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Radioactive wastes are stored in a spherical type 316 stainless steel tank of inner diameter \(1 \mathrm{~m}\) and \(1 \mathrm{~cm}\) wall thickness. Heat is generated uniformly in the wastes at a rate of \(3 \times 10^{4} \mathrm{~W} / \mathrm{m}^{3}\). The outer surface of the tank is cooled by air at \(300 \mathrm{~K}\) with a heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Determine the maximum temperature in the tank. Take the thermal conductivity of the wastes as \(2.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

Short Answer

Expert verified
The maximum temperature in the tank is approximately 422 K.

Step by step solution

01

Calculate the Volume of the Wastes

The inner diameter of the tank is given as 1 m. Therefore, the radius is \[ r_i = \frac{1}{2} = 0.5 \text{ m} \].The volume of the spherical tank is given by the formula:\[ V = \frac{4}{3} \pi r_i^3 \].Substitute for \( r_i \):\[ V = \frac{4}{3} \pi (0.5)^3 \approx 0.5236 \text{ m}^3 \].
02

Calculate the Heat Generated in the Wastes

The waste generates heat at a rate of \( 3 \times 10^4 \text{ W/m}^3 \).Hence, the total heat generated is:\[ Q_{ ext{gen}} = 3 \times 10^4 \times 0.5236 \approx 15707.96 \text{ W} \].
03

Determine Heat Loss through the Tank's Wall

Considering the small thickness of the wall, the tank behaves nearly as a sphere. The outside radius of the tank is approximately \[ r_o = 0.5 + 0.01 = 0.51 \text{ m} \].The heat loss through the tank wall is\[ Q_{ ext{loss}} = 4 \pi k_{ ext{waste}} (r_i)(r_o) \frac{(T_{ ext{max}} - T_{ ext{surface}})}{r_o - r_i} \], where \( k_{ ext{waste}} = 2.0 \text{ W/mK} \).
04

Apply Heat Balance

In a steady state, the heat generated equals the heat lost. Hence, \[ Q_{ ext{gen}} = Q_{ ext{loss}} \].This gives:\[ 15707.96 = 4 \pi (2.0) \cdot 0.5 \cdot 0.51 \frac{(T_{ ext{max}} - T_{ ext{surface}})}{0.51 - 0.5} \].Further, solve this equation to find \( T_{ ext{max}} - T_{ ext{surface}} \): \[ 15707.96 = 4\pi\cdot 0.51\cdot(2.0)(T_{ ext{max}} - T_{ ext{surface}}/0.01 ) \].
05

Solve for Maximum Temperature

Given \( h = 100 \text{ W/m}^2 \text{K} \) and air temperature as 300K, apply the surface temperature formula:\[ Q_{ ext{conv}} = 4 \pi (r_o^2) \cdot h(T_{ ext{surface}} - 300) \].Using this gives:\[ Q_{ ext{conv}} = 100 \cdot 4 \pi\cdot(0.51)^2(T_{ ext{surface}} - 300) \].This way equate this to the equation from step 4, and solve for maximum temperature.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a key concept in understanding heat transfer in materials. It measures how easily heat can pass through a material. In the context of the exercise, we are dealing with radioactive waste inside a stainless-steel tank. The thermal conductivity of the waste is given as \( 2.0 \text{ W/mK} \). This means that for every meter of thickness, the material can conduct 2 watts of heat per degree Kelvin difference in temperature.

This property is crucial because it determines how quickly the heat generated within the waste will spread throughout the tank. A higher thermal conductivity would imply faster heat distribution across the material, leading to potentially lower maximum temperatures. Conversely, a lower thermal conductivity means slower heat distribution, which can result in higher localized temperatures. Understanding this concept helps in predicting how the system reaches a thermal equilibrium and ensures safe operating conditions, especially when dealing with hazardous materials like radioactive waste.
Heat Generation
Heat generation refers to the process by which heat is produced within a system, typically due to chemical reactions, electrical resistance, or radioactive decay. In our exercise, the radioactive waste inside the tank generates heat at a rate of \( 3 \times 10^4 \text{ W/m}^3 \).

It is important to quantify this heat generation because it directly influences the system's thermal behavior. The amount of heat generated needs to be managed and dissipated to prevent excessive temperatures that could compromise the integrity of the tank and, by extension, the safety of its contents.
  • Understand that heat generation is the driving factor for heat transfer within the system.
  • Heat generation's rate must be determined accurately to assess how much heat needs to be managed.
Calculating the total heat generated involves multiplying the rate of heat generation by the volume of the waste. This step is critical for setting up the subsequent heat balance analysis.
Heat Balance
In thermal systems, achieving a heat balance is synonymous with attaining a steady state, where the amount of heat generated equals the amount of heat lost. This balance is integral to maintaining safe operation conditions, as it prevents heat build-up, which could lead to temperature increases beyond safe limits.

