Answer
a) $T_2=112°F$
b) $\nu_2=244\ ft/s$
Work Step by Step
From the energy balance:
$\dot{m}(h_1+\nu_1^2/2)=\dot{m}(h_2+\nu_2^2/2)$
From table A-17E (65°F): $h_1=125.40\ Btu/lbm$
Given $\nu_1=750\ ft/s,\ \nu_2\approx 0\ ft/s$
and solving for the outlet enthalpy: $h_2=136.63\ Btu/lbm$
From table A-17E ($h_2$): $T_2=571.6°R=112°F$
From the material balance:
$\frac{A_1\nu_1}{v_1}=\frac{A_2\nu_2}{v_2}$
$\frac{P_1A_1\nu_1}{RT_1}=\frac{P_2A_2\nu_2}{RT_2}$
Given $P_2=14.5\ psia, P_1=13\ psia,\ A_2/A_1=3$:
and solving for the outlet velocity: $\nu_2=244\ ft/s$
