10.1 Hydrogen has been considered as the fuel of the future. One way to produce hydrogen is by elec- trolysis of water with electricity generated by solar cells or wind energy. To be used in a variety of applications, this hydrogen must be purified by removing impurity gases such as O2, N2, and H2O. One way of doing this is by passing hydrogen gas through a polymer diffusion membrane in which hydrogen is soluble and diffuses relatively rapidly through the polymer while the other gases do not. This problem addresses the transport of hydrogen through polybutadiene tubes to purify it. H2 diffuses as the hydrogen molecule via interstitial diffusion in polymers.
a. For gases in polymers, the solubility or concentration is a function of pressure of the form, C = Sp, where S is the solubility constant, units [S] = mol/cm3 atm and is typically around 2.0 × 10−6 mol/cm3 atm at 300 K. If the p(H2) on the inside of the tube is 2 atm and p(H2) = 1 atm at the outside of the tube, calculate the concentrations of hydrogen in the inner and outer walls of the polybutadiene tube.
b. Calculate D(H2) at 27°C when D0 = 0.053 cm2/s and Q = 21.3 kJ/mol.
c. Hydrogen is diffusing through this tube that has an outer diameter of 5 mm and an inner diameter of 3 mm. The cylindrical symmetry cannot be ignored in this case because of the large wall thickness to tube radius ratio. Calculate and plot the concentration, C(r), in mol/cm3 as a function of radius for the steady-state concentration profile through the wall thickness in terms of the inner tube radius, R1, and the outer tube radius, R2.
d. From the result in c., calculate dC/dr in mol/cm4 at both the inside and outside surface of the tube.
e. Calculate the H2 flux density, mol/cm2 s at both the inside and outside surface of the tube.
f. For this cylindrical diffusion, develop an expression the hydrogen flux per unit length (mol/s cm)of the polybutadiene tube.
g. Calculate how many feet of polymer tube would be needed operating at 27°C and 2 atm H2 internal pressure to produce 100 ft3/h of H2 on the outside of the tube.
h. From the result in g., would you suggest any modifications to the tubing? If “No” why not?
If “Yes,” what might be modified and give some justification for your suggestions?
10.2 An alternative method of generating high purity hydrogen is by passing hydrogen gas through a palladium, Pd, diffusion membrane in which hydrogen is soluble and diffuses rapidly while the other gases do not. This problem addresses the transport of hydrogen through Pd tubes to purify it.
a. Hydrogen is soluble in Pd up to 10 a/o (atomic % as H) at 500°C at a hydrogen pressure of 100 atm (Brandes and Brook 1992). If the density of Pd is ρ(Pd) = 12.02 g/cm3 and its
Exercises 375 molecular weight is M(Pd) = 106.4 g/mol (Haynes 2013), calculate what 10 a/o is in moles
of H per cm3. The hydrogen forms an interstitial solid solution in Pd so that the number of Pd atoms per cm3 essentially does not change forming the solid solution.
b. The solubility of hydrogen in Pd follows Sievert’s law: that is, H2 (g) = 2H (solid solution) or C(H) = Sp1/2. Calculate S so that units [C] are mol/cm3 when units [p] = atm. Calculate C(H) in mol/cm3 at 500°C and p = 5 atm.
c. Calculate D(H) at 500°C when D0 = 2.9 × 10−3 cm2/s and Q = 22,200 J/mol.
d. Hydrogen is diffusing through a Pd tube with an outer diameter of 30 mils and a wall thickness of 5 mils. Assume that the cylindrical symmetry cannot be ignored again in this case. Calculate and plot the concentration, C(r), in mol/cm3 as a function of radius for the steady-state concentration profile through the wall thickness in terms of the inner tube radius, R1, and the outer tube radius, R2, where it can be assumed that pR2(H2)=1atm.
e. From the result in c., calculate dC/dr in mol/cm4 at both the inside and outside surface of the tube.
f. Calculate the H2 flux density, mol/cm2 s at both the inside and outside surface of the tube.
g. For this cylindrical diffusion, develop an expression the hydrogen (Note: H2) flux per unit length (mol/s cm)of the Pd tube.
h. Calculate how many feet of Pd tube would be needed operating at 500°C and 5 atm inter- nal H2 pressure to produce 100 ft3/h of H2 at STP on the outside of the tube.
i. From the result in h., would you suggest any modifications to the tubing? If “No” why not? If “Yes,” what might be modified and give some justification for your suggestions?
