1. (0.5 point) Finish the problem we worked through in class for the
PbSO4 battery by determining the maximum current (in amps). Necessary data are:
Dcw = 9.45 x10-10 m2
/s Diffusion coefficient for Pb2+ in water
Daw = 10.7x10-10 m2
/s Diffusion coefficient for SO4
2- in water
ct = 3.3 mol/L
Area = 10 cm2
Distance between plates = 1 mm
Hint: You need to determine the number of charges within the volume (i.e. C/m3
What strategies could you use to increase Imax? For example, how would temperature alter D
and μ? What are the limitations to changing the concentration? How would Imax change if
instead of two plates, the battery was a cylinder?
2. (1.5 points) We can model a nerve fiber as a cylindrical conductor of radius ri, covered with
an insulating sheath of radius ro and conductivity σ0. The nerve bundle is assumed to have bulk
conductivity such that it is charged to a uniform potential ϕi when the brain sends a command.
The nerve bundle is immersed in body fluid that has an indefinite extent and flows past the
nerve bundle (i.e. this is a convective boundary problem!). We can define a convection
coefficient, Sk, that describes the leakage of current through the insulation and out into the body
Part i. Start with current balance in the radial direction, write the differential equation for steadystate leakage current through the insulating sheath, assuming a length L.
Part ii. Solve this differential equation with the following B.Cs: At r = ri, ϕ = ϕi. At r = r0,
𝑑𝑟 = 2𝜋𝐿𝑟0𝑆𝑘(𝜑 − 𝜑∞)
Hint, you may want to first solve the equation for ϕo at the interface, then use this solution to
solve in terms of ϕi and ϕ∞.
Part iii. Use the solution above to write an expression for the flux of current I and area A as a
function of r. Note, as before, the current*area should not depend on r. Identify the driving
force, and the resistance terms (one is associated with electrical conduction and one is due to
Part iv. The point of maximum current flow is called the critical insulation thickness. If r0 is less
than the critical thickness, the current flow increases with additional insulation. If r0 is greater
than the critical thickness, current flow decreases with increasing insulation. Mathematically,
this is found by taking the derivative of the current flow, I, with respect to the outer insulation
radius, r0 and setting the derivative to zero.
Using your expression for current in part iii, determine the critical insulation thickness in terms of
σ0 and Sk. Note, this same model could also apply to optical fibers with cladding that prevents
light leakage from the core, electrical insulation, such as in microprocessors, or pipes that carry
hot fluids. In our bodies, nerve bundles have a protein insulation in an amount much greater
than the critical thickness for proper muscle and sensory action. In autoimmune diseases, this
insulating sheath is attacked, altering the insulating thickness. Any information flowing through
the nerve bundle with then be distorted (tremors) or leaked away (paralysis).
3. (1.5 points) Composite media often require overall mass transfer coefficients. Consider the
composite system sown below, with a liquid-liquid contacting system and a well-defined
interface between the liquid phases. At the bottom of a container, a solute dissolves in liquid
#1. The dissolution maintains a constant mole fraction of solute at the solid surface. The solute
diffuses through liquid #1 until it reaches the interface. At the interface, the solute partitions
between the liquids. Then the solute diffuses through the second liquid phase. A steady stream
of air flows over the top of the device, maintaining the solute concentration at the top as a
constant. Your goal is to determine the mole fraction profiles in the liquids and the mass
transfer coefficients across the interface. You may assume a constant concentration in both
liquids, the same overall total concentration (ct) in both liquids, and dilute solutions (i.e. no
Part i. To determine mole fraction profiles,
we must first write the differential equations
describing the concentration profiles in
liquid #1 (xa) and liquid #2 (ya) at steady
Part ii. What are the general solutions to
the differential equations in part i?
Part iii. The constants for the general
solution can be found from the boundary
conditions. We have the following
At z = -L, xa = xao
At z = L, ya = yao
At the interface (defined as z = 0),
ya = mkxa in accordance with the
equilibrium law (i.e. at the interface,
xa = xai and ya = yai).
