Consider a 2.65 m by 10.0 m housing wall, with a section shown in the diagram. A

Question

Consider a 2.65 m by 10.0 m housing wall, with a section shown in the diagram.
A builder has the choice of using 1.5 in. by 3.5 in. studs (at $3.80 each, with R-value R=2),
or 1.5 in. by 5.5 in. studs (at $5.90 each; R=7). The studs are spaced with a centre-to-centre distance of 40 cm.
The wall with the thinner studs can accommodate R=12 insulation (at $7.20 per m2), whereas the thicker wall can accommodate R=36 insulation (at $12.40 per m2).
Assume that the house is to be built in an area where the average difference between the inside and the outside temperature is 14.0 C. If the cost of heating is $0.15 per kWh, how long will it take for the homeowner to recoup the extra costs associated with building the better insulated wall?

Consider a 2.65 m by 10.0 m housing wall, with a section shown in the diagram. A builder has the choice of using 1.5 in. by 3.5 in. studs (at $3.80 each, with R-value R=2), or 1.5 in. by 5.5 in. studs (at $5.90 each; R=7). The studs are spaced with a centre-to-centre distance of 40 cm. The wall with the thinner studs can accommodate R=12 insulation (at $7.20 per m^2), whereas the thicker wall can accommodate R=36 insulation (at $12.40 per m^2). Assume that the house is to be built in an area where the average difference between the inside and the outside temperature is 14.0 C. If the cost of heating is $0.15 per kWh, how long will it take for the homeowner to recoup the extra costs associated with building the better insulated wall?

Explanation / Answer

The rate of heat transfer across an area through an R factor is

=A(tH-Tc)/R

Since there is no diagram, I will assume there are 25 studs.

The metric dimension of a stud edge is
.0381 m

since there are 25 of them, the wood area is
.9525 * 2.35
2.238 m^2

The balance of the area is insulation

=20.35-2.238
=18 m^2

The transfer of heat will be

12*(2.238/Rwood+18/Rinsulation)

for 2x4 studs (1.5 x 3.5)

=12*(2.238/4+18/12)
=24.714 W

for 2x6 studs (1.5 x 5.5)

=12*(2.238/7+18/38)

=9.5 W

(Note: I know from experience that the actual R factor in a 2x6 nominal wall is R18, not R 38. The conduction through this wall should be 16 W


The cost of the 2x4 wall is
=25*2.3+18*3.3
=$117

The cost of the 2x6 wall is
=25*4.45+18*6.25
=$224

the difference therefore is $107
at .10 per kw hour
$117 is
1,117 kw hours

Every day the difference in kw hours

(24-9.5)*24/1000

.348 kw hours per day
3,209.8 days

8.8 years

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