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Chapter 6: Problem 52

On a summer day the air temperature is \(27^{\circ} \mathrm{C}\) and the relative humidity is \(30 \%\). Water evaporates from the surface of a lake at a rate of \(0.10 \mathrm{~kg} / \mathrm{h}\) per square meter of water surface area. The temperature of the water is also \(27^{\circ} \mathrm{C}\). Determine the value of the convection mass transfer coefficient. 6.53 It is observed that a 230 -mm-diameter pan of water at \(23^{\circ} \mathrm{C}\) has a mass loss rate of \(1.5 \times 10^{-5} \mathrm{~kg} / \mathrm{s}\) when the ambient air is dry and at \(23^{\circ} \mathrm{C}\). (a) Determine the convection mass transfer coefficient for this situation. (b) Estimate the evaporation mass loss rate when the ambient air has a relative humidity of \(50 \%\). (c) Estimate the evaporation mass loss rate when the water and ambient air temperatures are \(47^{\circ} \mathrm{C}\), assuming that the convection mass transfer coefficient remains unchanged and the ambient air is dry.

Short Answer

Expert verified
The convection mass transfer coefficient for the lake situation is \(7.71 \times 10^{-9} \, \mathrm{kg/m^2s}\). To solve part a), b), and c) for the 230 mm diameter pan, follow the same steps as shown for calculating the convection mass transfer coefficient, taking into account the different temperatures and relative humidity values given.

Step by step solution

01

List the given information

In this problem, we have: 1. Air temperature: \(T_{air} = 27 ^\circ \mathrm{C}\) 2. Relative humidity: 30% 3. Evaporation rate: 0.10 kg/h per square meter 4. Water temperature: \(T_{water} = 27^\circ \mathrm{C}\) For the second situation, we are given: 1. Pan diameter: 230 mm 2. Mass loss rate: \(1.5 \times 10^{-5} \mathrm{~kg} / \mathrm{s}\) 3. Air temperature: \(T_{air} = 23^\circ \mathrm{C}\) 4. Water temperature: \(T_{water} = 23^\circ \mathrm{C}\)
02

Calculate the partial pressure of the water vapor

In order to calculate the convection mass transfer coefficient, we first need to determine the partial pressure of the water vapor (Pa) at the given temperature and humidity conditions. We can use the relation: \(P_v = \phi \times P_v^*\) where: - \(P_v\) is the partial pressure of water vapor (Pa), - \(\phi\) is the relative humidity (dimensionless), - \(P_v^*\) is the saturation pressure of dry air at the given temperature (Pa). Since the relative humidity is 30%, we have: \(P_v = 0.30 \times P_v^*\) Next, we need to find \(P_v^*\) at the given temperature. We will use Antoine's equation for this: \(P_v^* = 10^{A - \frac{B}{T + C}}\) where A, B, and C are constants at the given temperature. For the given air temperature of \(27^\circ \mathrm{C}\), the constants are: - A = 8.07131 - B = 1730.63 - C = 233.426 Now, we can calculate the saturation pressure: \(P_v^* = 10^{8.07131 - \frac{1730.63}{27 + 233.426}} = 3570.75 \, \mathrm{Pa}\)
03

Calculate the convection mass transfer coefficient

Now that we have the partial pressure, we can use Fick's law to calculate the convection mass transfer coefficient (\(h_m\)): \(h_m = \frac{dm}{dt} \times \frac{1}{A \times (P_v - P_v^*)}\) where: - \(h_m\) is the convection mass transfer coefficient (kg/m虏s), - \(dm/dt\) is the evaporation rate (kg/h), - A is the surface area of the lake (m虏), - \(P_v\) and \(P_v^*\) are the partial pressures of the water vapor calculated in Step 2. In our case, the evaporation rate is given to be 0.10 kg/h per square meter. Therefore: \(h_m = \frac{0.10 \, \mathrm{kg/h}}{3600 \, \mathrm{s/h}} \times \frac{1}{1 \times (3570.75 \, \mathrm{Pa})} = 7.71 \times 10^{-9} \, \mathrm{kg/m^2s}\) Hence, the convection mass transfer coefficient is \(7.71 \times 10^{-9} \, \mathrm{kg/m^2s}\). In part a), for 230 mm diameter pan, we repeat steps 2 and 3 with data for a 230 mm diameter pan, and we will find the mass transfer coefficient. In part b), with the mass transfer coefficient found in step a), we are now equipped to find the evaporation mass loss rate when the relative humidity is 50%. In part c), assuming the mass transfer coefficient remains unchanged, we can estimate the evaporation mass loss rate when the water and ambient air temperatures are 47 degrees Celsius.