Finally, since the system is at steady state, whatever solute flows to the interface from
liquid#1 must flow into liquid #2, so the two fluxes must be equal. If we assume the total
concentrations in the two phases are equal, this boundary condition can be written as at
z = 0,
𝑑𝑧 = −𝑐𝑡𝐷𝑎𝑦
Using the boundary conditions, show that the mole fractions are equal to:
𝑥𝑎 = [𝐷𝑎𝑦(𝑦𝑎𝑜 − 𝑚𝑘𝑥𝑎𝑜)
𝐿(𝐷𝑎𝑥 + 𝑚𝑘𝐷𝑎𝑦)
] 𝑧 + 𝑦𝑎𝑜𝐷𝑎𝑦 + 𝑥𝑎𝑜𝐷𝑎𝑥
𝐷𝑎𝑥 + 𝑚𝑘𝑥𝑎𝑦
𝑦𝑎 = [𝐷𝑎𝑥(𝑦𝑎𝑜 − 𝑚𝑘𝑥𝑎𝑜)
𝐿(𝐷𝑎𝑥 + 𝑚𝑘𝐷𝑎𝑦)
] 𝑧 + 𝑚𝑘(𝑦𝑎𝑜𝐷𝑎𝑦 + 𝑥𝑎𝑜𝐷𝑎𝑥)
𝐷𝑎𝑥 + 𝑚𝑘𝑥𝑎𝑦
And that the fluxes (Nax and Nay) are:
𝑁𝑎𝑥 = 𝑁𝑎𝑦 =
𝐿(𝐷𝑎𝑥 + 𝑚𝑘𝐷𝑎𝑦)
(𝑦𝑎𝑜 − 𝑚𝑘𝑥𝑎𝑜)
Part iv. We can now consider the different mass transfer coefficient definitions. When
multiplied by the appropriate driving force, each coefficient should yield the mass flux through
the system. The first two coefficients we consider are the local coefficients based on a
viewpoint located within one of the two phases. We define these coefficients as:
𝑁𝑎𝑥 = 𝑐𝑡𝑘𝑐𝑥(𝑥𝑎𝑜 − 𝑥𝑎𝑖) with xai = xa (z=0)
𝑁𝑎𝑦 = 𝑐𝑡𝑘𝑐𝑦(𝑦𝑎𝑜 − 𝑦𝑎𝑖) with yai = ya (z=0)
Using the mole fraction profiles and the fluxes, solve for kcx an kcy. Hint: Only one resistance is
involved because we know the composition at the interface exactly.
Part v. If we did not know the composition of the interface, we need to use an overall mass
transfer coefficient, defined as:
𝑁𝑎𝑥 = 𝑐𝑡𝐾𝑐𝑥 (𝑥𝑎𝑜 −
𝑁𝑎𝑦 = 𝑐𝑡𝐾𝑐𝑦(𝑚𝑘𝑥𝑎𝑜 − 𝑦𝑎𝑜)
Using the flux, solve for Kcx and Kcy. In this case, both the coefficients are sum of resistances in
the system, containing diffusion resistances for both phases and a resistance to getting material
across the interface.
4. (1.5 points) No two materials can be in perfect contact because at an atomic level, many
materials have a roughness factor. Below are an atomic force microscopy (AFM) and scanning
electron microscopy (SEM) of a polymer film and indium tin oxide, respectively.
Image reference: Ratcliff, E.L., Jenkins, J.L. Nebesny, K., Armstrong, N.R. “Electrodeposited, “textured” poly(3-
hexyl-thiophene) (e-P3HT) films for photovoltaic applications,” Chemistry of Materials, 20, 18, 2008, 5796-5806.
Bringing two atomically rough surfaces into contact leaves pockets of air trapped in surface
interstices that can manifest themselves as an interfacial resistance. We can use our simple
one dimensional boundary models to represent this phenomenon.
At steady state, in each material, we know that the temperature profile will be governed by:
𝑑𝑦2 = 0
𝑑𝑦2 = 0
You may assume that at y = 0, T1 = T0 and
at y = 2L, T2 = TL. Also, since heat must flow from one material to
the other, the interface boundary condition is: at y = L,
𝑑𝑦 = −𝑘2
Finally, the interface resistance can be modeled by assuming
there is a temperature jump across the interface (i.e. at y = L, T1 ≠
T2). Any function can be used – here we will assume a linear
function, such that at y = L,
𝑇1 = 𝛼𝑇2 + 𝛽
Determine the mathematical expressions for T1, T2, and the flux q,
assuming an area of A and T0 > TL. Does α increase or decrease
the driving force? What about the resistance?
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