Answer
$\mathcal{V}_1=29.9\ m/s$
$\mathcal{V}_2=66.1\ m/s$
Work Step by Step
From the material balance, for ideal gases:
$\dot{m}=\dfrac{P_1\pi D_1^2\mathcal{V}_1}{4RT_1}=\dfrac{P_2\pi D_2^2\mathcal{V}_2}{4RT_2}$
$\mathcal{V}_2=\mathcal{V}_1\times\dfrac{P_1D_1^2T_2}{P_2D_2^2T_1}$
Given $T_1=65°C,\ P_1=200\ kPa,\ T_2=60°C,\ P_2=175\ kPa,\ D_1/D_2=1.4$:
$\mathcal{V}_2=2.2104\ \mathcal{V}_1$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{Q}$
$c_pT_1+\mathcal{V}_1^2/2=c_pT_2++\mathcal{V}_2^2/2+q$
Given $q=3.3\ kJ/kg,\ c_p=1.005\ kJ/kg.K$
and solving for the inlet velocity:
$\mathcal{V}_1=29.9\ m/s$
hence $\mathcal{V}_2=66.1\ m/s$
