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andre [41]
3 years ago
15

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.
Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

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

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Answer:

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Explanation:

We have,

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Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon
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Explanation:

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