Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer:
Explanation:
Find attached the solution to the question
Answer:
The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)
Explanation:
The compressive stress in a cylindrical component can be calculated aby dividing the compressive force F to the cross sectional area A:
fc= F/A
If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.
For acylindrical component the cross sectional area A is:
A=πR^
If the diameter of the component is 81 mm, the radius is the half:
R=81mm /2 = 40.5 mm
Then A result:
A= 3.14 * (40.5 mm)^2 = 5150.4 mm^2
In square meters:
A= 3.14 * (0.0405 m)^2 = 0.005150 m^2
Replacing 708 kN to the force:
fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2
Using the force in Newtons:
F= 70800 N
Finally the compressive stress in Pa is:
fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa