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Consider a cylindrical shell of length L, inner radius r1, and outer radius r2 whose thermal conductivity varies in a specified temperature range as k(T)=k0(1+βT2) where k0 and β are two specified constants. The inner surface of the shell is maintained at a constant temperature of T1 while the outer surface is maintained at T2. Assuming steady one-dimensional heat transfer, obtain a relation for the heat transfer rate through the shell.

Short Answer

Expert verified
Answer: The final relation for the heat transfer rate is given by: dQdt=2πk0Lln(r2r1)[arctan(βT2)arctan(βT1)β]

Step by step solution

01

Write down Fourier's Law for heat conduction

Fourier's Law of heat conduction for a cylindrical shell can be written as: dQdt=k(T)AdTdr where dQdt is the heat transfer rate, k(T) is the thermal conductivity, A is the surface area of the cylindrical shell, and dTdr is the temperature gradient with the radius r.
02

Calculate surface area A

The surface area of a cylindrical shell of length L and radius r can be calculated as: A=2πrL
03

Substitute the given conductivity variation

Now we need to substitute the given conductivity variation k(T)=k0(1+βT2) in the Fourier's Law: dQdt=k0(1+βT2)2πrLdTdr
04

Solve the differential equation

The equation we derived in step 3 is an ordinary differential equation, to solve it, we can separate the variables as follows: T1T211+βT2dT=k0LdQdtr1r21rdr
05

Evaluate the integrals

Evaluate the left-hand side (LHS) and the right-hand side (RHS) of the equation obtained in step 4: T1T211+βT2dT=1βarctan(βT)|T1T2 and k0LdQdtr1r21rdr=k0LdQdtln(r)|r1r2
06

Obtain the final relation for heat transfer rate

Equate the LHS and RHS evaluation and simplify the equation obtained in step 5 to get the final relation for the heat transfer rate dQdt: dQdt=2πk0Lln(r2r1)[arctan(βT2)arctan(βT1)β]

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Key Concepts

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

Fourier's Law
Fourier's Law is fundamental to understanding heat conduction. It describes how heat energy is transferred through a material. In its basic form, the law states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the area through which it is being transferred.

In the context of a cylindrical shell, which is our focus here, Fourier's Law can be expressed specifically for radial heat conduction. This formulation considers the shape and nature of the material. For cylindrical geometry, the heat transfer rate along the radius is determined by the formula:
  • dQdt=k(T)AdTdr
where:
  • dQdt is the heat transfer rate.
  • k(T) represents temperature-dependent thermal conductivity.
  • A is the surface area of the cylindrical shell.
  • dTdr is the radial temperature gradient.
Understanding this law helps us predict how quickly heat flows through structures like pipes or tubes, which commonly have cylindrical geometries.
cylindrical shell
A cylindrical shell is a hollow cylinder with a specific thickness, characterized by two radii: the inner radius r1 and the outer radius r2. Understanding heat transfer in cylindrical shells is crucial for many practical applications, such as insulation of pipes in plumbing, chemical processes, and mechanical engineering.
The critical aspect of a cylindrical shell involves its surface area, which is calculated as:
  • A=2πrL
where:
  • r is the radius at which we're considering the surface area.
  • L is the length of the cylinder.
This surface area plays a crucial role in heat transfer calculations using Fourier's Law. When heat flows radially, the surface area changes with radius, impacting the rate of heat conduction.
By understanding these properties, engineers can design systems to control heat flow effectively, ensuring efficiency and safety in a variety of applications.
thermal conductivity
Thermal conductivity is a material property that measures a material's ability to conduct heat. It tells us how well heat moves through a substance. In the context of our problem, it varies with temperature, modeled by the equation:
  • k(T)=k0(1+βT2)
where:
  • k0 is the base thermal conductivity at a reference temperature.
  • β is a constant that adjusts how much the thermal conductivity changes with temperature squared.
  • T is the temperature of the material.
This dependence highlights that as temperature changes, so does the material's ability to transfer heat. For example, in our cylindrical shell, the conductivity increases as the temperature rises, due to the positive β term.
Engineers often need to consider these variations in thermal properties to predict and control heat transfer in devices effectively. It plays a vital role in the design of materials used for thermal management in engineering systems.
ordinary differential equation
An ordinary differential equation (ODE) is a mathematical equation that relates a function with its derivatives. In our context, it is a powerful tool to model the relationship between temperature and heat flow in systems like the cylindrical shell.
We arrive at an ODE while solving for heat transfer across the shell:
  • dQdt=k0(1+βT2)2πrLdTdr
To find the heat transfer rate, we need to solve this equation by integrating it over the given boundaries. This involves separating variables and considering both temperature and radius as factors.
The integration gives us a function that links the temperature difference across the shell with the heat transfer rate, which is calculated explicitly in specific boundary conditions. By evaluating these integrals, we obtain a comprehensive relationship for designing and analyzing heat transfer systems.

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Most popular questions from this chapter

What is heat generation? Give some examples.

Consider a large plane wall of thickness L=0.3 m, thermal conductivity k=2.5 W/mK, and surface area A= 12 m2. The left side of the wall at x=0 is subjected to a net heat flux of q˙0=700 W/m2 while the temperature at that surface is measured to be T1=80C. Assuming constant thermal conductivity and no heat generation in the wall, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the temperature of the right surface of the wall at x=L.

How is integration related to derivation?

Consider a water pipe of length L=17 m, inner radius r1=15 cm, outer radius r2=20 cm, and thermal conductivity k=14 W/mK. Heat is generated in the pipe material uniformly by a 25kW electric resistance heater. The inner and outer surfaces of the pipe are at T1=60C and T2=80C, respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

Consider a large 5-cm-thick brass plate (k= 111 W/mK ) in which heat is generated uniformly at a rate of 2×105 W/m3. One side of the plate is insulated while the other side is exposed to an environment at 25C with a heat transfer coefficient of 44 W/m2K. Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

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