Water enters a $50mm$ diameter thin$-$walled tube at $27C$ with a flow rate of $450k\frac{g}{h}.$ The heat

transfer from the tube wall to the fluid is given as $q_s^{\text{'}}(\frac{W}{m})=ax,$ where the coefficient $a$ is $20$

$\frac{W}{m}2$ and $x(m)$ is the axial distance from the tube entrance.

$($a$)$ Beginning with a properly defined differential control volume in the tube, derive an

expression for the temperature distribution $Tm(x)$ of the water. $0\mathrm{.}5$ points

$($b$)$ What is the outlet temperature of the water for a heated section $30m$ long? $0\mathrm{.}5$ points

$($c$)$ Plot the mean fluid temperature, $Tm(x),$ and the tube wall temperature, $Ts(x),$ as a function

of distance along the tube. Assume a convective coefficient $h=1000\frac{W}{m^2}K\mathrm{.}1$ point

$($d$)$ What value of a uniform wall heat flux, $q_s^{\text{'}\text{'}}($instead of $q_s^{\text{'}}=ax),$ would provide the same

fluid outlet temperature as that determined in part $($b$)?$ For this type of heating, plot the

temperature distributions requested in part $($c$).1$ points

$($e$)$ Let us now assume that the tube has a nonnegligible thickness of $1cm.$ The thermal

conductivity of the $316$ stainless steel tube is $16\mathrm{.}3\frac{W}{m}K.$ The tube is now surrounded by a

fluid at $T_{}=80C$ with a $h_o=5000\frac{W}{m^2}K.$ The internal convective coefficient can be assumed

as $h_i=1000\frac{W}{m^2}K.$ Plot the temperature of water as a function of the tube axis $($length$).$ Hint:

consider the thermal resistance for cylindrical coordinates. $2$ points Applying energy balance equation then,

$d{q}_{\text{c}\text{o}\text{n}\text{v}}=m{C}_{p}d{T}_m$

For part $($a$)$

$d{q}_{\text{c}\text{o}\text{n}\text{v}\text{e}}=q^{\text{'}}dx$

$ax.dx=m{C}_{p}d{T}_m$

$ax=m{C}_p\frac{d{T}_m}{dx}$

$a{\displaystyle \underset{0}{\overset{x}{}}}xdx=m{C}_p{\displaystyle \underset{T_m}{\overset{T}{}}}d{T}_m$

$T_m(x)=(T_m{)}_1+\frac{a{x}^2}{2T_m{C}_p}$

For part $($b$)$

The outlet temperature of the water for a heated section $30m$ long:

$T_m(30)=27+\frac{20(30{)}^2}{2(\frac{450}{3,600})4,180}=44\mathrm{.}2249C$ The outlet temperature of the water for a heated section $30m$ long:

$T_m(30)=27+\frac{20(30{)}^2}{2(\frac{450}{3,600})4,180}=44\mathrm{.}2249C$

Explanation:

Here, we derived the expression for temperature distribution with the help of energy balance equation in a control volume system that is $-$

$d{q}_{\text{c}\text{o}\text{n}\text{v}\text{e}}=q^{\text{'}}dx$

Step $2$

For part $($c$)$

$q_s^{\text{'}}=ax$

$q_s^{\text{'}}=q_{\text{c}\text{o}\text{n}\text{v}}=h(x)d[T_s(x)-T_m(x)]$

So$,$ fully developed flow is $h=$ constant and temperature varies linearly.

For developing flow $h(x)$ decrease with increasing in distance. For part $($d$)$

For uniform wall heating,

$q=q_s^{\text{'}\text{'}}DL=m{C}_p(T_{m}0-T_{m}1)$

$q_s^{\text{'}\text{'}}=\frac{\frac{450}{3,600}4,180(44\mathrm{.}2249-27)}{D30}$

$q_s^{\text{'}\text{'}}=\frac{95\mathrm{.}493}{D}\frac{W}{m^2}$

Explanation:

Here, we derived heat flux and and shows the mean fluid temperature for fully developed and developing flow conditions.

Answer

For part $($a$)-T_m(x)=(T_m{)}_1+\frac{a{x}^2}{2T_m{C}_p}$

For part $($b$)-44\mathrm{.}2249C$

For part $($c$)-$ The diagram is shown above.

For part $($d$)-q{s}^{\text{'}\text{'}}=\frac{95\mathrm{.}493}{D}\frac{W}{m^2}$

This are my answers. Could you write me a matlab code based on this questions and my answers