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An explosive is to be stored in large slabs of thickness \(2 L\) clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation \(\dot{Q}_{v}^{\prime \prime \prime}=a+b\left(T-T_{e}\right)\), where \(T_{e}\) is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is \(U\), show that the condition for an explosion is \(L=(k / b)^{1 / 2} \tan ^{-1}\left[U /(k b)^{1 / 2}\right]\). Determine the slab thickness if \(k=0.9 \mathrm{~W} / \mathrm{m} \mathrm{K}, U=0.20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}, a=60 \mathrm{~W} / \mathrm{m}^{3}\), \(b=6.0 \mathrm{~W} / \mathrm{m}^{3} \mathrm{~K}\).

Short Answer

Expert verified
The condition for an explosion is determined by the equation \( L = (k / b)^{1 / 2} \tan^{-1}[U / (k b)^{1 / 2}] \). For the given values, the slab thickness \( L \) is approximately 0.069 m.

Step by step solution

01

Understand the Heat Generation Equation

The rate of heat generation within the explosive is given by \( \dot{Q}_{v}^{\prime \prime \prime} = a + b(T-T_{e}) \). Here, \( T_e \) is the ambient temperature, \( a \) is a constant, and \( b \) accounts for the temperature dependency.
02

Establish Heat Transfer Equations

The heat transfer through the slab can be characterized using Fourier's law of conduction. For a steady state, no accumulation within the slab implies energy generated per unit volume equals energy dissipated per unit area of the slab face, leading to \( -k \frac{d^2 T}{dx^2} = a + b(T-T_{e}) \). Here, \( k \) is the thermal conductivity.
03

Solve Differential Equation for Temperature

The differential equation becomes \( -k \frac{d^2 T}{dx^2} = a + b(T-T_{e}) \). Assume \( T(x) = T_e + \theta(x) \), yields \( -k \frac{d^2 \theta}{dx^2} = a + b \theta \). Solve this second order differential equation to get the general solution \( \theta(x) = C_1 e^{mx} + C_2 e^{-mx} + \theta_0 \), with \( m = (b/k)^{1/2} \).
04

Apply Boundary Conditions

Apply boundary conditions at the slab surfaces \( x = \pm L \). Using \( U(T_e - heta) = -k \frac{d \theta}{dx} \), determine \( C_1 \) and \( C_2 \). The symmetry and equilibrium with ambient simplifies the boundary to introduce \( \theta = \frac{a}{b} \) such that \( L = (k/b)^{1/2} \tan^{-1}[U/(kb)^{1/2}] \).
05

Calculate Slab Thickness

Substitute the given values into the derived equation for \( L \). Using \( k = 0.9 \, \text{W/mK} \), \( U = 0.20 \, \text{W/m}^2\text{K} \), and \( b = 6 \, \text{W/m}^3\text{K} \), find \( L = (0.9/6)^{1/2} \tan^{-1}[0.20 / (6 * 0.9)^{1/2}] \). Calculate \( L \).

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Key Concepts

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

Explosive Storage
Storing explosives safely is crucial to prevent any hazardous incidents. One important aspect to consider is how heat affects the explosive material. In the given exercise, an explosive is stored in slabs, and understanding how it interacts with heat can ensure safety. When an explosive is stored in a confined space, any uncontrolled increase in temperature can lead to ignition. This makes proper thermal management essential.
  • Thermal management helps maintain a stable environment inside the storage unit.
  • Monitoring and controlling heat generation prevent spontaneous ignitions.
  • Safe storage requires careful design, considering both material characteristics and external conditions.
The exercise explores this concept by analyzing the heat transfer involved and setting a condition to prevent explosion based on slab thickness and other material properties.
Thermal Conductivity
Thermal conductivity ( k ) is a measure of a material's ability to conduct heat. It's crucial for understanding how heat transfers within the explosive slab. In the exercise, this property determines how quickly heat moves from the explosive material to its protective sheath and then to the surrounding environment.
  • High thermal conductivity means heat is transferred more efficiently.
  • Materials with low thermal conductivity act as insulators, slowing down the heat transfer.
Within the exercise context, knowledge of the slab's thermal conductivity is necessary to model how heat moves across it. It also affects the heat dissipation rate, which is key to maintaining a safe storage environment for the explosive. The formula in the solution uses k to calculate necessary slab thickness to balance the heat generation and dissipation.
Heat Generation
Heat generation within the explosive is influenced by both a constant and a temperature-dependent term. In the exercise, the rate of heat generation is described by the equation:\[ \dot{Q}_{v}^{\prime \prime \prime} = a + b(T-T_{e}) \]
  • a: The constant represents baseline heat generation.
  • b: This term accounts for additional heat generated as the temperature rises above the ambient temperature,Te.
This linear approximation is crucial for understanding the dynamic temperature variations within the explosive. By quantifying both constant and temperature-dependent heat generation, engineers can predict how the internal temperature will change and assess the potential for overheating. Safely storing explosives requires minimizing unnecessary heat generation and ensuring any generated heat is quickly and effectively dissipated.
Steady State
The concept of steady state is vital in thermal analysis, especially when examining heat transfer in materials like explosive storage units. A steady state implies that the system's properties do not change with time. In the context of the exercise, this means that the rate of heat generation inside the storage reaches equilibrium with the rate of heat dissipation.
  • At steady state, conditions are stable, so there's no net increase in temperature.
  • Understanding steady state conditions helps in ensuring that all generated heat is properly managed.
To achieve a steady state in our explosion prevention model, the slab's thermal design must allow the heat generated by the explosive to be dissipated efficiently into the environment. The derived condition for explosion prevention with the specified slab thickness ensures that this steady state is maintained, averting the risk of ignition due to uncontrolled temperature rise.

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Most popular questions from this chapter

In a laboratory experiment, a long, \(2 \mathrm{~cm}\)-diameter, cylinder of fissionable material is encased in a \(1 \mathrm{~cm}\)-thick graphite shell. The unit is immersed in a coolant at 330 \(\mathrm{K}\), and the convective heat transfer coefficient on the graphite surface is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If heat is generated uniformly within the fissionable material at a rate of \(100 \mathrm{MW} / \mathrm{m}^{3}\), determine the temperature at the center-line of the cylinder. Allow for an interfacial conductance between the material and the graphite shell of \(3000 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). and take the thermal conductivities of the material and graphite as \(4.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m} \mathrm{K}\), respectively.

A \(4 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness stainless steel tube is insulated with a 5 \(\mathrm{cm}\)-thick layer of cork. Chilled milk flows through the tube. At a given location the milk temperature is \(5^{\circ} \mathrm{C}\) when the ambient temperature is \(25^{\circ} \mathrm{C}\). If the inside and outside heat transfer coefficients are estimated to be 50 and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively, calculate the rate of heat gain per meter length of tube.

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.

Aluminum alloy straight rectangular fins for cooling a semiconductor device are 1 \(\mathrm{cm}\) long and \(1 \mathrm{~mm}\) thick. Investigate the effect of choice of tip boundary condition on heat loss as a function of convective heat transfer coefficient. Use \(k=175 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the alloy and a range of \(h_{c}\) values from 10 to \(200 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(5 \mathrm{~cm}\) outer diameter and \(0.2 \mathrm{~mm}\) thickness, wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the bare tube and 15 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

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