10.3 The equilibrium constants for the reaction: KCl (s) = K+ (soln) + Cl− (soln) are 19.76 at 100°C and 8.241 at 30°C where the standard state for the solution species is 1 molal/ (Roine 2002).
a. Calculate the concentration, mol/cm3, of K+ and Cl− in equilibrium with pure KCl at 100°C.
b. Do the same at 30°C.
c. If the density of solid KCl is 1.984 g/cm3 and its molecular weight M = 74.56 g/mol (Haynes 2013), calculate the concentration, mol/cm3, of KCl in solid KCl.
d. A Spherical particle of initial radius, a0, of KCl is dissolving into a large amount of pure water at 30°C. Develop an equation for the steady-state dissolution of a spherical particle of KCl of initial diameter of 2 mm; that is, develop an expression for the radius a as a function of time.
e. Plot the steady-state concentration, mol/cm3, of KCl in both the solid and liquid as a function of distance, r/a, from the center, of the dissolving sphere out to r/a = 10.
f. Calculate how long it takes to dissolve the 2 mm sphere of KCl if the diffusion coefficient for KCl in H2O is D = 2.1 × 10−5 cm2/s at 30°C.
g. Now suppose that a large amount of solution of KCl in equilibrium with pure KCl at 100°C were suddenly cooled to 30°C and that KCl solid particles now grow. Plot the steady-state concentration, mol/cm3, of KCl in both the solid and liquid as a function of distance, r/a, from the center of the now growing sphere out to r/a = 10.
10.4 a. Salt is often added to a glass of beer to increase the head by forming CO2 bubbles by het- erogeneous nucleation. As these carbon dioxide bubbles rise to the top of the glass, they grow by diffusion of dissolved CO2 into the bubble. The solubility at room temperature of CO2 gas is 7.9 × 10−5 mol/cm3 (Haynes 2013) and because of the pressure in the bottle, the actual amount of gas in solution is five times the solubility (i.e., the solution is supersatu- rated). If the diffusion coefficient of CO2 in the liquid is 1 × 10−4 cm2/s, calculate and plot the CO2 bubble size as a function of time up to 1 min. Assume that the bubble is growing by diffusion and that your glass is tall enough (a yard) so that the bubble takes that long to get to the top of the glass. Ignore any effects of the motion of the bubble or the motion of the growing bubble–liquid interface and assume steady-state conditions.
b. Plot the CO2 concentration versus r/a in mol/cm3 from the center of the bubble out to 10 times the bubble radius.
10.5 Manganese is being deposited by chemical vapor deposition by the following reaction:
MnCl g2
( )
+ H g2( )
Mn s( )
+ 2HCl g( )
For the surface reaction, the reaction rate constant, k, is given by k = ko exp(–Q/RT) with ko= 1010 cm/s and Q = 150,000 J/mol. For the same reaction controlled by diffusion through the gas phase, the following data are obtained: D = Do T3/2/p with Do = 0.0001 cm2 atm/s K3/2 and a gas boundary layer thickness of δ = 1 cm.
a. On the same graph plot log10hR, log10hD, and log10h versus 104/T from 200 to 1500°C, where hR = k is the mass transfer coefficient for the surface reaction, hD = D/δ is the mass transfer coefficient for diffusion through the gas phase, and h is the overall mass transfer coefficient and:
1 1 1
h hR hD
= +
b. From the following thermodynamic data, calculate the equilibrium constant for the reac- tion at 540°C given the following data (Units (T) = K):
MnCl2 (g): ΔGo = −295,400 – 263.6 T J/mol HCl (g) ΔGo = −92,300 – 186.8 T J/mol
c. If the initial MnCl2 pressure is 10−2 atm and the hydrogen pressure stays constant at essen- tially p(H2) = 1 atm, calculate the equilibrium pressures of MnCl2 and HCl at 540°C.
d. Calculate h (cm/s) at 540°C.
e. The density and molecular weight of manganese are 7.20 g/cm3 and 54.938 g/mol, respec- tively (Haynes 2013). Calculate the deposition rate (mils/h) of Mn at 540°C.
f. Calculate the MnCl2 pressure at the surface of the depositing Mn at 540°C.
