A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a fi
1 answer:
Answer:
procedure attached below
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
Explanation:
<em>Given data:</em>
Minimum tensile strength = 865 MPa
Ductility = 10%EL
Desired Final diameter = 6.0 mm
20% cold worked 7.94 mm diameter 1040 steel stock
<u> Describe the procedure you would follow to obtain this material.</u>
assuming 1040 steel experiences cracking at 40%CW
<em>attached below is a detailed procedure of obtaining the material</em>
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
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Answer:
Qx = 9.10 m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 )²× 9
× sin18 × cos18
Qd = 94.305 × m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × × π × 85
× ( 9
)³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × - 85.2 ×
Qx = 9.10 m³/s