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Introduction

The principle aim of this module is to introduce students to the basics of steady-state thermal analysis. Modelling thermal systems allows engineers to evaluate and predict the performance of a system affected by heat transfers, which can inform many design, operational and maintenance decisions faced by engineers in industry. Some examples of steady-state thermal analyses that may be used are listed below:

To inform the design of thermal systems.
To predict the thermal performance of a system
To analyse thermal losses in a system and evaluate potential solutions
To calculate heating and cooling loads of a system
To evaluate the thermal stresses on a system.

In this module, we shall cover the following topics:

A review of the numerical analyses of systems affected by conduction, convection and radiation.
The derivation of and application of the Energy Balance Equation.
The principles of Finite Element Analysis (FEA) and modelling thermal systems.
The methodology of conducting FEA using the ANSYS: Steady State Thermal software environment.

Review of Heat Transfer

Heat can be transferred through a system in three general modes:

Conduction in which energy flows through a solid from a high-temperature region to a low-temperature region in proportion to the temperature gradient.
Convection in which heat flows from a solid to its surrounding fluid environment. Heat energy is transferred from the solid to the fluid at the boundary and then is removed from the boundary area via mass transfer (i.e. fluid motion). The rate of heat transfer is contingent on the properties of the boundary and the fluid.
Radiation in which heat flows from a solid to its surrounding environment by electromagnetic radiation. In this case, heat transfer is not contingent on the properties of the surrounding substance, i.e.: radiation occurs even when the system is surrounded by a vacuum.

Notation

The quantities that will be used in this module are defined as follows:

$q$ - Rate of conductive heat transfer (W)

$f$ - Heat flux (W/m²) ‘Heat transfer per unit cross-sectional area’

$k$ - Thermal conductivity (W/m K)

$h$ - Convective heat transfer coefficient (W/m² K)

$c$ - Specific heat (J/kg K)

$α$ - Coefficient of thermal expansion (m/K)

$Q$ - Rate of heat generation per unit volume

$T$ - Temperature (°C or K)

$Δ T$ - Temperature difference (°C or K)

$d T / d x$ - Temperature gradient

$t$ - Time (s)

$A$ - Cross-sectional area (m²)

Table 1: Common symbols used in heat transfer calculations

Steady-State Conduction

Introduction to Conduction

‘Thermal conduction’ is the transfer of heat energy through a body by means of inter-particle collisions that are naturally occurring all the time.

Where neighbouring regions of particles have a consistent difference in internal energy (i.e.: a temperature gradient exists) energy is transferred from the higher energy region to the lower energy region.

What results is a spontaneous flow of heat from hotter to colder regions of an object and this flow is proportional to the temperature gradient.

In the absence of any external sources or sinks of energy, temperature differences through a body decay over time leading to a state of thermal equilibrium.

thermal conduction diagram

University Physics Volume 2 / Creative Commons Attribution 4.0

Fourier’s Law (Differential Form)

The basic equation describing heat conduction is Fourier’s Law:

“The rate of heat transfer per unit area (the heat flux) is proportional to the gradient of the temperature in the normal direction and flows in the direction of the decreasing temperature.”

Heat flux (f) is therefore given by the following formula:

$f = - k \frac{d T}{d x}$ eq.(1)

Flux is defined as the heat transfer by unit area; therefore this relationship can also describe the flux in terms of the rate of heat transfer (q):

$f = \frac{d q}{d A}$ eq.(2)

If we assume that a given cross-section has uniform heat transfer, eq. (2) simplifies to f = q/A. We are then able to write an equation for rate of heat transfer (q) in terms of the temp gradient:

$q = - k A \frac{d T}{d x}$ eq.(3)

Thermal Conductivity

$q = - k A \frac{d T}{d x}$ eq.(3)

Equation. 3 tells us that solids with a large value of ‘k’ (thermal conductivity) are good heat conductors and that with a low value of ‘k’ are poor conductors (i.e.: good insulators). Some examples of typical values for various materials are listed below:

Metal	Thermal Conductivity (W/m K)	Metal	Thermal Conductivity (W/m K)
Aluminium	230	Brick	0.6
Copper	390	Concrete	0.8
Steel	46	Glass	0.8

Table 2: Typical thermal conductivity values for various common materials

Note: we shall use these typical thermal conductivity values when referencing materials in worked examples within this module.

Fourier’s Law (Integral Form)

Eq. (1) is a first order differential equation and assuming k and f are constant, it can be integrated to give Fourier’s Law in integral form.

$f = - k \frac{d T}{d x}$ eq.(1)

$\int_{T_{x} = 0}^{T_{x}} d T = - \frac{f}{k} \int_{x = 0}^{x} d x$

$(T_{(x)} - T_{0}) = - \frac{f}{k} (x - x_{0})$

$T_{(x)} = - \frac{f}{k} x + T_{0}$ eq.(4)

T₀ is the temperature at position x = 0. Note that if the cross-sectional area throughout the solid is not uniform, the flux (f) will not be constant.

Example 1

steel rod diagram

A steel rod of length 35 cm and a uniform cross-section has one end in contact with a hot plate at temperature (T = 200 °C). Assuming that the perimeter of the rod is insulated and the flux through the rod is 3 kW/m, determine the temperature of the other end of the rod.

