Answer
$\nu_{1,15°C}= 6.37\ m/s$
$\nu_{1,40°C}= 6.42\ m/s$
Work Step by Step
Approximating by a saturated liquid:
$T_1=15°C$): $v_1=v_2=0.001001\ m³/kg$
Inlet :
$\nu_1=\dot{m}\frac{4v_1}{\pi D_1^2}$
With $\dot{m}= 0.5\ kg/s,\ D_1=0.01m$:
$\nu_1= 6.37\ m/s$
Outlet:
$\nu_2=\dot{m}\frac{4v_2}{\pi D_2^2}$
With $D_2=0.015m$:
$\nu_2=2.83\ m/s$
Second case
$T_1=40°C$: $v_1=0.001008\ m³/kg$
$\nu_1=\dot{m}\frac{4v_1}{\pi D_1^2}$
$\nu_1= 6.42\ m/s$
