Example 08a: Humidification: Drying lumber with air: required volumetric air flow rate

Example problem from: Example question (in abbreviated form):
There is an error in the statement of the dry bulb temperature (claims is 100°F) and the relative humidity (claims is 39.6%) at intermediate State0 (Node0) between State1 and State2 in the course PDF worked solution
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This is a more complex problem than the previous example questions, because the air (which dries the lumber through evaporating moisture) is undergoing humidification. There is water mass transfer to the air, so the absolute humidity of air doing the drying must increase!

It is assumed that where the example question is asking for 'how much outside air is needed per minute' it is asking for the volumetric flow rate of the humid air mixture. The result given is in fact for the "entry" State1:

The course PDF Answers (where available) from the worked example problem are given in the 3rd column in the custom table in the top of the image in IP units.

When handled by the `MPsy` class, we have 2 objects representing the before and after states. Note how the humidity ratio 'w' of the 1st object is used to set the humidity ratio of the 2nd object, because it is assumed that there is no change in the contained water mass within the humid air.

The 2nd table that shows the differences between them, as well as indicating the change in each main psychrometric variable, and the "direction" of change.

We see that for such a humidification process:

• The humidity ratio 'w' and the dew point temperature 'tdp' increase.
• The mass-specific enthalpy per dry air 'hda' increases.

There is a dedicated `Psy` library function for calculating the required volumetric humid air mixture flow rate for this humidification case.

The psychrometric chart function shows both states, as well as an arrow indicating the change.

In the next slide we see how such humidification can be represented as a 2-step (3-state) process.

Notes
Snippets (quotes/extracts)