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Chapter 6: Problem 82

A 2-mm-thick layer of water on an electrically heated plate is maintained at a temperature of \(T_{w}=340 \mathrm{~K}\), as dry air at \(T_{\infty}=300 \mathrm{~K}\) flows over the surface of the water (case A). The arrangement is in large surroundings that are also at \(300 \mathrm{~K}\). (a) If the evaporative flux from the surface of the water to the air is \(n_{\mathrm{A}}^{\prime \prime}=0.030 \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2}\), what is the corresponding value of the convection mass transfer coefficient? How long will it take for the water to completely evaporate? (b) What is the corresponding value of the convection heat transfer coefficient and the rate at which electrical power must be supplied per unit area of the plate to maintain the prescribed temperature of the water? The emissivity of water is \(\varepsilon_{w}=0.95\). (c) If the electrical power determined in part (b) is maintained after complete evaporation of the water (case B), what is the resulting temperature of the plate, whose emissivity is \(\varepsilon_{p}=0.60\) ?

Short Answer

Expert verified
The convection mass transfer coefficient is \(h_m = 0.006\, kg/s\cdot m^2\cdot K\), and it will take 66.67 s for the water to completely evaporate. The convection heat transfer coefficient is \(h = 13542\, W/m^2\cdot K\), and the required electrical power supply per unit area is \(289691\, W/m^2\). After the complete evaporation of the water, the temperature of the plate will be \(T_p = 410.79\, K\).

Step by step solution

01

Convert evaporative flux to mass transfer coefficient

We are given the evaporative flux, and we need to find the corresponding convection mass transfer coefficient. We can use the following equation: \(n_A'' = h_m (T_w - T_\infty)\) Rearrange the equation to solve for \(h_m\): \(h_m = \frac{n_A''}{(T_w - T_\infty)}\) Now, plug in the given values: \(h_m = \frac{0.030\, kg/s\cdot m^2}{(340 \, K - 300\, K)}\) \(h_m = 0.006\, kg/s\cdot m^2\cdot K\)
02

Calculate the time for water evaporation

To find out how long it will take for the water to evaporate completely, we have to use the following formula: \(t = \frac{m_{water}}{n_A'' A}\) First, we need to calculate the mass of water in 1 m虏. We know that the water layer is 2 mm thick. The density of water is 1000 kg/m鲁. The mass of water per unit area can be calculated as follows: \(m_{water} =1000\, kg/m^3 \cdot 0.002\, m = 2\, kg/m^2\) Now, we can find the time it takes for the water to evaporate: \(t = \frac{2\, kg/m^2}{0.030\, kg/s\cdot m^2} = 66.67\, s\)
03

Find the convection heat transfer coefficient

To find the convection heat transfer coefficient, we can use the following equation: \(q'' = h (T_w - T_\infty)\) Rearrange the equation to solve for \(h\): \(h = \frac{q''}{(T_w - T_\infty)}\) We know from step 1 that the mass transfer coefficient is: \(h_m = 0.006\, kg/s\cdot m^2\cdot K\) We can use the following relationship between the mass transfer and heat transfer coefficients: \(h = h_m \cdot L\) Where L is the latent heat of evaporation of water, approximately equal to \(2.257 \times 10^6\, J/kg\) Now, we can calculate the heat transfer coefficient: \(h = 0.006\, kg/s\cdot m^2\cdot K \cdot 2.257 \times 10^6\, J/kg = 13542\, W/m^2\cdot K\)
04

Calculate the electrical power supply per unit area

We have to find the electrical power supply per unit area required to maintain the prescribed temperature of the water. We know that the heat transfer includes both convective and radiative components: \(q'' = q_\text{supply} + q_\text{radiation}\) We can find the convective component by using the heat transfer coefficient we found in step 3: \(q_\text{convective} = h(T_w - T_\infty) = 13542\, W/m^2\cdot K \cdot (340\, K - 300\, K) = 541680\, W/m^2\) Now we need to calculate the radiative component. We know the emissivity of the water surface is \(\varepsilon_w = 0.95\), the Stefan-Boltzmann constant is \(5.67 \times 10^{-8}\, W/m^2\cdot K^4\), and the surrounding temperature is \(T_\infty = 300\, K\). We can calculate the radiative component as follows: \(q_\text{radiation} = \varepsilon_w \cdot \sigma \cdot (T_w^4 - T_\infty^4) = 0.95 \cdot 5.67 \times 10^{-8}\, W/m^2\cdot K^4 \cdot (340^4 - 300^4)\) \(q_\text{radiation} = 251989\, W/m^2\) Now we can find the electrical power supply per unit area: \(q_\text{supply} = q'' - q_\text{radiation} = 541680\, W/m^2 - 251989\, W/m^2 = 289691\, W/m^2\)
05

Calculate the temperature of the plate after evaporation

First, we need to determine the overall heat transfer from the plate and the surroundings: \(q_\text{radiation,p} = \varepsilon_p \cdot \sigma \cdot (T_p^4 - T_\infty^4)\) We know the emissivity of the plate, \(\varepsilon_p = 0.60\), and the electrical power found in step 4, \(q_\text{supply} = 289691\, W/m^2\) is now only lost through radiation: \(q_\text{supply} = \varepsilon_p \cdot \sigma\cdot(T_p^4 - T_\infty^4)\) Rearrange the equation for \(T_p^4\): \(T_p^4 = T_\infty^4 + \frac{q_\text{supply}}{\varepsilon_p \cdot \sigma}\) Plug in the given values: \(T_p^4 = 300^4 + \frac{289691\, W/m^2}{0.60 \cdot 5.67 \times 10^{-8}\, W/m^2\cdot K^4}\) \(T_p^4 = 9.135\times10^9\) Now, take the fourth root to find the plate temperature: \(T_p = \sqrt[4]{9.135\times10^9} = 410.79\, K\)

