/*! 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 145 A wet towel hangs on a clothes l... [FREE SOLUTION] | 91Ó°ÊÓ

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A wet towel hangs on a clothes line under conditions for which one surface receives solar irradiation of \(G_{S}=900 \mathrm{~W} / \mathrm{m}^{2}\) and both surfaces are exposed to atmospheric (sky) and ground radiation of \(G_{\text {tam }}=\) \(200 \mathrm{~W} / \mathrm{m}^{2}\) and \(G_{\mathrm{g}}=250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Under moderately windy conditions, airflow at a temperature of \(27^{\circ} \mathrm{C}\) and a relative humidity of \(60 \%\) maintains a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. The wet towel has an emissivity of \(0.96\) and a solar absorptivity of \(0.65\). As a first approximation the properties of the atmospheric air may be evaluated at a temperature of \(300 \mathrm{~K}\). Determine the temperature \(T_{s}\) of the towel. What is the corresponding evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long?

Short Answer

Expert verified
The temperature of the wet towel (\(T_s\)) is \(20.25^{\circ} \mathrm{C}\), and the evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long is \(1.16 \times 10^{-5} \text{ kg/s}\).

Step by step solution

01

Calculate net solar radiation received

Net solar radiation received can be calculated by multiplying the solar irradiation with the solar absorptivity.\[Q_{\text {solar }}=A_s G_{S}\]Where \(A_s\) is the solar absorptivity and \(G_S\) is the solar irradiation.\[Q_{\text {solar }}=0.65 \times 900=585 \text{ W/m}^2\]
02

Calculate long-wave radiation emitted by the towel

To calculate the long-wave radiation emitted by the towel, we use the Stefan-Boltzmann law.\[Q_{\text {Tlwr }}=e \sigma T_{s}^{4}\]Where, \(e\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \(\left( 5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)\), and \(T_s\) is the temperature of the towel (in Kelvin).
03

Calculate net long-wave radiation received

Net long-wave radiation received by the towel is the difference between atmospheric radiation and ground radiation.\[Q_{\text {net _{lwr }}}=G_{\text {tam }}+G_{g}\]Plugging in the given values,\[Q_{\text {net _{lwr }}}=200+250 = 450 \text{ W/m}^2\]
04

Calculate net radiation absorbed

The net radiation absorbed by the towel is the difference between the net solar radiation and net long-wave radiation.\[Q_{\text {net }}=Q_{\text {solar }}-Q_{\text {net _{lwr }}}\]Substitute the values,\[Q_{\text {net }}=585-450 = 135 \text{ W/m}^2\]
05

Calculate heat transfer by conduction (convection)

Heat transfer by conduction (convection) is given by\[Q_{c}=h * A (T_a - T_s)\]Where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area of the towel, and \(T_a\) is the ambient temperature. The surface area of the towel can be calculated as:\[A=2 \times 0.75 \times 1.50 = 2.25 \text{ m}^2\]We are given \(h = 20 \,\text{W/m}^2\cdot\text{K}\), thus the heat transfer by conduction (convection) is given by: \[Q_{c}=20 * (T_a - T_s)\]
06

Calculate heat balance

The heat balance is given by \[Q_{\text {net }}=Q_{c}\]We have two unknowns \(T_s\) and \(Q_c\), so replacing unknowns in the equation, we get: \[135=20 * (27 - T_s)\]
07

Find the temperature of the towel

Solve the equation for \(T_s\): \[T_s=27-\frac{135}{20}\]Therefore,\[T_s=20.25 ^{\circ} \mathrm{C}\]
08

