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For fully developed laminar Poiseuille flow in a tube, I came across this relation:

$$ \bar{u^2} = 2 u_b^2 $$

where $u_b$ is the bulk or mean velocity, defined by:

$$ u_b = \frac{1}{\rho A} \int_A \rho u dA$$

and the velocity profile $u(r)$ for Poiseuille pipe flow $u(r)$ is:

$$ u(r) = \frac{1}{4\mu} \frac{\Delta P}{L}(r^2 - R^2)$$

Using this velocity profile, the mean velocity for Poiseuille flow is:

$$ u_b = \frac{R^2}{8 \mu } \frac{\Delta P}{L}$$

How can we prove that $ \bar{u^2} = 2 u_b^2 $?


I tried to prove $ \bar{u^2} = 2 u_b^2 $ by first calculating

$$ \bar{u^2} \stackrel{?}{=} \frac{1}{\rho A} \int_A \rho u^2 dA$$

since this is similar in spirit to the definition of $u_b$. However, this resulted in $ \bar{u^2} = \frac{4}{3}u_b^2 $. Why is my above definition for $\bar{u^2}$ wrong?

I then came across a proper way to get $\bar{u^2}$, starting with the mass-average of kinetic energy energy the flowing fluid, $\bar{u^2}/2$:

$$ \dot{m} \frac{\bar{u^2}}{2} = \int_A \rho u (u^2/2)dA$$

which yields

$$ \bar{u^2} = \frac{1}{\dot{m}}\int_A \rho u^3dA = \frac{1}{\rho u_b A}\int_A \rho u^3dA = \frac{R^4}{32\mu^2} (\frac{\Delta P}{L})^2 = 2u_b$$

and this is the correct result!

Now, why does $\bar{u^2} = \frac{1}{\dot{m}}\int_A \rho u^3dA$ ? Is it satisfactory to say that it comes from the average kinetic energy?

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    $\begingroup$ In my judgment, the mass average kinetic energy is the correct quantity to use in carrying out a macroscopic energy balance on an open system. This is shown in Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ – Chet Miller Jan 14 at 13:13

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