Problem $1$

a$)$ Given the system shown in the Figure below, calculate the required power of the pump to

deliver water from the bottom to the upper tank at a rate of $0\mathrm{.}34\frac{m^3}{m}in.$ All of the piping is

$10\mathrm{.}16cm$ internal diameter smooth circular pipe. Consider the following localized viscous

dissipations along with the corresponding friction loss factors reported in parenthesis:

sudden fluid contraction $(0\mathrm{.}45),90$ deg elbow $(0\mathrm{.}5),$ and sudden fluid expansion $(1).2$ points

The density of water can be assumed to be equal to $1\frac{g}{c}{m}^3,$ and the viscosity of $1$mPa.$s$

b$)$ Calculate the power of the pump considering various pipe rugosity $(k)$ of the pipe: $0\mathrm{.}001016,$

$0\mathrm{.}004064,0\mathrm{.}01016,0\mathrm{.}04064cm.$ Plot the power of the pump as a function of the friction

factor $f$ and draw a trendline through the data. $2$ points

c$)$ Discuss interventions to the system in order to reduce the power of the pump. Motivate your

choice$/$s quantitatively using also the mechanical energy balance. $1$ 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$)-q{s}^{\text{'}\text{'}}=\frac{95\mathrm{.}493}{D}\frac{W}{m^2}$

Could you write $me$ a matlab code based $on$ the four questions and $my$ answers