Question

In a counter flow heat exchanger the oil is coooled from ${90}^{\circ }C$ to ${50}^{\circ }C$ by means of water entering the cooler at ${20}^{\circ }C$ and leaving the cooler at ${60}^{\circ }C$
The LMTD and AMTD for the heat exchanger are respectively

• A. 40 & 40
• B. None
• C. 40 & 30.8
• D. 30.8 & 40

Answer 1

The correct option is B None

$T_{hi}={90}^{o}C;T_{he}={50}^{o}C$
$T_{ci}={20}^{o}C;T_{ce}={60}^{o}C$

Energy balance:

Energy supplied by hot fluid = energy absorbed by cold fluid

$\dot{m_h}{c}_{ph}(90-50)=\dot{m_c}.c_{pc}(60-20)$

$\dot{m_h}.c_{ph}\times 40=\dot{m_c}.c_{pc}\times 40$

$\dot{m_h}.c_{ph}=\dot{m_c}.c_{pc}\left(equalheatcapacity\right)$


${\theta }_1={\theta }_2={30}^{o}C$

$Then,LMTD={\theta }_1={\theta }_2={30}^{o}C$
$LMTD={30}^{o}C$

Arithmetic mean temperature difference (AMTD):

$AMTD=\left(\frac{T_{hi}+T_{he}}{2}\right)-\left(\frac{T_{ci}+T_{cc}}{2}\right)$

$=\left(\frac{90+50}{2}\right)-\left(\frac{60+20}{2}\right)$

$=70-40={30}^{o}C$

For counter flow heat exchanger with equal heat capacity

${\theta }_1={\theta }_2=LMTD=AMTD$

And temperature profiles are linear and parallel to each other.