/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Starting with an energy balance ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Starting with an energy balance on a ring-shaped volume element, derive the two-dimensional steady heat conduction equation in cylindrical coordinates for \(T(r, z)\) for the case of constant thermal conductivity and no heat generation.

Short Answer

Expert verified
Question: Derive the 2D steady heat conduction equation in cylindrical coordinates for a temperature function, T(r, z), with constant thermal conductivity and no heat generation. Answer: The 2D steady heat conduction equation in cylindrical coordinates is given by: $$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^2 T}{\partial z^2}=0.$$

Step by step solution

01

Visualize the ring-shaped volume element in cylindrical coordinates

Consider a ring-shaped volume element in cylindrical coordinates \((r, \theta, z)\) with inner radius \(r\), outer radius \(r + \Delta r\), height \(\Delta z\), and thickness \(\Delta \theta\). The volume of the element is given by \(V = \Delta V = (r\Delta\theta)(\Delta r)(\Delta z)\).
02

Set up the energy balance equation

For steady-state heat conduction with no heat generation, the energy balance equation states that the amount of heat entering the volume element is equal to the amount of heat leaving it. We will consider heat transfer in the radial and axial directions only, since the problem is two-dimensional. Thus, we can write the energy balance as: $$q_r(r) - q_r(r+\Delta r) + q_z(z) - q_z(z+\Delta z) = 0.$$
03

Express heat flux in terms of temperature and thermal conductivity

The heat flux in the radial and axial directions can be expressed using Fourier's law of heat conduction: $$q_r = -k\frac{\partial T}{\partial r}\text{ and }q_z = -k\frac{\partial T}{\partial z},$$ where \(k\) is the constant thermal conductivity.
04

Substitute the expressions for heat flux and evaluate the limits

Substitute the expressions for \(q_r\) and \(q_z\) into the energy balance equation and evaluate the result in the limit as \(\Delta r\) and \(\Delta z\) approach zero: $$(-k\frac{\partial T}{\partial r})_{r} + (-k\frac{\partial T}{\partial r})_{r+\Delta r}+(-k\frac{\partial T}{\partial z})_{z} + (-k\frac{\partial T}{\partial z})_{z+\Delta z}=0.$$ Divide the entire equation by \(k\Delta r\Delta z\) and simplify, $$\frac{1}{\Delta r} \left(\frac{\partial T}{\partial r}|_{r+\Delta r} - \frac{\partial T}{\partial r}|_{r}\right) + \frac{1}{\Delta z} \left(\frac{\partial T}{\partial z}|_{z+\Delta z} - \frac{\partial T}{\partial z}|_{z}\right) =0.$$
05

Recognize the limits as partial derivatives and obtain the 2D steady-state heat conduction equation in cylindrical coordinates

