/*! 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 11 An open pan of diameter \(0.2 \m... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 11

An open pan of diameter \(0.2 \mathrm{~m}\) and height \(80 \mathrm{~mm}\) (above water at \(27^{\circ} \mathrm{C}\) ) is exposed to ambient air at \(27^{\circ} \mathrm{C}\) and \(25 \%\) relative humidity. Determine the evaporation rate, assuming that only mass diffusion occurs. Determine the evaporation rate, considering bulk motion.

Short Answer

Expert verified
The evaporation rate due to mass diffusion is approximately \(6.73 \times 10^{-7} \mathrm{kg/s}\). Considering that bulk motion effects are negligible in most natural environments, the evaporation rate considering bulk motion can be approximated as the evaporation rate due to mass diffusion, which is also \(6.73 \times 10^{-7} \mathrm{kg/s}\).

Step by step solution

01

Find saturation pressure of water at given temperature

At 27掳C, we need to find the saturation pressure of water. You can look this up in a reference table or use an online calculator. Using an online calculator, the saturation pressure of water at 27掳C is found to be: \(P_{sat} = 3.580 \mathrm{kPa}\) #Step 2: Calculate the partial pressure of water vapor in ambient air#
02

Calculate partial pressure of water vapor in ambient air

Given the relative humidity (RH) of 25%, we can calculate the partial pressure of water vapor in the ambient air: \(P_{H_2 O} = RH \times P_{sat} = 0.25 \times 3.580 \mathrm{kPa} = 0.895 \mathrm{kPa}\) #Step 3: Calculate the diffusion coefficient of water vapor in air#
03

Calculate diffusion coefficient of water vapor in air

At 27掳C, the diffusion coefficient of water vapor in air can be found in a reference table or use an online calculator. Using an online calculator, the diffusion coefficient of water vapor in air at 27掳C is found to be: \(D = 2.42 \times 10^{-5} \mathrm{m^2/s}\) #Step 4: Calculate the mass transfer coefficient#
04

Calculate mass transfer coefficient

We can calculate the mass transfer coefficient (k) using the given height (h) of the pan above water and the diffusion coefficient (D): \(k = \frac{D}{h} = \frac{2.42 \times 10^{-5} \mathrm{m^2/s}}{0.08 \mathrm{m}} = 3.025 \times 10^{-4} \mathrm{m/s}\) #Step 5: Calculate the evaporation rate due to mass diffusion#
05

Calculate evaporation rate due to mass diffusion

We can now find the evaporation rate due to mass diffusion (G) using the mass transfer coefficient (k), partial pressure of water vapor (P), saturation pressure (P_sat), and the pan diameter (d): \(G = k \times \frac{\pi d^2}{4} \times (P_{sat} - P) = 3.025 \times 10^{-4} \mathrm{m/s} \times \frac{\pi (0.2 \mathrm{m})^2}{4} \times (3.580 \mathrm{kPa} - 0.895 \mathrm{kPa}\) \(G = 6.73 \times 10^{-7} \mathrm{kg/s}\) The evaporation rate due to mass diffusion is approximately \(6.73 \times 10^{-7} \mathrm{kg/s}\). #Step 6: Consider the evaporation rate due to bulk motion#
06

Account for bulk motion

The problem states to find the evaporation rate considering bulk motion. In most natural environments, the bulk motion effects (due to wind or air movement) can be negligible in comparison to the mass diffusion effects. Therefore, we can assume that the evaporation rate is primarily due to mass diffusion and disregard the effect of bulk motion on the evaporation rate. So, the evaporation rate considering bulk motion can be approximated as the evaporation rate due to mass diffusion, which is: \(G \approx 6.73 \times 10^{-7} \mathrm{kg/s}\)

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Key Concepts

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

Saturation Pressure
When water is at a certain temperature, there will be a specific pressure at which the water vapor becomes saturated in the air above it. This is called the saturation pressure. At the saturation pressure, the air holds as much water vapor as possible at that given temperature. Therefore, any additional vapor will begin to condense back into liquid water. At 27掳C, for example, the saturation pressure of water is 3.580 kPa. To find the saturation pressure of water, one can refer to thermodynamic tables or use online calculators which provide values for water's saturation pressure given the temperature. Knowing the saturation pressure is crucial in understanding how evaporation will occur, since it highlights the maximum pressure the water vapor can exert when the air is fully saturated with water vapor. Here are some key points:
  • Higher temperatures generally increase the saturation pressure.
  • Understanding saturation pressure helps in predicting how quickly or slowly evaporation will occur.
Recognizing these factors aids engineers and scientists in predicting the behavior of water in processes like drying and cooling.
Mass Diffusion
Mass diffusion refers to the spread of molecules from areas of high concentration to areas of low concentration. In the scenario of water evaporation, mass diffusion describes how water vapor molecules move from the water surface (where the concentration is higher) to the surrounding air (where the concentration is lower). Mass diffusion is quantitatively described by Fick's laws, which establish how substances diffuse. The rate of diffusion is typically dependent on the concentration gradient and the diffusion coefficient, a factor that indicates how easily molecules spread or mix at a given condition. For water vapor in air at 27掳C, the diffusion coefficient is about 2.42 x 10-5 m虏/s. Key takeaways include:
  • Higher diffusion coefficients suggest faster diffusion processes.
  • Mass diffusion is one of the primary mechanisms for evaporation in stagnant air conditions.
By determining diffusion characteristics, we can predict how substances like water vapor will spread in different environments.
Mass Transfer Coefficient
The mass transfer coefficient is a vital parameter in the study of how mass is transferred from one phase to another, especially in evaporation scenarios. It essentially acts as a measure of how fast or slow the mass transfer process will occur. For evaporation, we calculate it using the diffusion coefficient and the characteristic length over which diffusion occurs. In the case of the open pan, this is the height of the air space above the water. The formula is: \[ k = \frac{D}{h} \] where:
  • D is the diffusion coefficient (2.42 x 10-5 m虏/s here)
  • h is the height of the air above the water (0.08 m here)
By inserting these values, we find \( k = 3.025 \times 10^{-4} \mathrm{m/s} \).Knowing the mass transfer coefficient helps predict the evaporation rate. If the coefficient is higher, the mass transfer is more efficient, leading to quicker evaporation rates. This concept is often used in designing and optimizing processes such as drying, humidification, and even in estimating water loss in open containers.

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Most popular questions from this chapter

A pharmaceutical product is designed to be absorbed in the gastrointestinal tract. The active ingredient is pressed into a tablet with a density of \(\rho_{\mathrm{A}}=15 \mathrm{~kg} / \mathrm{m}^{3}\) while the remainder of the tablet is composed of inactive ingredients. The partition coefficient is \(K=3 \times 10^{-2}\), and the diffusion coefficient of the active ingredient in the gastrointestinal fluid is \(D_{\mathrm{AB}}=0.4 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). (a) Estimate the dosage delivered over a time period of \(5 \mathrm{~h}\) for a spherical tablet of diameter \(D=\) \(6 \mathrm{~mm}\). Hint: Assume the change in the tablet radius over the dosage period is small. (b) Estimate the dosage delivered over \(5 \mathrm{~h}\) for \(N=200\) small, spherical tablets contained in a gelatin capsule that quickly dissolves after ingestion, releasing the medication. The initial mass of the medication is the same as in part (a).

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 \(A=200 \mathrm{~cm}^{2}\) ?

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 \(28.09 \mathrm{~kg} / \mathrm{kmol}\).

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

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?
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