/*! 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 117 Dry air at atmospheric pressure ... [FREE SOLUTION] | 91Ó°ÊÓ

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Dry air at atmospheric pressure and \(350 \mathrm{~K}\), with a free stream velocity of \(25 \mathrm{~m} / \mathrm{s}\), flows over a smooth, porous plate \(1 \mathrm{~m}\) long. (a) Assuming the plate to be saturated with liquid water at \(350 \mathrm{~K}\), estimate the mass rate of evaporation per unit width of the plate, \(n_{\mathrm{A}}^{\prime}(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\). (b) For air and liquid water temperatures of 300,325 , and \(350 \mathrm{~K}\), generate plots of \(n_{\mathrm{A}}^{\prime}\) as a function of velocity for the range from 1 to \(25 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
To summarize the solution to the given problem, we first calculated the Reynolds number, Prandtl number, and Schmidt number for the given conditions using the properties of dry air at atmospheric pressure. Then, we estimated the heat transfer coefficient using the Chilton-Colburn analogy. Next, we determined the Sherwood number and mass transfer coefficient. Finally, we estimated the mass rate of evaporation per unit width of the plate, \(n_A'\), using the mass transfer coefficient and the vapor pressure difference. In part (b), we repeated these calculations for different temperatures and generated plots of \(n_A'\) as a function of velocity.

Step by step solution

01

Calculate Reynolds Number

To find the Reynolds number, use the formula: \[Re = \frac{\rho V L}{\mu}\] Where \(Re\) is the Reynolds number, \(\rho\) is the density of the air, \(V\) is the free-stream velocity, \(L\) is the length of the plate, and \(\mu\) is the dynamic viscosity. We have the velocity, \(V = 25 \mathrm{m/s}\), and the length of the plate, \(L = 1 \mathrm{m}\). The properties of dry air at atmospheric pressure are: \(\rho = 1.09 \mathrm{kg/m^3}\), and \(\mu = 1.81 \times 10^{-5} \mathrm{Pa}\cdot \mathrm{s}\). We can calculate the Reynolds number as follows: \[Re = \frac{1.09~ \mathrm{kg/m^3} \cdot 25~ \mathrm{m/s} \cdot 1~ \mathrm{m}}{1.81 \times 10^{-5} ~\mathrm{Pa}\cdot \mathrm{s}}\]
02

Calculate Prandtl and Schmidt Numbers

Calculate the Prandtl number, \(Pr\) and the Schmidt number, \(Sc\) using the following formulas: \[Pr = \frac{\mu C_p}{k}\] and \[Sc = \frac{\mu}{\rho D}\] Where \(C_p\) is the heat capacity, \(k\) is the thermal conductivity, and \(D\) is the mass diffusivity. The properties of dry air at atmospheric pressure are: \(C_p = 1007 ~\mathrm{J} / \mathrm{kg}\cdot \mathrm{K}\), \(k = 0.02558 ~\mathrm{W} / \mathrm{m}\cdot \mathrm{K}\), and \(D = 2.44 \times 10^{-5} ~\mathrm{m^2/s}\). Calculate the Prandtl and Schmidt numbers as follows: \[Pr = \frac{1.81 \times 10^{-5} ~\mathrm{Pa} \cdot \mathrm{s} \cdot 1007 ~\mathrm{J} / \mathrm{kg}\cdot \mathrm{K}}{0.02558 ~\mathrm{W} / \mathrm{m}\cdot \mathrm{K}}\] \[Sc = \frac{1.81 \times 10^{-5} ~\mathrm{Pa} \cdot \mathrm{s}}{1.09 ~\mathrm{kg/m^3} \cdot 2.44 \times 10^{-5} ~\mathrm{m^2/s}}\]
03

Estimate Heat Transfer Coefficient

With the Reynolds, Prandtl, and Schmidt numbers, we can estimate the heat transfer coefficient, \(h\), using the Chilton-Colburn analogy. This is given by: \[St = \frac{h}{\rho V C_p} = 0.664 Re^{-1/2} Pr^{1/3} = \frac{k_{ave}}{L}\] Calculate the heat transfer coefficient, \(h\), from this formula: \[h = St \rho V C_p\]
04

Determine Sherwood Number

Using Chilton-Colburn analogy for mass transfer, calculate the Sherwood number, \(Sh\): \[Sh = 0.664 Re^{-1/2} Sc^{1/3}\]
05

Estimate Mass Transfer Coefficient

To estimate the mass transfer coefficient, \(k_A\), use the formula: \[k_A = \frac{Sh \cdot D}{L}\]
06

