I am an academic in the UK and I have constructed a kinetic instrument that uses a small flow channel 0.25 mm square. This channel is imaged during steady state flow conditions using a CCD. I know the volume comming through the channel per unit time (around 1 mL per second and so I can calculate the linear velocity of the solution in the flow channel. When I use this linear velocity to generate my time base, I get an error of 1.3 in my kinetic rates compared with other measurements. This assumes the flowing channel cross section is 0.25 x 0.25 mm, i.e. 0.0625 mm ^3. However if I assume a flowing channel of diameter 0.25 mm I get the correct answer!! Is there any way of simulating flow in this channel and showing that the corner quadrants are static, i.e. that the true flow is down a 0.25 mm diameter cylinder in the middle?

Any help would be appreciated as I am a biologist!!

Don't know that I can answer your question - so I'll start with a few of my own!

What do you mean by 'linear velocity'? The most obvious velocity is usually called the 'bulk velocity' and is calculated from V = (volume flow rate)/(cross sectional area).

The maximum velocity in a cross section will be greater than V and will occur somewhere away from the walls of the channel. I think you could calculate the exact magnitude and location for a fluid of constant viscosity in laminar flow using an undergraduate fluids text and a calculator. However, I can't think of many 'biological' fluids with a constant viscosity.

The maximum occurs because the fluid 'sticks' to the walls - that is, it has zero velocity right at the wall. So the corners of your channel have areas of very low velocity because two walls have a major slowing effect. Note as well that you'll have a different flow for an open (3-walled) channel than from a closed (4-walled) channel.

If your calibration/correlation coefficient depends on the maximum velocity on a cross section and you're using a bulk velocity to calculate the coefficient, an error of 1.3 seems plausible. [For an academic 2-dimensional case, Vmax/Vbulk = 1.5].

To answer more directly, assuming you know the viscosity of your fluid, you can calculate the details of the velocity in the channel, analytically for an 'easy' viscosity, numerically for a known viscosity function.

However, if you have a friend on the engineering faculty at your institution, you might consider buying him a pint and asking him about your problem! : )

Good luck!

Dear Jim

Thanks for your response.

Yes, by linear velocity I mean bulk velocity. I call it linear because I know the field of view calibration in my image from the CCD camera (i.e., 1 pixel = 0.01 mm for example). Then knowing my bulk velocity I can caculate the time domain, i.e. the solution has zero age at the begining of the channel and age 1 millisecond or so at the end. This time domain allows me to calculate the rate of the chemical reaction which is out by the factor of 1.3.

The channel is closed , i.e 4 walled and I believe I have conditions for turbulent flow (approx 1 mL/second down a channel 0.25 x 0.25 mm).

It sounds as if it is likely that the flow is static in the corner quadrants although it is not a trivial calculation?? I will buy you that pint (somehow, even if i have to FEDEX it) if it can be done. (The viscosity of the solution will be that of water since it is all done in dilute aqueous phase).

thanks for your help

chris