Question 6.P.19: Derive a relationship between the pressure difference record...

An energy balance for an incompressible fluid in turbulent flow is given by:

\Delta u^2 / 2+g \Delta z+v \Delta P+F=0                 (equation 2.55)

Ignoring functional losses and assuming the pitot tube to be horizontal,

\left(u_2^2-u_1^2\right) / 2=-v\left(P_2-P_1\right)

If the fluid is brought to rest of plane 2, then:

-u_1^2 / 2=-v\left(P_2-P_1\right)

and:                                  u_1=\sqrt{2 v\left(P_2-P_1\right)}=\underline{\underline{\sqrt{2 g h}}}          (equation 6.10)

If the duct radius is r, the velocity u_y at a distance y from the wall (and s from the centreline) is given by the one-seventh power law as:

u_y=u_s\left(\frac{y}{r}\right)^{1 / 7}                     (equation 3.59)

where u_s is the velocity at the centreline.
The flow, d Q, through an annulus of thickness d y_1 distance y from the axis is:

d Q=2 \pi s d y u_s\left(\frac{y}{r}\right)^{1 / 7}

Multiplying and dividing through by r² gives:

d Q=2 \pi r^2 u_s \frac{s}{r}\left(\frac{y}{r}\right)^{1 / 7} d \left(\frac{y}{r}\right)

or, since s = (r – y):      =2 \pi r^2 u_s\left(1-\frac{y}{r}\right)\left(\frac{y}{r}\right)^{1 / 7} d \left(\frac{y}{r}\right)

The total flow is:     Q=2 \pi r^2 u_s \int_0^1\left[\left(\frac{y}{r}\right)^{1 / 7}-\left(\frac{y}{r}\right)^{8 / 7}\right] d \left(\frac{y}{r}\right)

=2 \pi r^2 u_s\left[\frac{7}{8}\left(\frac{y}{r}\right)^{8 / 7}-\frac{7}{15}\left(\frac{y}{r}\right)^{15 / 7}\right]_0^1=0.817 \pi r^2 u_s

The average velocity, u_{ av }=Q / \pi r^2=0.817 u_s

Thus:                                           u_y=u_{ av }, 0.817 u_s=u_s(y / r)^{1 / 7}

∴                                             (y / r)=0.243 and s / r=\underline{\underline{0.757}}