Therefore 45.77 x W = L x A (written more simply as 45.77W = LA)

L in this is, of course, the same as the speed in the column in inches/second, ie. L = Speed

So Speed S = 45.77W/A inches per second (Equation 1) In terms of the diameter of the column .... if the diameter of the column is D, then A = Pi x D x D / 4

( or A =PiDD/4)

So Speed S = 45.77W / [PiDD /4]

ie. S = 58.3W/DD (Equation 2)

or W = SDD/58.3 (Equation 3)

[Where S is in inches/second, W is in kilowatts, and D is in inches.]

Take your pick which equation you use ... they are all equivalent.

Sorry about the 'DD's etc, but I think it is perhaps a bit clearer to write

DD for D squared than D^2

(source = M. Nixon)

So, let's get down to business...

Let's work out a 2" column first, so we've got a yardstick to go by.

The standard 2" x 40" packed column requires approx 1.2KW power input to achieve good separation.

This is a known standard, however let's prove it to be absolutely sure of what we are doing...

HETP means (H)eight (E)quivalent of a (T)heoretical (P)late.

Therefore, for a 2" x 40" column, using Tony Ackland's S/S packed, we do this...

HETP = 0.13m

Change it to inches (no confusion)...

1 HETP (in inches) = 0.13 x 40

Therefore 1 HETP = 5.2 inches of S/S packing.

Thus for our 40" tall packed column, we calculate getting approx. 7 separations or distillations (40 / 5.2).

That's good separation. It's not perfect but if you want better, then just increase the column height by multiples of 5.2 inches (1 plate).

Now we need to work out the power requirements...

If ideal speed = 18"/sec, then to find out what power is required, we need to start with this Mike Nixon's vapor speed equation and plug in the parameters

for our column.

Speed is 18"/sec, which is 18 = 45.77W/A.

Or, it can be expressed as 18 = 45.77W / [PiDD /4] if that is easier. They are the same thing (as Mike said).

Now plug in the parameters for the 2" column, and do one step at a time...

18 = 45.77W / [PiDD /4]

18 = 45.77W / 3.1416 x 2 x 2 /4

18 = 45.77W / 3.1416

18 = 14.57W (rounded)

So far, so good. But how do we get the figure for W (our power requirement). To continue..

As we know in math, whatever we do to one side of an equation, we must do to the other side, to isolate the unknown (in this case the unknown is W, the power

we require).

.

From the last step, we have,

18 = 14.57W

18 / 14.57 = 14.57W /14.57 (both sides divided)

1.2354152367879203843514070006863 = W

Rounded, we get

W = 1.24 Kilowatts

And now at last we have our answer.

We have PROVED what we set out to do,

ie establish the power input for a STANDARD 2" x 40" mesh-packed column.

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Using the same math , it's simple to work out power for a 3" x 40" (7 plate) column...

Plug in the 3" diameter.

18 = 45.77W / [PiDD /4]

18 = 45.77W / 3.1416 x 3 x 3 /4

Now do the math...

18 = 45.77W / 7.0686

18 = 6.5W (rounded)

Continuing...

18 / 6.5 = 6.5W / 6.5 (both sides divided)

2.7692307692307692307692307692308 = W

Rounded, we get

W = 2.78 Kilowatts

We can see clearly that the power required to run a 3" column is more than DOUBLE that of a 2" column.