Question

Air enters a 16-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows, and it leaves the pipe at 180 kPa and 43°C. The gas constant of air is 0.287 kPa·m3/kg·K. Whats the volumetric flow rate of the inlet/outlet, mass flow rate and velocity & volume flow rate at the exit?

Answer 1

Explanation:

(a)  We will determine the mass flow rate as follows.

                m = $\rho_{1} V_{1}$

                    = $\frac{P_{1}}{RT_{1}}A_{1}v_{1}$

                    = $\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}$

Putting the given values into the above formula as follows.

      m = $\frac{P_{1}}{RT_{1}} \times \frac{D^{2}}{4} \pi v_{1}$

          = $\frac{200}{0.287 \times 293 K} \times \frac{(0.16)^{2}}{4} \pi \times 5$                          

          = 0.239 kg/s

Hence, the mass flow rate of the inlet/outlet is 0.239 kg/s.

(b)  Now, we will determine the final volume rate as follows.

            $V_{2} = \frac{m}{\rho_{2}}$

                        = $\frac{RT_{2}m}{P_{2}}$

                        = $\frac{0.287 \times 313 \times 0.239}{180}$

                        = 0.119 $m^{3}/s$

And, the final velocity will be determined as follows.

               $v_{2} = \frac{V_{2}}{A}$

                         = $\frac{4V_{2}}{D^{2} \times \pi}$

                         = $\frac{4 \times 0.119}{(0.16)^{2} \times \pi}$

                         = 5.92 m/s

Therefore, the volumetric flow rate is 0.119 $m^{3}/s$ and velocity rate is 5.92 m/s.