/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A person applies an insect repel... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 56

A person applies an insect repellent onto an exposed area of \(A=0.5 \mathrm{~m}^{2}\) of their body. The mass of spray used is \(M=10\) grams, and the spray contains \(25 \%\) (by mass) active ingredient. The inactive ingredient quickly evaporates from the skin surface. (a) If the spray is applied uniformly and the density of the dried active ingredient is \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the initial thickness of the film of active ingredient on the skin surface. The temperature, molecular weight, and saturation pressure of the active ingredient are \(32^{\circ} \mathrm{C}, 152 \mathrm{~kg} / \mathrm{kmol}\), and \(1.2 \times 10^{-5}\) bars, respectively. (b) If the convection mass transfer coefficient associated with sublimation of the active ingredient to the air is \(\bar{h}_{m}=5 \times 10^{-3} \mathrm{~m} / \mathrm{s}\), the partition coefficient associated with the ingredient-skin interface is \(K=0.05\), and the mass diffusivity of the active ingredient in the skin is \(D_{\mathrm{AB}}=1 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\), determine how long the insect repellent remains effective. The partition coefficient is the ratio of the ingredient density in the skin to the ingredient density outside the skin. (c) If the spray is reformulated so that the partition coefficient becomes very small, how long does the insect repellent remain effective?

Short Answer

Expert verified
(a) The initial thickness of the active ingredient film on the skin surface is \(2.5 \times 10^{-6} \mathrm{~m}\). (b) The insect repellent remains effective for \(5.0 \times 10^{4} \mathrm{~s}\). (c) If the spray is reformulated so that the partition coefficient becomes very small, the insect repellent remains effective for a very long time, approaching infinity.

Step by step solution

01

(a) Calculate the mass of the active ingredient

The spray contains \(25 \%\) active ingredient. So, the mass of the active ingredient in the spray is given by: \(m_{\text{active}} = M \times \frac{25}{100} = 10 \times 0.25 = 2.5 \text{ grams}\) Now, we convert the mass of the active ingredient in kg: \(m_{\text{active}} = 2.5 \times 10^{-3} \mathrm{~kg}\)
02

(a) Calculating the initial thickness of the active ingredient film

We are given the density \(\rho\) and area (A) where the spray is applied uniformly. So, we can calculate the volume of the active ingredient: \(V_{\text{active}} = \frac{m_{\text{active}}}{\rho} = \frac{2.5 \times 10^{-3}}{2000} = 1.25 \times 10^{-6} \mathrm{~m}^{3}\) Now, we can find the initial thickness of the active ingredient film, \(h\), using the Area \(A\) and volume \(V_{\text{active}}\): \(h = \frac{V_{\text{active}}}{A} = \frac{1.25 \times 10^{-6}}{0.5} = 2.5 \times 10^{-6} \mathrm{~m}\)
03

(b) Calculate the time for the insect repellent to remain effective

The time for the insect repellent to remain effective can be calculated using the equation: \(t = \frac{K \times h \times \rho}{\bar{h}_{m} \times D_{\mathrm{AB}}}\) Plugging in the given values, we get: \(t = \frac{0.05 \times 2.5 \times 10^{-6} \times 2000}{5 \times 10^{-3} \times 1 \times 10^{-13}}\) \(t = 5.0 \times 10^{4} \mathrm{~s}\)
04

(c) Calculate the time for the insect repellent to remain effective with a small partition coefficient

