3
$\begingroup$

I'm building something to dry glass tubes. It's raised a question which I'd like some background knowledge on before I draw up requirements for the design.

Say I want to dry something with heat alone, then obviously in a room environment (50% humidity, relatively free airflow, atmospheric pressure) heating something above 100°c would quickly evaporate fairly small amounts of water.

Similarly, a powerful enough jet of air would blast off most water, quickly evaporating any remains.

But say I use a heated airflow. Say the heater power is fixed, what would the relationship be between speed of drying and airflow? I think this is a hard question to answer without specifics, but perhaps one question in isolation could be answered:

In principle, for small amounts of water, and a fixed heater power: if airflow is doubled, then the difference in temperature between that airflow and room temperature would be halved. Similarly, if the airflow is halved, the difference in temperature would be doubled. What would the effect on rate of evaporation be?

Perhaps this is impossible to answer without more information, in which case, how might this problem be solved? A compromise would have to be established on criteria like cost of higher pressure air flow generation, noise levels, power consumption, energy efficiency, rate of drying, surface area of objects to be dried, etc etc etc. Might this be better off just developed with experimentation?

Any level of guidance would be appreciated

$\endgroup$
  • $\begingroup$ What equations have you considered to describe the physical phenomena here? $\endgroup$ – willpower2727 Jun 1 '16 at 16:56
  • 2
    $\begingroup$ Requirements are things that you should be drafting to figure out what you want, before you know exactly how you are going to solve it. Write down what type of glassware, how fast do you need the process to be, how much space is it allowed to take, how many items per day, do the items need to be safe to touch after drying, how wet are the items to start with, are they wet from distilled water or tap water, how much manual intervention, what investment cost, what operation cost... $\endgroup$ – Han-Kwang Nienhuys Jun 1 '16 at 17:20
  • $\begingroup$ There are all sorts of industrial dryers (tray dryers?) used in industry, these might work here. How big a system are you planning? Are you drying a big roomful of glassware each day, or maybe a cupboard/cabinet full? $\endgroup$ – Andy Jun 2 '16 at 7:38
1
$\begingroup$

If you are only interested in temperature, flow rate, and drying rate, then you could consider the following quantities for a given air temperature $T$:

  • How much heat you need for evaporation (based on how much water is on the glassware)?
  • How much heat you need to bring the entire mass of glassware to that temperature?
  • What volume of air do you need to carry the moisture of one batch, assuming saturated vapor and how much heat do you need to heat all that air?

This will give you the the minimum amount of energy needed to dry the glassware. To increase the rate of evaporation, I would put priority on increasing the water-carrying capacity of the resulting air, since evaporation rate is primarily driven by the difference between the actual water content of the air and the maximum water content. Specifically, the rate of evaporation is $$Q = K (c_{\mathrm{max}} - c), $$ where $K$ is a proportionality constant, $c_{\mathrm{max}}$ is the temperature-dependent maximum water content of the air (e.g., in kg/m3 units) and $c$ is the actual water content. For your system, you could start by taking $c$ as the final water content of the air after all the glassware has dried.

Next, consider increasing the flow velocity of the air without increasing the volume of air. This will increase the proportionality constant $K$. Turbulent flow, as opposed to laminar flow, is a good thing here.

$\endgroup$
  • $\begingroup$ @Jodes - I have updated my answer. The earlier formulation about "decrease RH" was rather sloppy. $\endgroup$ – Han-Kwang Nienhuys Jun 2 '16 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.