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How can I calculate/measure/estimate the drag coefficient of a square, vertical plate slowly (nor sure how slowly is slow) moving up and down in a liquid other than water?

Any input is welcome.

Thanks!

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You really need to know the Reynolds number of the plate to get a decent estimate. For a smooth flat plate, we generally assume the critical Reynolds number of the plate is roughly 500,000. This simply means that the boundary layer transitions from laminar, at some region of the plate, to turbulent over the remaining portion of the plate. The critical Reynolds number is dependent on the roughness of the plate, but you can take 500,000 since you are interested in an estimate. Now, the estimate of drag is highly dependent whether or not the boundary layer flow is laminar or turbulent, or a mix of both. You can assume the plate has a length $L$. The Reynolds number for the full length of the plate is given by, $$ Re_L = \frac{\rho V L}{\mu}$$ where $\rho$ is the density, $V$ is the velocity, and $\mu$ is the dynamic viscosity of the fluid (liquid or gas). If this value is less than 500,000 then it is safe to assume the boundary layer flow is completely laminar over the entire plate. In this case, the famous Blasius solution in boundary layers will provide a very good estimate for the skin friction over the plate. As a function of distance along the plate, the skin friction coefficient as given by the Blasius solution is, $$ C_f = \frac{0.664}{Re_x^{1/2}} $$

Notice that in this case the Reynolds number is local and now a function of $x$, which just simply means replace $L$ with $x$ in the Reynolds number definition. In this case, the drag coefficient is given by the integrated skin friction coefficient over the entire length of the plate. $$ C_D = \frac{1}{L} \int_0^L \frac{0.664}{Re_x^{1/2}} \ dx $$ $$ C_D = \frac{1.328}{Re_L^{1/2}} $$

Note, the above result is only true if the flow is completely laminar over the full length of the plate, namely $Re_L < 500,000$. In the event that $ Re_L > 500,000$, you should use a combined laminar and turbulent analysis to properly compute the drag coefficient. For instance, you can determine the estimated transition distance (transition from laminar to turbulent) based on the critical Reynolds number, $$ x_{cr} = \frac{500,000 \mu}{\rho V} $$ This is the distance to which laminar flow in the boundary layer exist. Beyond this distance you will have turbulent boundary layer flow. There are many empirical correlations for the local skin friction for a turbulent boundary layer. Here are a few,

$$ C_f \approx \frac{0.027}{Re_x^{1/7}} $$ or $$ C_f \approx \frac{0.059}{Re_x^{1/5}} $$

Therefore, using the second expression (more commonly used in engineering), our skin friction distribution over a smooth flat plate can be expressed in the following piecewise form:

$$ C_f(x) = \begin{cases} \frac{0.664}{Re_x^{1/2}} & \text{if } \quad x \leq x_{cr} \quad \text{(laminar)} \\ \\ \frac{0.059}{Re_x^{1/5}} & \text{if } \quad x > x_{cr} \quad \text{(turbulent)} \end{cases} $$

Finally, as one would expect, the drag coefficient is simply the integrated skin friction coefficient over the length of the plate. Hence,

$$ C_D = \frac{1}{L} \int_0^L C_f \ dx = \frac{1}{L} \left[ \int_0^{x_{cr}} \frac{0.664}{Re_x^{1/2}} \ dx + \int_{x_{cr}}^L \frac{0.059}{Re_x^{1/5}} \ dx \right]$$

I presume you can follow the math from here. However, the average skin friction (usually a good estimate) for combined laminar and turbulent flow over a flat plate takes the form, $$ C_D = \frac{0.031}{Re_L^{1/7}} $$

Now the final thing to keep in mind is that all of the above analysis assumes flow over the top of a plate. If the plate is submerged in a fluid, i.e flow on both sides of the plate, then you will need to double the $C_D$ to determine the total drag coefficient.

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  • $\begingroup$ Does 1.328 in the formula after integration come from the fact that the plate has two sides in contact with the fluid? $\endgroup$ – Pompilia Dec 18 '16 at 16:32
  • $\begingroup$ Nope, if you carry out the integration you should get the 1.328, which accounts for one side of the plate. $\endgroup$ – TRF Dec 19 '16 at 15:07
  • $\begingroup$ I had already done that when I asked but failed to integrate properly. :D It's been a while since I did that. redone it now and got the 1.328. $\endgroup$ – Pompilia Dec 21 '16 at 18:25
  • $\begingroup$ I have calculated Reynolds numbers and drag coefficients and the results are a bit misleading. Basically, I have very low Reynolds no. (0.003) because I am moving the plate slowly (2 mm/s) - that's how I want it. The drag coefficient then results high (29), which gives the impression that the drag is the cause for which the speed is so low, which is not the case. Or does it give this impressions to ME only? Due to the way the drag coefficient is related to speed (through Re) it looks as if the plate cannot move faster because the drag is so high. Comments? Thanks! $\endgroup$ – Pompilia Dec 22 '16 at 5:38
  • $\begingroup$ The drag coefficient may be high for a very low Reynolds number, but the actually drag force is proportional to the square of the velocity. Remember, $ D = \frac{1}{2} \rho V^2 A C_D$. In which case, squaring a velocity on the order of mm/s will yield a very low drag force. $\endgroup$ – TRF Dec 22 '16 at 5:41
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The drag coefficient of a flat plate moving normal to its plane is approximately $C_D=\frac{D}{\frac{1}{2}\rho\,V^2 A}\approx1.17$, with $D$ the drag, $\rho$ the density of the fluid, $V$ the speed of the plate, and $A$ its area. This is for motion of a thin plate at moderate speed, so the flow is fully separated.

If we are talking about flow at very slow speed (Stokes flow), then the drag will depend on viscosity, and things become more involved. In particular, the exact shape of that plate, its thickness and the shape of its edges become important. There is no simple formula for the general case. In these situations (slow, so-called "creeping flow"), the drag is often expressed as $D=6\pi\,\eta\,r_s V$, where $\eta$ is the kinetic viscosity of the fluid, and $r_s$ is the so-called "Stokes radius".

For the case of a circular disk of radius $R$ normal to the flow, one finds $r_s/R=\frac{8}{3\pi}\approx0.85$ or $D=16\,\eta\,R\,V$.

If the plate is not circular, you can get a decent estimate using the area $A$ of the plate, estimating an "effective Stokes radius" $r_e$ as $r_e\approx \sqrt{A/\pi}$ and plugging that into the above formula for the drag, so we use $D\approx6\pi\,\eta\,r_e V$.

Clarification: The exposition above assumes that the plate is moving normal to its plane, which is how I understood the original post. However, I do realize that this is not clear. If the plate moves tangential to its plane, then the relations given in TRF's answer are the pertinent ones.

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  • $\begingroup$ The square plate is vertical and moves up and down, which I think means it's moving tangential to its plane. Plus, I need the drag coefficient in order to calculate the drag force, so... I cannot calculate the coefficient from the drag. $\endgroup$ – Pompilia Dec 7 '16 at 10:23
  • $\begingroup$ In that case TRF's answer gives you all you need. $\endgroup$ – Pirx Dec 7 '16 at 12:23

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