Respuesta :
Explanation:
(a) Â We will determine the mass flow rate as follows.
        m = [tex]\rho_{1} V_{1}[/tex]
          = [tex]\frac{P_{1}}{RT_{1}}A_{1}v_{1}[/tex]
          = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
Putting the given values into the above formula as follows.
   m = [tex]\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}[/tex]
     = [tex]\frac{200}{0.287 \times 293 K} \times \frac{(0.16)^{2}}{4} \pi \times 5[/tex]             Â
     = 0.239 kg/s
Hence, the mass flow rate of the inlet/outlet is 0.239 kg/s.
(b) Â Now, we will determine the final volume rate as follows.
      [tex]V_{2} = \frac{m}{\rho_{2}}[/tex]
            = [tex]\frac{RT_{2}m}{P_{2}}[/tex]
            = [tex]\frac{0.287 \times 313 \times 0.239}{180}[/tex]
            = 0.119 [tex]m^{3}/s[/tex]
And, the final velocity will be determined as follows.
        [tex]v_{2} = \frac{V_{2}}{A}[/tex]
             = [tex]\frac{4V_{2}}{D^{2} \times \pi}[/tex]
             = [tex]\frac{4 \times 0.119}{(0.16)^{2} \times \pi}[/tex]
             = 5.92 m/s
Therefore, the volumetric flow rate is 0.119 [tex]m^{3}/s[/tex] and velocity rate is 5.92 m/s.
