The Equation of Continuity — Question 4

Question 4

A centerline speed is not an area-average speed

A steady axial flow in a circular pipe of radius $R$ has the prescribed profile $v(r)=v_{\max}(1-r^2/R^2)$. Density is uniform. This profile may describe a viscous flow; continuity does not require an inviscid model.

1. Integrate over circular rings to find $Q$ and the mean speed.

2. What fraction of the volume flow passes through the inner half-radius disk? Compare it with that disk's area fraction.

Question 4 - Solution

A centerline speed is not an area-average speed

(a) Derivation and calculation

$$dA=2\pi r\,dr,\quad Q=2\pi v_{\max}\int_0^R\left(r-\frac{r^3}{R^2}\right)dr.$$$$\boxed{Q=\frac12\pi R^2v_{\max}},\qquad\boxed{\bar v=Q/(\pi R^2)=v_{\max}/2}.$$Using centerline speed as mean speed doubles the flow estimate.

(b) Result and interpretation

$$Q(a)=2\pi v_{\max}\int_0^a\left(r-\frac{r^3}{R^2}\right)dr=\pi v_{\max}\left(a^2-\frac{a^4}{2R^2}\right).$$$$\boxed{\frac{Q(R/2)}{Q(R)}=2\left(\frac14-\frac1{32}\right)=\frac7{16}=43.75\%}.$$$$A(R/2)/A(R)=1/4=25\%.$$The central region carries disproportionate flow because it moves faster.

Physical check

The flux integral works for the prescribed viscous profile; incompressible continuity does not require inviscid flow.