Ch9.5 Heat Conduction Through Wall

• We apply concepts of heat flux and Fourier's law to model conduction of heat in one direction through a wall.
  
• We formulate a differential equation with respect to spatial variable \( x \) using heat flux \( \small{J(x)} \) for equilibrium temperature \( \small{U(x)} \) in wall.
• The analytic solution of resulting second order boundary value problem is derived and explored in Ch11.2.

I wouldn't buy anything with velcro.

It's a total rip-off.

• Walls are everywhere in our lives and provide an essential role in meeting our basic needs for shelter.
  
• Investigating how heat conducts through a wall gives us technical insight into a long running part of human history.
  
• Walls and a roof are defining aspects of shelter.
  
• Shelter is foundational in Maslow's hierarchy of needs [2].
  

• Shelter has enabled individual survival.
• Shelter has helped foster the development of human society and culture.
• Shelter at a basic level can include clothing.
• Shelter has enabled human adaptation to varying climates across millions of years.
  
• Shelter has supported population expansion worldwide under varying conditions.
  
• Indirect methods have been used to identify important periods in human clothing and shelter usage together with associated physiological evolution [3,4].

Parameter values for numerical solution are listed below.

• Wall has thickness \( \small{ L = 1 \, m} \), made of unspecified material.
• Conductivity of material is \( \small{k = 1 \, W/(m \,^\circ C)} \).
  
• Initial temperature is \( \small{U(0) = 10 \,^\circ C} \)
• Initial slope is \( \small{U'(0) = (-1/k)J(0) = -1 \,^\circ C /m} \)

Formulate differential equation for equilibrium temperature inside wall as function of distance from interior side of wall.

• Wall comprised of homogeneous material.
• Both sides of wall held uniformly constant at specified temperatures.
• One side of wall is warmer than other side of wall.
• Heat flows through material from warmer side to cooler side.
• Heat flux through wall is by conduction only.
• Heat is conducted according to Fourier's law.
• Conductivity of material is constant.
• Temperature inside wall is in equilibrium state.

We will model temperature at a point inside wall by considering thin section and then letting thickness of section tend to zero.

Using the balance law, the word equation for our assumptions and compartment diagram is shown below.

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{change \, in \, heat } \\ \mathrm{content \, of \, section } \end{Bmatrix} &= \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{into \, section } \end{Bmatrix} - \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{out \, of \, section } \end{Bmatrix} \\ &= 0 \end{aligned}} \]

• Distance \( x \) of a point within wall is measured from warmer side \( \small{x=0} \) to cooler side \( \small{x=L} \) meters.
• Thin section of wall modeled ranges from \( x \) to \( \small{x+ \Delta x} \).
• Surface area of face of section is \( \small{A \, m^2} \).
• \( \small{J(x)=} \) flux of heat at \( x \), in Watts/\( \small{m^2} \).
• \( \small{U(x)=} \) temperature of wall at \( x \), in Celcius C.
• \( \small{k>0} \) is the conductivity of material within wall.

Recall the following unit considerations:

• Heat conduction is expressed in terms of Joules/sec.
• The units of heat flux \( \small{J(x)} \) is \( \small{W/m^2} \).
• The units of \( \small{J(x)A} \) are \( \small{\left(\frac{W}{m^2}\right)(m^2) = W = \frac{Joules}{sec}} \)

Word equation for thermal equilibrium becomes

\[ \small{0 = J(x)A - J(x+ \Delta x)A } \]

Let \( \small{h = \Delta x} \), and divide both sides by \( \small{(- A h)} \):

\[ \small{ \frac{J(x+ h) - J(x)}{h} = 0 } \]

Apply limit operator with \( h \rightarrow 0 \) to both sides:

\[ \small{ \begin{aligned} \lim_{h \rightarrow 0} \frac{J(x+ h) - J(x)}{h} & = \lim_{h \rightarrow 0} (0) \\ \frac{dJ}{dx} & = 0 \end{aligned} } \]

Using Fourier's law:

\[ \small{ \begin{aligned} \frac{d}{dx} \left(-k \frac{dU}{dx} \right) &= 0 \\ \frac{d^2 U}{dx^2} &= 0 \end{aligned} } \]

• Since ODE is simple, we take a quick look at exact solution.

\[ \small{\frac{d^2 U}{dx^2} = 0 \Rightarrow U(x) = a x + b} \]

• Note that time \( t \) is not present in our solution.
• Coeffs \( a \) and \( b \) determined within context of problem.

