Matthew Schrandt
MES 601 Final Test
Physical Chemistry Test
1. The contact angle between hexadecane and n-hexatriacontane (C36H78, solid) is 46 degrees. LV for
hexadecane is 27.6 mJ/m2. What is the work of adhesion between hexadecane and n-hexatriacontane?
(hexadecane-hexatriacontane) = 46°
(hexadecane-air) = 27.6 mJ/m2
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Assuming the sl between hexadecane and n-hexatriacontane is equivalent to that of decane (lv = 23.9
mJ/m2) and n-hexatriacontane, what is the decane - n-hexatriacontane contact angle?
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Do you think this is a reasonable assumption? Why? Would it be a reasonable assumption if water
were the liquid rather than decane?
Yes, this is a reasonable assumption, the surface tensions (γ) between the liquids is close and the
angles used are fairly close. This means that decane and hexadecane are fairly close. However,
water would be a bad substitute since the surface tension
which is a lot higher. This
makes sense as well since water has a completely different chemical structure and properties
whereas hexadecane and decane have similar properties due to their chemically similar structures.
2. From the data below, compare the solid surface energy components for automobile parts painted with
an acrylic paint and a urethane-based paint using two different furnace types. Which paint and furnace
will give the best adhesion between the paint and metal surface.
Contact Angle
gas
urethane
acrylic
electric
urethane
acrylic
Work
gas
Bake time
0.75
2
4
0.75
2
4
0.75
2
4
0.75
2
4
CH2I2
48.3
50.5
53.8
49.3
50.3
53.5
39.7
40.1
41.1
19.7
30.8
44.3
H2O
88.1
97.4
98.1
92.7
94.5
96.9
81.6
83.5
90.3
82
84.1
87
C2H6O2
74.2
75.5
78.6
73.6
75.9
78.9
69.2
70.5
72.6
69.9
70.5
71.4
γ
50.8
72.8
48
urethane
acrylic
electric
urethane
acrylic
CH2I2
84.5937
83.11277
80.80277
83.9266
83.24941
81.017
89.8855
89.65801
89.08102
98.6267
94.43516
87.15719
H2O
75.2137
63.42368
62.54239
69.37065
67.08818
64.05404
83.43484
81.04119
72.41882
82.9318
80.2833
76.61006
C2H6O2
61.06945
60.01824
57.48755
61.55239
59.69352
57.24105
65.04513
64.02273
62.35396
64.49567
64.02273
63.31005
Maximum work is best paint. The maximum work occurs for the acrylic paint in the electric oven for
CH2I2 (w=98.6267 mJ/m2). From the trends observed in the chart above, the values for the electric ovens
vs those for the gas oven show that the electric oven in general has greater work required to remove
adhesion. The urethane paints seem to be overall more adhesive for shorter bake times (with the
exception of the acrylic paint in the electric oven in CH2I2) but the acrylic paints seem to be more
adhesive over a longer bake time.
Overall, the acrylic paint and the electric oven seem to be the best.
3. A steel part was painted many years ago. How much of the surface is now covered with paint if the
measured water contact angle of the part is 93.4 degrees? The water contact angle of the steel prior to
initial painting was 43.8 degrees. After initial painting the contact angle was found to be 114 degrees.
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What would the paint coverage be if the surface roughness of the painted surface (after the years of
use) 1.15?
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Transport Phenomena Test
1. Doctors are studying the different flow rates throughout the human vasculature system. After
years of eating fries and hamburgers, four out of ten Americans are obese, and most Americans
have a decreased vessel diameter due to fat buildup, increasing the risk of heart attack and
stroke.
(a) Use Buckingham Pi theory to develop the dimensionless numbers for this system.
Representative variables:
ρ, P, L, V, μ
P
V
μ
L
ρ
Length
Kg/m·s2
m/s
Kg/m·s
M
Kg/m3
Mass
Kg/m·s2
Kg/m·s
Time
Kg/m·s2
m/s
Kg/m·s
Kg/m3
Kg/m3
5 q’s
3 u’s
2 non-primary q’s (π’s)
Choose the π corresponding to:
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Kg : 1 – a = 0
a=1
m : -1 + a – b – c = 0
b=1
s : -1 + 2a + c = 0
c = -1
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(b) Assume that the growth of fatty deposits on the inside of blood vessels will be limited by the
shear stress imposed on the deposit by the flow of blood through the vessel; that is, assume that
the growth of fatty deposits on the inside wall of a blood vessel decreases with increasing shear
stress τrz. As the deposits grow and the tube narrows, will the growth of fatty deposits slow down
over time, or accelerate? Answer this question assuming: (i) constant pressure and (ii) constant
flow rate.
