Specific heat of hot and cold air is 1.05 kJ/KgK, U=80W⁄(m^2 K), find the area of the heat exchanger for parallel and counter flow configuration.

By energy balance equation outlet temperature of cold air

$m_h C_ph (T_hi-T_ho )$=$m_c C_pc (T_Co-T_ci)$

1.25×1.05(66-38)=1.6×1.05×($T_co$-15)

$T_co$=36.875℃

When the flow is counter

$T_lm$=$\frac{((66-36.875)-(38-15))}{ln\frac{((66-36.875)}{(38-15))}}$

$T_lm$=25.94℃

Q=$m_c C_pc (T_Co-T_ci )$

=1.6×1.05×(36.875-15)

Q=36.75KW=36750W

Q=UA$∆T_lm$

36750=80×A×25.94

A=17.71$m^2$

When the flow is parallel $∆T_lm$=$\frac{((66-15)-(38-36.875))}{ln\frac{((66-15)}{(38-36.875))}}$

$T_lm$=13.077

Q=UA$∆T_lm$ 36750=80×13.077×A

A=35.12$m^2$