A cylindrical specimen of a metal alloy 45.8 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 378 MPa ca
uses the specimen to plastically elongate to a length of 54.2 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 45.8 mm to a length of 55.8 mm
1 answer:
You might be interested in
Answer:
The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is
Explanation:
Given
x1 = 0 mm
x2 = 6 mm = 6 * m
c1 = 2 kg/
c2 = 0.4 kg/
T = 600 °C
Area = 0.25
D = 1.7 *
First equation
J = - D
Second equation
J =
To find the J (flux) use the First equation
J = - 1.7 * *
To find M use the Second equation
=
M =
Answer:
70.66 cm^3
The specific volume for P = 70.66 is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )
Explanation:
Viral equation : Z = 1 + Bp + Cp^2 + Dp^3 + -----
Viral equation can also be rewritten as :
Z = 1 + B ( P/RT )
B ( function of time )
Temperature = 310 K
P1 = 8 bar
P2 = 75 bar
<u>Determine the specific volume in cm^3 </u>
V = 70.66 cm^3
<u>b) comparing the specific volumes to the experimental values </u>
70.58 and 3.90
The specific volume for P = 70.66 is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )
attached below is the detailed solution
Answer:
v=82 m/s
s=116m
Explanation:
a=20t
using condition given at t=0
c=-8
now equation becomes
v=10t²-8
v at t= 3s v=82 m/s
again
now using condition given s=50 at t=0
b=50
now equation becomes
calculating s at t=3s
s=116m
Answer:
exit temperature 285 K
Explanation:
given data
temperature T1 = 270 K
velocity = 180 m/s
exit velocity = 48.4 m/s
solution
we know here diffuser is insulated so here heat energy is negleted
so we write here energy balance equation that is
0 = m (h1-h2) + m × .....................1
so it will be
.....................2
put here value by using ideal gas table
and here for temperature 270K
h1 = 270.11 kJ/kg
solve it we get
h2 = 285.14 kJ/kg
so by the ideal gas table we get
T2 = 285 K