What are the benefits of using the engineering design process? combining scientific knowledge and creativity in a rigid and stru
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Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj -
a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N
Answer:
= 2.6906 ×
lb-s/in²
Explanation:
given data
velocity V = 64 in/sec
separated by a gap x = 0.41 in
relative motion by shear stress = 0.42 lb/in²
solution
we know that shear stress is directly proportional to rate of change of velocity as per newton's law of viscosity.
....................1
so here coefficient of dynamic viscosity and
is velocity gradient
and
put here value and we get
0.42 =
= 2.6906 ×
lb-s/in²