A jet airplane is flying horizontally with its nose up at a 17° angle. The gage
ID: 1410597 • Letter: A
Question
A jet airplane is flying horizontally with its nose up at a 17° angle. The gage pressure at the inlet of the jet engine is -4.1 psig and the gage pressure at the exit is 8.0 psig. The inlet and exit area are 2.75 ft² and 3.70 ft² respectively. If we fix the control volume on the airplane and consider the incoming velocity positive, the vertical component of the pressure force acting on the inlet of the jet in lbf is
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A jet airplane is flying horizontally with its nose up at a 17° angle. The gage pressure at the inlet of the jet engine is -4.1 psig and the gage pressure at the exit is 8.0 psig. The inlet and exit area are 2.75 ft² and 3.70 ft² respectively. If we fix the control volume on the airplane and consider the incoming velocity positive, the vertical component of the pressure force acting on the inlet of the jet in lbf is
a. -1623.6 b. 10.8 c. 3.8 d. -4086.9 e. 12.0 f. 474.7 A jet airplane is flying horizontally with its nose up at a 17 degree angle. The gage pressure at the inlet of the jet engine is -4.1 psig and the gage pressure at the exit is 8.0 psig. The inlet and exit area are 2.75 ft^2 and 3.70 ft^2 respectively. If we fix the control volume on the airplane and consider the incoming velocity positive, the vertical component of the pressure force acting on the inlet of the jet in lbf isExplanation / Answer
Angle =17°
Inlet pressure P1=-4.1psig
Exit pressure P2=8 psig
Inlet area A1=2.75 ft^2
Exit area A2=3.7 ft^2
This concept belongs to mechanical properties of fluids
According to the pascal law
Pressure =F/A
P cos(theta)=F/A
Now we find the force applied on inlet
4.1cos(17°)=F/2.75
F=4.1×2.75×cos(17°)
=4.1×2.75×0.956304756
=10.8 IBF
Therefore the force applied on inlet =10.8
The correct option is B
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