Derive the kinetic energy $()$ correction factor for the following pipe flow velocity distributions $($see Roberson et al$.,$ eqn. $5-2).$ You can use an integrating calculator or computer program to check your answer, but show your work.

Laminar flow:

Turbulent flow:

$v=v_{\text{m}\text{a}\text{x}}(1-(\frac{r}{R}{)}^2)$

$v=v_{\text{m}\text{a}\text{x}}(\frac{y}{R}{)}^{\frac{1}{7}}$

Do the following $($show your work$).$

a$.$ Find the volumetric flow rate $(Q)$ for a pipe of radius $R$ by integrating each equation:

$Q={\displaystyle \underset{0}{\overset{R}{}}}$vdA

let $dA=2rdr.$ Note that for the turbulent velocity distribution you'$1$l need to work with the variable $y($distance from the wall$)$ instead of $)=(R-r.$ Differentiate this to get $dr=-dy.$ Plug into the equation for $dA$ and integrate from $y=R$ to $y=0)$

b$.$ Find the average velocity $V=\frac{Q}{A}$

c$.$ Find $\frac{V}{v_{\text{m}\text{a}\text{x}}}($answer $=\frac{1}{2}$ for laminar flow$)$

d$.$ Find $,$ the kinetic energy correction factor, $=\frac{1}{A}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(\frac{V}{V}{)}^{3}dA($answer $=2$ for laminar flow, $1\mathrm{.}06$ for turbulent flow$)$