The code below tells the computer that each of the variables (watt, ft, ...) represent a unit.

In [1]:
from pint import UnitRegistry
ureg = UnitRegistry()
K = ureg.degK
m = ureg.m
watt = ureg.watt
ft = ureg.ft
hour = ureg.hour
BTU = ureg.BTU
F = ureg.degR

This is the representation in mathematics

$$k_{wood} = 0.15 \frac{watt}{K \cdot m}$$$$R_{wood} = \frac{t_{wood}}{k_{wood}}$$$$R_{total} = R_{wood} + R_{foam} + R_{wood}$$

Below is an implementation of the calculation in Python code with units.

We define the conductivities of wood and foam using values from a table in Wikipedia. We also define the units.

In [2]:
k_wood = 0.15 * watt / K / m
k_foam = 0.02 * watt / K / m

We define the thicknesses of the layers of wood and foam in the panel.

In [3]:
t_wood = 0.01 * m
t_foam = 0.14 * m

We can then calculate the R-values for the a wood layer and a foam layer.

In [4]:
r_wood = t_wood / k_wood
r_foam = t_foam / k_foam

The R-value is the sum of the layers. One wood, one foam, and then another wood layer.

In [5]:
r_SIP_SI = r_wood + r_foam + r_wood
r_SIP_SI
Out[5]:
7.13333333333 kelvin meter2/watt

We can multiply by a conversion factor to find the english R-value.

In [6]:
print(r_SIP_SI.magnitude * 5.68) 

40.5173333333

The software we are using allows the computer to convert between these units. We see the numbers agree to 4 significant figures.

In [7]:
r_SIP_SI.to(ft**2 * hour / BTU * F)
Out[7]:
40.5049451666 degR foot2 hour/btu

We can also calculate the power lost by a large area of this wall.

$$ Q = UA\Delta T = \frac{A \Delta T}{R_{total}} $$
In [8]:
length = 50 * m
height = 3 * m
t_inside = 20
t_outside = 5
height * length / r_SIP_SI * (t_inside - t_outside) * K
Out[8]:
315.420560748 watt