Estimate Mass Rate of Evaporation

Now, we can estimate the mass rate of evaporation per unit width of the plate, \(n_A'\), using the mass transfer coefficient and the difference in vapor pressures, \(\Delta P_v\): \[n_A' = k_{A\infty} \Delta P_v = k_A \frac{P_v^W - P_v^A}{R T}\] Where \(P_v^W\) is the vapor pressure of the liquid water, \(P_v^A\) is the vapor pressure of the air, \(R\) is the gas constant, and \(T\) is the temperature. Calculate \(n_A'\) for part (a) and following this approach, generate plots of \(n_A'\) as a function of velocity for the given temperature range in part (b).

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Key Concepts

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

Understanding the Reynolds Number
The Reynolds number (Re) is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. It helps distinguish between laminar flow, which is smooth and orderly, and turbulent flow, which is chaotic and mixed. The textbook example provides the formula for calculating the Reynolds number as

\[\begin{equation}Re = \frac{\rho V L}{\mu}\end{equation}\]

where \(\rho\) is the fluid density, \(V\) is the velocity, \(L\) the characteristic length (such as the length of a plate), and \(\mu\) the dynamic viscosity of the fluid. A high Reynolds number indicates turbulent flow, whereas a low number suggests laminar flow—each carrying implications for mass and heat transfer processes.

In practical terms, when air flows over a smooth, porous plate as described in the exercise, the Reynolds number helps predict how the air will interact with the plate surface, which is critical for determining evaporation rates and heat transfer characteristics.
The Significance of Prandtl and Schmidt Numbers
The Prandtl and Schmidt numbers are two more dimensionless numbers that play a vital role in the study of heat and mass transfer. The Prandtl number (\(Pr\)) correlates the momentum diffusivity (viscosity) to the thermal diffusivity. It is calculated using the formula

\[\begin{equation}Pr = \frac{\mu C_p}{k}\end{equation}\]

Where \(C_p\) is the specific heat at constant pressure, and \(k\) is the thermal conductivity of the fluid. A high Prandtl number suggests that the fluid conducts momentum better than heat.
  • Air typically has a Prandtl number around 0.7, signifying that heat and momentum diffusivities are approximately the same, a fact that simplifies the analysis of convective heat transfer.
  • For the Schmidt number (\(Sc\)), the focus shifts to mass transfer, where it compares momentum diffusivity to mass diffusivity (\(D\)), expressed as:

    \[\begin{equation}Sc = \frac{\mu}{\rho D}\end{equation}\]

    It indicates how a substance diffuses through another substance.
  • A high Schmidt number implies that momentum diffuses much faster than mass, meaning a thin boundary layer for momentum than for mass transfer.
These numbers are especially important in problems involving phase changes, like evaporation, where proper understanding of both heat and mass transfer is crucial for accurate predictions and calculations as in the textbook exercise scenario.
Calculating the Mass Transfer Coefficient
The mass transfer coefficient (\(k_A\)) is a measure of how rapidly a substance will transfer from one phase to another through a unit area and is part of the formula to estimate the mass rate of evaporation, \(n_A'\). It is intrinsically linked to the Sherwood number (\(Sh\)), which is the mass transfer equivalent of the Nusselt number for heat transfer. Following the textbook steps, it's calculated by:

\[\begin{equation}Sh = 0.664 Re^{-1/2} Sc^{1/3}\end{equation}\]

The mass transfer coefficient then derives from the Sherwood number as:

\[\begin{equation}k_A = \frac{Sh \cdot D}{L}\end{equation}\]

with \(D\) being the mass diffusivity and \(L\) the characteristic length. Using this coefficient, one can quantify the mass rate of evaporation per unit width of the plate, an essential step for industrial applications such as drying processes in food or chemical industries. This coefficient is pivotal in the provided exercise, where understanding the evaporation from a wetted surface into air is the centerpiece of the analysis.

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Most popular questions from this chapter

Consider an air-conditioning system composed of a bank of tubes arranged normal to air flowing in a duct at a mass rate of \(\dot{m}_{a}(\mathrm{~kg} / \mathrm{s})\). A coolant flowing through the tubes is able to maintain the surface temperature of the tubes at a constant value of \(T_{s}

An \(L=1\)-m-long vertical copper tube of inner diameter \(D_{i}=20 \mathrm{~mm}\) and wall thickness \(t=2 \mathrm{~mm}\) contains liquid water at \(T_{w}=0^{\circ} \mathrm{C}\). On a winter day, air at \(V=3 \mathrm{~m} / \mathrm{s}, T_{\infty}=-20^{\circ} \mathrm{C}\) is in cross flow over the tube. (a) Determine the heat loss per unit mass from the water (W/kg) when the tube is full of water. (b) Determine the heat loss from the water (W/kg) when the tube is half full.

