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dalvyx [7]
3 years ago
15

The rings of Saturn occupy the region inside Saturn's Roche limit.

Physics
1 answer:
Oliga [24]3 years ago
7 0
I suppose this is a true or false question and that sentence is true.
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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

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