The overall heat transfer coefficient is expected to be 1450 W/m2C. Using NTU method, calculate the following:

i) The mass flow rate of water

ii) The effectiveness of the heat exchanger

iii) The surface area required.

Given:

$m_h$=0.55Kg⁄s,$C_p$=2.45KJ⁄KgC,$T_(h,1)$=115℃,$T_(h,2)$=40℃,$T_c1$=15℃,$T_(c,2)$=75℃,

U=1450$\frac{W}{(m^2 C)}$

The mass flow rate of water:

Q=$ϵC_min T_(h.1)-T_(c,1)$

To find $m_c$,use energy balance equation

0.55× 2450 (115 - 40)= $m_c$× 4186 (75 - 15)

mc = 0.40kg/s

Effectiveness of heatexchanger

Heat capacity rate of hot fluid: $C_h$=$m_h$ $C_ph$=0.55×2.45=1.35KW⁄K

Heat capacity rate of cold fluid:$C_c=m_c C_pc$=0.40×4.186=1.67KW/K

$C_h \lt C_c$

$C_h$= $C_min$

ϵ=$\frac{(m_h C_ph (T_(h,in)-T_(h,out) ))}{(C_min⁡ (T_(h,in)-T_(c,in) ) )}$=$\frac{(115-40)}{(115-15)}$=0.75

∈=75%

Q=0.75×1350×(115-15)

Q=101.250W

Area Of heat exchanger

Q=$UA∆T_lm$

$∆T_lm$=$\frac{(∆T_1-∆T_2)}{(ln⁡((∆T_1)/(∆T_2 )) )}$

=$\frac{((115-75)-(40-15))}{ln⁡((115-75)/(40-15))}$

=31.9℃

A=$\frac{101.25}{(31.9×1450)}$

A=$2.19m^2$