Answer:
The area of heat exchanger [tex]A=0.00218\ m^2[/tex]
Explanation:
Given that
Brine heated from 10°C to 35°C and water cools from 60°C to 45°C.
[tex]U=950\ \frac{W}{m^2K}[/tex]
Lets take these are act as counter flow heat exchanger.
So
[tex]\Delta T_1=35^{\circ}C,\Delta T_2=25^{\circ}C[/tex]
As we know that
[tex]Q=m_wC_p\Delta T_w=U\times A\times LMTD[/tex]
Here
[tex]LMTD=\dfrac{\Delta T_1-\Delta T_2}{\ln \frac{\Delta T_1}{\Delta T_2}}[/tex]
[tex]LMTD=\dfrac{35-25}{\ln \frac{35}{25}}[/tex]
LMTD=29.7°C.
[tex]m_wC_p\Delta T_w=U\times A\times LMTD[/tex]
[tex]C_p=4.187\frac{KJ}{kg.K}[/tex] for water
0.3 x 4.187 x 15 = 950 x A x 29.7
[tex]A=0.00218\ m^2[/tex]
So the area of heat exchanger [tex]A=0.00218\ m^2[/tex]