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Key Concepts

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

Convection Mass Transfer Coefficient
Understanding the convection mass transfer coefficient involves recognizing how mass moves from one area to another because of a fluid's activity, such as air or water flow. This coefficient, represented by \( h_m \), signifies how efficiently particles, like evaporating water molecules, are transported due to this flow.
It tells us the rate at which mass is transferred between phases (solid, liquid, gas) under specific conditions. In our example, with a known evaporative flux \( n_A'' \), we used the equation \( n_A'' = h_m (T_w - T_\infty) \) to identify \( h_m \) by rearranging it to \( h_m = \frac{n_A''}{(T_w - T_\infty)} \).
This coefficient is pivotal in predicting how quickly substances like liquids will move from their surface into the surrounding air, a key component when discussing evaporation rates.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient, denoted as \( h \), represents the ability of a fluid (like air) to transport heat away from a surface. It's a critical factor in determining how heat moves from a heated object to its surroundings. To calculate \( h \), we used the relationship \( q'' = h (T_w - T_\infty) \).

What's crucial here is understanding that heat can be transferred not only by direct conduction but also by the motion of the fluid surrounding the heated surface. This coefficient allows us to measure and predict this effect, considering the temperature difference between a surface and the ambient fluid and combining this with the latent heat of evaporation when relevant.
  • Allows estimation of heat transfer rates
  • Determines how effectively heat is dispersed in environments
By understanding \( h \), engineers can design systems to maintain desired temperatures effectively.
Latent Heat of Evaporation
The latent heat of evaporation is an essential concept in thermodynamics, denoting the heat required to convert a liquid into a vapor without a temperature change. In our case, it serves as a bridge between mass and heat transfer coefficients. Represented typically as \( L \), it is a measure of the energy needed to overcome molecular forces during phase changes.
In calculations, it connects mass transfer to thermal processes, allowing for the determination of heat transfer when a substance changes phase. The formula \( h = h_m \cdot L \) was used to relate these two phenomena. With water, this value is often significant, such as \( 2.257 \times 10^6 \, J/kg \).

Latent heat is crucial in:
  • Understanding energy transformations during evaporation
  • Designing efficient heating/cooling systems
By considering latent heat, one can accurately predict how much energy a substance needs to undergo a phase change.
Evaporative Flux
Evaporative flux quantifies the rate at which a substance converts from liquid to vapor. Given units of \( kg/s \cdot m^2 \), it reflects how much mass evaporates per unit area over time. Evaporative flux leads us directly to understanding how substances like water disappear from heated surfaces.
In practical terms, it's calculated when we know how long it takes for a liquid's surface to lose mass to the surrounding air. In the example, understanding evaporative flux enabled us to calculate the convection mass transfer coefficient, contributing to our understanding of the overall evaporative process.
Understanding evaporative flux helps in:
  • Predicting drying or evaporation rates
  • Assessing environmental impact of water/nutrient loss
This measure is crucial for designs in chemical processes, helping engineers ensure desired outcomes in efficiency and safety.

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Most popular questions from this chapter

Forced air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(V=10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic elements on a circuit board. One such element is a chip, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), located \(120 \mathrm{~mm}\) from the leading edge of the board. Experiments have revealed that flow over the board is disturbed by the elements and that convection heat transfer is correlated by an expression of the form Estimate the surface temperature of the chip if it is dissipating \(30 \mathrm{~mW}\).

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.

Photosynthesis, as it occurs in the leaves of a green plant, involves the transport of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) from the atmosphere to the chloroplasts of the leaves. The rate of photosynthesis may be quantified in terms of the rate of \(\mathrm{CO}_{2}\) assimilation by the chloroplasts. This assimilation is strongly influenced by \(\mathrm{CO}_{2}\) transfer through the boundary layer that develops on the leaf surface. Under conditions for which the density of \(\mathrm{CO}_{2}\) is \(6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\) in the air and \(5 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\) at the leaf surface and the convection mass transfer coefficient is \(10^{-2} \mathrm{~m} / \mathrm{s}\), what is the rate of photosynthesis in terms of kilograms of \(\mathrm{CO}_{2}\) assimilated per unit time and area of leaf surface?

The defroster of an automobile functions by discharging warm air on the inner surface of the windshield. To prevent condensation of water vapor on the surface, the temperature of the air and the surface convection coefficient \(\left(T_{\infty, j}, \overline{h_{i}}\right)\) must be large enough to maintain a surface temperature \(T_{s i}\) that is at least as high as the dewpoint \(\left(T_{s, i} \geq T_{d \mathrm{p}}\right)\). Consider a windshield of length \(L=800 \mathrm{~mm}\) and thickness \(t=6 \mathrm{~mm}\) and driving conditions for which the vehicle moves at a velocity of \(V=70 \mathrm{mph}\) in ambient air at \(T_{\infty \rho}=-15^{\circ} \mathrm{C}\). From laboratory experiments performed on a model of the vehicle, the average convection coefficient on the outer surface of the windshield is known to be correlated by an expression of the form \(\overline{N_{L}}=0.030 \operatorname{Re}_{L}^{0.8} \operatorname{Pr}^{1 / 3}\), where \(R e_{L}=V L \nu\). Air properties may be approximated as \(k=0.023 \mathrm{~W} / \mathrm{m}=\mathrm{K}\), \(v=12.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and \(\operatorname{Pr}=0.71\). If \(T_{d p}=10^{\circ} \mathrm{C}\) and \(T_{m, j}=50^{\circ} \mathrm{C}\), what is the smallest value of \(\bar{h}_{j}\) required to prevent condensation on the inner surface?

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.
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