In the training video the at 1:30 the instructor starts talking about combustion air for the two appliances. 36000bthu for the water heater and 80000btuh for the furnace. He adds them together for a total 116000btuh. He then says that you require 50ft3 for each 1000bthuh. So he divides 116000btuh by 50ft3. He comes to the conclusion that he requires 6000ft3 of space for combustion air, but 116000/50=2320ft3.
If it was 6000ft3 then divide by 8ft high room = 750ft2 room for this not to be a confined space?
If it was 2320ft3 then divide by 8f high room = 290ft2 room, which is approx 17ftx17ft room which is still crazy huge.
What am I missing?
The contractor license to do the calculations. Unless you can perform these calculations at multiple elevations I’d stay away from any reference in my report.
The InterNACHI reference is very general. The requirements in Colorado will be different than Miami.
Factor in wood or metal louvers and calculations change. You need to understand not only room volume but methods of bringing combustible air into the room. This is a rabbit hole most will not enter. Not part of the SOP.
I am in the process of becoming a home inspector and I am completing the courses. I was asking because I did not understand the course material.
“To burn 1 cubic foot of gas, approximately 1,000 Btu for natural gas, requires 10 cubic feet of air for perfect combustion (Figure 2). Ten cubic feet of air contains 2 cubic feet of oxygen and 8 cubic feet of nitrogen and other gases combined with water vapor.”