Question

A process industry employ a counter flow heat exchanger to cool 0.8 kg/s of oil from${120}^{0}Cto{40}^{0}C$ by the use of water entering at ${20}^{0}C$. The overall heat transfer coefficient is estimated to be$1600W/m^2.K.$Assume exit temperature and mass flow rate of water ${80}^{0}C$, 0.75kg/s, respectively. The effectiveness of the given heat exchanger is

• A. 0.6
• B. 0.65
• C. 0.8
• D. 0.70

Answer 1

The correct option is C 0.8

(B)


Energy balance:
Energy released by oil = Energy received by water

${\stackrel{\cdot }{m}}_o.c_{po}.(T_{hi}-T_{he})={\stackrel{\cdot }{m}}_w.c_{pw}(T_{ce}-T_{ci})$

${\stackrel{\cdot }{m}}_o.c_{po}\times (120-40)={\stackrel{\cdot }{m}}_w.c_{pw}(80-20)$

$80\times {\stackrel{\cdot }{m}}_o.c_{po}=60\times {\stackrel{\cdot }{m}}_w.c_{pw}$

${\stackrel{\cdot }{m}}_o.c_{po}<{\stackrel{\cdot }{m}}_w.c_{pw}$

$C_{max}={\stackrel{\cdot }{m}}_w.c_{pw}\&C_{min}={\stackrel{\cdot }{m}}_o.c_{po}$

Effectiveness$(\in )=\frac{C_{min}(T_{hi}-T_{he})}{C_{min}(T_{hi}-T_{ci})}=\left(\frac{120-40}{120-20}\right)=\frac{80}{100}=0.8$