To establish heat balance in the exercise, we equate the heat generated by the radioactive waste to the heat dissipated through the tank's walls. This is derived from the principle that in a steady state, the rate of heat entering or generated within the system must equal the rate of heat leaving the system.
  • The equation for heat balance is:
  • \( Q_{\text{gen}} = Q_{\text{loss}} \)
Solving for the temperature difference helps in calculating the maximum temperature in the system. This process ensures that the waste storage remains under safe control and prevents structural failure or hazardous conditions.
Convective Heat Transfer
Convective heat transfer involves the movement of heat between a solid surface and a fluid, like air or liquid, that is in motion relative to the solid. In our tank scenario, convective heat transfer occurs on the outer surface of the tank, which is cooled by air at \( 300 \text{ K} \).

The efficiency of this cooling process depends on the heat transfer coefficient, given as \( 100 \text{ W/m}^2 \text{K} \), and the surface area of the tank. This coefficient indicates how effectively heat is transferred from the solid (tank wall) to the moving fluid (air).
  • Convective heat transfer helps in dissipating excess heat generated inside the tank.
  • Calculating the convective heat transfer involves using the formula: \( Q_{\text{conv}} = 4 \pi (r_o^2) \cdot h(T_{\text{surface}} - 300) \)
This ensures the waste inside maintains a safe temperature by effectively removing heat from the outer surface of the tank, thus preventing overheating through effective cooling.

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Most popular questions from this chapter

A \(4 \mathrm{~mm}\)-diameter, \(25 \mathrm{~cm}\)-long aluminum alloy rod has an electric heater wound over the central \(5 \mathrm{~cm}\) length. The outside of the heater is well insulated. The two \(10 \mathrm{~cm}\)-long exposed portions of rod are cooled by an air stream at \(300 \mathrm{~K}\) giving an average convective heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the power input to the heater is \(10 \mathrm{~W}\), determine the temperature at the ends of the rod. Take \(k=190\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) for the aluminum alloy.

Two air flows are separated by a \(2 \mathrm{~mm}\)-thick plastic wall. A \(20.2 \mathrm{~cm}\)-long, \(2 \mathrm{~cm}\)-diameter aluminum rod transfers heat from one flow to the other as shown. The hot air flow is at \(70^{\circ} \mathrm{C}\), and the convective heat transfer coefficient to the rod is \(48 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\); the cold air flow is at \(20^{\circ} \mathrm{C}\) and is at a lower velocity, giving a heat transfer coefficient of only \(24 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Determine the rate of heat transfer and the temperature of the midsection of the rod. Take \(k=190 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

On the flight of Apollo 12, plutonium oxide \(\left(\mathrm{Pu}^{238} \mathrm{O}_{2}^{16}\right)\) was used to generate electrical power. Heat was generated uniformly through the loss of kinetic uniformly through the loss of kinetic energy from alpha particles emitted by the \(\mathrm{Pu}^{238}\). Consider a sphere of plutonium oxide of \(3 \mathrm{~cm}\) diameter covered with thermo-electric elements for converting heat to electricity. The physical properties of these elements (tellurides) and heat rejection considerations suggest that the surface of the sphere be at \(200^{\circ} \mathrm{C}\). On the other hand, the ceramic nature of the plutonium oxide allows a maximum temperature of \(1750^{\circ} \mathrm{C}\). With these constraints, determine (i) the maximum allowable volumetric heating rate. (ii) the electrical power generated, assuming a thermal efficiency of \(4 \%\). Take \(k_{\mathrm{PuO}_{2}}=4 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

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The absorber of a simple flat-plate solar collector with no coverplate consists of a \(2 \mathrm{~mm}\)-thick aluminum plate with \(6 \mathrm{~mm}\)-diameter aluminum water tubes spaced at a pitch of \(10 \mathrm{~cm}\), as shown. On a clear summer day near the ocean, the air temperature is \(20^{\circ} \mathrm{C}\), and a steady wind is blowing. The solar radiation absorbed by the plate is calculated to be 680 \(\mathrm{W} / \mathrm{m}^{2}\), and the convective heat transfer coefficient is estimated to be \(14 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If water at \(20^{\circ} \mathrm{C}\) enters the collector at \(7 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) per meter width of collector, and the collector is \(3 \mathrm{~m}\) long, estimate the outlet water temperature. For the aluminum take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(\varepsilon=0.20\). (Hint: Evaluate the heat lost by reradiation using Eq. (1.19) with a constant value of \(h_{r}\) corresponding to a guessed average plate temperature. Then apply the steady-flow equation to an elemental length of the collector, and so derive a differential equation governing the water temperature increase along the collector.)

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