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Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the density is 0.762 kg/m3 and the velocity is 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flowrate through the nozzle and (b) the exit area of the nozzle.

Answer :

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To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

[tex]\dot{m}=\rho A V[/tex]

Where,

[tex]\dot{m} =[/tex] mass flow rate

[tex]\rho =[/tex] Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

[tex]\rho_1 = 2.21kg/m^3[/tex]

[tex]V_1 = 20m/s[/tex]

[tex]A_1 = 60*10^{-4}m^2[/tex]

[tex]\rho_2 = 0.762kg/m^3[/tex]

[tex]V_2 = 160m/s[/tex]

PART A) Applying the flow equation we have to

[tex]\dot{m} = \rho_1 A_1 V_1[/tex]

[tex]\dot{m} = (2.21)(60*10^{-4})(20)[/tex]

[tex]\dot{m} = 0.2652kg/s[/tex]

PART B) For the exit area we need to arrange the equation in function of Area, that is

[tex]A_2 = \frac{\dot{m}}{\rho_2 V_2}[/tex]

[tex]A_2 = \frac{0.2652}{(0.762)(160)}[/tex]

[tex]A_2 = 2.175*10^{-3}m^2[/tex]

Therefore the Area at the end is [tex]21.75cm^2[/tex]

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