The tube surface temperature is maintained at 70℃ and the flowing water is heated from the inlet 15℃ to an outlet temperature of 45℃. Calculate the following.

(i)The heat transfer coefficient from the tube surface of the water

(ii)The heat transfer and

(iii)The length of the tube. Take the physical properties of water at its mean bulk temperature of 30℃.ρ=995.7 Kg/$m^3$ ,$C_p$=(4.174KJ)/Kgk,k=61.718×$10^(-2)$ W/mK,v=0.805×$10^(-6)$ $m^2⁄s$,Pr=5.42

Let us first determine the type of flow

Re=$\frac{UD}{v}$=$\frac{(12×0.06)}{(0.805×10^(-6) )}$=8.94×$10^5$

Since Re > 8.94×$10^5$, the flow is turbulent

Nu=0.023 $Re^.8$ $Pr^(1/3)$

Nu=$\frac{hD}{k}$=0.023×$(8.94×10^5 )^.8$× $5.42^(1/3)$=2330.58

h=$\frac{2330.58×61.718}{P0.06=23.973×$10^3$}$ $\frac{W}{(m^2 K)}$ h=23973 $\frac{W}{(m^2 K)}$ Heat Transfer, Q=$mC_P (T_out-T_in )$=$UρAC_P (T_out-T_in )$

=12×995.7×$\frac{π}{4}$×$0.06^2$×4174(45-15)

Q=4.2303×$10^6$ W

We know that

Q=hA∆T

=h×π×D×L×($T_s-T_mean$)

4.2303×$10^6$=23973×π×0.06×L×(70-30)

L=23meter