In the production of sheet metals or plastics, it is customary to cool the material before it leaves the production process for storage or shipment to the customer. Typically, the process is continuous, with a sheet of thickness \(\delta\) and width \(W\) cooled as it transits the distance \(L\) between two rollers at a velocity \(V\). In this problem, we consider cooling of an aluminum alloy (2024-T6) by an airstream moving at a velocity \(u_{\infty}\) in counter flow over the top surface of the sheet. A turbulence promoter is used to provide turbulent boundary layer development over the entire surface. (a) By applying conservation of energy to a differential control surface of length \(d x\), which either moves with the sheet or is stationary and through which the sheet passes, derive a differential equation that governs the temperature distribution along the sheet. Because of the low emissivity of the aluminum, radiation effects may be neglected. Express your result in terms of the velocity, thickness, and properties of the sheet \(\left(V, \delta, \rho, c_{p}\right)\), the local convection coefficient \(h_{x}\) associated with the counter flow, and the air temperature. For a known temperature of the sheet \(\left(T_{i}\right)\) at the onset of cooling and a negligible effect of the sheet velocity on boundary layer development, solve the equation to obtain an expression for the outlet temperature \(T_{a}\). (b) For \(\delta=2 \mathrm{~mm}, V=0.10 \mathrm{~m} / \mathrm{s}, L=5 \mathrm{~m}, W=1 \mathrm{~m}\), \(u_{\infty}=20 \mathrm{~m} / \mathrm{s}, T_{\infty}=20^{\circ} \mathrm{C}\), and \(T_{i}=300^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{a}\) ?

Fluid velocities can be measured using hot-film sensors, and a common design is one for which the sensing element forms a thin film about the circumference of a quartz rod. The film is typically comprised of a thin \((\sim 100 \mathrm{~nm})\) layer of platinum, whose electrical resistance is proportional to its temperature. Hence, when submerged in a fluid stream, an electric current may be passed through the film to maintain its temperature above that of the fluid. The temperature of the film is controlled by monitoring its electric resistance, and with concurrent measurement of the electric current, the power dissipated in the film may be determined. Proper operation is assured only if the heat generated in the film is transferred to the fluid, rather than conducted from the film into the quartz rod. Thermally, the film should therefore be strongly coupled to the fluid and weakly coupled to the quartz rod. This condition is satisfied if the Biot number is very large, \(B i=\bar{h} D / 2 k \geqslant 1\), where \(\bar{h}\) is the convection coefficient between the fluid and the film and \(k\) is the thermal conductivity of the rod. (a) For the following fluids and velocities, calculate and plot the convection coefficient as a function of velocity: (i) water, \(0.5 \leq V \leq 5 \mathrm{~m} / \mathrm{s}\); (ii) air, \(1 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\). (b) Comment on the suitability of using this hot-film sensor for the foregoing conditions.

The thermal pollution problem is associated with discharging warm water from an electrical power plant or from an industrial source to a natural body of water. Methods for alleviating this problem involve cooling the warm water before allowing the discharge to occur. Two such methods, involving wet cooling towers or spray ponds, rely on heat transfer from the warm water in droplet form to the surrounding atmosphere. To develop an understanding of the mechanisms that contribute to this cooling, consider a spherical droplet of diameter \(D\) and temperature \(T\), which is moving at a velocity \(V\) relative to air at a temperature \(T_{\infty}\) and relative humidity \(\phi_{\infty}\). The surroundings are characterized by the temperature \(T_{\text {sur }}\) Develop expressions for the droplet evaporation and cooling rates. Calculate the evaporation rate \((\mathrm{kg} / \mathrm{s})\) and cooling rate \((\mathrm{K} / \mathrm{s})\) when \(D=3 \mathrm{~mm}, V=7 \mathrm{~m} / \mathrm{s}, T=40^{\circ} \mathrm{C}, T_{\infty}=25^{\circ} \mathrm{C}\), \(T_{\text {sar }}=15^{\circ} \mathrm{C}\), and \(\phi_{\infty}=0.60\). The emissivity of water is \(\varepsilon_{w}=0.96\).

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