(i) constant pressure:
decelerates due to increased shear stress (τ) with increased speed.
Despite keeping higher Cmax, the shear stress causes less growth.
(ii) constant flow rate: decelerates since
. Since Vx is constant, as dx shrinks
(diameter shrinks), the shear stress increases and thus the growth slows.
(c) The diameter of the carotid artery, the blood vessel that supplies blood to the brain, is 7 mm, and
the average velocity and density of blood is 0.4 m/s and 1060 kg/m3 respectively. Across the
blood vessel length of 1m and with a blood viscosity of 0.0027 Pa·s, calculate: (i) the Reynolds
number of blood through this healthy vessel; (ii) the friction factor; (iii) the pressure difference
across the vessel.
d = 0.007 m
V = 0.4 m/s
ρ = 1060 kg/m3
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L=1m
µ = 0.0027 Pa·s
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(d) In one such person, fat has built up on the interior vessel wall over a period of years, restricting
blood flow and endangering the person’s life. If there is a 0.75 mm layer of fat on the carotid arterial
wall resulting in a decreased vessel radius Rf, what is the new average velocity which gives the
same flow rate Q as in part c?
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2. It has been proposed that some MEMS (Micro Electro-Mechanical Systems) device parts for high
temperature applications could be made out of magnesium oxide by water etching, due to the slight
solubility of magnesium oxide in water. The etching reaction is:
MgO + H2O → Mg2+ + 2OH−
A magnesium oxide plate 5 cm (0.05 m) long partially covered by a polymer etching mask is placed in a
water tank near an impeller which generates a free stream water velocity of 1 m/s past the plate.
Data:
• Mg2+ diffusivity in water: D = 10−5 cm2/s (= 10−9 m2/s ).
• Mg2+ solubility in this particular solution: Cs = 10−2 mol/cm3.
(a) Sketch the shapes of the momentum and diffusion boundary layers above the plate on one sketch,
clearly indicating which of the two boundary layers will be larger.
Mg2+
H2O
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𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
𝛿
𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝛿𝑇
𝑥
)
)
)
𝑉
𝑥
𝐿
(b) If the magnesium oxide dissolution is limited by mass transfer of Mg2+ ions away from the interface,
will the dissolution rate be uniform across the plate? If not, indicate on your sketch from part 2a
where on the plate you expect to see the highest and lowest dissolution rates.
No, the dissolution rate will not be linear across the plate. The rate will be highest at x=0 and
slowest at x=L as shown in the graph below.
𝑥
𝐿
𝑥
(c) Calculate the local Sherwood number Shx at the back end of the plate (x = L).
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(d) If the plate were at a different temperature from the fluid, how would the energy transfer boundary
layer compare to the momentum and mas transfer boundary layers.
The thermal energy boundary barrier is always smaller than both the diffusion and the momentum
boundary layers for this system.
3.
Two immiscible liquids are flowing in laminar flow between two parallel plates. Is the velocity
distribution shown below ever possible? Please explain your answer, assuming that the velocity
profile given in the figure is parabolic for both Liquid A and B.
Liquid A
Liquid B
The proposed velocity distribution is not possible. To have the different maximum velocity profiles
shown, the faster liquid (B) must have a smaller viscosity than the other liquid (A). On top of that, at the
interface of the two liquids, there must be no difference in the shear stress. By which, it can be seen that
there should be no difference in the velocity of the two liquids at the interface (as displayed in the
diagram). However, since there is no τ change at the interface, the relative τ for liquid A should dictate
that liquid A reaches the speed at the boundary equal to the speed of liquid B at the boundary. So liquid
A then feels the boundary of liquid B as a faster liquid and thus does not slow to the boundary speed,
but instead stays sped up. Liquid B, while meeting the boundary speed, must decrease its maximum
speed to match the drag felt by liquid A. Thus, the velocity distribution should instead follow the profile
drawn in to the left.