Question

In a condenser, water enters at ${30}^{\circ }C$ and flows at the rate 1500 kg/hr. The condensing steam is at a temperature of ${120}^{\circ }C$ and cooling water leaves the condenser at ${80}^{\circ }C$. Specific heat of water is 4.187 kJ/kgK. If the overall heat transfer coefficient is 2000 W/m${}^2$K, the heat transfer area is

• A.
  $0.707m^2$
• B.
  $7.07m^2$
• C.
  $70.7m^2$
• D.
  $141.4m^2$

Answer 1

The correct option is A

$0.707m^2$


$\therefore {\theta }_1={120}^{\circ }-{30}^{\circ }={90}^{\circ }C$
${\theta }_2={120}^{\circ }-{80}^{\circ }={40}^{\circ }C$

Mass flow rate,

$m=\frac{1500}{3600}=0.4167kg/s$

Log mean temperature difference,

${\theta }_m=\frac{{\theta }_1-{\theta }_2}{ln\left(\frac{{\theta }_1}{{\theta }_2}\right)}=\frac{90-40}{ln\left(\frac{90}{40}\right)}={61.65}^{\circ }C$

From energy balance,

$Q=m{c}_{pw}\Delta T=\cup A{\theta }_m$

$0.4167\times 4.187\times 1000\times 50$

= $2000\times A\times 61.65$

$orA=0.707m^2$