The following data were obtained when a cold-worked metal was annealed. (a) Estimate the recovery, recrystallization, and grain
1 answer:
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Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter = 12.8 mm
Final diameter = 10.7
Engineering stress = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4( )² ; Ag = π/4(
)²
% = π/4 [ ( ( )² - (
)²) / ( π/4 (
)²) ]
= ( ( )² - (
)²) / (
)² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress =
( 1 +
)
is engineering strain
= dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
= 0.431
so we substitute the value of into our initial equation;
True stress = 460 ( 1 + 0.431)
True stress = 460 (1.431)
True stress = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa
Answer:
The answers to the question are as follows
(a) W = -175.6 MJ
(b) W = -329.256 MJ
The peak temperature of the isentropic compression process is 886.974 K
Explanation:
(a) We are given the initial conditions as
v₂ = 10 m³
T₂ = 15 °C
p₂ (gauge) = 4.5 MPa gauge → 4.5 MPa + 1 atm = 4.5 MPa + 101325 Pa = 4.601 MPa
p₁ = 1 atm
Therefore isothermal compression we have the work done given by
per unit mass of the given gas, hence
From the relation
p₁·v₁ =p₂·v₂ therefore v₁ = p₂·v₂/p₁ = 4.6 MPa× 10 m³/(1 atm) = 4.6 MPa× 10 m³/(101325 Pa) = 454.115 m³
Therefore W₁₂ = 101325 Pa × 454.11 m³× ㏑((10 m³)/(454.115 m³)) = 46013250×(-3.82) = -175575813.855 J = -175.6 MJ
W = -175.6 MJ
(b) For isentropic compression we have
W = m×cv×(T₂ -T₁)
for air we put K = 1.4
therefore we have from which
v₁ = 152.65 m³
We also have or
from which we find the value of T₂ as
= 886.974 K (peak temperature)
Therefore from pv = RT and R =cp -cv = 1.005 -0.718 = 0.287 kJ/kg·K
Therefore number of moles = pv/(RT) = (4601325×10)(287×288.15) = 556.394 kg
m = 556.394 kg
Therefore work done at constant pressure = m·cp·(T₂-T₁) gives
556.394 kg × 1.005 kJ/kg⋅K×(298.15 K-886.974 K ) = -329256.19 kJ or -329.256 MJ
The peak temperature of the isentropic process = 886.974 K