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A chemically reacting mixture is stored in a thin-walled spherical container of radius \(r_{1}=200 \mathrm{~mm}\), and the exothermic reaction generates heat at a uniform, but temperaturedependent volumetric rate of \(\dot{q}=\dot{q}_{o} \exp \left(-A / T_{o}\right)\), where \(\dot{q}_{o}=5000 \mathrm{~W} / \mathrm{m}^{3}, A=75 \mathrm{~K}\), and \(T_{o}\) is the mixture temperature in kelvins. The vessel is enclosed by an insulating material of outer radius \(r_{2}\), thermal conductivity \(k\), and emissivity \(\varepsilon\). The outer surface of the insulation experiences convection heat transfer and net radiation exchange with the adjoining air and large surroundings, respectively. (a) Write the steady-state form of the heat diffusion equation for the insulation. Verify that this equation is satisfied by the temperature distribution $$ T(r)=T_{s, 1}-\left(T_{s, 1}-T_{s, 2}\right)\left[\frac{1-\left(r_{1} / r\right)}{1-\left(r_{1} / r_{2}\right)}\right] $$ Sketch the temperature distribution, \(T(r)\), labeling key features. (b) Applying Fourier's law, show that the rate of heat transfer by conduction through the insulation may be expressed as $$ q_{r}=\frac{4 \pi k\left(T_{s, 1}-T_{s, 2}\right)}{\left(1 / r_{1}\right)-\left(1 / r_{2}\right)} $$ Applying an energy balance to a control surface about the container, obtain an alternative expression for \(q_{r}\), expressing your result in terms of \(\dot{q}\) and \(r_{1}\). (c) Applying an energy balance to a control surface placed around the outer surface of the insulation, obtain an expression from which \(T_{s, 2}\) may be determined as a function of \(\dot{q}, r_{1}, h, T_{\infty}, \varepsilon\), and \(T_{\text {sur }}\) (d) The process engineer wishes to maintain a reactor temperature of \(T_{o}=T\left(r_{1}\right)=95^{\circ} \mathrm{C}\) under conditions for which \(k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, r_{2}=208 \mathrm{~mm}, h=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, \varepsilon=0.9, T_{\infty}=25^{\circ} \mathrm{C}\), and \(T_{\text {sur }}=35^{\circ} \mathrm{C}\). What is the actual reactor temperature and the outer surface temperature \(T_{s, 2}\) of the insulation? (e) Compute and plot the variation of \(T_{s, 2}\) with \(r_{2}\) for \(201 \leq r_{2} \leq 210 \mathrm{~mm}\). The engineer is concerned about potential burn injuries to personnel who may come into contact with the exposed surface of the insulation. Is increasing the insulation thickness a practical solution to maintaining \(T_{s, 2} \leq 45^{\circ} \mathrm{C}\) ? What other parameter could be varied to reduce \(T_{s, 2}\) ?

Step by step solution

01

(Step 1: Heat diffusion equation for insulation)

The steady-state heat diffusion equation for the insulation and the given temperature distribution need to be verified. The heat diffusion equation is: \[ \frac{\partial^2 T}{\partial r^2} + \frac{2}{r}\frac{\partial T}{\partial r} = 0 \] Differentiating \((T_s_1 - T_s_2)\left[\frac{1 - \left(\frac{r_1}{r}\right)}{1 - \left(\frac{r_1}{r_2}\right)}\right]\) twice w.r.t r, we need to show that it satisfies the heat diffusion equation.

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Key Concepts

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

Heat Diffusion Equation

The heat diffusion equation plays a central role in understanding how heat moves through materials like the insulation in our problem. In steady-state scenarios, where temperatures do not change over time, the equation simplifies to: \[\frac{\partial^2 T}{\partial r^2} + \frac{2}{r} \frac{\partial T}{\partial r} = 0\] This equation reveals that the rate at which temperature changes with respect to the radius, combined with the radial symmetry of the sphere, must sum to zero.
In essence, it is telling us that, in a stable condition, heat flow into a material must equal heat flow out. This principle, properly applied, ensures temperature uniformity across the material, making it vital for getting the temperature distribution equation correct.
For our problem, the equation is shown to be satisfied by substituting the specified temperature distribution into the diffusion equation, confirming the steady-state condition.

Fourier's Law

Fourier's Law provides the basis for calculating how heat flows through materials, such as the insulation around our spherical container. It is represented by the formula: \[ q = -k \frac{dT}{dr} \]where \( q \) is the heat transfer rate, \( k \) is the thermal conductivity of the material, and \( \frac{dT}{dr} \) is the temperature gradient.

• This tells us that heat transfer through a material is directly proportional to how quickly temperature changes with distance.
• The negative sign indicates heat moves from hot to cold regions.

In our case, the insulation uses Fourier's Law to calculate the rate of heat transfer by conduction, which can be expressed as\[ q_r = \frac{4 \pi k(T_{s, 1} - T_{s, 2})}{\left(1 / r_1\right) - \left(1 / r_2\right)} \]This restatement in spherical coordinates accounts for the geometry of the problem and how it affects heat transfer.

Energy Balance

In thermal systems, maintaining an energy balance is crucial to understanding how energy is conserved and transferred. This balance ensures that all the energy entering a system equals the energy leaving it plus any changes within the system.
For the spherical container and its surroundings, applying energy balance calculations allows us to derive expressions for the heat transfer rate through the insulation.
To use this principle in practice, a control surface—an imaginary boundary—is drawn around the system. The conservation of energy is then examined across this surface.

• In the problem, this concept helps find expressions for the reactor surface temperature \( T_{s, 2} \).
• By evaluating the energy entering and leaving different parts of the system, insights into the control of temperatures can be gained.

Using these balances provides a clearer understanding of alternative ways heat can be managed, sometimes suggesting potential design changes or operational adjustments.

Thermal Conductivity

Thermal conductivity is a material property dictating how easily heat can pass through a material. Represented by the symbol \( k \), it is a constant in Fourier's Law and tells us the efficiency of heat transmission within the insulation material.
High thermal conductivity materials allow heat to pass through them quickly, making them less effective as insulators.

• Conversely, low thermal conductivity materials slow down heat flow, making them ideal for insulation as they better retain heat within a bounded area.
• In the problem, thermal conductivity crucially impacts how heat is managed through the insulating layer on the sphere.

By choosing materials with the appropriate thermal conductivity, engineers can optimize the temperature behavior, minimizing excessive cooling or heating. Understanding this property is essential for making informed decisions about which materials to use in a given thermal application.