Question

A counter-flow heat exchanger is used to cool oil $(c_p=2.20kJ/kgK)$ from ${110}^{\circ }Cto{85}^{\circ }C$ at a rate of $0.75kg/s$ by cold water $(c_p=4.18kJ/kgK)$ that enters the heat exchanger at ${20}^{\circ }C$ at a rate of $0.6kg/s.$ If the overall heat transfer coefficient is $800W/m^{2}K,$ the heat transfer area of the heat exchanger will be

• A. $0.810m^2$
• B. $0.745m^2$
• C. $0.790m^2$
• D. $0.770m^2$

Answer 1

The correct option is B $0.745m^2$



$\text{Energy balance:}\text{Heat released by hot oil = Heat received by cold water}\begin{array}{cc}\dot{m_o}{c}_{po}(T_{hi}-T_{he})& ={\dot{m}}_w{c}_{pw}(T_{ce}-T_{ci})\\ 0.75\times 2.2(110-85)& =0.6\times 4.18(T_{ce}-20)\\ \Rightarrow T_{ce}={36.44}^{\circ }C& \\ LMTD& =\frac{{\theta }_1-{\theta }_2}{ln\left(\frac{{\theta }_1}{{\theta }_2}\right)}=\frac{73.55-65}{ln\left(\frac{73.55}{65}\right)}={69.18}^{\circ }\\ \Rightarrow {\theta }_m& ={69.18}^{\circ }C\\ \text{Heat transfer rate}& ={\dot{m}}_o{c}_{po}(T_{hi}-T_{he})\\ UA{\theta }_m& =\dot{m_o}{c}_{po}(T_{hi}-T_{he})\\ 800\times A\times 69.18& =0.75\times 2200(110-85)\\ \Rightarrow A& =0.745m^2\end{array}$