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Chapter 14: Problem 35

Referring to Problem 14.34, a more representative model of respiration in a spherical organism is one for which oxygen consumption is governed by a firstorder reaction of the form \(\dot{N}_{\mathrm{A}}=-k_{1} C_{\mathrm{A}}\). (a) If a molar concentration of \(C_{\mathrm{A}}\left(r_{o}\right)=C_{\mathrm{A}, o}\) is maintained at the surface of the organism, obtain an expression for the radial distribution of oxygen, \(C_{\mathrm{A}}(r)\), within the organism. Hint: To simplify solution of the species diffusion equation, invoke the transformation \(y \equiv r C_{\mathrm{A}}\). (b) Obtain an expression for the rate of oxygen consumption within the organism. (c) Consider an organism of radius \(r_{o}=0.10 \mathrm{~mm}\) and a diffusion coefficient of \(D_{\mathrm{AB}}=10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). If \(C_{\mathrm{A}, o}=5 \times 10^{-5} \mathrm{kmol} / \mathrm{m}^{3}\) and \(k_{1}=20 \mathrm{~s}^{-1}\), estimate the corresponding value of the molar concentration at the center of the organism. What is the rate of oxygen consumption by the organism?

Short Answer

Expert verified
The radial distribution of oxygen within the organism is given by the expression \(C_A(r) = \frac{B}{r}\), where \(B\) is the integration constant. The rate of oxygen consumption within the organism is \(\dot{N}_A = \frac{D_{AB}B}{r_o^2}\). Given the specific values, the molar concentration at the center of the organism is undefined, indicating a steep concentration gradient. The rate of oxygen consumption by the organism is \(5\times10^{-10}\text{ kmol/(m}^2\text{s)}\).

Step by step solution

01

Set up the species diffusion equation.

First, let's set up the species diffusion equation, which is: \[\frac{d^2 C_A(r)}{dr^2}+ \frac{2}{r}\frac{d C_A(r)}{dr} = \frac{\dot{N}_A}{D_{AB}}\] And the given first-order reaction: \[\dot{N}_A = -k_1 C_A\] We will use these equations to find an expression for the radial distribution of oxygen, \(C_A(r)\).
02

Use the transformation y = r C_A.

Now, let's use the given transformation to simplify the species diffusion equation. Applying the transformation \(y = r C_A\), we can differentiate with respect to \(r\): \[\frac{dy}{dr} = \frac{dr C_A(r) + r dC_A(r)}{dr^2} = C_A(r) + r \frac{dC_A(r)}{dr}\] Now differentiate with respect to \(r\) again: \[\frac{d^2 y}{dr^2} = \frac{dC_A(r)}{dr} + \frac{dC_A(r)}{dr} + r \frac{d^2 C_A(r)}{dr^2} = 2\frac{dC_A(r)}{dr} + r \frac{d^2 C_A(r)}{dr^2}\]
03

Substitute into the species diffusion equation and solve.

Substitute the transformation into the species diffusion equation: \[2\frac{dC_A(r)}{dr} + r\frac{d^2 C_A(r)}{dr^2} = -\frac{k_1 C_A}{D_{AB}}\] Now, integrate with respect to \(r\): \[2\int \frac{dC_A(r)}{dr}dr + \int r\frac{d^2 C_A(r)}{dr^2}dr = -\int \frac{k_1 C_A}{D_{AB}}dr\] \[2\frac{dC_A(r)}{dr} + r \frac{d^2 C_A(r)}{dr^2} = -\frac{k_1}{D_{AB}}y + B\] where \(B\) is the integration constant. Now, substitute the transformation \(y = r C_A\) back into the equation: \[2\frac{dC_A(r)}{dr} + r\frac{d^2 C_A(r)}{dr^2} = -\frac{k_1}{D_{AB}}rC_A + B\] \[C_A(r) = A\frac{e^{-\sqrt{k_1/D_{AB}}r}}{r} + \frac{B}{r}\] This is the expression for the radial distribution of oxygen within the organism.
04

Apply boundary conditions and find A and B.

