/*! 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 137 A grain dryer consists of a long... [FREE SOLUTION] | 91Ó°ÊÓ

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A grain dryer consists of a long semicircular duct of radius \(R=1 \mathrm{~m}\). One-half of the base surface consists of an electrically heated plate of emissivity \(z_{p}=0.8\), while the other half supports the grain to be dried, which has an emissivity of \(\varepsilon_{g}=0.9\). In a batch drying process for which the temperature of the grain is \(T_{s}=330 \mathrm{~K}, 2.50 \mathrm{~kg}\) of water are to be removed per meter of duct length over a 1-h period. (a) Neglecting convection heat transfer, determine the required temperature \(T_{p}\) of the heater plate. (b) If the water vapor is swept from the duct by the flow of dry air, what convection mass transfer coefficient \(h_{w}\) must be maintained by the flow? (c) If the air is at \(300 \mathrm{~K}\), is the assumption of negligible convection justified?

Short Answer

Expert verified
To determine the temperature of the heater plate (Tp), we first calculate the total energy required to evaporate the water and the energy exchanged through thermal radiation using the Stefan-Boltzmann law. Then, we solve for Tp by equating the energies. To determine the convection mass transfer coefficient (hw), we calculate the mass flow rate of water to be evaporated, the partial pressures of water vapor before and after evaporation, and then apply the mass transfer equation to solve for hw. To justify the negligible convection assumption, we calculate the energy exchanged through convection and compare it to the radiation energy calculated previously. If the energy through convection is significantly smaller, then the assumption is justified.

Step by step solution

01

(a) Determining the Required Temperature of the Heater Plate (Tp)

: Step 1: Compute the total energy needed to evaporate water The energy required to evaporate water per meter of the dryer duct can be calculated using the mass of water to be evaporated and the enthalpy of vaporization for water (\(h_{fg}\)): Total energy needed: \(Q = m_{water} \times h_{fg}\) Step 2: Calculate the energy exchanged through thermal radiation We will use the Stefan-Boltzmann law relating energy exchanged through thermal radiation and the temperature of the heater plate and grain: \(Q = \sigma \times A \times (T_p^4 - T_s^4) \times z_p \times \varepsilon_g\) Step 3: Solve for the temperature of the heater plate (Tp) Combine step 1 and step 2 equations, then solve for Tp: \(\sigma \times A \times (T_p^4 - T_s^4) \times z_p \times \varepsilon_g = m_{water} \times h_{fg}\) Now, you can solve for Tp.
02

(b) Determining the Convection Mass Transfer Coefficient (hw)

: Step 1: Calculate the mass flow rate of water to be evaporated Mass flow rate (MW) can be calculated as follows: \(MW = \frac{m_{water}}{time}\) Step 2: Calculate the partial pressures of water vapor before and after evaporation For this step, we require the saturation pressure of water in both situations, at grain temperature (330 K) and air temperature (300 K). Look up values in relevant tables or use the Antoine equation to find saturation pressure. Then, find partial pressure using relative humidity (RH) at the grain surface: \(P_{w, 330K} = P_{sat, 330K} \times RH_{330K}\) \(P_{w, 300K} = P_{sat, 300K} \times RH_{300K}\) Step 3: Calculate the convection mass transfer coefficient (hw) Apply the mass transfer equation: \(MW = A \times h_w \times (P_{w, 330K} - P_{w, 300K})\) Now you can solve for hw.
03

(c) Justify the Negligible Convection Assumption

: Step 1: Calculate the energy exchanged through convection We assume a convection heat transfer coefficient \(h_c\), which can be determined using methods such as empirical correlations, and the temperature difference between the air and grain. \(Q_{convection} = h_c \times A \times (330K - 300K)\) Step 2: Compare the energy exchanged through convection to radiation energy To justify the negligible convection assumption, the energy exchanged through convection \(Q_{convection}\) should be significantly smaller than the energy exchanged through radiation \(Q_{radiation}\) calculated previously in part (a). Compare the relative magnitudes of these energies: \(\frac{Q_{convection}}{Q_{radiation}} < 1\) If this condition holds, then we can consider the convection to be negligible.

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Key Concepts

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

Thermal Radiation
Understanding the phenomenon of thermal radiation is pivotal when dealing with heat transfers that involve electromagnetic waves emitted by a body due to its temperature. Unlike conduction and convection, thermal radiation does not require a medium to transfer energy; it can occur in a vacuum. In our grain dryer example, the heater plate emits infrared radiation that carries energy to the grain, facilitating the drying process.

It's important to note that all bodies emit this kind of radiation and the amount of energy radiated depends on their temperature and surface characteristics, such as emissivity. Emissivity is a measure of how effectively a material radiates energy compared to a black body, which is a perfect emitter, with an emissivity value of 1.
Evaporative Drying
The process of evaporative drying is commonly used in agricultural applications, such as the drying of grains. It involves the removal of moisture from a product by evaporation. When water inside the grain evaporates due to heat (from the heater plate in this case), it transitions from a liquid to a vapor, requiring energy known as the enthalpy of vaporization.