Calculate evaporation rate

We know that the heat transfer due to evaporation \(Q_e\) is given by:\[Q_e=mL\]Where \(m\) is the evaporation rate and \(L\) is the latent heat of vaporization. For water, \(L = 2.26 \times 10^{6} \mathrm{J / kg}\). The heat transfer due to evaporation is equal to the heat transfer by conduction (convection). Thus, we have:\[Q_c=mL\]Plugging in given values,\[20 * (27 - 20.25)=m * (2.26 \times 10^{6})\]Therefore, the evaporation rate is: \[m=\frac{20 * (27 - 20.25)}{(2.26 \times 10^{6})}=1.16 \times 10^{-5} \text{ kg/s}\] So, the temperature \(T_s\) of the wet towel is \(20.25^{\circ} \mathrm{C}\), and the evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long is \(1.16 \times 10^{-5} \text{ kg/s}\)

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Key Concepts

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

Solar Irradiation
Solar irradiation is a critical factor in various heat transfer calculations, including those related to drying a wet towel outdoors. It refers to the power per unit area received from the Sun in the form of electromagnetic radiation. In our problem, the solar irradiation (\(G_{S}\)) was given as 900 W/m², and the solar absorptivity (\(A_s\)) of the towel determined how much of this irradiation was actually absorbed by the towel to contribute to its drying. In general, higher solar irradiation and higher absorptivity result in greater heat input and faster drying times.
Understanding sunlight's role in heat transfer is crucial for applications ranging from solar panel design to climate control in buildings. When considering solar irradiation in heat calculations, it's also important to factor in the angle of sunlight, duration of exposure, and the surface's reflective properties. In our exercise, the straightforward multiplication of solar irradiation by solar absorptivity provided an easy way to gauge the net solar heat gain by the towel.
Solar irradiation is often a variable condition, dependent on weather, geographic location, and time of day. Therefore, mastering these calculations is essential for accurately predicting the behavior in real-world scenarios, such as drying laundry or harnessing solar power.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a cornerstone of thermal physics, describing how the amount of thermal radiation emitted by a black body in space is related to its temperature. Specifically, the law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (\(Q_{\text{Tlwr}}\) in our exercise) is directly proportional to the fourth power of the black body's absolute temperature (\(T_s\) in the exercise).
In mathematical terms, it is expressed as:
\[\begin{equation} Q_{\text{Tlwr}} = e \times \text{σ} \times T_{s}^{4} \text{, where \text{\text{e}} is the emissivity, and \text{σ} is the Stefan-Boltzmann constant.} \br\br\br\br\end{equation}\] In the context of our towel-drying problem, we used this law to estimate the amount of long-wave radiation emitted by the towel. The emissivity value provided (\(e = 0.96\)) indicates that the towel is a near-perfect emitter, coming close to the ideal black body assumption. This law is pivotal for understanding how objects emit heat through radiation and is widely used in fields such as engineering, astrophysics, and environmental science.
Convection Heat Transfer
Convection heat transfer plays a significant role in the drying process and is driven by the movement of fluids (in this case, air) over the towel's surfaces. The convection heat transfer coefficient (\(h\)), which was given as 20 W/m²⋅K for our problem, quantifies the rate of heat transfer per unit area per degree of temperature difference between the towel and the surrounding air.
In a practical scenario, convection is impacted by factors such as the properties of the fluid, the flow regime (laminar or turbulent), the temperature differential, and the physical configuration of the system. Windy conditions, described as 'moderately windy' in the exercise, enhance convection by moving more air over the towel's surfaces, which in turn can increase the evaporation rate and hasten drying.
When calculating heat transfer by convection, the formula is usually represented as \[\begin{equation}Q_c = h \times A \times (T_a - T_s)\text{. Here, \text{A} is the surface area, and \text{T_a} and \text{T_s} are the ambient and surface temperatures, respectively.}\br\br\br\br\end{equation}\] The larger the temperature differential and the higher the convection heat transfer coefficient, the greater the rate of heat transfer. Understanding convection is essential for a broad range of applications, from designing efficient cooling systems to improving weather prediction models.
Evaporation Rate
The evaporation rate is a measure of how quickly a liquid turns into vapor, which is central to drying processes such as a wet towel on a clothesline. In the heat transfer context, evaporation involves the removal of heat from a surface due to the phase change of water from liquid to vapor, requiring energy known as the latent heat of vaporization (\(L\)).
To determine the evaporation rate (\(m\)), we use the relationship \[\begin{equation}Q_e = mL\text{. In our exercise, we equated the heat transfer due to evaporation to the heat transfer by convection to the towel, leading to \text{m} being the variable of interest.}\br\br\br\br\end{equation}\] The rate of evaporation increases with greater surface area exposure, higher air temperatures, lower humidity, and greater airflow—all of which apply to our towel-drying scenario. These factors can be manipulated in industrial processes to optimize drying times, such as in the design of dehydrators or the scheduling of agricultural practices. The concept of evaporation rate is not only important in drying clothes but also pivotal in fields such as meteorology, environmental engineering, and agriculture.

