 # Measuring Energy Efficiency in Windows: How much can I save?

It’s really easy to say ‘our product is more energy efficient’.  That really doesn’t say anything.  That statement could literally just mean the product is .01% more efficient.  So today I’m going to break down how much more efficient our Indow Window inserts are.   If you want to skip reading the thorough explanation and just get to ‘how much money could I save’ you can check out this spreadsheet I put together.  If that’s also more work than you want to do then the to put it simply: compared to standard single pane windows in Boston MA a 1,000 sqft home could save an estimated \$377.22 per year.

note* the spreadsheet assumes gas heating is used @ a rate of \$1.60 cents per therm for gas and 0.226 for electricity (what I found to be the average in the Boston area)

The spreadsheet calculates monetary savings from using window inserts in your home, however,  saving money is not the only benefit.  The energy savings from indow window inserts also have a positive impact on the environment.

## How do we calculate this number?

To get started we’ll need to understand a bit of vocabulary.  Let’s take a look at the first part of the spreadsheet in the image below and identify what information we need. R-Value:  The R-value is the resistance to heat conduction rating of the window.  Essentially it’s how resistant is the window to losing heat.  The higher the value the better.  A standard single pane window is somewhere between 0.8-1

The R-value for indow window inserts stacked with a single pane is 1.94

To keep our estimated savings strict we’ll assume a value of 1 but you can change this if you know the value for your specific windows.

Area (in sqft): This is simple, just the area of your home.  length * width.  For the calculation we’ll use the square footage of 1000, this will make it easy to guestimate what the area should be from that base.

DegreeDays: degreedays is a much less commonly known metric we use to measure the average difference between your desired temperature and the temperature outside.  In the equation, we will multiply this by 24 to get the amount of hot or cold loss throughout the day.  This is a fairly complicated metric you don’t really need to understand but if you’re interested in learning more check out this link.  This is probably the most important part of the equation to calculate the energy efficiency of your windows.

Heating DegreeDays: Heating degree days (HDD) is the degreedays measurement in relation to having to heat your home.  Meaning a positive HDD value at any time is colder than the desired room temperature. For more information check out the link above.

Cooling DegreeDays: Cooling degree days (CDD) is the degreedays measurement in relation to having to heat your home.  Meaning a positive CDD value at any time is warmer than the desired room temperature. For more information check out the link above.

Heating and Cooling COP: This is the coefficient of performance.  Here’s a brief excerpt from the Wikipedia article:

“[the COP is the] ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs.”

Therms: A therm is a non-standard unit of heat energy equal to 100,000 BTU.  For more information on this check out the Wikipedia page for Therms

kWh(kilowatt-hour): The kilowatt-hour is 3.6 megajoules or one kilowatt sustained over the course of 1 hour.

BTU (British thermal unit):  the BTU is the quantity of heat required to raise the temperature of one pound of liquid water by 1 degree Fahrenheit at the temperature that water has its greatest density (approximately 39 degrees Fahrenheit). (source)

Alright, now we’re done with the vocab required to start doing our calculations and find out how much energy we can save with Indow Window Inserts!

The Equation:

step 1

HDD = heating degree days, R1= initial R-value, R2 = new R-value with indow window inserts.

Heating day BTU savings= H(HDD, R1, R2) = [(HDD * 24 * Area) / R1] – [(HDD * 24 * Area) / R2]

Cooling day BTU savings= H(CDD, R1, R2) = [(CDD * 24 * Area) / R1] – [(CDD * 24 * Area) / R2]

\$ Saved = (BTUsaved * ThermsPerBTU * [gas or electricity price] ) / [COP]

\$ saved on heating =  5411 * 0.00001 * 1.6 / 0.8 = \$331.14 per year

\$ saved on cooling = 1423.8 * 0.00001 * 0.08 / 3.1 = \$46.08 per year