# Mean squared velocity in Poiseuille flow

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?

• 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. – Chet Miller Jan 14 at 13:13