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:
Explanation:
General solution
in equation (iii)
Therefore,
Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number () for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
where is the density of the fluid,
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
Temperature at 28 mm distance after t time = = 50 degree C
we know that
from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
solving fot t we get
t = 2244.3 sec