1.
Ertl was awarded the Nobel prize for researching gas-solid interface interactions; more specifically, the
adsorption of gas into a solid and the subsequent chemical reactions, such as oxidation, that occur. It
was noted that the use of surface coatings can hinder unwanted, damaging corrosion from occurring
due to this adsorption. Previous nobel prizes had been given out for chemical processes involving
hydrogenation, and these works were built on by examining hydrogen near metal surfaces. Hydrogen
and nitrogen gas added to an iron metal surface will create ammonia, the iron acting as a catalyst. In
order to reach this conclusion, Ertl analyzed the surface composition of the metal using Auger Electron
Spectroscopy.
He found that Potassium on the iron surface also aided in the formation of Ammonia in vacuum
conditions. Also, in ambient conditions, the surface composition is quite complex. At high pressures,
the surface composition was close to that modeled from the low pressure measurements.
From his experiments, he was able to determine the reaction mechanisms between nitrogen and
hydrogen. With those mechanisms, he also measured the energies for the different reactions. His
energy diagram of N2 and H2 showed the progression to ammonia. It is shown that nitrogen “cleaving” is
the limiting energy in the steps, and that potassium reduces the necessary excitation. Potassium would
“donate” electrons to the surrounding iron elements to allow for better adsorption.
The surface chemistry was demonstrated to be an important factor when modeling interface
interactions. This becomes important when analyzing material properties near the surface since bulk
material properties are less applicable.
2.
a) Mean height= 2.96 µm
b) Ra=0.820, Rq=3.12
c)
Histogram
12
45.00%
40.00%
35.00%
30.00%
25.00%
20.00%
15.00%
10.00%
5.00%
0.00%
Frequency
10
8
6
4
2
0
Diff from Mean
Normal Dist
Diff from mean
The Gaussian (normal distribution) equation for this plot is
f x  
1
e
 2

x2
2 2
2
Where σ=0.981. For the other quantities asked for: mean of difference   0.0 , variance   0.962 ,
Skewness = -0.197, Kurtosis = 2.41 which is close to the normal distribution kurtosis value of 3. The
histogram is moderately conforming to the normal distribution, but lacking toward the 0 mean distance.
Using
, the ACF=-0.499 which shows anti-correlation.
d) The bearing area steadily decreases as h* increases.
1.2
1
tp
0.8
0.6
h*
0.4
0.2
0
0
1
2
3
4
5
6
h*
3.
At 20° C, water-air surface tension ϒ is 0.0728 J/M2. Using Dupre’s equation:
Wsl Platinum= 0.128 J/M2
Wsl Parrafin = 0.048 J/M2
It is seen that water adhesion to paraffin is very low compared to water adhesion with platinum.
4.
The equation for adhesive force is
At 20° C, water-air surface tension ϒ is 0.0728 J/M2, R=0.01, h=0.0001, angles =40°
This results in Fa = 0.350 N
5.
flat punch with sphere indenting:
See Comsol for FEA model of interaction.
Parameters were assumed from previous courses: P=1e3 N, E=2e11 Pa, v=0.3, R=0.001, and ϒ=0.5
(assume compatible metals)
This leads to the following calculated values:
E*=1.1e11, a=0.000190, P0,1=1.33e10, P0,2=1.92e7.
finally, these values were used to calculate the pressure distribution
the contact area, against a hertz pressure distribution
in
where
=1.33e10 (same as hertz-punch)
Plotting in excel results in:
1.4E+10
1.2E+10
1E+10
8E+09
hertz
6E+09
hertz-punch
4E+09
2E+09
0
0
0.00005
0.0001
0.00015
0.0002
This shows that results are slightly higher than hertz on its own, but still very close.
6.
From http://content.lib.utah.edu/utils/getfile/collection/etd3/id/1960/filename/1946.pdf and
http://deepblue.lib.umich.edu/bitstream/handle/2027.42/31515/0000437.pdf?sequence=1 ,
where E’= E/(2(1-v^2)), E is modulus of elasticity, v is
poisson’s ratio, H is Brinell hardness, σs is standard deviation of the asperity heights, and ρ is the average
summit radius R. For Al-7075 assuming grade t6, E=70 GPa, H=3*500 MPa yield=1500 MPa.
It is noted in the reference above that almost universally, plastic deformation occurs in solid-solid
contact.
Ψ=14.1 which is greater than 1 predicting plasticity.
For elasticity only, the standard deviation would need to be reduced to make psi =1 or less.
 H 
s  
  =7.07 nm.
 E' 
2
 H 
  =13.9 nm
 E' 
2
Using material properites from previous courses gives Ψ=10.1 and  s  