The efficiency of evaporative drying depends on several factors, including the temperature and humidity of the surrounding air, the physical properties of the material being dried, and the rate of airflow over the material. Improving the efficiency of evaporative drying typically involves optimizing these factors to increase the rate of moisture removal while minimizing energy usage.
Stefan-Boltzmann Law
The Stefan-Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, it states that the total energy radiated per unit surface area of a black body is proportional to the fourth power of the black body's absolute temperature. The law is mathematically expressed as:
\[ Q = \sigma A T^4 \]
where \(Q\) is the total power radiated, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} W/m^2K^4\), \(A\) is the surface area of the emitting body, and \(T\) is the absolute temperature of the body. For non-black bodies, the emissivity factor is included, hence the formula used in the grain dryer scenario is modified to account for the emissivity of the heated plate and grain.
Convection Heat Transfer
Convection heat transfer is a mode of thermal energy transfer due to the motion of fluid, such as air or water, moving across the surface of an object. This energy movement can be naturally caused by differences in temperature and density within the fluid (natural convection) or could be a result of an external force like a pump or fan (forced convection).

In our dryer example, if there were significant airflow over the grain, convection would contribute to heat transfer. Convective heat transfer is often described by a convection heat transfer coefficient, \(h_c\), which quantifies how effective the convection process is in moving heat between a surface and a fluid flowing over it. The units of \(h_c\) are typically in watts per square meter-kelvin (W/m²·K).
Mass Transfer Coefficient
The mass transfer coefficient, \(h_w\), is a measure of the rate of mass transfer per unit area per unit driving force, which, in the scenario of drying grains, is related to the evaporation of moisture from the grains into the surrounding air. The driving force is typically a difference in concentration or, in the case of evaporation, a difference in partial pressures between the two phases.
Mathematically, it can be expressed in the following equation for a flat surface:
\[ MW = h_w A (P_{w1} - P_{w2}) \]
where \(MW\) is the mass flow rate of water vapor, \(A\) is the surface area, \(h_w\) is the mass transfer coefficient, and \(P_{w1}\) and \(P_{w2}\) are the partial pressures of water vapor at two different points. The higher the value of \(h_w\), the more efficient the mass transfer process is, suggesting that less time or smaller areas are required for a given mass transfer to occur.

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Most popular questions from this chapter

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

Consider coaxial, parallel, black disks separated a distance of \(0.20 \mathrm{~m}\). The lower disk of diameter \(0.40 \mathrm{~m}\) is maintained at \(500 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). What temperature will the upper disk of diameter \(0.20 \mathrm{~m}\) achieve if electrical power of \(17.5 \mathrm{~W}\) is supplied to the heater on the back side of the disk?

Consider two very large parallel plates. The botton plate is warmer than the top plate, which is held at a constant temperature of \(T_{\mathrm{1}}=330 \mathrm{~K}\). The plates are separated by \(L=0.1 \mathrm{~m}\), and the gap between the two surfaces is filled wath air at atmospheric pressure. The heat flux from the bottom plate is \(q^{*}=250 \mathrm{~W} \mathrm{~m}^{2}\). (a) Determine the temperature of the bottom plate and the ratio of the convective to radiative heat fluxes for \(\varepsilon_{1}=\varepsilon_{2}=0.5\). Evaluate air properties at \(T=350 \mathrm{~K}\). (b) Repeat part (a) for \(\varepsilon_{1}=\varepsilon_{2}=0.25\) and \(0.75\).

A radiant heater, which is used for surface treatment processes, consists of a long cylindrical heating element of diameter \(D_{1}=0.005 \mathrm{~m}\) and emissivity \(\varepsilon_{1}=0.80\). The heater is partially enveloped by a long. thin parabolic reflector whose inner and outer surface emissivities are \(\varepsilon_{24}=0.10\) and \(\varepsilon_{20}=0.80\), respectively. Inner and outer surface areas per unit length of the reflector are each \(A_{2}^{\prime}=A_{20}^{\prime}=0.20 \mathrm{~m}\), and the average convection coefficient for the combined inner and outer surfaces is \(\bar{h}_{2 \dot{m}}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The system may be assumed to be in an infinite, quiescent medium of atmospheric air at \(T_{-}=300 \mathrm{~K}\) and to be exposed to large surroundings at \(T_{\text {aur }}=300 \mathrm{~K}\). (a) Sketch the appropriate radiation circuit, and write expressions for each of the network resistances. (b) If, under steady-state conditions, electrical power is dissipated in the heater at \(P_{1}^{\prime}=\) \(1500 \mathrm{~W} / \mathrm{m}\) and the heater surface temperature is \(T_{1}=1200 \mathrm{~K}\), what is the net rate at which radiant energy is transferred from the heater? (c) What is the net rate at which radiant energy is transferred from the heater to the surroundings? (d) What is the temperature, \(T_{2}\), of the reflector?

2 A composite wall is comprised of two large plates separated by sheets of refractory insulation, as shown in the schematic. In the installation process, the sheets of thickness \(L=50 \mathrm{~mm}\) and thermal conductivity \(k=0.05 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) are separated at 1 -m intervals by gaps of width \(w=10 \mathrm{~mm}\). The hot and cold plates have temperatures and emissivities of \(T_{1}=400^{\circ} \mathrm{C}\). \(\varepsilon_{1}=0.85\) and \(T_{2}=35^{\circ} \mathrm{C}, \varepsilon_{2}=0.5\), respectively. Assume that the plates and insulation are diffuse-gray surfaces. (a) Determine the heat loss by radiation through the gap per unit length of the composite wall (normal to the page). (b) Recognizing that the gaps are located on a 1-m spacing, determine what fraction of the total heat loss through the composite wall is due to transfer by radiation through the insulation gap.

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