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CHE 499 (Spring 04)
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Problem set #6
1. (p. 3.161) During hibernation of warm-blooded animals (homoisotherms), the heart beat and
the body temperature are lowered and in some animals the body waste is recycled to
reduce energy consumption. Up to 40% of the total weight may be lost during the
hibernation period. The nesting chamber of the hibernating animals is at some distance
from the ground surface, as shown in Figure 1(i). The heat transfer from the body is
reduced by the reduction in the body temperature T1 and by the insulating effects of the
body fur and the surrounding air (assumed stagnant). A simple thermal model for the
steady-state, spherical, one-dimensional heat transfer is given in Figure 1(ii). The thermal
resistances of air and soil can be determined from
Ra
2 L , Rk,soil = L  Ra
Rk,air =
4k s Ra L
4k a Ra
1
An average temperature T2 is used for the ground surrounding the nest. The air gap size
Ra  Rf is an average taken around the animal body.
Determine the heat loss from the body for (i) L = 2.5Ra, and (ii) L = 10Ra.
R1 = 10 cm, Rf = 11 cm, Ra = 11.5 cm, T1 = 20oC, T2 = 0oC.
For air ka = 0.0267 W/mK, for fur kf = 0.036 W/mK, for soil ks = 0.52 W/mK.
Figure 1. Conduction heat transfer from a warm-blooded animal during hibernation.
2. (p. 3.562) A fermentation broth consists of an aqueous solution of nutrients and cells. As the
cells grow, they cluster into spherical pellets of radius R(t). On average, the cell density
inside a pellet is 0.02 mg of cell mass per cubic millimeter of pellet volume. The dissolved
oxygen concentration in the broth is 5 g/cm3. The cells utilize oxygen at a rate of 1.2 mmol
of oxygen per hour per gram of cell mass, via a zero-order reaction. Assume that the diffusion
coefficient DAB of oxygen within the pellet is 1.810-5 cm2/s. How large can R(t) become
before the oxygen concentration becomes zero at the center of a pellet? Assume that the broth
external to the pellets is well mixed. The cells and broth have densities close to that of water.
1
2
Kaviany, Principles of Heat Transfer, Wiley, 2002, p. 350
Middleman, An Introduction to Mass and Heat Transfer, Wiley, 1998, p.107
3. Repeat problem (2), but allow for the presence of a finite convective resistance to mass
transfer to the surface of the pellet, such that the flux to the surface is given by
molar flux = kc(CP  CB)
where CB is the broth oxygen concentration and CP is the (unknown) concentration of oxygen
at the pellet surface. Calculate the critical value of R(t) (at which the oxygen concentration
becomes zero at the center of a pellet) for two cases: 2kcR/DAB = 2, and 2kcR/DAB = 20.
4. (p. 26.163) The “drug patch” shown in the figure below releases a water-soluble epidermal
growth factor (species A) to repair a specific region of wounded tissue on the human body. A
slow release of the drug is critical for regulating the rate of tissue repair. The drug layer (pure
solute A) rests on top of a diffusion barrier. The diffusion barrier is essentially a microporous polymer material consisting of tiny parallel pores filled with liquid water (species B).
The diffusion barrier controls the rate of drug release. The thickness (L), pore size (dpore), and
porosity of the diffusion layer determine the dosage rate of the drug to the tissue directly
beneath it. The maximum solubility of the drug in water is 1 mole/m3 at 25oC. The pore
diameter (dpore) is 110-8 m, and the equivalent molecular diameter of the drug is 0.2510-8
m. The total surface area of the patch is 4 cm2, but the cross-sectional area of the pores
constitute only 25% of the surface area for flux. The diffusion coefficient of the drug in water
at infinite dilution is 110-10 m2/s at 25oC.
Drug patch
Drug reservoir
Impermeable
barrier
Diffusion barrier
(water-filled micropores)
L
Skin surface
Infected body tissue
(sink for drug)
(a) Determine the effective diffusion coefficient of the drug in the diffusion barrier using the
Renkin equation.
Dm/D = (1 

a/r)2 1 

3
a
a
a
2.1   2.09   0.95 
r
r
r
5



(b) Estimate the thickness of the diffusion barrier (L) necessary to achieve a maximum possible
dosage rate of 0.05 mole per day, assuming that the drug is instantaneously consumed once
it exits the diffusion barrier and enter the body tissue.
3
Welty, J. R., Fundamentals of Momentum, Heat, and Mass Transfer, Wiley, 2001, p. 548.
(c) The above calculations assume a temperature of 25oC. However, the human body is actually
at 37oC, and it is likely that the diffusion barrier equilibrates to that temperature. If all other
process parameters are constant, what is the percentage change in dosage rate if the
temperature is increased to 37oC? You may assume that maximum solubility of the drug in
water is unchanged at 1 mole/m3 in the range 25oC to 37oC.
2-9, 2-11, 2-12, 2-13 (Text)