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Chapter 3: Problem 95

Radinactive wastes \(\left(k_{1 w}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{m}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{e}=0.6 \mathrm{~m}\). Heat is genersed volumetrically within the wastes at a unifom rate of \(\hat{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and \(T_{\mathrm{s}}=25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The inner surface temperature of the sphere is approximately 1040345 K.

Step by step solution

01

Calculate Thermal Resistance Components

There are three forms of heat transfer in this system: conduction through the stainless steel container and convection to the surrounding water.1. **Conduction Resistance of the Sphere (R_{cond}):**The conduction resistance through the spherical shell can be calculated using the formula for a spherical shell:\[R_{cond} = \frac{1}{4\pi k_{m}} \left(\frac{1}{r_{i}} - \frac{1}{r_{e}}\right)\]Where:- \(k_m = 15 \text{ W/m}\cdot\text{K}\)- \(r_i = 0.5 \text{ m}\)- \(r_e = 0.6 \text{ m}\)Substituting these values, we get:\[R_{cond} = \frac{1}{4\pi \times 15} \left(\frac{1}{0.5} - \frac{1}{0.6}\right) \approx 0.00177 \text{ K/W}\]2. **Convection Resistance (R_{conv}):**The convection resistance from the outer surface to the surroundings is calculated by:\[R_{conv} = \frac{1}{h \cdot 4\pi r_{e}^{2}}\]Where:- \(h = 1000 \text{ W/m}^2\cdot\text{K}\)Substituting, we get:\[R_{conv} = \frac{1}{1000 \times 4\pi \times 0.6^2} \approx 0.00022 \text{ K/W}\]
02

Total Thermal Resistance

The total resistance to heat transfer is the sum of the conduction and convection resistances:\[R_{total} = R_{cond} + R_{conv} \approx 0.00177 + 0.00022 = 0.00199 \text{ K/W}\]
03

Calculate the Heat Generation

Heat is generated within the waste material at a rate given by:\[\dot{Q}_{gen} = \hat{q} \cdot V\]Where \(\hat{q} = 10^8 \text{ W/m}^3\) and \(V\) is the volume of the sphere filled with waste:\[V = \frac{4}{3} \pi r_i^3 \approx \frac{4}{3} \pi (0.5)^3 \approx 0.5236 \text{ m}^3\]Substitute into the equation to find the total heat generation:\[\dot{Q}_{gen} = 10^8 \times 0.5236 \approx 5.236 \times 10^7 \text{ W}\]
04

Calculate Temperature of Inner Sphere Wall

Using the heat generation and total thermal resistance, the inner surface temperature can be found. First, use the energy balance equation that represents heat flow from the inner to outer surface:\[T_i - T_s = \dot{Q}_{gen} \cdot R_{total}\]Where \(T_s = 25^\circ \text{C}\).\[T_i - 25 = 5.236 \times 10^7 \times 0.00199\]Solving for \(T_i\):\[T_i \approx 25 + 1040320\;\text{K} \approx 1040345 \;\text{K}\]
05

Conclusion

Applying the thermal resistances and the calculated heat generation, we determine the temperature at the inner surface of the sphere to account for the substantial heat generation due to radioactive waste.

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Key Concepts

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

Conduction
Conduction is a mode of heat transfer occurring predominantly in solids, where heat energy is transferred by the vibration of particles, without any movement of the material itself. In the context of the exercise, conduction occurs through the stainless steel container. The thermal conduction resistance, denoted as \(R_{cond}\), measures how well the stainless steel resists the flow of thermal energy. This resistance is calculated using the formula for a spherical shell:
  • \[R_{cond} = \frac{1}{4\pi k_{m}} \left(\frac{1}{r_{i}} - \frac{1}{r_{e}}\right)\]
where \(k_m\) is the thermal conductivity of the material, and \(r_i\) and \(r_e\) are the inner and outer radii, respectively. Stainless steel, with its notable conductivity, allows heat generated inside to gradually move outward. Recognizing the conductive path and calculating the corresponding resistance enables us to better manage heat dissipation effectively.
Convection
Convection is another mode of heat transfer that involves the bodily movement of fluid particles, such as water or air, transferring heat. In our scenario, the outer surface of the container is exposed to a water flow which facilitates convective heat transfer. Convection resistance, \(R_{conv}\), quantifies the resistance against this heat movement. It's determined by:
  • \[R_{conv} = \frac{1}{h \cdot 4\pi r_{e}^{2}}\]
where \(h\) represents the heat transfer coefficient and \(r_e\) is the outer radius of the container. Convection is integral in dissipating heat from the container's surface. The flowing water around the sphere ensures the spread of heat into the surrounding environment, making convection an efficient method for cooling down the spherical container, especially needed in this case due to high heat generation by the radioactive wastes.
Heat Generation
Heat generation is a process where energy is introduced into a system, typically as heat. In the exercise, it's detailing the radioactive materials stored within the spherical container, which generate heat at a constant volumetric rate, \(\hat{q}\). This internal heat generation is important because it constantly supplies energy that must be dissipated to prevent overheating.To determine the quantity of generated heat, we multiply the volumetric heat generation rate by the volume of the inner spherical region containing the material:
  • \[\dot{Q}_{gen} = \hat{q} \cdot V\]
where \(V\) is the volume of the sphere. This tells us how much heat needs to be managed through conduction and convection. Assessing heat generation accurately is essential for designing the containment material and thickness necessary to safely store waste, ensuring that the structure can handle and transfer heat efficiently without reaching critical temperatures.

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Most popular questions from this chapter

Stearn at a temperature of \(250^{\circ} \mathrm{C}\) flows through a steel pipe (AISI 1010) of 60-mm inside diameter and 75-mm autside diameter. The convection coefficient between the team and the inner surface of the pipe is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while that between the outer surface of the pipe and the wrroundings is \(25 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), The pipe emissivity is \(0.8\). and the temperature of the air and the surroundings is 20 C. What is the heat loss per unit length of pipe?

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W})\) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?

A thin flat plate of length \(L\) thickness \(t\), and width \(W=L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{e}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q^{*}\) - (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

Consider two long. slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocoeple, it was observed that the temperatures of the rods were equal at the positions \(x_{A}=0.15 \mathrm{~m}\) and \(x_{\mathrm{A}}=\) \(0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(A\) is known to be \(k_{A}=\) \(70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{m}}\) for rod \(\mathrm{B}\).

A hollow aluminem sphere, with an clectrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are \(0.15\) and \(0.18 \mathrm{~m}\), respectively. and testing is done under steady-state conditions with the inner surface of the aluminum maintained at \(250^{\circ} \mathrm{C}\). In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of \(0.12 \mathrm{~m}\). The system is in a roorn for which the air temperature is \(20^{\circ} \mathrm{C}\) and the convection cocfficient at the outer surface of the insulation is \(30 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). If \(80 \mathrm{~W}\) are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?
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