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Most popular questions from this chapter

Two small surfaces, \(A\) and \(B\), are placed inside an isothermal enclosure at a uniform temperature. The enclosure provides an irradiation of \(6300 \mathrm{~W} / \mathrm{m}^{2}\) to each of the surfaces, and surfaces A and B absorb incident radiation at rates of 5600 and \(630 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Consider conditions after a long time has elapsed. (a) What are the net heat fluxes for each surface? What are their temperatures? (b) Determine the absorptivity of each surface. (c) What are the emissive powers of each surface? (d) Determine the emissivity of each surface.

The \(50-\mathrm{mm}\) peephole of a large furnace operating at \(450^{\circ} \mathrm{C}\) is covered with a material having \(\tau=0.8\) and \(\rho=0\) for irradiation originating from the furnace. The material has an emissivity of \(0.8\) and is opaque to irradiation from a source at room temperature. The outer surface of the cover is exposed to surroundings and ambient air at \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that convection effects on the inner surface of the cover are negligible, calculate the heat loss by the furnace and the temperature of the cover.

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of \(30-\mathrm{mm}\) diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The energy flux associated with solar radiation incident on the outer surface of the earth's atmosphere has been accurately measured and is known to be \(1368 \mathrm{~W} / \mathrm{m}^{2}\). The diameters of the sun and earth are \(1.39 \times 10^{9}\) and \(1.27 \times 10^{7} \mathrm{~m}\), respectively, and the distance between the sun and the earth is \(1.5 \times 10^{11} \mathrm{~m}\). (a) What is the emissive power of the sun? (b) Approximating the sun's surface as black, what is its temperature? (c) At what wavelength is the spectral emissive power of the sun a maximum? (d) Assuming the earth's surface to be black and the sun to be the only source of energy for the earth, estimate the earth's surface temperature.

A manufacturing process involves heating long copper rods, which are coated with a thin film, in a large furnace whose walls are maintained at an elevated temperature \(T_{w}\). The furnace contains quiescent nitrogen gas at 1-atm pressure and a temperature of \(T_{x}=T_{w}\). The film is a diffuse surface with a spectral emissivity of \(\varepsilon_{\lambda}=0.9\) for \(\lambda \leq 2 \mu \mathrm{m}\) and \(\varepsilon_{\lambda}=0.4\) for \(\lambda>2 \mu \mathrm{m}\). (a) Consider conditions for which a rod of diameter \(D\) and initial temperature \(T_{i}\) is inserted in the furnace, such that its axis is horizontal. Assuming validity of the lumped capacitance approximation, derive an equation that could be used to determine the rate of change of the rod temperature at the time of insertion. Express your result in terms of appropriate variables. (b) If \(T_{w}=T_{\mathrm{o}}=1500 \mathrm{~K}, T_{i}=300 \mathrm{~K}\), and \(D=10 \mathrm{~mm}\), what is the initial rate of change of the rod temperature? Confirm the validity of the lumped capacitance approximation. (c) Compute and plot the variation of the rod temperature with time during the heating process.

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