Problem $1(20$ pts$)$

Oil with the density of $900k\frac{g}{m^3}$ and the kinematic viscosity of $4\mathrm{.}2{10}^{-4}\frac{m^2}{s}$ is discharged by a $3-mm-$diameter and $2-m$ long horizontal pipe from a reservoir to the atmosphere. The height of the liquid level above the center of the pipe is $2\mathrm{.}5m.$ Neglect the minor losses and determine,

a$)$ The type of the flow in the pipe

b$)$ The entrance length, $L_e$

c$)$ The equation of the velocity profile in the pipe

d$)$ The equation of the shear stress profile in the pipe

e$)$ The shear stress at $r=1mm$ from the center

FORMULAS

$Re=\frac{VD}{}=\frac{VD}{\text{v}\text{a}\text{r}},=\frac{du}{dy}={}_w\frac{2r}{D}$

$Q=VA,=(+{}_t)\frac{du}{dy}=\frac{du}{dy}-\frac{}{b}ar(u^{\text{'}}{v}^{\text{'}})$

$\frac{P_1}{}+z_1+\frac{V_1^2}{2g}+H_P-H_T-{\displaystyle \underset{?}{\overset{?}{}}}{h}_L=\frac{P_2}{}+z_2+\frac{V_2^2}{2g},{}_w=\frac{pD}{4L}=\frac{8V}{D}=\frac{f}{8}{V}^2$

$h_f=f(\frac{L}{D})\frac{V^2}{2g}=\frac{8fL{Q}^2}{g{}^2{D}^5}=\frac{4{}_{w}L}{D},Q=\frac{D^{4}p}{128L}$

$p=\frac{2L}{r}=\frac{4L{}_w}{D}=\frac{32LV}{D^2}=h_L,V_c=\frac{p{D}^2}{16L}$

$\frac{p-Lsin}{L}=\frac{2}{r},u(r)=(\frac{p{D}^2}{16L})(1-(\frac{2r}{D}{)}^2)=V_c(1-(\frac{2r}{D}{)}^2)$

$\frac{L_e}{D}=0\mathrm{.}06Re$ for Laminar flow $\frac{V_c-(\frac{?}{\text{b}\text{a}\text{r}}(u))}{u^{**}}=2\mathrm{.}5ln(\frac{R}{y})or\frac{(\frac{?}{\text{b}\text{a}\text{r}}(u))}{v_c}=(1-\frac{r}{R}{)}^{\frac{1}{n}}$

$\frac{L_e}{D}=4\mathrm{.}4R{e}^{\frac{1}{6}}$ for Turbulent flow $\frac{V}{V_c}=\frac{2n^2}{(n+1)(2n+1)}$

$=\frac{du}{dy}={}_w\frac{2r}{D}$

$=(+{}_t)\frac{du}{dy}=\frac{du}{dy}-\frac{}{b}ar(u^{\text{'}}{v}^{\text{'}})$

${}_w=\frac{pD}{4L}=\frac{8V}{D}=\frac{f}{8}{V}^2$

$Q=\frac{D^{4}p}{128L}$

$V_c=\frac{p{D}^2}{16L}$

$u(r)=(\frac{p{D}^2}{16L})(1-(\frac{2r}{D}{)}^2)=V_c(1-(\frac{2r}{D}{)}^2)$

$\frac{V_c-(\frac{?}{\text{b}\text{a}\text{r}}(u))}{u^{**}}=2\mathrm{.}5ln(\frac{R}{y})or\frac{(\frac{?}{\text{b}\text{a}\text{r}}(u))}{V_c}=(1-\frac{r}{R}{)}^{\frac{1}{n}}$

$\frac{V}{V_c}=\frac{2n^2}{(n+1)(2n+1)}$