Reducing Evaporation Rates With Rippled Surfaces

A theory that I want to test out is that adding a ripple to a surface makes it easier to clean in certain situations.  I work in retail, and sometimes the cleaning of non food safety related issues are delayed due to more important tasks.  In particular I'm talking about drips.  You might see these on metal trays under bottles of milk in the fridge, or on plastic sheets under the shelves in the meat department.  In both cases drips are meant to be caught on a surface that is easy to remove and clean.  The problem is that if you leave a spill too long it dries and become hard to remove and requires vigorous cleaning which may damage the surface and waste time.

To avoid this, you want to slow the rate of evaporation.  This can be done by decreasing the surface area by increasing the depth.  A rippled surface is perfect for this.  You essentially make little cups to hold the spill.  Another requirement that would be nice to include is no tight internal corners.  Anything that is designed to be cleaned shouldn't contain a concave surface that you can't get a finger into.  Anyone that knows me well understands that I think in equations, and the one below matches our criteria perfectly.

$$$z=10 - 8 \left(\dfrac{1-\cos(2\pi x / 20)}{2}\right)\left(\dfrac{1-\cos(2\pi y / 20)}{2}\right)$$$

Let's talk about what this equation means.  The two large sections in the brackets are periodic terms that create ripples in the x and y directions.  The 20 means that this ripple will repeat every 20mm.  By subtracting the cos term from one and then dividing that result by 2, the term in the bracket will range from 0 to 1.  Multiplying the two brackets together will also give a result between 0 and 1.  By multiplying this by 8 we now have a function that ranges from 0 to 8.  The 10 describes the maximum of the equation.  By subtracting the rest of the equation from 10 we now have a surface that ranges from 2 to 10 above zero.  The important things to remember are, 20 specifies how wide the ripples are, and 8 describes how deep they are.

Sometimes though it helps to have the real thing in your hand to test so I created a 3d model to send away for 3d printing.  There doesn't seem to be anything out there to create a 3d model from an equation so I had to write some software to do that.  I'll post that when I tidy it up and comment it properly.

I'm not made of money so the model is only 100mm x 100mm x 10mm (a volume of 100mL) with 20mm wide ripples that are 8mm deep.  By a stroke of luck, I happened to write the software is a way that easily calculates the volume of the model.  In this case the model is 80 mL.  As the bounding box of the model is 100 mL this means the volume of the 25 little cups is 20mL.  Each one holding 0.8 mL.

Rippled Drip Tray

I don't have the print yet, but it has been done and photos sent to me.  In theory this surface should hold 20 mL of liquid and covers an area of 100 square centimeters, so as a test I poured 20 mL of water on a flat surface and it spread to cover 180 square centimeters.  So already the ripple pattern has reduced the surface area by 45%

3D Printed Drip Tray Top

That may not sound like much, but the ripples can be made deeper.  If they were 3x times deeper (24mm) the surface would hold 60mL.  Once they get too deep though, you would need to make the ripples wider to make cleaning easier.  Changing the width of the ripples doesn't effect the volume that the surface would hold though.

3D Printed Drip Tray Top

In this demonstration I've shown the surface as a solid block with depressions.  In reality you'd use something like a polypropylene sheet moulded to this shape.  It would give an object shaped similar to an egg carton.

3D Printed Drip Tray Base

Anyway, this is just a thought that I wanted to explore.  Maybe it'll work out, maybe it won't.  Either way the process was enjoyable.