Ch11.2 Heat Loss Through a Wall

Background

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We solve an ODE and apply boundary conditions to find equilibrium temperature inside a brick wall.
The boundary condition on inside of wall is kept at a constant temperature.
The boundary condition on outside of wall follows from Newton's law of cooling.

Background

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Consider a building with walls of thickness \( L = 20 ~cm. \)
Inside of house is kept at \( u_i = 20^\circ C. \)
Outside of wall is exposed to surrounding air at a temperature \( 5^\circ C \).
Subscripts \( i \) and \( o \) denote inside and outside of wall, respectively.

Background

Inside wall temperature is fixed at room temperature, \( u_i = 20^\circ C. \)
Outside wall temperature is higher than \( 5^\circ C \), and loses heat to surroundings, at temperature \( u_o = 5^\circ C. \)

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Model Assumptions and Approach

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Let x-axis measure distance from inside of wall.
For \( x \) nearer to 0, temperature is closer to \( 20^\circ C. \)
In general, temperature is a function of the two variables x and t (Ch12).
For equilibrium, temperature will be a function of x only.

Model Assumptions and Approach

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Neglect complicated heat flow at edges, top and bottom.
Assume heat flow will be in x-direction only.
Assume no windows.

Model Assumptions and Approach

Since inside and outside temperatures are constant once thermal equilibrium is established, temperature at each point inside wall will be constant.
Outside temperature of wall satisfies Newton's law of cooling.
Temperature of outer surface of wall can be determined from solution for temperature in wall (using boundary conditions).

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Differential Equation

From Ch9.5, the differential equation is

\[ \frac{d^2 U}{dx^2} = 0 \]

ODE arose from heat balance in thermal equilibrium.
ODE does not depend on conductivity of the material.
Conductivity incorporated through boundary conditions.

\[ \begin{aligned} U''(x) &= 0 \\ U'(x) &= C_1 \\ U(x) &= C_1 x + C_2 \end{aligned} \]

Boundary Conditions

Inside: Fixed temperature

\[ U(0) = C_2 = u_i \]
Solution so far:

\[ \begin{aligned} U(x) &= C_1 x + C_2 \\ &= C_1 x + u_i \end{aligned} \]

Outside: Newton's Law
- Use to find \( C_1 \)
- Note also \( ~U'(x)=C_1 \)

Rate of heat exchange: \( Q = hA\Delta U \) (p. 228)
Rate of heat conduction: \( J(x)A \) (p. 237)

\[ \begin{aligned} J(L)A &= h A\left(U(L)- u_o \right) \\ J(L) &= h \left(U(L)- u_o \right) \end{aligned} \]

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Boundary Conditions

Outside Wall:

\[ \begin{aligned} J(L) &= h \left(U(L)- u_o \right) \\ U(L) &= \frac{J(L)}{h} + u_o \end{aligned} \]

Fourier's Law:

\[ \begin{aligned} J(x) &= -k U'(x) \\ J(L) &= -k C_1 \\ \end{aligned} \]

From \( ~U(x)=C_1x+u_i \):

\[ \begin{aligned} U(L) &= C_1 L + u_i \end{aligned} \]

Thus

\[ \begin{aligned} C_1 L + u_i &= \frac{J(L)}{h} + u_o \\ C_1 L &= \frac{-k C_1}{h} + u_o - u_i \\ (hL + k) C_1 &= h(u_o - u_i) \\ C_1 &= \frac{u_o - u_i}{L + \frac{k}{h}} \\ \therefore U(x) &= \left(\frac{u_o - u_i}{L + \frac{k}{h}}\right) x + u_i \end{aligned} \]

Solution to Boundary Value Problem

Solution

\[ U(x) = \left(\frac{u_o - u_i}{L + \frac{k}{h}}\right) x + u_i \]

Code chunk:

Ch11.2Ex1 <- function(h) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
L <- 0.15 #Wall thickness (meters)
k <- 0.5  #Conductivity of brick
#h <- 10  #Convective heat transfer coeff
U <- function(x){(uo-ui)/(L+k/h)*x+ui}

Plot Code in R

Ch11.2Ex1 <- function(h) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
L <- 0.15 #Wall thickness (meters)
k <- 0.5  #Conductivity of brick
#h <- 10  #Convective heat transfer coeff
U <- function(x){(uo-ui)/(L+k/h)*x+ui}

curve(U,xlab ="x (meters)",
ylab ="Temperature (C)", 
type = "l",col="blue",xlim=c(0,L),ylim=c(0,ui))

abline(h = uo, type="l",col="red")

legend("topright", legend = c("Wall Temperature", "Outside Temperature"), col = c("blue","red"), 
lty = c(1,1)) }

Solution to Boundary Value Problem

Ch11.2Ex1(10)

plot of chunk unnamed-chunk-3

\[ U(x) = \left(\frac{u_o - u_i}{L + \frac{k}{h}}\right) x + u_i \]

Code chunk:

Ch11.2Ex1 <- function(h) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
L <- 0.15 #Thickness 
k <- 0.5  #Conductivity
#h = heat transfer