If the partition coefficient, K, becomes very small, the time for the insect repellent to remain effective can be calculated using the same equation as in (b): \(t = \frac{K' \times h \times \rho}{\bar{h}_{m} \times D_{\mathrm{AB}}}\) Since K' is very small, K' x h x 蟻 becomes close to zero. Therefore: \(t \approx \frac{0}{\bar{h}_{m} \times D_{\mathrm{AB}}}\) Because any number divided by 0 is infinity, the insect repellent will remain effective for a very long time, approaching infinity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection Mass Transfer Coefficient
In the context of mass transfer processes, the convection mass transfer coefficient \(\bar{h}_{m}\) represents the ease with which a substance can transfer from the surface of a solid into a moving fluid, such as air. This coefficient is crucial when studying substances like an active ingredient in a skin-applied insect repellent.
Convection mass transfer involves the transportation of mass through a fluid due to the movement of the fluid itself. Here are some key points:
  • The convection process is dependent on factors, including fluid velocity and the nature of the fluid.
  • In our example, the coefficient is given as \(5 \times 10^{-3} \hspace{2mm} \mathrm{m/s}\). This suggests that the active ingredient is transported from the skin surface into the air at this rate.
  • A higher coefficient would mean faster sublimation of the active ingredient into the surrounding air.
Understanding this concept is essential when determining how substances evaporate from surfaces.
Partition Coefficient
The partition coefficient \(K\) is an essential factor in the realm of mass transfer, especially when considering interfaces between two phases, like skin and air in our example. It reflects the propensity of a chemical substance to favor one medium over another.
The partition coefficient is defined as the ratio of the concentration of a compound in one phase to its concentration in another phase at equilibrium:
  • In our scenario, \(K\) was given as \(0.05\). This indicates the active ingredient's tendency to be more concentrated on the outer air phase than inside the skin.
  • When the partition coefficient is low, it implies that the ingredient prefers to remain in the air rather than being absorbed into the skin. This can significantly affect how long the repellent remains effective.
  • In practical applications, formulators can adjust \(K\) to optimize the effectiveness and longevity of products like insect repellents.
Mass Diffusivity
Mass diffusivity \(D_{AB}\) is a measure of how quickly molecules of one substance (A) diffuse through another substance (B). In our exercise, \(D_{AB}\) represents the diffusion of the active ingredient within the skin matrix.
This concept can be described as follows:
  • The given mass diffusivity is \(1 \times 10^{-13} \mathrm{~m}^{2}/\mathrm{s}\), a relatively low value which implies slow diffusion within the skin.
  • Mass diffusivity depends on several factors, including temperature, pressure, and the molecular properties of the active ingredient.
  • A higher diffusivity indicates faster spreading of the active ingredient, potentially leading to a shorter effective time for the insect repellent.
Understanding mass diffusivity is crucial for designing products that rely on controlled release of active ingredients, such as pharmaceutical patches or long-lasting repellents.

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Most popular questions from this chapter

Nitric oxide (NO) emissions from automobile exhaust can be reduced by using a catalytic converter, and the following reaction occurs at the catalytic surface: $$ \mathrm{NO}+\mathrm{CO} \rightarrow \frac{1}{2} \mathrm{~N}_{2}+\mathrm{CO}_{2} $$ The concentration of NO is reduced by passing the exhaust gases over the surface, and the rate of reduction at the catalyst is governed by a first- order reaction of the form given by Equation 14.66. As a first approximation it may be assumed that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness \(L\) that adjoins the surface. Referring to Figure 14.7, consider a situation for which the exhaust gas is at \(500^{\circ} \mathrm{C}\) and \(1.2\) bars and the mole fraction of \(\mathrm{NO}\) is \(x_{\mathrm{A}, L}=0.15\). If \(D_{\mathrm{AB}}=10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), \(k_{1}^{\prime \prime}=0.05 \mathrm{~m} / \mathrm{s}\), and the film thickness is \(L=1 \mathrm{~mm}\), what is the mole fraction of \(\mathrm{NO}\) at the catalytic surface and what is the NO removal rate for a surface of area \(A=200 \mathrm{~cm}^{2}\) ?

Hydrogen at a pressure of 2 atm flows within a tube of diameter \(40 \mathrm{~mm}\) and wall thickness \(0.5 \mathrm{~mm}\). The outer surface is exposed to a gas stream for which the hydrogen partial pressure is \(0.1 \mathrm{~atm}\). The mass diffusivity and solubility of hydrogen in the tube material are \(1.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) and \(160 \mathrm{kmol} / \mathrm{m}^{3}\) *atm, respectively. When the system is at \(500 \mathrm{~K}\), what is the rate of hydrogen transfer through the tube per unit length \((\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\) ?

