Recently I visited a sericulture facility in India and I saw that people there 'degummed' (https://www.youtube.com/watch?v=h13fctiTylI) silk in a batch process.

As a Chemical Engineer, I thought of this as an unsteady state unit operation and I have decided to work on making this process a continuous or steady-state operation as a summer project.

So far this is my schematic enter image description here

The silk thread will be pulled over a large drum and will pass over this drum a number of times in a helical fashion. This drum is immersed in a bath of boiling water with added chemicals so that the silk is degummed. The thread will come out of the other end another end of the cylinder and will be wound on another bobbin.

Now I have already determined what sort of chemicals and mass transfer I am gonna get and what should be my temperature of the bath. I am not able to work out on the residence time any part of the silk thread should spend in the bath ? for this I need to determine the RPM of the rotating drum? how do I calculate that? standard degumming time is about 60 minutes. So if I say that the length of the drum is L and its diameter is D and say it is rotating at an angular velocity of w. How do I develop a set of equations to determine my rpm so that I can change my drum dimension if the residence time is too long or too short.

  • 1
    $\begingroup$ Actually this link may quite help. It seems an easy question to me. If you want more detailed calculation, you must supply more details as diameters and gear ratios etc. $\endgroup$
    – kfb
    Commented Jun 22, 2017 at 10:38
  • $\begingroup$ Have you tried to derive an equation? You should be able to find a length of the line and the velocity. Divide length by velocity to find the time. $\endgroup$
    – JMac
    Commented Jun 22, 2017 at 10:47
  • 1
    $\begingroup$ Why change the drum dimension - just change the rpm either with a controller or two cones and a belt or gears - whatever comes to hand really. $\endgroup$
    – Solar Mike
    Commented Jun 22, 2017 at 16:07
  • $\begingroup$ @F.Bek thanks for the input... my diameter is 32 cms for my cylinder and it will be immersed in the bath partially. I intend to use a servo motor to rotate the drum. $\endgroup$
    – user11882
    Commented Jun 29, 2017 at 6:23
  • $\begingroup$ @JMac i did but my RPM was way too high for a given residence time... so i thought i was going wrong somewhere. $\endgroup$
    – user11882
    Commented Jun 29, 2017 at 6:24

1 Answer 1


Length $l$ of silk on the immersed part of the drum $m$ divided by feed rate $f$ is the time the thread is immersed T.

$l / f = T$.

Feed rate is same as linear velocity of the surface of the drum (assuming negligible silk thickness comparing to drum radius). $f = v = \omega r$.

The amount of thread on the drum ($l'$) is the drum circumference ($C$) times number of loops ($n$); number of loops is the usable length of the drum ($L$) divided by pitch of the helix ($p$; which must be no smaller than the silk thickness, preferably at least 2x as large).

$l' = nC$; $n = L/p$ ; $C = 2 \pi r$ ; $l' = 2\pi r L/p$

If the whole drum is immersed, $ l = l'$ so,

$$ {{2\pi r L/p} \over {\omega r}} = T$$

or, $$ \omega = {{ 2 \pi L } \over {pT}} $$

where the drum of length L and helix pitch p rotating at $\omega$ holds the thread immersed for time T.

(Interestingly, makes the radius non-issue).

If your drum is partially immersed, you'll need to use the Circular Segment length to determine percentage of the circumference that remains submerged, and adjust the time accordingly.

  • $\begingroup$ assuming that i take a helix angle of 70 degrees my pitch would be pi*diameter *tan(90-70) right so diameter will surely come into play right ?? $\endgroup$
    – user11882
    Commented Jun 29, 2017 at 6:43
  • $\begingroup$ @YasirSayyad: I assumed a very narrow helix, where the circumference is much larger than pitch - the timing would benefit from packing silk more densely on the drum. If your helix is loose, start off with $l' = n \sqrt{C^2+p^2}$. (and no, tan(20 deg) = 0.36, which would make the length less than circumference, this is definitely wrong.) $\endgroup$
    – SF.
    Commented Jun 29, 2017 at 7:59
  • $\begingroup$ alright, i think I got it @SF. thanks a lot for your help !!! $\endgroup$
    – user11882
    Commented Jun 30, 2017 at 6:11

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