$90,000$ kg h$-1$ of crude oil is to be heated from $40$oC to $120$oC in the tubes of a shell$-$ and$-$tube heat exchanger. The exchanger has $110$ tubes per pass, and the tube I$/$D is $14\mathrm{.}83$ mm$.$ The total heat exchange area is $65\mathrm{.}8$ m$2.$ Select an appropriate correlation for Nusselt number, and calculate the tube$-$side heat transfer coefficient, assuming a tube inner wall mean temperature of $132$C$.$Crude oil properties:densityspecific heat capacity thermal conductivity viscosity$800$ kg m$-32090$ J kg$-1$K$-10\mathrm{.}133$ W m$-1$K$-1(14\mathrm{.}8-2\mathrm{.}84$lnT $)$ x $10-3$ N s m$-2($T in oC

Tube side velocity

Tube side Re

$=$ mass flow $\frac{?}{(\text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}\text{t}\text{u}\text{b}\text{e}\frac{c}{s}\text{a}\text{r}\text{e}\text{a}\text{t}\text{u}\text{b}\text{e}\text{s}\text{p}\text{e}\text{r}\text{p}\text{a}\text{s}\text{s})}$

$=\frac{90000}{3600(8001\mathrm{.}727{10}^{-4}110)}=1\mathrm{.}65m{s}^{-1}$

$=$ density $$ velocity $$ tube$\frac{\text{I}}{D}\frac{?}{?}$ viscosity

$Re$

$=8001\mathrm{.}65\frac{0\mathrm{.}01483}{2\mathrm{.}36{10}^{-3}}$

$=8294\mathrm{.}75$

Note at room temperature, $T=20C,$ viscosity $=6\mathrm{.}3{10}^{-3}Ns{m}^{-2}.$ It is higher than $2\mathrm{.}0{10}^{-3}Ns{m}^{-2}.$ So$,$ treat as viscous. Therefore choose general correlation for turbulent flow and viscous fluid:

$=0\mathrm{.}027R{e}^{0\mathrm{.}8}P{r}^{0\mathrm{.}33}(\frac{}{{}_W}{)}^{0\mathrm{.}14}$

Pr

$=$ specific heat $$ viscosity $\frac{?}{?}$ thermal conductivity

$=20902\mathrm{.}36\frac{{10}^{-3}}{0\mathrm{.}133}$

$=37\mathrm{.}1$

Ignoring wall effect term initially,

Nu

$=0\mathrm{.}027{8294\mathrm{.}75}^{0\mathrm{.}8}{37\mathrm{.}1}^{0\mathrm{.}33}$

$=121\mathrm{.}4$

$=$ Nusselt number $$ thermal conductivity $\frac{?}{?}$ tube$\frac{\text{I}}{D}$

$=121\mathrm{.}4\frac{0\mathrm{.}133}{0\mathrm{.}01483}W{m}^{-2}{K}^{-1}$

$=1088\mathrm{.}95Wm{m}^{-2}{K}^{-1}$

h

Then correct for wall effect.

h

$=$ uncorrected $h($ bulk $to$ wall viscosity ratio ${)}^{0\mathrm{.}14}$

$=1088\mathrm{.}95(2\mathrm{.}36\frac{{10}^{-3}}{(14\mathrm{.}8-2\mathrm{.}84ln(132)){10}^{-3}}{)}^{0\mathrm{.}14}$

$=1088\mathrm{.}95(\frac{2\mathrm{.}36{10}^{-3}}{9\mathrm{.}33{10}^{-4}}{)}^{0\mathrm{.}14}$

$=1088\mathrm{.}951\mathrm{.}14$

$=1241\mathrm{.}4W{m}^{-2}{K}^{-1}$