A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffus
ion condition. The diffusion coefficient for nitrogen in steel at this temperature is 5.9 × 10^(-11) m^2/s, and the diffusion flux is found to be 4.7 × 10^(-7) kg/m^2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.5 kg/m^3.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
1 answer:
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Answer:
The value of v2 in each case is:
A) V2=3v for only Vs1
B) V2=2v for only Vs2
C) V2=5v for both Vs1 and Vs2
Explanation:
In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.
Also, what the problem asks is the value V2 in each case, where:
If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.
In the first case we can use an equivalent resistance between R2 and R3:
And
In the second case we can use an equivalent resistance between R2 and (R1+R4):
And
If we consider both batteries:
The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)