Shear rate in rectangular "channel"

AI Thread Summary
The discussion focuses on calculating the shear rate in a rectangular channel to simulate blood flow conditions. The user seeks to understand how to compute shear rate at the wall, particularly in a channel with mixed materials, and whether to assume homogeneity for simplification. Key equations for calculating average wall shear rate and volumetric flow rate are provided, with corrections made regarding unit discrepancies. The user initially miscalculated the volumetric flow rate, leading to confusion about its plausibility compared to blood flow rates in arteries. Ultimately, the conversation clarifies the correct approach and equations, leading to a more accurate understanding of the shear rate in the given channel setup.
Ekaekto
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Hi everyone,

Non-physics student here, and slightly in over my head :D

I am slightly stumped for ideas regarding the calculation of the shear rate in a rectangular type channel, which I need to effectively simulate a blood flow type condition in a channel that has to be rectangular due to the specimen that I am testing.

Of course I could just go ahead and use velocity as the only measure, but it seems to me to be better if I had the shear rate and could compare that to blood flow.

The channel that I have is 10 mm wide, 2 mm high (and 90 mm long). The velocity can be changed at will, the maximum I can reach (but don't want to, it's too high :D) is 0.7 m/s. How would I calculate this correctly? Do I need to use Stokes equation or am I thinking too complicated?

Another problem that arises is that I have a mix of materials in this channel, because it is basically plates screwed together with a silicone seal in the middle. The bottom of the channel is made from stainless steel, the sides of the channel are made from silicone and the top is made from PP.
Would it make sense to approximate having a homogenous material? The most important area (where the test specimen is placed) is the stainless steel bottom, so would it make sense to assume stainless steel everywhere?

Thanks for the help :)
 
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You have axial fluid flow down a rectangular channel, and you would like to know the shear rate at the wall, correct?
 
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Yes, its fully developed laminar flow. That far I have managed to calculate :)

If possible it would be best for me to know how the shear rate behaves across the cross section, however at the walls would suffice also, as the shear rate for blood that is usually found (edit: in literature) is also given as the one at the wall (I'd guess because it's non-Newtonian). So if I wish to use a similar shear rate, I'd really only need the one at the wall.

If there's any type of publication on the topic that would also be of help, I can try understanding it myself, calculating and coming back if any questions come up
 
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For a rectangular channel, the average wall shear rate is given by $$\gamma=\frac{QP\lambda}{8A^2}$$where Q is the volumetric throughput rate, A is the cross sectional area of the channel and P is the wetted perimeter, and where the shape factor ##\lambda## is given by:$$\lambda=\left[(1-0.351b/a)(1+b/a)\right]^2$$with b representing the short side of the rectangle and a representing the long side.

For more details on this, see "Predicting Non-Newtonian Flow Behavior in Ducts of Unusual Cross Section," Miller, C., Industrial and Engineering Chemistry Fundamentals, 11, 1972, pages 524-528.
 
Thank you very much! That's a great help!
 
After calculating, the ##Q## I am getting seems rather small to me.

I based the calculation on the wall shear rate of the carotid artery that is ##γ = 333,3 s^{-1}##

##a = 0.0104 m## and ##b = 0.002 m##

So, ##A^2 = 4.3264*10^{-10} m^2##
##λ = 1.253##
##P = 0.0248 m ## (as the entire channel is wetted, I calculated a regular ##2a+2b## - is this assumption correct?)

So when solving for ##Q ## I would get the following formula: $$Q = \frac {8A^2γ} {Pλ},$$

and adding in all calculated numbers and SI-Units I reached the following conclusion: $$Q = \frac {8*4.3264*10^{-10}m^2*333.3s^{-1}} {0.0248m*1.253} = 3,71*10^{-5} \frac {m} {s},$$

Does that seem plausible? Are the SI-units even correct? I always thought ##Q## is given as ##\frac {m^3}{s}## but I got ##\frac {m}{s}##. It seems very slow to me, for my specific cross-section it would mean using a volumetric flow slower than the one I have currently, which is already rather small at ##800 µL/min = 1.333333333336*10^{-8} \frac{m^3}{s}## which yields a velocity of ##9*10^{-4}\frac{m}{s}## at the point of interest.

I probably have a slight misconception here, because I am comparing it with the volumetric flow rate of blood through the carotid artery which is much higher at ##238.84 \frac {mL}{min} = 3.981 \frac{m^3}{s}##

There's probably some mistake in my calculation that I am missing.

Cited papers:
S.O. Oktar, C. Yücel, D. Karaosmanoglu, K. Akkan, H. Ozdemir, N. Tokgoz, T. Tali (2006) Blood-Flow Volume Quantification in Internal Carotid and Vertebral Arteries: Comparison of 3 Different Ultrasound Techniques with Phase-Contrast MR Imaging American Journal of Neuroradiology Feb , 27 (2) 363-369
Wu, S. P., Ringgaard, S. , Oyre, S. , Hansen, M. S., Rasmus, S. and Pedersen, E. M. (2004), Wall shear rates differ between the normal carotid, femoral, and brachial arteries: An in vivo MRI study. J. Magn. Reson. Imaging, 19: 188-193. doi:10.1002/jmri.10441
 
In your below calculation the units for A^2 should be m^4 not m^2 and that will result in m^3/s which resolves your units issue for Q

Ekaekto said:
Q=8A2γPλ,Q=8A2γPλ,​
Q = \frac {8A^2γ} {Pλ},

and adding in all calculated numbers and SI-Units I reached the following conclusion:
Q=8∗4.3264∗10−10m2∗333.3s−10.0248m∗1.253=3,71∗10−5ms,​
 
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JBA said:
In your below calculation the units for A^2 should be m^4 not m^2 and that will result in m^3/s which resolves your units issue for Q

D'oh, that was daft of me. Thanks :)
Ok, but then I see no other issue with the calculation...other than intuitively it seems very small to me. But science doesn't care about intuition I suppose :D
 
$$238.84\ mL/min=3.981\ mL/s = 3.981\times 10^{-3}\ m^3/s$$If the carotid artery were a circular tube with this volumetric flow rate, and the shear rate in the tube were 333.3 1/s, I would calculate a value of the tube diameter of 0.5 cm (using a value of ##\lambda=16## for a circular tube).

I'm very sorry. In my post #4, I gave you the wrong equation for ##\lambda##. The correct equation is:
$$\lambda=\frac{24}{\left[(1-0.351b/a)(1+b/a)\right]^2}$$
With this equation, for your geometry, b/a = 0.192, I calculate ##\lambda=19.4##. Now, using the equation I gave with the correct value of ##\lambda## would yield a flow rate of 2.4 cc/sec with a shear rate of 333.3 1/s. So this is comparable to the value of 4 cc/sec you gave for a carotid artery.
 
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Oh yes, I also made a mistake above leaving out the ##10^{-3}## for the volumetric flow in a blood vessel. I just forgot it when typing up my manual notes.

Now it all makes more sense :)
 
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