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Key Concepts

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

Understanding the Evaporation Rate
The evaporation rate is the speed at which water transitions from a liquid to a vapor. It plays a critical role in phenomena like drying clothes, cooking, and the water cycle. Calculating the rate of evaporation can be crucial when trying to understand weather patterns, designing HVAC systems, or even when working on the intricacies of chemical processes.

In the context of the textbook exercise, the evaporation rate is given per unit area (square meter) and time (hour). This rate can be influenced by several factors such as the air temperature, water temperature, relative humidity, and the convection mass transfer coefficient. When relative humidity is low, the air can accept more water vapor, and thus evaporation occurs more rapidly, which is why knowing the evaporation rate can help predict how quickly a lake may lose water on a hot, dry day.

It's also important to note that evaporation cools the surface from which the water is evaporating. This is because the molecules with the highest energy are more likely to escape into the vapor phase, leaving the cooler ones behind. This effect is experienced on a hot day when you feel cooler as sweat evaporates from your skin.
The Role of Relative Humidity
Relative humidity is the amount of moisture that air contains compared to the maximum it could hold at a certain temperature. It鈥檚 expressed as a percentage. The higher the relative humidity, the less capacity the air has to absorb additional moisture, resulting in a lower evaporation rate. It鈥檚 intimately connected with people's comfort levels, preservation of materials, and many industrial processes.

In the given exercise, we see scenarios where the evaporation rate is affected by two different relative humidity levels: 30% and 50%. The drier the air (lower relative humidity), the faster water can evaporate into it. Understanding relative humidity is essential for predicting the evaporation rate under different environmental conditions.

For practical purposes like dehumidification in buildings or in drying technologies in industry, controlling relative humidity is vital to achieve the desired outcome. This is because the ability of the air to pick up moisture heavily depends on its relative humidity, and by extension, so does the efficiency of the evaporation process.
Fick's Law of Diffusion Explained
Fick's law of diffusion is fundamental in understanding how substances like water vapor move from regions of high concentration to areas of low concentration. In simpler terms, this law tells us that a substance will naturally spread out to fill the available space, moving from areas where there's a lot of it to areas where there's not so much.

Mathematically, Fick's first law states that the rate of transfer of a mass through a unit area is proportional to the negative gradient of concentration, often applied to find the diffusion coefficient in various materials. It's crucial for calculating spreading rates of gases, and in this case, the rate of evaporation.

When we discuss the convection mass transfer coefficient in our textbook problem, we're using the principles of Fick's law but with a focus on the convective movement of vapor rather than just molecular diffusion. This coefficient tells us how effectively mass is transferred in a fluid, which is, in this case, the water vapor in the air above the lake. The calculated coefficient is a measure of how efficiently the water molecules are being transported away from the surface of the lake.

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Most popular questions from this chapter

An industrial process involves evaporation of a thin water film from a contoured surface by heating it from below and forcing air across it. Laboratory measurements for this surface have provided the following heat transfer correlation: $$ \overline{N u_{L}}=0.43 R e_{L}^{0.58} P r^{0.4} $$ The air flowing over the surface has a temperature of \(290 \mathrm{~K}\), a velocity of \(10 \mathrm{~m} / \mathrm{s}\), and is completely dry \(\left(\phi_{\infty}=0\right)\). The surface has a length of \(1 \mathrm{~m}\) and a surface area of \(1 \mathrm{~m}^{2}\). Just enough energy is supplied to maintain its steady-state temperature at \(310 \mathrm{~K}\). (a) Determine the heat transfer coefficient and the rate at which the surface loses heat by convection. (b) Determine the mass transfer coefficient and the evaporation rate \((\mathrm{kg} / \mathrm{h})\) of the water on the surface. (c) Determine the rate at which heat must be supplied to the surface for these conditions.

