Answer
9.96 kg/sec at the inlet and 4.98 kg/sec at each exit
Work Step by Step
In the given question as it is in steady state so we directly write ,
$dm_{1}$/dt=$dm_{2}$/dt+$dm_{3}$/dt
also mass flow rate can be written as density$\times$area$\times$velocity.
for 2 and 3 outlets mass flow rate will be same because the conditions are exactly the same ,
therefore ,
$dm_{1}$/dt=2$dm_{2}$/dt
now we just need to find the density for the 1 inlet , then we will automatically get mass flow rate for both the asked sections.
for finding the density , we need to find the specific density at the inlet, but for that the data that is given is not given exactly so we need to interpolate in this case,
we will interpolate between 350 C and 400C to get for 360C
putting the values
(360 - 350)/(400-360)= (x-0.57015)/(0.613731-x)
solving for x we get the specific volume as 0.60127
therefore density = 1/specific volume = 1.66 kg/$m^{3}$
now by the value of density we get mass flow rate as 9.96kg/sec at the inlet and 4.98 kg/sec at each exit.
