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.

1-D form of 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