Solution to Boundary Value Problem

Ch11.2Ex1(5)

plot of chunk unnamed-chunk-5

\[ U(x) = \left(\frac{u_o - u_i}{L + \frac{k}{h}}\right) x + u_i \]

Code chunk:

Ch11.2Ex1 <- function(h) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
L <- 0.15 #Thickness 
k <- 0.5  #Conductivity
#h = heat transfer

Solution to Boundary Value Problem

Ch11.2Ex1(100)

plot of chunk unnamed-chunk-7

\[ U(x) = \left(\frac{u_o - u_i}{L + \frac{k}{h}}\right) x + u_i \]

Code chunk:

Ch11.2Ex1 <- function(h) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
L <- 0.15 #Thickness 
k <- 0.5  #Conductivity
#h = heat transfer

Rate of Heat Loss from Wall

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The rate of heat loss from wall can be calculated using Fourier's Law

\[ \begin{aligned} J(x) &= -k~ U'(x) \\ J(L) &= -k ~C_1 \\ & = -k \frac{u_o - u_i}{L + \frac{k}{h}} \\ & = \frac{u_i - u_o}{\frac{L}{k} + \frac{1}{h}} \end{aligned} \]

Rate of Heat Loss from Wall

Rate of heat loss from wall:

\[ J = \frac{u_i - u_o}{\frac{L}{k} + \frac{1}{h}} \]

\( J \) is directly proportional to temperature difference but not directly proportional to the conductivity \( k \) or to the thickness of the wall \( L \).
Thus doubling wall thickness will not halve the heat loss from the wall, and nor would doubling the conductivity.
However, for decreased \( k \) and \( h \) there is a reduction in the heat flux \( J \).

Rate of Heat Loss from Wall

The rate of heat loss from wall:

\[ J = \frac{u_i - u_o}{\frac{L}{k} + \frac{1}{h}} \]

Ch11.2Ex2 <- function(L,k) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
#L = thickness (meters)
#k =  Conductivity 
h <- 10  #Convective heat transfer coeff
(U <- (ui-uo)/(L/k+1/h)) }

Examples: Rate of Heat Loss

Brick wall (\( W/m^2 \))

Ch11.2Ex2(0.15,0.5)

[1] 37.5

Window (\( W/m^2 \))

Ch11.2Ex2(0.005,0.8)

[1] 141.1765

Heat loss through window is significant

Heat Resistance R-Values

R-values represent combined thermal resistance:

\[ J = \frac{u_i - u_o}{R}, ~R = \frac{L}{k} + \frac{1}{h} = R_1 + R_2 \]

Here
- \( ~R_1 = L/k \) = resistance to heat flow through material
- \( ~R_2 = 1/h \) = resistance to heat flow through surface
Resistance to heat flow through material, \( L/k \), increases with width of the material \( L \) and decreases with the conductivity \( k \), as we might expect.

Heat Resistance R-Values

If we include a surface resistance for both inside and outside surfaces of a wall, we obtain

\[ J = \frac{u_i - u_o}{\frac{1}{h_i} + \frac{L}{k} + \frac{1}{h_o}} \]

Ch11.2Ex3 <- function(L,k) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
#L = thickness (meters)
#k =  Conductivity 
hi <- 10  #Convective heat transfer coeff
ho <- 10  #Convective heat transfer coeff
(U <- (ui-uo)/(1/hi + L/k + 1/ho)) }

Examples: Rate of Heat Loss

Brick wall (\( W/m^2 \))

Ch11.2Ex3(0.15,0.5)

[1] 30

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Insulation R-Values

Companies often mention R-value of insulation batts.
R2.0 batts have a thermal resistance of \( L/k = 2.0 W^\circ C^{-1} \).
We can calculate rate of heat loss for insulated wall of house.
Here, the total resistance is

\[ R = \frac{1}{h_i} + \frac{L_1}{k_1} + \frac{L}{k} + \frac{1}{h_o} \]

where \( k_1 \) and \( L_1 \) are conductivity and width of insulation batts, and \( k \) and \( L \) are those values corresponding to brick.
We assume same surface resistance on both sides of wall.

Insulation R-Values

For R2.0 batts, we use the following code.

Ch11.2Ex4 <- function(L,k) {
ui <- 20  #20C = 68F
uo <- 5   #5C = 41F
#L = thickness (meters)
#k =  Conductivity 
hi <- 10  #Convective heat transfer coeff
ho <- 10  #Convective heat transfer coeff
L1 <- 2   #R2.0 batts have L1/K1 = 2.0
k1 <- 1   #R2.0 batts have L1/K1 = 2.0
(U <- (ui-uo)/(1/hi + L1/k1 + L/k + 1/ho)) }

Insulation R-Values

Brick wall with R2.0 batts

Ch11.2Ex4(0.15,0.5)

[1] 6

Thus insulated wall has much less heat flux than a single brick wall, for which \( J = 30 W/m^2 \).

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Double Glazed Windows

For windows, double or triple glazing is used to reduce the loss of heat and is remarkably effective.
Over 90% of heat loss can be prevented (Ch11.3).

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