16.4.7 Flow Rate

As for the other cross-sections we will now derive the flow rate for the rectangular cross-section according to Eq. 10.9. Obviously, any solution to the Navier-Stokes equation we found in Eq. 16.53, Eq. 16.58, and Eq. 16.62 could be used for deriving the flow rate. In the following, we will limit ourselves to Eq. 16.53 since this is the most commonly found equation, which gives rise to some convenient simplifications. Using Eq. 10.9 and Eq. 16.53 we find the flow rate as

$\begin{array}{lll}Q\hfill & =\hfill & {\displaystyle {\int }_{A}udA}\hfill \\ \hfill & =\hfill & -\frac{4h^2}{\eta {\pi }^3}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^3}{\displaystyle {\int }_{-\frac{w}{2}}^{\frac{w}{2}}\left(1-\frac{\cosh \left(\left(2n+1\right)\pi \frac{y}{h}\right)}{\cosh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}\right)dy{\displaystyle {\int }_0^h\sin \left(\left(2n+1\right)\pi \frac{z}{h}\right)dz}}}\hfill \\ \hfill & =\hfill & -\frac{4h^2}{\eta {\pi }^3}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^3}{\left[\left(y-\frac{h}{\left(2n+1\right)\pi }\frac{\sinh \left(\left(2n+1\right)\pi \frac{y}{h}\right)}{\cosh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}\right)\right]}_{-\frac{w}{2}}^{\frac{w}{2}}{\left[-\frac{h}{\left(2n+1\right)\pi \cos \left(\left(2n+1\right)\pi \frac{z}{h}\right)}\right]}_0^h}\hfill \\ \hfill & =\hfill & \frac{4h^2}{\eta {\pi }^3}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^3}\left(\left(\frac{w}{2}+\frac{w}{2}\right)-\left(\frac{h}{\left(2n+1\right)\pi }\frac{\sinh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}{\cosh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}-\frac{h}{\left(2n+1\right)\pi }\frac{\sinh \left(\left(2n+1\right)\pi \frac{-w}{2h}\right)}{\cosh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}\right)\right)\left(-\frac{h}{\left(2n+1\right)\pi }\left(\frac{1}{\cos \left(\left(2n+1\right)\pi \right)}-1\right)\right)}\hfill \\ \hfill & =\hfill & \frac{4h^2}{\eta {\pi }^3}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^3}\left(w-2\frac{h}{\left(2n+1\right)\pi }\tanh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)\right)}2\frac{h}{\left(2n+1\right)\pi }\hfill \\ \hfill & =\hfill & -\frac{4h^2}{\eta {\pi }^3}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^4}\frac{2h}{\pi }}\left(w-\frac{2h}{\left(2n+1\right)\pi }\tanh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)\right)\hfill \\ \hfill & =\hfill & -\frac{8h^3}{\eta {\pi }^4}\frac{dp}{dx}{\displaystyle \sum _{n=0}^{\infty }\left(\frac{w}{{\left(2n+1\right)}^4}-\frac{2h}{{\left(2n+1\right)}^5\pi }\tanh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)\right)}\hfill \end{array}$

We can further simplify this expression using the fact that Eq. 16.63 contains a series with a known convergence value given by

${\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(2n+1\right)}^4}=\frac{1}{96}{\pi }^4}$

See Tab. 4.1 for details on this series. We can therefore rewrite Eq. 16.63 to

$Q=-\frac{h^{2}w}{12\eta }\frac{dp}{dx}\left(1-\frac{192h}{{\pi }^{5}w}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^5}\tanh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)}\right)$

Eq. 16.64 gives the general case of the flow rate in a rectangular channel.

Channels With Square Cross-Sections. If setting h = w in Eq. 16.64, we obtain

$Q=-\frac{w^2}{12\eta }\frac{dp}{dx}\left(1-\frac{192}{{\pi }^5}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^5}\tanh \left(\left(2n+1\right)\pi \frac{w}{2}\right)}\right)$

Approximation for Channels With Low Aspect Ratios. As you can see from Fig. 3.4b the hyperbolic tangent tanh x converges to 1 for values of x > 1. Therefore, for channels with aspect ratios r 1 we will find

$\begin{array}{lll}\frac{w}{h}\hfill & \to \hfill & \infty \hfill \\ \tanh \left(\left(2n+1\right)\pi \frac{w}{2h}\right)\hfill & \to \hfill & 1\hfill \end{array}$

in which case we can simplify Eq. 16.64 to

$Q=-\frac{h^{3}w}{12\eta }\frac{dp}{dx}\left(1-\frac{192h}{{\pi }^{5}w}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{{\left(2n+1\right)}^5}}\right)$

where we can apply a second known series convergence value given by

${\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(2n+1\right)}^5}=\frac{31}{32}\zeta \left(5\right)\approx 1.0045}$

Again, see Tab. 4.1 for details on this series. Using this series we can rewrite Eq. 16.66 to

$\begin{array}{lll}Q\hfill & =\hfill & \frac{h^{3}w}{12\eta }\frac{dp}{dx}\left(1-\frac{192h}{{\pi }^{5}w}\frac{1}{1.0045}\right)\hfill \\ \hfill & =\hfill & \frac{h^{3}w}{12\eta }\frac{dp}{dx}\left(1-0.63\frac{h}{w}\right)\hfill \\ \hfill & \hfill & \hfill \end{array}$

For $\frac{h}{w}\to 0$ we obtain the flow rate calculated for the infinitesimally elongated channel, Eq. 16.14. Obviously, we make mistakes if applying Eq. 16.67 for channels with aspect ratios in the range of 1 and above. Fig. 16.15 shows the relative error we obtain when using Eq. 16.67 instead of Eq. 16.64 for calculating the flow rate. As you can see, the error is in the range of 13 % for r = 1, whereas it drops below 1 % at r < 0.65. For smaller aspect ratios, the error is negligible. For aspect ratios bigger than 1 the simplification should not be used since the error introduced is significant.