Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
Answer:
The spring was compressed the following amount:
Explanation:
Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount ), and the total final energy is the addition of the kinetic energies of both masses:
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as
Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,
The letters have the same meaning as before.
Then he hoop stress would be,
And the longitudinal stress would be
The Mohr's circle is attached in a image to find the maximum shear stress, which is given as
Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