We're given that at the surface of the organism, the molar concentration is \(C_A(r_o) = C_{A,o}\). Apply this boundary condition: \[C_{A,o} = A\frac{e^{-\sqrt{k_1/D_{AB}}r_o}}{r_o} + \frac{B}{r_o}\] Now we need to find another boundary condition to solve for \(A\) and \(B\). Since the concentration must be finite at the center of the organism, we know that as \(r \to 0\), \(C_A(r) \to C_A(0)\). This implies that: \[A = 0\] So the radial distribution of oxygen within the organism is: \[C_A(r) = \frac{B}{r}\]
05

Find the rate of oxygen consumption within the organism.

To find the rate of oxygen consumption within the organism, let's first find the molar flux at the surface of the organism: \[\dot{N}_{A} = -D_{AB}(\frac{dC_A}{dr})_{r= r_o}\] Since we found that \(C_A(r) = \frac{B}{r}\), we can differentiate with respect to \(r\): \[\frac{dC_A}{dr} = -\frac{B}{r^2}\] Now, plug in the boundary condition \(r=r_o\): \[\dot{N}_A = - D_{AB}\left(-\frac{B}{r_o^2}\right)\] The rate of oxygen consumption within the organism is: \[\dot{N}_A = \frac{D_{AB}B}{r_o^2}\]
06

Calculate the molar concentration at the center of the organism and the rate of oxygen consumption.

Now we have all the expressions needed to find the molar concentration at the center of the organism, and the rate of oxygen consumption. We have \(r_o = 0.10\text{ mm}\), \(D_{AB} = 10^{-8} \text{m}^2\text{/s}\), \(C_{A,o} = 5 \times 10^{-5} \text{kmol/m}^3\), and \(k_1 = 20\text{ s}^{-1}\). First, find \(B\) using the boundary condition: \[C_{A,o} = \frac{B}{r_o} \Rightarrow B = C_{A,o} r_o\] Now find the molar concentration at the center of the organism: \[C_A(0) = \frac{B}{0} = \infty\] The molar concentration at the center of the organism is undefined, implying that the concentration gradient is very steep at the center. Now, find the rate of oxygen consumption by the organism: \[\dot{N}_A = \frac{D_{AB}B}{r_o^2} = \frac{D_{AB}C_{A,o} r_o}{r_o^2} = \frac{10^{-8}(5 \times 10^{-5})(0.10\times10^{-3})}{(0.10\times10^{-3})^2}\] \[\dot{N}_A = 5\times10^{-10} \text{kmol/(m}^2\text{s)}\] Therefore, the molar concentration at the center of the organism is undefined, and the rate of oxygen consumption by the organism is \(5\times10^{-10}\text{ kmol/(m}^2\text{s)}\).

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Key Concepts

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

Oxygen Consumption
Oxygen consumption in biological contexts refers to the process by which living organisms use oxygen to metabolize organic compounds, typically to produce energy. In the aforementioned exercise, oxygen consumption is described using a first-order reaction rate equation.

The reaction rate equation is given by \( \dot{N}_{A} = -k_{1} C_{A} \), where \( \dot{N}_{A} \) represents the rate of oxygen consumption, \( k_{1} \) is the first-order rate constant, and \( C_{A} \) is the concentration of oxygen. The negative sign indicates that oxygen is being consumed, so its concentration is decreasing over time. The equation suggests that the rate of oxygen consumption is directly proportional to the concentration of oxygen within the organism; as the availability of oxygen drops, so does the rate at which it is consumed.

In order to maintain life functions, organisms must efficiently distribute and consume oxygen. If the oxygen concentration on the organism's surface is known, with the help of diffusion equations, scientists can predict how oxygen is distributed within the organism and at what rate it is being consumed, essential knowledge for understanding respiratory systems and designing medical treatments.
Radial Distribution of Oxygen
The radial distribution of oxygen within an organism is a measure of how the concentration of oxygen varies as a function of distance from the organism's exterior towards its center. This concept is central in the given exercise, where we need to determine how oxygen is spread within a spherical organism.

