Chapter 4
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} <heat from the oil received by the water> | • Constant properties | |
| • Negligible Δ\rm E_{kin} and Δ\rm E_{pot} |
Step-by-Step
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:
