Answer:
The strength coefficient is K = 591.87 MPa
Explanation:
We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by
where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.
Solving for strength coefficient
From the strain hardening equation we can solve for K
And we can replace values
Thus we get that the strength coefficient is K = 591.87 MPa
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
Explanation:
From the question we are told that:
Initial Pressure
Initial Temperature
Final Pressure
Final Temperature
Work Output
Generally Specific Energy from table is
At initial state
With
Specific Volume
At Final state
Generally the equation for The Process is mathematically given by
Assuming Mass to be Equal
Where
Therefore