Answer to Question #109154 in Mechanical Engineering for Lukas v

Question

air enters a compressor with a density of 1.2 kg/m3 at mean velocity of 4 m/s in the 6 cm x 6 cm square inlet duct. air is discharged from the compressor with a mean velocity of 3 m/s in a 5 cm diameter circular pipe. determine the mass flow rate and the density at outlet?

Answer 1

Given:-

$\\Intial\,density\,(\rho _{1})=1.2\times kg/m^{3}\\[5pt] Intial\,velocity\,(v_{1})=4m/sec\\[5pt] Inlet\, Area(A_{1})=6\times 6\times 10^{-4}m^{2}\\[5pt] Inlet\, Area(A_{1})=36\times 10^{-4}m^{2}\\[5pt] Outlet \,velocity(v_{2})=3m/sec\\[5pt] Diameter(D)=5\times 10^{-2}m^{2}\\[5pt]$

Find the mass flow rate and the density at outlet.

Now,

$\\Outlet\,area\,(A_{2})=\frac{\pi }{4}\times D^{2}\\[5pt] Outlet\,area\,(A_{2})=\frac{\pi }{4}\times (5\times 10^{-2})^{2}\\[5pt] Outlet\,area\,(A_{2})=19.63\times 10^{-4}m^{2}\\[5pt] The\,mass\,flow\,rate\,is\\[5pt] m=\rho _{1}A_{1}V_{1}\\[5pt] m =1.2\times 36\times 10^{-4}\times 4\\[5pt] m=17.28\times 10^{-3}kg/s$

Therefore,

Conservation of mass between section (1) and (2).

so,

$\rho _{1}A_{1}V_{1}=\rho _{2}A_{1}V_{2}$

Hence, the density at section (2) .

$m_{2}=\rho _{2}A_{1}V_{2}\\[10pt] therefore.\\ m=m_{1}=m_{2}\\[5pt] Now,$

$\\\rho_{2} =\frac{m}{A_{2}\times V_{2}}\\[10pt] \rho_{2} =\frac{17.28\times 10^{-3}}{19.63\times 10^{-4}\times3}\\[10pt] \rho_{2} =2.933\,kg/m^{3}$

Answer:-

The mass flow rate and the density at outlet.

$\rho_{2} =2.933\,kg/m^{3}$