Question

In a double pipe counter flow heat exchanger oil of heat capacity rate $2.1kJ/s-K$ is cooled from ${120}^{\circ }C$ to ${60}^{\circ }C$ with help of cooling water 4.2 J/kgK which is flowing at a rate of 1.5 kg/s. The heat transfer surface area is $22m^2$ and cooling water enters at ${40}^{\circ }C$. What is the value of appropriate overll heat transfer coefficient?

• A. $123.3W/m^{2}K$
• B. $167.3W/m^{2}K$
• C. $157.3W/m^{2}K$
• D. $181.3W/m^{2}K$

Answer 1

The correct option is C $157.3W/m^{2}K$

$C_{ph}=2.1kJ/K$

$T_{hi}={120}^{\circ }C{T}_{ho}={60}^{\circ }C$

$T_{ci}={40}^{\circ }C$

$m_c=1.5kJ/s,C_{pc}=4.2kJ/kgK$

From energy balance,

$Q_1=C_{pj}(T_{hi}-T_{hv})=m_c.c_{pc}(T_{co}-T_{ci})$

$2.1\times (120-60)=1.5\times 4.2\times (T_{\infty }-40)$

$T_{\infty }={60}^{\circ }C$

${\theta }_1=60{\theta }_0=20$

$LMTD=\frac{{\theta }_1-{\theta }_0}{\ln \left(\frac{{\theta }_1}{{\theta }_0}\right)}=\frac{60-20}{\ln \left(\frac{60}{20}\right)}={36.41}^{\circ }C$

$Q=UALMTD$

$C_{ph}(T_{hi}-T_{\infty })=U\times 22\times 36.41$

$2.1\times (120-60)=U\times 22\times 36.41\times {10}^{-3}$

$U=157.3W/m^{2}K$