When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is
a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).Answer: 3.05 x 10-8 kg/m2-s
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Answer:
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Explanation:
Given that;
College Engineering 10+5 pts
When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).
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Answer:
Explanation:
Given:-
- The diameter of the cylinder, d = 50 mm.
- The compressive load, F = 80 KN.
Solution:-
- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.
- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).
εx = - [ σx - ν( σy + σz ) ] / E
εy = - [ σy - ν( σx + σz ) ] / E
εz = - [ σz - ν( σy + σx ) ] / E
- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:
εx = [ ν*σz ] / E
εy = [ ν*σz ] / E
εz = - [ σz ] / E .... Eq 1
- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.
- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:
εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx
εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E
εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy
εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E
- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.
- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).
Note: σx' = σy', The cylinder is radially enclosed around the entire surface.
Therefore,
- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2
σx' ( 1 - v ) = [ ν*σz ] / 2
σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]
- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):
εz' = - [ σz - ν( σy' + σx' ) ] / E
εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E
εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2
- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:
... Answer