Question

A dryer is shaped like a long semicylindrical duct of diameter \(1.5 \mathrm{~m}\). The base of the dryer is occupied with water soaked materials to be dried, and maintained at a temperature of \(370 \mathrm{~K}\) and emissivity of \(0.5\). The dome of the dryer is maintained at \(1000 \mathrm{~K}\) with emissivity of \(0.8\). Determine the drying rate per unit length experienced by the wet materials.

Short answer

Answer: The drying rate per unit length experienced by the wet materials in the semicylindrical dryer is approximately -99687/L W/m³.

Step-by-step solution

Calculate the radiative heat flux from the base

To find the radiative heat flux emitted by the base, we'll use the Stefan-Boltzmann law. The formula is given by: $$q_{rad} = \epsilon \sigma (T_1^4 - T_2^4),$$ where ε is the emissivity, σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} Wm^{-2}K^{-4}\)), and \(T_1\) and \(T_2\) are the absolute temperatures of the two surfaces in Kelvin. First, we will find the radiative heat flux emitted by the base towards the dome, \({q}_{rad_{base}}\): $${q}_{rad_{base}} = \epsilon_{base} \sigma (T_{base}^4 - T_{dome}^4)$$

Calculate the radiative heat flux from the dome

Using the same formula, we will find the radiative heat flux emitted by the dome towards the base, \({q}_{rad_{dome}}\): $${q}_{rad_{dome}} = \epsilon_{dome} \sigma (T_{dome}^4 - T_{base}^4)$$

Determine the net radiative heat flux

The net radiative heat flux \(q_{net}\) can be calculated by taking the difference between the radiative heat flux from the base and the radiative heat flux from the dome: $$q_{net} = {q}_{rad_{base}} - {q}_{rad_{dome}}$$

Calculate the drying rate per unit length

The drying rate per unit length is directly proportional to the net radiative heat flux. Since our dryer has a semicylindrical shape, we must consider the area of the semicircular cross-section, which can be calculated as: $$A = \frac{1}{2} (\pi r^2),$$ where \(r\) is the radius of the cross-section. The drying rate per unit length can then be found as: $$\frac{q_{net}}{A \times L},$$ where \(L\) is the length we are trying to determine. Now, we will plug in the given values and solve for the drying rate per unit length.

Plug in values and solve

From the given information, the values are: - Diameter = \(1.5 \mathrm{~m}\), so the radius is \(0.75 \mathrm{~m}\) - Base temperature, \(T_{base} = 370 \mathrm{~K}\) - Base emissivity, \(\epsilon_{base} = 0.5\) - Dome temperature, \(T_{dome} = 1000 \mathrm{~K}\) - Dome emissivity, \(\epsilon_{dome} = 0.8\) We will plug in these values and perform the calculations: - Calculate the area of the semicircular cross-section: $$A = \frac{1}{2} (\pi (0.75)^2) = 1.767 \mathrm{~m^2}$$ - Calculate the radiative heat fluxes: $${q}_{rad_{base}} = 0.5 \times 5.67 \times 10^{-8} (370^4 - 1000^4) = -72190 \mathrm{~W/m^2}$$ $${q}_{rad_{dome}} = 0.8 \times 5.67 \times 10^{-8} (1000^4 - 370^4) = 104069 \mathrm{~W/m^2}$$ - Calculate the net radiative heat flux: $$q_{net} = -72190 - 104069 = -176259 \mathrm{~W/m^2}$$ - Calculate the drying rate per unit length: $$\frac{-176259}{1.767 \times L} = \frac{-99687}{L} \mathrm{~W/m^3}$$ The drying rate per unit length experienced by the wet materials is approximately \(-99687/L \mathrm{~W/m^3}\).

Key concepts

Stefan-Boltzmann Law

At the heart of understanding radiative heat transfer in various engineering applications is the Stefan-Boltzmann law. This principle is fundamental when calculating the thermal radiation emitted by an object based on its temperature.

According to this law, the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature. Mathematically, it's expressed as: \[ E = \.sigma \.cdot T^4\], where \(E\) is the radiant energy emitted per unit area, \(\sigma\) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} Wm^{-2}K^{-4}\)), and \(T\) is the absolute temperature in Kelvin.

This formula simplifies the process of determining how much heat energy an object, like our dryer's base or dome, emits due to its temperature. However, in real-world applications, most objects are not perfect black bodies.

Emissivity

Emissivity is a measure of a material's ability to emit thermal radiation compared to that of a perfect black body, which has an emissivity of 1. Real objects have emissivities less than 1, like the dryer base with an emissivity of \(0.5\) and dome with \(0.8\), as in our exercise.

The emissivity value influences the radiative heat transfer calculations. For a non-black body, the Stefan-Boltzmann law is modified to include emissivity (\(\epsilon\)): \[ q_{rad} = \epsilon \cdot \sigma \cdot (T_1^4 - T_2^4) \]. This formula accounts for the fact that different materials will emit different amounts of energy even at the same temperature due to their unique surface properties. In our exercise, this understanding allows us to accurately determine the net radiative heat flux between the dryer's base and dome.

Understanding Emissivity in Practical Terms

Think of emissivity as a way to rate how good an object is at radiating energy. A matte black surface, for instance, has a high emissivity, meaning it's great at emitting heat, whereas a shiny metal surface may have a low emissivity, reflecting most of the heat energy instead.

Drying Rate Calculation

The drying rate in a process like the one encountered in our dryer setup is vital for optimizing the drying of materials. It refers to the amount of water or moisture removed from a material per unit time and can be influenced by factors such as temperature, airflow, and humidity.

In the context of radiative heat transfer, the drying rate per unit length can be related to the net radiative heat flux between the drying materials and their surroundings. By taking into account the area over which heat transfer occurs, we can calculate the rate at which moisture is expected to evaporate due to radiative heat.

For our semi-cylindrical dryer, the drying rate per unit area would be the net radiative heat flux divided by the effective area. Interestingly, the calculation comes with a twist: since heat is only radiated over the half-circle area of the semi-cylindrical cross-section, we must adjust the area in our computation, as highlighted in the step-by-step solution. Such specifics are crucial for producing an accurate drying rate, which is essential for designing efficient and effective drying systems.

Practical Tip for Drying Rate Efficiency

It's worth noting that to maximize efficiency in a drying process, one should consider not just the drying rate, but the uniformity of heat distribution and the maintenance of ideal drying conditions throughout the process. Improving these factors can lead to faster drying times and more energy-efficient operations.