Ultra-pure hydrogen is required in applications ranging from the manufacturing of semiconductors to powering fuel cells. The crystalline structure of palladium allows only the transfer of atomic hydrogen (H) through its thickness, and therefore palladium membranes are used to filter hydrogen from contaminated streams containing mixtures of hydrogen and other gases. Hydrogen molecules \(\left(\mathrm{H}_{2}\right)\) are first adsorbed onto the palladium's surface and are then dissociated into atoms \((\mathrm{H})\), which subsequently diffuse through the metal. The H atoms recombine on the opposite side of the membrane, forming pure \(\mathrm{H}_{2}\). The surface concentration of H takes the form \(C_{\mathrm{H}}=K_{s} p_{\mathrm{H}_{2}}^{05}\), where \(K_{s} \approx\) \(1.4 \mathrm{kmol} / \mathrm{m}^{3} \cdot \mathrm{bar}^{0.5}\) is known as Sieverts constant. Consider an industrial hydrogen purifier consisting of an array of palladium tubes with one tube end connected to a collector plenum and the other end closed. The tube bank is inserted into a shell. Impure \(\mathrm{H}_{2}\) at \(T=\) \(600 \mathrm{~K}, p=15\) bars, \(x_{\mathrm{H}_{2}}=0.85\) is introduced into the shell while pure \(\mathrm{H}_{2}\) at \(p=6\) bars, \(T=600 \mathrm{~K}\) is extracted through the tubes. Determine the production rate of pure hydrogen \((\mathrm{kg} / \mathrm{h})\) for \(N=100\) tubes which are of inside diameter \(D_{i}=1.6 \mathrm{~mm}\), wall thickness \(t=75 \mu \mathrm{m}\), and length \(L=80 \mathrm{~mm}\). The mass diffusivity of hydrogen \((\mathrm{H})\) in palladium at \(600 \mathrm{~K}\) is approximately \(D_{\mathrm{AB}}=7 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

If an amount of energy \(Q_{o}^{\prime \prime}\left(\mathrm{J} / \mathrm{m}^{2}\right)\) is released instantaneously, as, for example, from a pulsed laser, and it is absorbed by the surface of a semi-infinite medium, with no attendant losses to the surroundings, the subsequent temperature distribution in the medium is $$ T(x, t)-T_{i}=\frac{Q_{o}^{\prime \prime}}{\rho c(\pi \alpha t)^{1 / 2}} \exp \left(-x^{2} / 4 \alpha t\right) $$ where \(T_{i}\) is the initial, uniform temperature of the medium. Consider an analogous mass transfer process involving deposition of a thin layer of phosphorous (P) on a silicon (Si) wafer at room temperature. If the wafer is placed in a furnace, the diffusion of \(\mathrm{P}\) into Si is significantly enhanced by the high-temperature environment. A Si wafer with \(1-\mu \mathrm{m}\)-thick P film is suddenly placed in a furnace at \(1000^{\circ} \mathrm{C}\), and the resulting distribution of \(P\) is characterized by an expression of the form $$ C_{\mathrm{P}}(x, t)=\frac{M_{\mathrm{P}, o}^{\prime \prime}}{\left(\pi D_{\mathrm{P}-\mathrm{Si}} t\right)^{1 / 2}} \exp \left(-x^{2} / 4 D_{\mathrm{P}-\mathrm{Si}} t\right) $$ where \(M_{\mathrm{P}, o}^{\prime \prime}\) is the molar area density \(\left(\mathrm{kmol} / \mathrm{m}^{2}\right)\) of \(\mathrm{P}\) associated with the film of concentration \(C_{\mathrm{P}}\) and thickness \(d_{o}\). (a) Explain the correspondence between variables in the analogous temperature and concentration distributions. (b) Determine the mole fraction of \(P\) at a depth of \(0.1 \mu \mathrm{m}\) in the Si after \(30 \mathrm{~s}\). The diffusion coefficient is \(D_{\mathrm{P}-\mathrm{Si}}=1.2 \times 10^{-17} \mathrm{~m}^{2} / \mathrm{s}\). The mass densities of \(\mathrm{P}\) and \(\mathrm{Si}\) are 2000 and \(2300 \mathrm{~kg} / \mathrm{m}^{3}\), respectively, and their molecular weights are \(30.97\) and \(28.09 \mathrm{~kg} / \mathrm{kmol}\).

A 1-mm-thick square \((100 \mathrm{~mm} \times 100 \mathrm{~mm})\) sheet of polymer is suspended from a precision scale in a chamber characterized by a temperature and relative humidity of \(T=300 \mathrm{~K}\) and \(\phi=0\), respectively. Suddenly, at time \(t=0\), the chamber's relative humidity is raised to \(\phi=0.95\). The measured mass of the sheet increases by \(0.012 \mathrm{mg}\) over \(24 \mathrm{~h}\) and by \(0.016 \mathrm{mg}\) over \(48 \mathrm{~h}\). Determine the solubility and mass diffusivity of water vapor in the polymer. Preliminary experiments have indicated that the mass diffusivity is greater than \(7 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).
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