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Chapter 14: Problem 140

Consider a 15-cm-internal-diameter, 10-m-long circular duct whose interior surface is wet. The duct is to be dried by forcing dry air at \(1 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\) through it at an average velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through a chilled room, and it remains at an average temperature of \(15^{\circ} \mathrm{C}\) at all times. Determine the mass transfer coefficient in the duct.

Short Answer

Expert verified
Based on the calculations, the mass transfer coefficient (k_m) in the wet circular duct through which dry air is being forced is approximately 9.3 x 10^-6 m/s. This value is crucial for evaluating the drying process in the duct.

Step by step solution

01

Calculate the Reynolds number

The Reynolds number is a dimensionless quantity that helps to predict fluid flow patterns. It is the ratio of inertial forces to viscous forces and can be calculated using the formula: $$ Re = \frac{蟻VD}{渭} $$ Where \(蟻\) is the fluid density, \(V\) is the average velocity of fluid flow, \(D\) is the hydraulic diameter, and \(渭\) is the dynamic viscosity of the fluid. Given, average velocity \(V = 3~m/s\), duct diameter \(D = 0.15~m\). To find the density and dynamic viscosity of dry air at \(1~atm\) and \(15掳C\), we can refer to standard air property tables. For dry air at \(1~atm\) and \(15掳C\), the density (\(蟻\)) is approximately \(1.225~kg/m^3\), and the dynamic viscosity (\(渭\)) is around \(1.8 脳 10^{-5}~Pa.s\) Now, we can calculate the Reynolds number: $$ Re = \frac{(1.225~kg/m^3)(3~m/s)(0.15~m)}{1.8 \times 10^{-5}~Pa.s} 鈮� 30625 $$
02

Calculate the Schmidt number

Schmidt number is a dimensionless number representing the ratio between molecular momentum transport and mass transport. It can be calculated using the formula: $$ Sc = \frac{渭}{蟻D_m} $$ Where \(D_m\) is the mass diffusivity of the fluid. For air-water vapor, the mass diffusivity \(D_m\) at \(15掳C\) is approximately \(2.5 脳 10^{-5}~m^2/s\). Now, we can calculate the Schmidt number: $$ Sc = \frac{1.8 \times 10^{-5}~Pa.s}{(1.225~kg/m^3)(2.5 脳 10^{-5}~m^2/s)} 鈮� 0.59 $$
03

Determine the mass transfer coefficient using Chilton-Colburn analogy

The Chilton-Colburn analogy is a useful tool to determine mass transfer coefficients for turbulent flow while knowing the Reynolds and Schmidt numbers. The analogy suggests that: $$ Sh = \frac{k_mD}{D_m} = 0.023 Re^{0.83} Sc^{0.44} $$ Where \(Sh\) is the Sherwood number and \(k_m\) is the mass transfer coefficient. Now, we can calculate the mass transfer coefficient \(k_m\): $$ k_m = \frac{Sh 脳 D_m}{D} = \frac{0.023 脳 30625^{0.83} 脳 0.59^{0.44} 脳 2.5 \times 10^{-5}~m^2/s}{0.15 ~m} 鈮� 9.3 \times 10^{-6}~ m/s $$ The mass transfer coefficient (\(k_m\)) in the duct is approximately \(9.3 \times 10^{-6}~m/s\).

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Key Concepts

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

Reynolds Number
In fluid dynamics, the Reynolds number (Re) is a key concept to understand. It helps us predict the behavior of fluid flow. This dimensionless number is essentially a ratio between inertial and viscous forces in a fluid.

A low Reynolds number indicates that viscous forces are dominant, resulting in laminar or smooth flow. Conversely, a high Reynolds number suggests inertial forces take precedent, causing turbulent or chaotic flow.

For the exercise involving a circular duct, the formula used to calculate the Reynolds number is: \[ Re = \frac{蟻VD}{渭} \]
  • \( 蟻 \) is the fluid density.
  • \( V \) represents the average velocity.
  • \( D \) is the hydraulic or pipe diameter.
  • \( 渭 \) is the dynamic viscosity.
Using the given values, the calculated Reynolds number was approximately 30,625, suggesting turbulent flow. Understanding this number provides insights into the flow dynamics and helps in predicting whether the flow will be smooth or turbulent under given conditions.
Schmidt Number
The Schmidt number (Sc) is another important dimensionless quantity in the study of mass transfer. It links the momentum diffusivity (viscosity) and mass diffusivity of a fluid. Simply put, it compares the rate at which momentum is transferred throughout a fluid versus the transfer rate of mass by diffusion.

The Schmidt number formula is:\[ Sc = \frac{渭}{蟻D_m} \]
  • \( 渭 \) denotes the dynamic viscosity.
  • \( 蟻 \) is the fluid density.
  • \( D_m \) stands for mass diffusivity.
In our duct problem, we calculated the Schmidt number to be approximately 0.59. This number indicates how effectively momentum diffuses compared to mass within the fluid. Scalars like this are pivotal in studies involving both heat and mass transfer as they provide a comparative understanding of these processes.
Chilton-Colburn Analogy
The Chilton-Colburn analogy is a fascinating concept that draws a parallel between heat, mass, and momentum transfer in fluid flows. Particularly relevant in turbulent flow conditions, this analogy allows engineers to estimate mass transfer coefficients using known parameters like Reynolds and Schmidt numbers.

According to the analogy, the Sherwood number (Sh) can be tied to both Reynolds and Schmidt numbers as follows:\[ Sh = 0.023 Re^{0.83} Sc^{0.44} \]The Sherwood number bears resemblance to the Nusselt number used in heat transfer and denotes the ratio of convective to diffusive mass transfer.

From the exercise, after substituting the calculated values for Reynolds and Schmidt numbers, the mass transfer coefficient (\(k_m\)) was determined using the equation:\[ k_m = \frac{Sh \cdot D_m}{D} \]By understanding the Chilton-Colburn analogy, we can bridge the behavior of mass transfer with well-studied heat and momentum transfer theories. This analogy is crucial for efficiently optimizing systems involving all three transfer phenomena.

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Most popular questions from this chapter

Dry air whose molar analysis is \(78.1\) percent \(\mathrm{N}_{2}\), \(20.9\) percent \(\mathrm{O}_{2}\), and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) during the process, determine (a) the molar analysis of the saturated air and \((b)\) the density of air before and after the process. What do you conclude from your results?

Consider a 20 -cm-thick brick wall of a house. The indoor conditions are \(25^{\circ} \mathrm{C}\) and 50 percent relative humidity while the outside conditions are \(50^{\circ} \mathrm{C}\) and 50 percent relative humidity. Assuming that there is no condensation or freezing within the wall, determine the amount of moisture flowing through a unit surface area of the wall during a \(24-\mathrm{h}\) period.

Under what conditions will the normalized velocity, thermal, and concentration boundary layers coincide during flow over a flat plate?

Using Henry's law, show that the dissolved gases in a liquid can be driven off by heating the liquid.

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.
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