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vladimir1956 [14]
3 years ago
8

Consider a subsonic engine inlet (i.e., a diffuser) with an inlet area of 1.5 m2 and exit-to-inlet area ratio of 1.44. Air enter

s the diffuser flowing at 5 kg/s with a static pressure of 0.350 bar and a velocity of 110 m/s. The gases exit the diffuser at a static pressure of 0.375 bar and a velocity of 80 m/s. Assume that the ambient (external) pressure on the walls of the diffuser is 0.340 bar.a.) For these conditions, what is the force (in units of lbf and N) transmitted to the structure
holding the diffuser? Provide both direction and magnitude of the force, and use a
picture to help present your answer.
b.) Based solely on your results, comment on whether the force acting on the diffuser would
help speed up or slow down a vehicle that used this diffuser as part of a jet propulsion
system.
Physics
1 answer:
Kisachek [45]3 years ago
8 0

Ai=2300 in2 , Ao=Ai*1.44=3312 in2

m=10 lbm/s

Pi=5 psia , Po=5.4 psia , Pa=5.5 psia

Vi=120 m/s , Vo=78 m/s

a) Force =m(Vo-Vi) = -190.5 N = -42.82 lbf (towards the inlet)

b) since force is negative it will slow down the system.

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F=8.0*10^{-10}N

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See the attached file for the masses distributions

The force between two masses at distance r is expressed as

F=\frac{Gm_{1}m_{2}  }{r^{2} }\\ G=Gravitional constant \\

since the masses are of the same value, the above formula can be reduce to

F=\frac{Gm^{2}}{r^{2} }\\

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

F_{12} =\frac{Gm^{2}}{r^{2} }j\\

The force on the lower left corner of the mass due to the lower right side of the mass is

F_{14} =\frac{Gm^{2}}{r^{2} }i\\

The force on the lower left corner of the mass due to the upper right side of the mass is

F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\

The net force can be express as

F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}

if we insert values we arrive at

F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j

if we solve for the magnitude, we arrive at

F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}

Hence the net force on one of the masses is

F=8.0*10^{-10}N

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