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 k

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.

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Nho 1 year 2021-08-03T10:17:59+00:00 1 Answers 14 views 0

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    2021-08-03T10:19:56+00:00

    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

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

    Where,

    [tex]\dot{m} = mass flow rate\\\rho = Density\\V = Velocity[/tex]

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

    [tex]\rho_1 = 2.21kg/m^3V_1 = 40m/s[/tex]

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

    PART A) Applying the flow equation

    [tex]\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/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}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2[/tex]

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