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at \(100 \mathrm{~m} / \mathrm{s}\) in air that is at a temperature of \(-23^{\circ} \mathrm{C}\). If the characteristic length of the airfoil is \(L=2 \mathrm{~m}\) and wind tunnel measurements indicate an average friction coefficient of \(\bar{C}_{f}=0.0025\) for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of \(T_{s}=5^{\circ} \mathrm{C}\) ?

Experiments have been conducted to determine local heat transfer coefficients for flow perpendicular to a long, isothermal bar of rectangular cross section. The bar is of width \(c\) parallel to the flow, and height \(d\) normal to the flow. For Reynolds numbers in the range \(10^{4} \leq R_{d} \leq 5 \times 10^{4}\), the face-averaged Nusselt numbers are well correlated by an expression of the form The values of \(C\) and \(m\) for the front face, side faces, and back face of the rectangular rod are found to be the following: \begin{tabular}{llll} \hline Face & cld & \(\boldsymbol{C}\) & \(\boldsymbol{m}\) \\ \hline Front & \(0.33 \leq\) cld \(51.33\) & \(0.674\) & \(1 / 2\) \\ Side & \(0.33\) & \(0.153\) & \(2 / 3\) \\ Side & \(1.33\) & \(0.107\) & \(2 / 3\) \\ Back & \(0.33\) & \(0.174\) & \(2 / 3\) \\ Back & \(1.33\) & \(0.153\) & \(2 / 3\) \\ \hline \end{tabular} Determine the value of the average heat transfer coefficient for the entire exposed surface (that is, averaged over all four faces) of a \(c=40\)-mm-wide, \(d=30\)-mm-tall rectangular rod. The rod is exposed to air in cross flow at \(V=10 \mathrm{~m} / \mathrm{s}, T_{x}=300 \mathrm{~K}\). Provide a plausible explanation of the relative values of the face-averaged heat transfer coefficients on the front, side, and back faces.

For flow over a flat plate of length \(L\), the local heat transfer coefficient \(h_{x}\) is known to vary as \(x^{-1 / 2}\), where \(x\) is the distance from the leading edge of the plate. What is the ratio of the average Nusselt number for the entire plate \(\left(\overline{N u}_{L}\right)\) to the local Nusselt number at \(x=L\left(N u_{L}\right)\) ?

The naphthalene sublimation technique involves the use of a mass transfer experiment coupled with an analysis based on the heat and mass transfer analogy to obtain local or average convection heat transfer coefficients for complex surface geometries. A coating of naphthalene, which is a volatile solid at room temperature, is applied to the surface and is then subjected to airflow in a wind tunnel. Alternatively, solid objects may be cast from liquid naphthalene. Over a designated time interval, \(\Delta t\), there is a discernible loss of naphthalene due to sublimation, and by measuring the surface recession at locations of interest or the mass loss of the sample, local or average mass transfer coefficients may be determined. Consider a rectangular rod of naphthalene exposed to air in cross flow at \(V=10 \mathrm{~m} / \mathrm{s}, T_{\mathrm{s}}=300 \mathrm{~K}\), as in Problem 6.10, except now \(c=10 \mathrm{~mm}\) and \(d=30 \mathrm{~mm}\). Determine the change in mass of the \(L=500\)-mm-long rod over a time period of \(\Delta t=30 \mathrm{~min}\). Naphthalene has a molecular weight of \(M_{\mathrm{A}}=128.16 \mathrm{~kg} / \mathrm{kmol}\), and its solid-vapor saturation pressure at \(27^{\circ} \mathrm{C}\) and \(1 \mathrm{ltm}\) is \(p_{\text {A, } a t}=1.33 \times 10^{-4}\) bar.
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