Question

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 1

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

$\dot{m}=\rho A V$

Where,

$\dot{m} =$ mass flow rate

$\rho =$ Density

V = Velocity

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

$\rho_1 = 2.21kg/m^3$

$V_1 = 20m/s$

$A_1 = 60*10^{-4}m^2$

$\rho_2 = 0.762kg/m^3$

$V_2 = 160m/s$

PART A) Applying the flow equation we have to

$\dot{m} = \rho_1 A_1 V_1$

$\dot{m} = (2.21)(60*10^{-4})(20)$

$\dot{m} = 0.2652kg/s$

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

$A_2 = \frac{\dot{m}}{\rho_2 V_2}$

$A_2 = \frac{0.2652}{(0.762)(160)}$

$A_2 = 2.175*10^{-3}m^2$

Therefore the Area at the end is $21.75cm^2$