10.6 A Hollow amorphous silica (SiO2 glass) laser fusion sphere 100 μm in diameter is filled with hydrogen gas at a total initial pressure of 1 atm. The diffusion coefficient of hydrogen in SiO2 is given by D = D0 exp(−Q/RT), where D0=5 65 10. × −4cm s2/ and Q = 43,027 J/mol. The solubility of hydrogen in the silica is given by C(H2) = Sp(H2), where Units (C) = mol/cm3, Units (p) = atm, and S = 4.1×10−7 mol/cm3 atm.
wt. % Al2O3
0 10 20 30 40 50 60 70 80 90 100
Temperature (°C)
600 800 1000 1200 1400 1600 1800 2000 2200
Liquid
Liquid + Al2O3
Na3AlF6+ Al2O3 Na3AlF6+ liquid
Na3AlF6 Al2O3
FIGURE E.1 The Al2O3–Na3AlF6 phase diagram. (After Levin et al. 1964.)
Exercises 377
a. From Fick’s second law, develop the steady-state model for the concentration as a func- tion of distance, C(x), through the sphere wall assuming that the wall thickness is small relative to the sphere radius: the linear solution. Let Ci = the concentration at the inside of the sphere wall, x = 0, and p(H2) = 0 at the outer wall, x = L.
b. Develop an expression for the H2 pressure in the sphere as a function of time, S, D, T, L, and sphere diameter, d. Assume only steady state and, because the wall thickness is much less than the sphere radius, ignore spherical coordinates. Take into consideration that Ci will decrease as the pressure drops that leads to an exponential decrease in p(H2) with time.
c. Calculate the value of D (cm2/s) at −100°C.
d. Calculate how long it takes (days) for the pressure to drop 10% at −100°C if the sphere has a wall thickness of 10 μm and the hydrogen content of the ambient external atmosphere can be considered to be zero.
10.7 Aluminum metal is made by electrolyzing a cryolite, Na3AlF6, melt containing Al2O3 dissolved in liquid solution. A schematic of the Na3AlF6–Al2O3 phase diagram is given in Figure E.1. The overall cryolite melt has 10 w/o Al2O3 dissolved. Two hundred micron diameter spherical par- ticles of Al2O3 are dumped into the melt for dissolution at 1300°C and the melt is sufficiently large so that the particles do not interact nor do they change the overall concentration of the melt.
a. The density and molecular weight of molten cryolite are 2.084 g/cm3 and 209.94 g/
mol and that of molten Al2O3 are 2.89 g/cm3 and 101.96 g/mol, respectively (Haynes 2013). Molten Na3AlF6 and molten Al2O3 form an ideal liquid solution. Calculate the Al2O3 concentration in mol/cm3 in the 10 w/o melt far from the dissolving Al2O3 and calculate the equilibrium Al2O3 concentration (mol/cm3) in the melt at the solid Al2O3– melt interface.
b. Plot the Al2O3 concentration (mol/cm3) from the center of the Al2O3 particles to three times the particle radius into the melt. The density of solid Al2O3 is 4.0 g/cm3.
c. Calculate how long it takes the Al2O3 particles to dissolve in the cryolite.