To solve this, the exercise leverages a transformation of the species diffusion equation using \( y \equiv r C_{A} \), where \( r \) is the radial distance from the center of the organism. The species diffusion equation, combined with the first-order reaction, forms the basis for determining how the concentration changes with \( r \). After the transformation and implementing appropriate boundary conditions, such as a known concentration at the organism's surface (\( C_{A}(r_{o}) = C_{A, o} \) ), the distribution equation is found to be \( C_{A}(r) = \frac{B}{r} \), where \( B \) is a constant determined by these conditions.

It is critical that solutions to the diffusion equation exhibit a physically realistic behavior鈥攃oncentration cannot be infinite, and at the center of a spherically symmetrical organism, the concentration gradient must approach zero. This distribution helps us to visualize and calculate how efficiently an organism can transport and utilize oxygen, directly affecting its metabolic processes and overall health.
First-order Reaction
A first-order reaction is one in which the rate of the reaction is directly proportional to the concentration of a single reactant. In the context of our exercise, the consumption of oxygen within an organism is modeled as a first-order reaction, with respect to the oxygen concentration.

The first-order kinetics described by the equation \( \dot{N}_{A} = -k_{1} C_{A} \) imply that as the concentration of oxygen decreases, the rate of its consumption diminishes as well. It's important to note that the rate constant \( k_{1} \) can vary widely depending on the species and environmental conditions, as it encapsulates the biological and chemical complexities of how organisms utilize oxygen.

Understanding the implications of first-order kinetics is fundamental in biological systems since it helps in quantifying the removal or consumption rates of substances, including pollutants or drugs within biological organisms. Additionally, when paired with diffusion equations, it provides insightful information for bioengineers, environmentalists, and pharmacologists in their respective fields.

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Most popular questions from this chapter

A He-Xe mixture containing \(0.75\) mole fraction of helium is used for cooling of electronics in an avionics application. At a temperature of \(300 \mathrm{~K}\) and atmospheric pressure, calculate the mass fraction of helium and the mass density, molar concentration, and molecular weight of the mixture. If the cooling system capacity is \(10 \mathrm{~L}\), what is the mass of the coolant?

A large sheet of material \(40 \mathrm{~mm}\) thick contains dissolved hydrogen \(\left(\mathrm{H}_{2}\right)\) having a uniform concentration of \(3 \mathrm{kmol} / \mathrm{m}^{3}\). The sheet is exposed to a fluid stream that causes the concentration of the dissolved hydrogen to be reduced suddenly to zero at both surfaces. This surface condition is maintained constant thereafter. If the mass diffusivity of hydrogen is \(9 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), how much time is required to bring the density of dissolved hydrogen to a value of \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\) at the center of the sheet?

A thick plate of pure iron at \(1000^{\circ} \mathrm{C}\) is subjected to a carburization process in which the surface of the plate is suddenly exposed to a gas that induces a carbon concentration \(C_{C, s}\) at one surface. The average diffusion coefficient for carbon and iron at this temperature is \(D_{\mathrm{C}-\mathrm{Fe}}=3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). Use the correspondence between heat and mass transfer variables in addressing the following questions. (a) Consider the heat transfer analog to the carburization problem. Sketch the mass and heat transfer systems. Show and explain the correspondence between variables. Provide the analytical solutions to the heat and mass transfer problems. (b) Determine the carbon concentration ratio, \(C_{\mathrm{C}}(x, t) / C_{\mathrm{C}, s}\), at a depth of \(1 \mathrm{~mm}\) after \(1 \mathrm{~h}\) of carburization. (c) From the analogy, show that the time dependence of the mass flux of carbon into the plate may be expressed as \(n_{\mathrm{C}}^{\prime \prime}=\rho_{\mathrm{C}, s}\left(D_{\mathrm{C}-\mathrm{Fe}} / \pi t\right)^{1 / 2}\). Also, obtain an expression for the mass of carbon per unit area entering the iron plate over the time period \(t\).

Beginning with a differential control volume, derive the diffusion equation, on a molar basis, for species A in a three-dimensional (Cartesian coordinates), stationary medium, considering species generation with constant properties. Compare your result with Equation 14.48b.

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 \((\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\) ?
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