Originally posted by: Marlin1975
And i disagree with dullard in that it will save money as the rooms you stay in the most will warm up faster and those are the rooms that have the thermostats so the heat will turn off sooner as those rooms get warm faster without really heating up the utility room.
Crude math time. I will make some assumptions of course, feel free to attack the assumptions if you feel that they are incorrect. But at least give a good scientific justification as to why you are attacking them.
Assumptions:
1) I assume the hot vents line the utility room ceiling.
2) I assume that the utility room is centralized in the basement. Thus, heat that flows through the utility room ceiling is not wasted heat.
3) I assume that the basement rooms around the utility room are rooms in use or that they at least allow heat to flow from them to the rooms in use above them. Thus heat lost through the utility room walls is not wasted heat. I can see people unhappy with this assumption, but then the best solution is to insulate these interior walls, not the ductwork.
4) I assume that there isn't any strong forced convection in the utility room. That is, I assume there aren't strong drafts within an interior centralized room. If this isn't the case, it probably comes from holes in the ductwork allowing hot air to flow into the utility room and that is your real problem - not insulation. Fix the holes and you are done.
5) I assume no other complications: just concrete below, ducts above, and air in-between.
6) I assume average concrete of 1 W/m/K thermal conductivity and 4" thick.
7) I assume the ducts are 6' above the floor with air of 0.026 W/m/K thermal conductivity between the concrete and the ducts.
8) I assume the "hot enough to burn" means the ducts are at 150°F and that the ground is at 50°F.
Results:
1) With those assumptions, this is a 1-dimensional thermal conductivity problem. We only need to calculate the temperature of the top of the concrete floor. This concrete floor is not the air temperature. Step on it with bare feet, and you'll feel that it is cold.
2) The temperaure of the top of the concrete floor is calculated by:
(Tfloor) = (Tducts - Tground*alpha) / (alpha + 1)
where
alpha = (Kconcrete * Dair) / (Kair * Dconcrete)
Kconcrete = conductivity of concrete = 1 W/m/K
Dair = distance of the air between concrete and ducts = 6 feet
Kair = 0.026 W/m/K
Dconcrete = 1/3 feet
Tducts = temperature of ducts = 150°F
Tground = temperature of ground below concrete = 50°F
Thus, alpha = 692.3 and Tfloor = 50.14°F. Yep, the concrete floor feels cold because it is cold. This is true even if the utility room air feels hot (The air actually would be dangerously hot at the top and gradually becoming cooler towards the floor if these assumptions are true).
3) The heat loss is through the ground and is calculated by:
Ploss = Power lost = Kconcrete * (Tfloor - Tground) * Afloor / Dconcrete
where
Afloor = 200 ft^2.
Thus, Ploss = 14.9 watts. You lose about the same amount of energy as 1/4th of a typical lightbulb because the utility room is hot. Or just a bit more than one CFL bulb.
4) What does this cost you? Where I live, electricity goes as high as 10 cents per kilowatt-hour. If you go by gas heating, it'll be even cheaper.
Cost = 14.9 W * 24 hr * $0.10/KwHr = 3.6 cents per day
Now, you tell me if saving 3.6 cents per day is worth the labor and expense to insulate the ducts. Hint: if it were worth it, it'll be in the building codes. If you really can't stand that massive 3.6 cent loss, throw down an old rug in the utility room (ie insulate where the heat is lost).