Question

Consider a spherical reactor of \(5-\mathrm{cm}\) diameter operating at steady condition has a temperature variation that can be expressed in the form of \(T(r)=a-b r^{2}\), where \(a=850^{\circ} \mathrm{C}\) and \(b=5 \times 10^{5} \mathrm{~K} / \mathrm{m}^{2}\). The reactor is made of material with \(c=\) \(200 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=9000 \mathrm{~kg} / \mathrm{m}^{3}\). If the heat generation of reactor is suddenly set to \(9 \mathrm{MW} / \mathrm{m}^{3}\), determine the time rate of temperature change in the reactor. Is the heat generation of reactor suddenly increased or decreased to \(9 \mathrm{MW} / \mathrm{m}^{3}\) from its steady operating condition?

Short answer

Question: In a spherical reactor, with given temperature distribution, material properties, and heat generation rate, determine the time rate of temperature change and whether the heat generation is increased or decreased from the steady operating condition. Answer: The time rate of temperature change in the reactor is 7.222 K/s, which is positive. This indicates that the temperature is increasing over time, meaning that the heat generation has increased from its steady operating condition.

Step-by-step solution

Write down the given parameters

The problem provides the following information: Diameter of the reactor: \(d = 0.05\) m (converted from cm to meters) Temperature distribution: \(T(r) = a - br^2\) (with \(a = 850^{\circ}C\) and \(b = 5 \times 10^5 K/m^2\)) Material properties: - Specific heat: \(c = 200 \frac{J}{kg \cdot{ }^{\circ}C}\) - Thermal conductivity: \(k = 40 \frac{W}{m \cdot K}\) - Density: \(\rho = 9000 \frac{kg}{m^3}\) Heat generation rate: \(q_g = 9 \frac{MW}{m^3} = 9 \times 10^6 \frac{W}{m^3}\) (converted from MW to W)

Set up the heat diffusion equation for spherical coordinates

Recall the heat diffusion equation for spherical coordinates: \(\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 k \frac{\partial T}{\partial r}\right) + \rho c \frac{\partial T}{\partial t} = q_g\) Plug the given values: \(\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \cdot 40 \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial r}\right) + 9000 \cdot 200 \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = 9 \times 10^6\)

Calculate the partial derivative with respect to r

We need to calculate the radial gradient of the temperature distribution: \(\frac{\partial T}{\partial r} = \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial r} = -10^6 r\) Now, we plug this back into the heat diffusion equation: \(\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \cdot 40 \cdot(-10^6 r)\right) + 9000 \cdot 200 \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = 9 \times 10^6\)

Solve for the time rate of temperature change

Now, we have: \(\frac{1}{r^2} \frac{\partial}{\partial r}\left(- 4 \times 10^7 r^3\right) + 9000 \cdot 200 \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = 9 \times 10^6\) Taking the radial derivative: \(\frac{1}{r^2} \cdot (-12 \times 10^7 r^2) = -12 \times 10^7\) Now, solving for the time rate of temperature change: \(-12 \times 10^7 + 9000 \cdot 200 \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = 9 \times 10^6\) Divide by 9000 * 200 (converting from power to energy): \(-12 \times 10^7 / 9000 \cdot 200 + \frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = 9 \times 10^6 / 9000 \cdot 200\) Rearrange and evaluate the time rate of temperature change: \(\frac{\partial (850 - 5 \times 10^5 r^2)}{\partial t} = \frac{9 \times 10^6}{9000\cdot200} + \frac{12 \times 10^7}{9000\cdot200} = \frac{13 \times 10^7}{9000\cdot200} = 7.222 K/s\) Thus, the time rate of temperature change is 7.222 K/s.

Analyze whether the heat generation is increased or decreased

Since the time rate of temperature change is positive, it means that the temperature is increasing over time. Therefore, we can conclude that the heat generation has increased to 9 MW/m³ from its steady operating condition.

Key concepts

Understanding Temperature Distribution

The temperature distribution within an object, like a spherical reactor, can be mapped as a function of position. In the given exercise, the temperature distribution within the reactor is expressed using the equation T(r) = a - br^2. This equation allows us to determine the temperature at any point within the sphere by simply substituting the radial distance r from the center into the equation. The constants a and b are parameters that describe how the temperature varies radially within the sphere.

Understanding how temperature varies within the reactor is crucial for analyzing energy transfer, chemical reactions, and physical properties, among other factors. For instance, regions closest to the heat source typically exhibit higher temperatures, whereas temperatures may decrease with distance from the source due to heat dissipation. The equation implies a parabolic temperature profile with the maximum temperature at the center of the sphere (where r = 0). Evaluating this distribution provides insights into the reactor's thermal behavior during operation, which is paramount for safety and efficiency.

Deciphering Heat Generation Rate

The heat generation rate (represented by q_g in the exercise) refers to the amount of heat produced per unit volume of the material over time. In this case, the reactor has a heat generation rate of 9 MW/m³, indicating that significant energy is produced in a given volume of the reactor every second. This variable is central to the study of thermal systems, particularly when assessing the temperature rise due to internal heat generation, as is common in exothermic reactions or when electrical energy is converted into heat within the material.

When engineers or scientists design systems, knowing the heat generation rate is essential to maintain temperature within safe or functional limits and to calculate cooling requirements. This information is vital in preventing overheating and ensuring the reliability and longevity of the equipment. Therefore, accurately determining and managing the rate of heat generated is a foundational aspect of thermal management in systems like reactors.

Time Rate of Temperature Change

The time rate of temperature change in the exercise is a measure of how quickly the temperature changes with respect to time, given in units like Kelvin per second (K/s). By employing the heat diffusion equation, specifically tailored for a spherical coordinate system due to the shape of the reactor, it's possible to calculate how fast the temperature within the reactor is changing over time after a new heat generation rate is introduced.

The calculation factors in the material's specific heat capacity (c), density (ρ), and other relevant parameters to solve for the rate at which the temperature is rising or falling. In this example, after computing the necessary derivatives and simplifying the equation, we find the time rate of temperature change to be 7.222 K/s, symbolizing a rapid increase in temperature. Consequently, such a figure indicates that after the heat generation rate is set to 9 MW/m³, the reactor's temperature begins to climb quickly, which can be attributed to an increase in the heat being produced inside the reactor compared to its previous steady state.