## Q. 4.6

Basic Shell-and-Tube Heat Exchanger
Consider cooling of hot oil ($\rm\dot m_{oil}$ = 5.5 kg / s; $\rm T_{\dot m}$ = 380K, $\rm T_{out}$ = 320K; $\rm c_p$ = 2.122kJ /(kg ⋅ K))  by water ($\rm T_{in}$ = 280K, $\rm\dot m_{water}$ = 5.65kg / s; $\rm c_p$ = 4.18kJ /(kg ⋅  K))

 Assumptions: Model: System: • No losses • Steady turbulent uniform streams • Δ$\rm\dot Q_{oil}$ = Δ$\rm\dot Q_{water}$ • Constant properties • Negligible Δ$\rm E_{kin}$ and Δ$\rm E_{pot}$

## Verified Solution

$\begin{array}{l}\rm{{\Delta\dot{ Q}_{{oil}}=\Delta\dot{ Q}_{{water}}; \quad\Delta\dot{ Q}=\mathrm{\dot m}\Delta{h}={\dot m}\,{c_{p}}~\Delta{T}}}\\\\ \rm{{({\dot m}{}{c_{p}})_{{oil}}\left({T_{in}-T_{o u t}}\right)_{{oil}}=(\dot m c_p)_{water}\left({T_{in}-T_{o u t}}\right)_{{water}}}}\end{array}$

Thus,

$\begin{matrix}\rm T_{{out,w}}={T}_{{in,w}}+{\frac{({\dot mc_{p}})_{0}\;({T}_{{in}}-{T}_{{out}})_{0}}{({\dot m\,c_{p}})_{{w}}}}\\\\\rm{\underline{{{T}_{{out,w}}}}}=280{K}+30{K}={\underline{{310{K}}}}\end{matrix}$

Comments: The basic “heat-exchange balance”, Δ$\rm\dot Q_ A$ = Δ$\rm\dot Q_ B$ , holds for other configurations as well. Here are two basic examples depicted: