The basement of my house has concrete walls that are 025 m

The basement of my house has concrete walls that are ~ 0.25 m thick. The total surface area of the walls is ~ 125 m^2. Lowe\'s sells insulation sheets that are 2.5 m long times 1.25m wide times 0.01m thick, and cost $13 each. They have a stated R-value of R = 7.7 F ft^2 hr/BTU. If heating my house using wood has an equivalent heating cost of $0.05/kW hr, should I insulate my basement with the Lowe\'s insulation, or will it be a waste of money? If I insulate it, how long will it take me to recoup the cost of the insulation through heat savings?

Solution

Note: Temperature of the room and the surroundings, R Value or Thermal conductivity of the wall  are required to arrive at a numerical value of the problem. However, an attempt is being made to derive the equations where the above values if known can be plugged in to get the answers.

Area of one insulation sheet = 2.5m x 1.25m =3.5 m²

Number of insulation sheets required to cover 125 m² area = 125/3.5 = 40 sheets

Cost of 40 sheets = 40 x Cost of one insualtion sheet = 40 x $ 13 = $ 520

Given, R value of insulation sheet = 7.7 F ft² hr/BTU = 0.176 x 7.7 K m²/W = 1.35 K m²/W

( R value in SI = R value in US x 0.176)

R value of 0.25m (equal to 10 inch) thick basement concrete = 1.2 ft² hr/BTU or 0.21 K m²/W (Source internet google books)

Now, energy savings in a year is calculated by following formula

E = {24 x (Total surface area of walls) x (U_w - U_i) x DD_H } kW hr / yr

where w and i stand for wall and insulation respectively

U_w = 1/R_w & U_i = 1/R_i  

DD_H = Heating day per year in deg C-day/yr (VALUE MISSING IN PROBLEM !)

DD_H can be calculated if average temperature difference inside & outside of basement is known multiplied by the number of heating days in a year.

Payback =Cost of sheets / Total heat saving cost in a year

Total heat saving cost in a year =  (E x c)

where c, equivalent heating cost = $ 0.05/ kW hr (Given)

E = energy saving as calculated above

Therefore, Payback = $ 520 / (E x $ 0.05)