As \(\Delta r \to 0\) and \(\Delta z \to 0\), the expression on the left-hand side of the equation represents the sum of two second-order partial derivatives: $$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^2 T}{\partial z^2}=0.$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Cylindrical Coordinates
Cylindrical coordinates are an alternative to the more common Cartesian coordinates but are especially useful in problems involving symmetry around a central axis. These coordinates are represented by three variables:
  • \( r \) - the radial distance from a central axis,
  • \( \theta \) - the angular position around the axis,
  • \( z \) - the height or axial position along the axis.
This system is ideal for scenarios where we deal with objects that are circular or cylindrical in shape, such as pipes or cables. It simplifies the analysis of such systems by aligning with their natural geometry.
In this particular problem, we reduce the complexity by focusing on radial and axial directions, which is aligned with a two-dimensional approach in cylindrical coordinates. This helps in applying the processes of heat conduction and energy analysis efficiently.
Exploring Thermal Conductivity
Thermal conductivity, denoted as \( k \), is a material property that indicates how well a material can conduct heat. It's a key factor in determining how quickly temperatures will equalize throughout a material.
This constant value is assumed in many heat conduction problems to simplify analysis and is given in units of \( \, W/(m\cdot K) \, \) (watts per meter per Kelvin).
To understand its role, imagine a metal rod heated at one end. High thermal conductivity ensures that heat spreads quickly along the rod, balancing temperatures. In the analyzed problem, assuming constant thermal conductivity simplifies our equations, making it easier to predict temperature distributions when using Fourier's law.
Knowing thermal conductivity is crucial when designing systems where heat transfer efficiency is critical, ensuring materials perform as expected under thermal loads.
Diving into Energy Balance
Energy balance is a fundamental principle in thermal dynamics, which states that energy entering a system should equal the energy leaving, if there's no accumulation within the system.
In the realm of steady heat conduction, where there is no heat generation or time-dependent change, we achieve balance by equating incoming with outgoing heat energy flows.
For our example, this energy consideration accounts for radial and axial flows, ignoring the circumferential direction as we're working in a two-dimensional radial-axial plane. It leads us to a simplified equation \[q_r(r) - q_r(r+\Delta r) + q_z(z) - q_z(z+\Delta z) = 0\]This equation describes how heat moves through a small volume element of our cylindrical system, ensuring that what enters and exits is balanced. This balance is pivotal for accurately applying Fourier's law and ultimately deriving the steady-state heat conduction equation in cylindrical coordinates.
Applying Fourier's Law of Heat Conduction
Fourier's Law is fundamental for understanding how heat flow relates to temperature gradients within a material. It tells us that the heat transfer rate is proportional to the negative of the temperature gradient and is accompanied by the thermal conductivity.
In mathematical terms for one-dimensional heat conduction, this relation for any direction \( x \) is expressed as: \[q_x = -k\frac{\partial T}{\partial x}\]
This implies that heat flows from high to low temperature, and greater thermal gradients result in greater heat transfer rates.
In the problem, Fourier's law aids in expressing heat flux (\( q_r\) and \( q_z\) ) in terms of temperature derivatives, simplifying the energy balance into something solvable. It is critical for integrating the concept of thermal conductivity into our understanding, thereby linking material properties with the mechanical challenges of heat movement.
Applying Fourier’s law effectively allows us to develop usable models for predicting temperature distributions – vital for thermal management in real-world applications like insulation and electronic cooling systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a steam pipe of length \(L=35 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(250^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=\) \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=160^{\circ} \mathrm{F},(a)\) 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.

Consider steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly. The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is \((a)\) none, \((b)\) small, or \((c)\) significant.

Consider a large plane wall of thickness \(L\) and constant thermal conductivity \(k\). The left side of the wall \((x=0)\) is maintained at a constant temperature \(T_{0}\), while the right surface at \(x=L\) is insulated. Heat is generated in the wall at the rate of \(\dot{e}_{\text {gen }}=a x^{2} \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) by solving the differential equation, obtain a relation for the variation of temperature in the wall \(T(x)\) in terms of \(x, L, k, a\), and \(T_{0}\), and (c) what is the highest temperature \(\left({ }^{\circ} \mathrm{C}\right)\) in the plane wall when: \(L=1 \mathrm{ft}, k=5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ}{ }^{\circ} \mathrm{F}, a=1200 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{5}\), and \(T_{0}=700^{\circ} \mathrm{F}\).

A spherical communication satellite with a diameter of \(2.5 \mathrm{~m}\) is orbiting around the earth. The outer surface of the satellite in space has an emissivity of \(0.75\) and a solar absorptivity of \(0.10\), while solar radiation is incident on the spacecraft at a rate of \(1000 \mathrm{~W} / \mathrm{m}^{2}\). If the satellite is made of material with an average thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the midpoint temperature is \(0^{\circ} \mathrm{C}\), determine the heat generation rate and the surface temperature of the satellite.

A spherical vessel is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the vessel. The inner diameter of the vessel is \(5 \mathrm{~m}\) and its inner surface temperature is at \(120^{\circ} \mathrm{C}\). The wall of the vessel has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0018 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). The vessel is situated in a surrounding with an ambient temperature of \(15^{\circ} \mathrm{C}\), the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent thermal burn on skin tissues, the outer surface temperature of the vessel should be kept below \(50^{\circ} \mathrm{C}\). Determine the minimum wall thickness of the vessel so that the outer surface temperature is \(50^{\circ} \mathrm{C}\) or lower.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.