Answer:
The velocity at exit of the nozzle is 175.8 m/s
Solution:
As per the question:
Air density inside the rocket, [tex]\rho_{i} = 5.25\ kg/m^{3}[/tex]
Speed, v = 1.20 m/s
The inner diameter of the rocket, [tex]d_{i} = 15.0\ cm = 0.15\ m[/tex]
The inner radius of the rocket, [tex]r_{i} = \frac{0.15}{2} = 0.075\ m[/tex]
The exit diameter of the nozzle, [tex]d_{e} = 2.50\ cm = 0.025\ m[/tex]
The exit radius of the nozzle, [tex]r_{e} = 0.0125\ m[/tex]
Air density inside the nozzle, [tex]\rho_{n} = 1.29\kg/m^{3}[/tex]
Now,
To calculate the air speed when it leaves the nozzle:
Mass rate in the interior of the rocket, [tex]\frac{dM}{dt} = \rho_{i}Av[/tex]
Mass rate in the outlet of the nozzle, [tex]\frac{dm}{dt} = \rho_{n}A'v'[/tex]
Now,
[tex]A = \rho_{i}r_{i}^{2}[/tex]
Now,
We know that:
[tex]Av = A'v'[/tex]
[tex]\rho_{i}r_{i}^{2}v = \rho_{n}r_{e}^{2}v'[/tex]
[tex]5.25\times 0.075^{2}\times 1.2 = 1.29\times 0.0125^{2}v'[/tex]
v' = 175.8 m/s