10.8 Aluminum alloys containing small amounts of copper (5 w/o or less) are very important alloys as they can be heat treated to precipitate CuAl2 as a second phase, which greatly increases the yield and tensile strengths of the alloy. The relevant part of the aluminum copper phase diagram is given in Figure E.2. These are called “age-hardening” alloys, whose development at the turn of the previous century, enabled the construction of large airships
600 500 400 300 200 100
0 1 2 3 4 5 6 7
Alpha
Alpha + CuAl2
Temperature (°C)
w/o Copper Al
FIGURE E.2 Part of the aluminum-copper phase diagram. (After Brandes and Brook 1992.)
such as the Hindenburg and, today, modern aircraft. To obtain the maximum properties, the particles must be kept small. If they become too large, the strength is not maximized, this is called “over aging.” A 4 w/o Cu alloy is “solution” heat treated to single phase alpha at 550°C, quenched, and cooled to 250°C and precipitation heat treated (aging). Assume that the precipitate particles are spherical. The density and molecular weight of aluminum are 2.70 g/cm3 and 26.98 g/mol, respectively, and assume that the density of the Al solid solution does not change with the small copper concentration. The density and molecular weight of CuAl2 are 4.73 g/cm3 and 117.51 g/mol and that of copper are 8.92 g/cm3 and 63.55 g/mol, respectively (Haynes 2013).
a. Calculate the copper concentrations (mol/cm3) in the original 4 w/o alloy, the alloy in equilibrium with CuAl2 at 250°C, and CuAl2.
b. Plot the concentration of Cu (mol/cm3) as a function of distance from the center of the growing spherical precipitates out to three times the radius into the alloy.
c. The diffusion coefficient of Cu in Al is given by D = Do exp(−Q/RT), where Do = 0.137 cm2/s and Q = 123.5 kJ/mol (Brandes and Brook 1992). Calculate D at 250°C.
d. What is the maximum time of heat treatment (hours) at 250°C so the precipitate par- ticles are not larger than 0.2 μm in diameter.
e. Make a plot of (C0 – Ce) × D horizontally (which is proportional to the rate of precipitate particle growth) versus temperature vertically from Te (the temperature at which the precipitate phase becomes stable) down to 250°C and determine the temperature at which the growth rate is a maximum.
10.9 a. With the iron–carbon phase diagram, Figure 7.4, calculate the time it takes for 10 μm diameter spherical cementite particles to dissolve in austenite at 1000°C assuming steady-state diffusion of the carbon in the austenite. Assume that the austenite com- position is constant at 0.77 weight% carbon; the density and molecular weights of the cementite, austenite, and carbon are ρ(Fe3C) = 7.694 g/cm3, M(Fe3C) = 179.55 g/mol, ρ(austenite) = 7.86 g/cm3, M(austenite) = 55.85 g/mol, ρ(C) = 2.2 g/cm3, and M(C) = 12.01 g/mol; and Do = 0.20 cm2/s and Q = 142.3 kJ/mol for carbon diffusion in austenite (Brandes and Brook 1992).
b. Plot the carbon concentration (g atoms/cm3) as a function of distance from the center of a spherical particle to a distance five times the particle radius from the particle’s center at the start of the dissolution process.
10.10 The oxidation rate of silicon can be given by x2=Bt, where x is the oxide thickness and B is the parabolic rate constant and has the following experimentally determined values in 1 atm of dry oxygen (Mayer and Lau 1990):
Temperature (°C) B (μ2/h)
1200 0.0450
1100 0.0270
1000 0.0117
920 0.0049
800 0.0011
a. Plot the SiO2 thickness (nm) as a function of time (0 to 2 h) for each of these tempera- tures on the same plot.
b. Make a plot of log10B versus 1/T (K−1) and determine the apparent activation energy for the oxidation process.
c. The diffusion coefficient for O2 in SiO2 is given by D D exp= 0 −Q/RT with D0 = 2.7 × 10−4 cm2/s and Q = 1.16 eV (Mayer and Lau 1990). How does the calculated activation energy from oxidation data compare with that for diffusion oxygen given in this reference?
d. Assuming that the diffusion coefficient for oxygen is that given in the literature in part c above, calculate the solubility coefficient, S, in C(O2) = Sp(O2), where Units (C) = mol/cm3 and Units (p) = atm at 1000°C.