Many calculators (example) can obtain heating load or the whole house or of a single room with room based on area, room height and other factors (windows, insulation etc).

My 1927 home has existing ductwork which I'd like to confirm whether the existing ducts/vents are properly sized. If not, I’d like to do adjustments (e.g., replace a 6” with an 8” duct).

However, this is not as straight forward:

  1. There is a single common return in the hallway which implies that all doors should be open anyway for proper airflow
  2. The house has two floors with heating only in the lower floor
  3. Lower floor features 2 bedrooms but otherwise is open floor plan with hallway, dining room, living room, kitchen connected (no doors)
  4. All rooms have standard ceiling heights of 8", except the living room which has a heigh ceiling (12") ... noteworthy they are connected via large opening.
  5. Next to the dining room are the stairs the the second floor (unheated). The rooms in the second floor can be closed but the upstairs corridor is still connected to the rest.

Room sizes first floor:

  • Bedroom 1: 149sqft
  • Bedroom 2: 176sqft
  • Bathroom: 58sqft
  • Hallway: 50sqft
  • Diningroom: 184sqft
  • Kitchen: 105sqft
  • Nook (next to kitchen): 71sqft
  • LivingRoom: 217sqft

Further info:

  • 1927 home in Bay Area, California
  • gas furnace was recently replaced with 3-ton heat pump using existing duct work and vents.
  • No cooling required, only heating in winter

First floor layout:

enter image description here

How do I go about calculating the heat load in such a layout/setup?

Do I calculate the heat load of each room?

How do I account for the fact that the rooms are connected with either no doors at all or doors that need to be open (Bedroom1/2+Bathroom) due to common return?

1 Answer 1


(Exterior) Wall area * wall U-value. If you have walls with different levels of insulation, they are added up separately after their separate U-values are multiplied by each area.

(Exterior) Window area * window U-value.

(Exterior) Door area * door U-value.

(Exterior) Ceiling area * ceiling U-value.

Add them all up and multiply by your design day differential temperature.

Open plan doesn't matter to load. It might matter to distribution, but not load. Ceiling between first and second floor doesn't matter. Walls that are internal don't matter if both sides of the wall are conditioned.

If you have a heated basement or the first floor is a slab on grade, floor and wall area of that, times U-value, times design differential (soil temperature outside it, typically higher than air temperature outside walls and ceiling when heating.)

U-value is 1/R-value and the units work out so you get heat flow when you multiply that by temperature differential.

Returns from closed doors are generally handled by the door having a gap (usually on the bottom) sufficient for airflow, unless the situation is such that an explicit vent is required. Still has nothing to do with heating load.

There is some additional load associated with ventilation and air leakage, which is most of the ventilation in most houses nearing 100 years old like yours. You can do a blower door test to quantify that, or fudge-factor it, or realize that it may be why the above seems to be smaller than your current unit if your contractor was competent.

  • Thanks; I calculated the total heating load already using the calculator (~40kBTU). But I will revise it with your suggestion with U-value. However, my question is really about distribution (in other words, I was referring to heating loads of each room). For each room, I’d like to know if the vent size is sufficient and if the duct leading to it is sufficient.
    – divB
    Jun 23, 2023 at 15:39
  • Process is the same by room - interior partitions don't count unless the other side is not conditioned, exterior envelope components do count by area and heat flow.
    – Ecnerwal
    Jun 23, 2023 at 15:44
  • Ok great, let me try this! One more thing, it’s very likely that the sum of CFM for each room is different than the CFM of the whole house. What do I do then? If it’s larger I guess I need to increase tonnage? If it’s lower it’s not critical?
    – divB
    Jun 23, 2023 at 16:57

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