Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
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


Answer:

Explanation:
solution:
from this below equation (1)
σ/2εo
...........(1)
we obtain:


Saturated Pressure Temperature chart for R-22 shows 45 degF at 76 psig
65-45= 20 degF superheat