# Measuring Home Heat Loss

I am interested in testing/measuring my house to see how leaky it is and how bad the insulation is. If my house is poorly sealed and insulated then I will know how to improve it and I can retest it afterwards and make sure I fixed the problem. Two local energy companies wanted too much money to do a blower door test and going around with fancy thermal camera. Purchasing the equipment myself is even more expensive. Recouping the cost of these official tests with future energy savings would be near impossible.

I was wondering if there was a low-tech way to do this. My idea is this:

1. Wait for a cold day and measure the outside temp.
2. Heat the house to a certain temp and then turn off the heat.
3. Time how long it takes for the inside temp to drop by a certain amount.
4. Calculate some number that tells me something about the thermal efficiency of my home.

By just doing the test, then making some home improvements and then repeating the test under similar conditions I know that I could gauge my relative improvement, but it wouldn't tell me if my home is poor/average/good compared to other homes.

I googled around and didnt find any low-tech way to ballpark this stuff. I did find these two equation:

Total Heat Loss = (Surface Area of Home) x (Delta T) x (U-Value)
Air Energy = (Mass of Air) x (Heat Capacity of Air) x (Temperature)

I can measure surface area with a tape measure. I can measure the mass of my air with a tape measure and the density of air constant (1.2kg/m^3). I can measure temp with a thermometer. Heat capacity of air is a constant (~1kJ/kg*K)

So, if I calculate the Air Energy at steps 2 and 3 of my experiment then I can subtract to find the heat loss that my home sustained. Plug this into equation 1 and I can then calculate the u-value of my entire house as a whole. However, Im not sure where time plays into the equation since waiting longer would mean more heat loss, and time is not in any of the variables. So thats something I dont know.

Even if I do all this, how would I know if my U-value is low, average or high?

So, back to the original question. Is there an established way to do a cheap/low-tech test of a house to see how efficient it is with regard to heat loss? Is there some rule of thumb like "An average 2000sqft home will lose 50k BTUs per hour"

• There's two numbers: thermal resistance, and air changes per hour at a given pressure. Those account for both radiative+conductive losses, and convective/physical movement losses. So you need to do your test on a day with absolutely no wind for the resistive measurement. But again on a day with a constant wind from a known direction so you can guess at how permeable your house is.
– Móż
Mar 14 at 22:14
• The cheap option would be a DIY blower test, you could possibly even use cardboard to block up the door with a hole for the fan to do the blow and suck tests. That wouldn't give you a repeatable or generalisable number, but that plus a few sticks of incense would let you find and fix the worst of the air leaks.
– Móż
Mar 14 at 22:16
• From your profile, looks like you are in SDG&E territory. There's a number at the bottom of this page you can call to schedule a free energy assessment.
– LShaver
Mar 15 at 12:51
• Time actually is in your first equation, Q = U A (T2-T1). The heat transfer coefficient, U, has units of watts/m^2 K, where a watt is 1 Joule per second. The resulting value, Q, is a heat loss rate such as Joules/second.
– Mark
Mar 16 at 1:55