View Solution

Steel has a thermal conductivity of k = 46 W/m K.

Eq.(4) tells us:

$T_{(x)} = - \frac{f}{k} x + T_{0}$ eq.(4)

$T_{(0.35)} = - \frac{3000}{46} 0.35 + 200$

$T_{(0.35)} = 177.2 ° C$

Example 2

A house is constructed with 250 mm solid brick exterior walls. The internal temperature is 20 °C. The external temperature is 10 °C.

Assuming that the internal and external wall surfaces are at the temperature of their surroundings:

What is the temperature gradient (dT/dx) of the wall?
What is the heat flux (f) of the wall?
What is the temperature (T) within the wall at a point 50 mm from the internal surface?

house diagram

View Solution

What is the temperature gradient (dT/dx) of the wall?

     $\frac{d T}{d x} = \frac{(T_{1} - T_{0})}{Δ x}$

     $\frac{d T}{d x} = \frac{(10 - 20)}{0.250}$

     $\frac{d T}{d x} = - 40 ° C / m (or k/m)$
What is the heat flux (f) of the wall?
• Brick has a thermal conductivity of k = 0.6 W/m K.
• Eq. (1) tells us:

     $f = - k \frac{d T}{d x}$

     $f = - 0.6 \times 40$

     $f = - 24 W/ m^{2}$
What is the temperature (T) within the wall at a point 50 mm from the internal surface?
• Eq. (4) tells us:

     $T_{(x)} = - \frac{f}{k} x + T_{0}$

     $T_{(0.05)} = - \frac{24}{0.6} 0.05 + 20$

     $T_{(0.05)} = 18 °C$

Conduction Through a Cylindrical Pipe

In a cylindrical pipe of length (l) with internal and external radii of r1 and r2 respectively we are interested in the conduction of heat through the pipe wall.

Considering eq. (3) for heat flux:

$q = - k A \frac{d T}{d x}$

As illustrated, the area for conduction is not fixed, i.e., the area increases as we move towards the external surface of the pipe (r1 → r2).

The area at any given distance from the centre can be expressed in terms of radius (r):

$A = 2 π r l$ eq.(5)

And the heat flux equation can therefore be rewritten as:

$q = - k (2 π r l) \frac{d T}{d r}$ eq.(6)

cylindrical pipe diagram

Eq. (6) is a first order differential equation and assuming $k$ and flux ( $f$ ) are constant once again, it can be integrated to describe heat flow through a cylindrical pipe wall:

Rearranging eq. (6) and integrating both sides: Note T_r1 and T_r2 are the internal and external surface temperatures respectively. (T1 and T2).

$- \frac{q}{2 π k l} \int_{r_{2}}^{r_{2}} \frac{1}{r} d r = \int_{T_{r 2}}^{T_{r 1}} d T$ eq.(7)

Using the log function identity: $(ln x - ln y) = (ln x / ln y)$

$- \frac{q}{2 π k l} (ln r_{2} - ln r_{1}) = T_{2} - T_{1}$

$- \frac{q}{2 π k l} (ln (\frac{r_{2}}{r_{1}})) = T_{2} - T_{1}$ eq.(8)

And rearranging in terms of heat flow:

$q = \frac{2 π k l (T_{2} - T_{1})}{ln (r_{2} / r_{1}}$ eq.(9)

Example 3

Given that there is no heat storage or generation within the pipe wall (i.e.: heat flux (q) is constant). Find the temperature 10 mm from the external surface given the following system description:

Internal radius of pipe = 100 mm
External radius of pipe = 150 mm
Internal surface temp = 100 C
External surface temp = 20 C

View Solution

As the heat flux through the pipe is constant, we can equate the heat flux to the outside wall with the heat flux to any internal point of radius (r):

$q = \frac{2 π k l (T_{1} - T_{2})}{ln (r_{2} / r_{1})} = \frac{2 π k l (T_{1} - T_{r})}{ln (r / r_{1})}$

Cancelling out common terms:

$\frac{(T_{1} - T_{2})}{ln (r_{2} / r_{1})} = \frac{(T_{1} - T_{r})}{ln (r / r_{1})}$

And rearranging for T_(r):

$T_{(r)} = T_{1} - \frac{(T_{1} - T_{2})}{ln (r_{2} / r_{1})} ln (r / r_{1})$ Note: This equation will be useful in the future.

$T_{(140)} = 100 - \frac{(100 − 20)}{ln (150/100)} ln (140/100) = 33.6 ° C$

References

Annaratone D. (2010) Steady Conduction (pp 13-27). In: Engineering Heat Transfer. Springer, Berlin, Heidelberg. 30 October 2009. Available as an EBOOK in the UHI library.

Modelling the Energy Balance Equation

Introduction

Review of Heat Transfer

Notation

Steady-State Conduction

Introduction to Conduction

Fourier’s Law (Differential Form)

Thermal Conductivity

Fourier’s Law (Integral Form)

Example 1

Example 2

Conduction Through a Cylindrical Pipe

Example 3

References