• To numerically solve 2nd order IVP, recall that IVP is first expressed as system of two first order IVPs.
• Consider

\[ \small{y'' + ay' + cy = g(x), \,\,\, y(0) = y_{10}, \,\,\, y'(0) = y_{20}} \]

• Let \( y_1 = y, \, y_2 = y_1' \), and note \( y'_2 = y'' \).
• Then IVP above is equivalent to system below:

\[ \small{ \begin{aligned} y_1' & = y_2, \, y_1(0) = y_{10} \\ y_2' & = g(x) - a y_2 - b y_1, \, y_2(0) = y_{20}\\ \end{aligned}} \]

The IVP for thermal equilibrium in the wall is

\[ \small{y'' = 0, \, y(0) = y_{10}, \, y'(0) = y_{20}} \]

From Fourier's law, \( \small{J = -ky'} \). Thus the IVP becomes

\[ \small{ \begin{aligned} y_1' &= y_2,\,\,\, y_1(0) = y_{10} \\ y_2' &= 0, \,\,\,\,\, y_2(0) = y_{20} = (-1/k)J(0) \end{aligned} } \]

Our goal is to find \( \small{y_1 = y} \), but we also find \( \small{y_2} \) along the way.

Recall the parameter values specified in the Background section:

• Thickness of wall: \( \small{L = 1 \, m} \)
• Conductivity of wall material: \( \small{k = 1 \, W/(m \,^\circ C)} \)
  
• Initial value of temperature: \( \small{U(0) = 10 \, ^\circ C} \)
• Initial slope: \( \small{U'(0) = (-1/k)J(0) = -1 \,^\circ C/m} \)

\[ \small{\begin{aligned} y'' &= 0, \, y(0) = y_{10}, \, y'(0) = y_{20} \\ y_1' &= y_2,\,\,\, y_1(0) = y_{10} \\ y_2' &= 0, \,\,\,\,\, y_2(0) = y_{20} = (-1/k)J(0) \end{aligned}} \]

#System of ODEs
a <- 0
b <- 0
g <- 0
f1 <- function(y2) {y2}   
f2 <- function(y1,y2) {g-a*y2-b*y1}

#Place additional graph in plot   
lines(x,-k*y2, type="l",col="red")

• For numerical solution, we converted an associated 2nd order IVP to a system of first order IVPs.
  

\[ \small{\begin{aligned} y'' &= 0, \, y(0) = y_{10}, \, y'(0) = y_{20} \\ y_1' &= y_2,\,\,\, y_1(0) = y_{10} \\ y_2' &= 0, \,\,\,\,\, y_2(0) = y_{20} = (-1/k)J(0) \end{aligned}} \]

• The initial conditions are adapted from physical observation.
  
• Instead, could have boundary conditions \( \small{y(0) = y_1} \) and \( \small{y(L) = y_2} \), or \( \small{y'(0) = f_1} \) and \( \small{y'(L) = f_2} \).
  
• Numerical methods for solving BVPs different than for IVPs.
  

• Assuming that the wall is in thermal equilibrium allows us to model temperature \( \small{U(x)} \) as a function of distance \( x \) only.
• This leads to an ordinary differential equation.
  
• Otherwise temperature would have the form \( \small{U(x,y,z,t)} \), and model will require a partial differential equation.
  

• Our textbook [1] points out that we must also account for the amount of heat “used up” in raising the temperature.
• More complicated differential equations can also occur when we include volumetric heat sources, as is when heat is produced by an electric current or chemical reaction, see [1].

[1] Mathematical Modeling with Case Studies, Barnes and Fulford, CRC Press, 2015.

[2] Shelter (building), https://en.wikipedia.org/wiki/Shelter_(building), retrieved on 10/2/2022.

[3] The Technology Of Clothing And Shelter, https://gohighbrow.com/the-technology-of-clothing-and-shelter/, retrieved on 10/2/2022.

[4] Origin of Clothing Lice Indicates Early Clothing Use by Anatomically Modern Humans in Africa, https://academic.oup.com/mbe/article/28/1/29/984822, retrieved on 10/2/2022.