Starting with an energy balance on a disk volume element, derive the one- dimensional transient heat conduction equation for \(T(z, t)\) in a cylinder of diameter \(D\) with an insulated side surface for the case of constant thermal conductivity with heat generation.

Short Answer

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Question: Derive the one-dimensional transient heat conduction equation for a cylinder with an insulated side surface, constant thermal conductivity, and heat generation. Answer: The one-dimensional transient heat conduction equation for a cylinder with an insulated side surface, constant thermal conductivity, and heat generation can be derived as: \(\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial z^2} + \frac{Q}{\rho c_p}\), where \(T(z, t)\) represents the temperature distribution within the cylinder as a function of position along the axial direction \(z\) and time\(t\).

Step by step solution

01

Identify the problem variables

Given a cylinder of diameter \(D\), we are tasked with finding the one-dimensional transient heat conduction equation for temperature \(T(z, t)\), with insulated side surfaces, constant thermal conductivity, and heat generation. The coordinate for the axial direction is represented by \(z\), while time is represented by \(t\).
02

Perform an energy balance on a disk volume element

Consider a disk-shaped volume element of thickness \(\Delta z\) at distance \(z\) from the bottom of the cylinder. The energy balance on this element can be expressed as: \(\text{Energy In} - \text{Energy Out} + \text{Heat Generation} = \text{Energy Storage}\) Energy In (from the lower face) = \(- kA \frac{\partial T}{\partial z}\) Energy Out (from the upper face) = \(- kA\left(\frac{\partial T}{\partial z} + \frac{\partial ^2T}{\partial z^2}\Delta z\right)\) Heat Generation (per unit volume \(Q\)) = \(Q \cdot A \Delta z\) Energy Storage (increase in internal energy) = \(c_p m \frac{\partial T}{\partial t} = \rho c_p A\Delta z \frac{\partial T}{\partial t}\) Here, \(k\) is the thermal conductivity, \(c_p\) is the specific heat capacity, \(\rho\) is the density, and \(A\) is the cross-sectional area of the cylinder (\(\frac{\pi D^2}{4}\)).
03

Apply the energy balance equation

Substituting the energy terms into the energy balance equation: \(-kA\frac{\partial T}{\partial z} - \left(-kA\left(\frac{\partial T}{\partial z} + \frac{\partial ^2T}{\partial z^2}\Delta z\right)\right) + QA\Delta z = \rho c_p A\Delta z \frac{\partial T}{\partial t}\) Simplifying the equation: \(-kA\frac{\partial T}{\partial z} + kA\frac{\partial T}{\partial z} + kA\frac{\partial^2 T}{\partial z^2}\Delta z + QA\Delta z = \rho c_p A\Delta z \frac{\partial T}{\partial t}\) Canceling and rearranging terms: \(kA\frac{\partial^2 T}{\partial z^2}\Delta z + QA\Delta z = \rho c_p A\Delta z \frac{\partial T}{\partial t}\) Divide both sides by \(A\Delta z\): \(k\frac{\partial^2 T}{\partial z^2} + Q = \rho c_p \frac{\partial T}{\partial t}\)
04

Obtain the one-dimensional transient heat conduction equation

We finally obtain the one-dimensional transient heat conduction equation for a cylinder with an insulated side surface and constant thermal conductivity with heat generation: \(\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial z^2} + \frac{Q}{\rho c_p}\) This equation gives the temperature distribution within the cylinder as a function of position along the axial direction \(z\) and time\(t\).

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Key Concepts

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

Energy Balance
Understanding the concept of energy balance is crucial when analyzing problems related to heat transfer. It is the principle that the energy entering a system minus the energy leaving the system equals the change in energy stored within the system. This might seem complex, but it's similar to balancing a checkbook, ensuring what comes in and goes out is recorded to find out the current balance.

In the context of the problem, the energy entering the disk volume is due to heat conduction from the lower face, while the energy leaving is from the upper face of the disk. Any heat that is generated within the volume element itself must also be considered as part of the energy that contributes to the change in storage. The balance can be metaphorically seen as a 'thermal account' of the disk element, tracking all the heat transactions to determine the resulting temperature changes over time. The equation for energy balance ensures that all heat contributions and withdrawals are accounted for in predicting the temperature distribution inside the cylinder.
Thermal Conductivity
Thermal conductivity, represented by the symbol 'k', is a material property that illustrates how well a material conducts heat. It's like the thermal equivalent to electrical conductivity; just as copper wires conduct electricity more effectively than rubber, different materials have different efficiencies when it comes to conducting heat.

The value of thermal conductivity plays a pivotal role in determining the rate at which heat transfers through a material. It influences how quickly a material can absorb heat from its surroundings or release it, essentially defining the speed of the thermal 'current'. In our exercise, the constant thermal conductivity simplifies the problem. It allows us to predict that the rate at which heat moves through the cylinder is consistent throughout, not affected by temperature changes or variations in material composition. This constancy is beneficial in deriving the transient heat conduction equation because it eliminates the need to account for changing thermal properties as heat propagates through the cylinder.
Heat Generation
In some systems, heat is not only transferred but also generated within the material itself. This internal heat generation can arise from chemical reactions, electrical energy, or even radioactive decay. It's similar to having a miniature heater inside the material, continuously adding warmth.

In the given exercise, heat generation per unit volume is symbolized by 'Q' and it's a source term in the energy balance equation. This term alters the temperature profile within the cylinder by adding extra energy that must be dissipated or conducted. The scenario changes from simply understanding how heat moves through an object to also incorporating how the object itself contributes to the heat within the system. This internal heat generation is crucial in various applications, such as electronic devices that generate heat during operation, thereby affecting their thermal management and design.

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Most popular questions from this chapter

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

A long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

Consider a steam pipe of length \(L=30 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(300^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=12.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=175^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.

A stainless steel spherical container, with \(k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used for storing chemicals undergoing exothermic reaction. The reaction provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner radius of \(50 \mathrm{~cm}\) and a wall thickness of \(5 \mathrm{~cm}\) and is situated in a surrounding with an ambient temperature of \(23^{\circ} \mathrm{C}\). The container's outer surface is subjected to convection heat transfer with a coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For safety reasons to prevent thermal burn to individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine the variation of temperature in the container wall and the temperatures of the inner and outer surfaces of the container. Is the outer surface temperature of the container safe to prevent thermal burn?

The conduction equation boundary condition for an adiabatic surface with direction \(n\) being normal to the surface is (a) \(T=0\) (b) \(d T / d n=0\) (c) \(d^{2} T / d n^{2}=0\) (d) \(d^{3} T / d n^{3}=0\) (e) \(-k d T / d n=1\)

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