Question

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

Answer 1

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

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

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

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

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

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

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$