Question

In a food processing plant, a brine solution is heated from $-{12}^{o}C$ to $-{6.5}^{o}C$ in a double pipe parallel-flow heat exchanger by water entering at ${35}^{o}C$ and leaving at ${20.5}^{o}C$ at the rate 9 kg/min. What is the heat exchanger area for an overall heat transfer coefficient of 860 W/$m^2$K?
Given Cp water = 4.186 kJ/kgK

• A. $0.293m^2$
• B. $0.286m^2$
• C. $0.327m^2$
• D. $0.271m^2$

Answer 1

The correct option is A $0.293m^2$

Heat transfer from the water,

$Q=m{c}_p\Delta T$

$=9\times 4.186\times 1000[35-20.5]$

$=546.27\times {10}^{3}J/min=9104.5J/s$

LMTD, ${\theta }_m=\frac{{\theta }_1-{\theta }_2}{ln\left(\frac{{\theta }_1}{{\theta }_2}\right)}$

In this case,
${\theta }_1=t_{h1}-t_{c1}=35-(-12)={47}^{o}C$

${\theta }_2=t_{h2}-t_{c2}=20.5-(-6.5)={27}^{o}C$

$\therefore {\theta }_m=\frac{47.27}{ln\left(\frac{47}{27}\right)}=36.08$

Area,
$A=\frac{Q}{U{\theta }_m}=\frac{9104.5}{860\times 36.08}=0.293m^2$