Conduction

Conduction

Concepts

• R-value

• U-value

• Fourier's Law

Thought experiment

Given a block of material sitting between two temperatures, what heat flow do you expect?

Fourier's Law

$q_x = -kA\frac{\Delta T}{\Delta x}$

• $q_x$ dimensions of energy per time or power

• $A$ dimensions of area

• $k$ dimensions of power per distance per degree

• $\Delta T$ is the temperature difference

• $\Delta x$ is the thickness of the material

Fourier's Law Differential Form

$q_x = -k \frac{dT}{dx}$

$q = -k \nabla T$

• $q_x$ dimensions of power per unit area

• $k$ dimensions of power per distance per degree

Fourier's Law, Buildings Form

$Q = U A \Delta T$

• $Q$ heat transfer dimensions of power

• $U$ dimensions of power per area per degree temperature

• $A$ dimension of area

• $T$ dimension of temperature

Conductivity

R value and U value

• Building materials publish an R-value

• Sometimes published as a U-value

Conduction

Conduction

Conduction

Units

• in US units R value is $ft^2 \circ F / BTU per hour$

• in SI units R value is $m^2 K / watt$

Parallel

• If you have two conducting surfaces in parallel, the U-values add

• In parallel, the heat can take either path

Series

• If you have two conducting surfaces in series, the U-values add

  according to

  $U_{total} = \left(\frac{1}{U_1} + \frac{1}{U_2}\right)^{-1}$

• In series, the heat must take both paths

Bathtub model of heat flow

• What is the input?

• Now the drain is faster with greater temperature

Typical R and U values

• Window range (US R-values)

  • R-1 for single pane

  • R-12.5 for more advanced windows

• Wall range

  • R-3.4 (2x4 no inssulation)

  • R-12.7 (2x4 R-13 insulation)

  • R-34.6 (2x6 R-21 insulation)

Cylindrical Insulator

Note that this R does not have the same dimensions as R-value. It has dimensions of temperature difference per unit power.

Spherical Insulator

Note that this R does not have the same dimensions as R-value. It has dimensions of temperature difference per unit power.

Typical UA values

• Residential Home ?

• Commercial Building ?

• Cooler ?

• Down Jacket ?

Activities

R-Value of SIP

• Are the components of the SIP in parallel or series?

• How do we find the properties of each?

• Wood 0.15 watts per kelvin per meter

• Polyurethane foam 0.02 watts per kelvin per meter

• Conductivities

• 4.5 inch panel 13.8 R value

k_wood = 0.15 W/K/m
k_foam = 0.02 W/K/m
t_wood = 0.01 m
t_foam = 0.08 m

r_wood = t_wood/k_wood => 0.0667 m^2*K/W
r_foam = t_foam/k_foam => 4 m^2*K/W

r_SIP_SI = 2 * r_wood + r_foam => 4.1333 m^2*K/W

r_SIP_US = r_SIP_SI * 5.68 ft^2/m^2*F/K*W/BTU => 23.4773 ft^2*F/BTU

Convert R-values

Convert between a US R-value and a metric R-value.

R-Value Conversion

Once you learn how to do this, you can use the value you calculate as a conversion factor. This will help you convert more quickly.

Estimate Wall Loss in ETC

• What is the R-value?

• What is the total area of walls?

• How much power do we need to maintain the ETC one degree above the

  outside temperature?

Calculus Approach to Heat Loss through conduction

We calculate the temperature as a function of time of a heated object that loses heat to its surroundings through an insulation. We start with a lumped mass approximation with conductive heat loss through an insulator. Using the heat capacity of the object we have the relation

$C = \frac{Q}{\Delta T}$

Where $C$, the heat capacity is the product of the density, $\rho$, the volume of the object $V$, and the specific heat capacity of the material $c$.

$C = \rho V c$

Over a small time interval $dt$, the heat lost by the object as heat conducts away is the product of the temperature difference $T - T_C$, the thermal conductivity of the insulation, $K$ and the time interval $dt$.

$Q = K (T - T_C) dt$

Substituting, we get

$\rho V c = K (T - T_C) dt/dT$

which we rearrange and integrate

$\int_{0}^{t} \frac{K}{\rho V c} dt = \int_{T_0}^{T} \frac{dT}{T - T_C}$

with initial conditions $T = T_0$ at $t=0$ and $T=T$ at $t=t$. Integrating, we get

$\frac{K}{\rho V c} t \bigg|_0^t = \ln(T - T_C) \bigg|_{T_0}^{T}$

$\frac{K}{\rho V c} t = \ln(\frac{T - T_C}{T_0 - T_C})$

We exponentiate both sides and get

$\exp(-\frac{K}{\rho V c} t) = \frac{T - T_C}{T_0 - T_C}$

$T = (T_0 - T_C) \exp (-\frac{K}{\rho V c} t) + T_C$

The equation shows an exponential decay in temperature starting at $T_0$, the initial temperature of the object, decaying to $T_C$.

Discrete approach

We can also do this numerically with a discrete time period.