Why is solar gain such a large share of this example's load?

Because this example has a relatively large window area (300 sq ft) using an average solar gain factor applied uniformly — real houses with less glazing, better shading, or more north/south-facing (vs. east/west) window orientation would see a meaningfully smaller solar gain contribution. This is exactly why window design decisions have an outsized effect on cooling load.

Would this house really need a 6.8-ton system?

Based on this simplified calculation with these specific assumptions, yes — but a load this high for a 2,500 sq ft house is a signal worth double-checking, since it suggests either unusually large glazing, poor shading, or an overly conservative solar gain assumption. A full Manual J with orientation-specific solar factors would likely show meaningfully lower total load for the same house.

Can I use this exact worked example as my project's calculation?

No — this is an educational walkthrough with illustrative assumptions, not a substitute for a project-specific Manual J calculation using your actual building's envelope details, window orientations, and local design conditions.

Engineering·10 min read·August 14, 2026

🏠 Manual J Worked Example: 2,500 Sq. Ft. House Cooling Load

A complete, step-by-step Manual J-style cooling load calculation for a representative 2,500 square foot single-story house, showing every input and how each load component is derived.

By EngineersUniverse Editorial Team

Engineering Editorial & Technical Review Team

Published August 14, 2026

The Scenario

This worked example walks through a simplified, Manual J-style cooling load calculation for a representative house: 2,500 sq ft, single-story, 9 ft ceilings, standard (code-minimum) insulation, 300 sq ft of double-pane low-E windows, climate zone 4 (mixed climate, moderate design temperature difference), 4 occupants, 1,000 W of equipment load, and 2 W/sq ft lighting density. The goal is to show every input and intermediate step so the calculation is fully auditable, not just a final tonnage number.

Step 1 — Envelope Conduction (Walls and Ceiling)

Approximate wall area for a roughly square single-story house: perimeter ≈ 4 × √(2,500) ≈ 200 ft, wall area ≈ 200 ft × 9 ft = 1,800 sq ft, minus 300 sq ft of window area = 1,500 sq ft of net wall. At standard insulation (assume a conduction factor of roughly 0.05 BTU/hr·ft²·°F for this simplified method) and a zone-4 design ΔT of 30°F: wall load ≈ 1,500 × 0.05 × 30 = 2,250 BTU/hr. Ceiling: 2,500 sq ft × 0.03 × 30 ≈ 2,250 BTU/hr.

Step 2 — Window Conduction and Solar Gain

Double-pane low-E glazing has a U-factor around 0.28: window conduction = 300 sq ft × 0.28 × 30°F = 2,520 BTU/hr. Solar heat gain through the same glazing, using a simplified average gain factor of roughly 150 BTU/hr per sq ft of glass: 300 × 150 = 45,000 BTU/hr — note this is intentionally a rough average across orientations; a full Manual J would apply different solar gain factors per window based on which direction each one actually faces, since a west-facing window gains far more in late afternoon than a north-facing one.

Step 3 — Internal Gains

Occupants: 4 people × 250 BTU/hr sensible each = 1,000 BTU/hr sensible (plus 4 × 200 = 800 BTU/hr latent, accounted for separately). Equipment: 1,000 W × 3.412 BTU/hr per watt ≈ 3,412 BTU/hr. Lighting: 2,500 sq ft × 2 W/sq ft × 3.412 ≈ 17,060 BTU/hr.

Step 4 — Summing Sensible Load

Adding every sensible component: wall (2,250) + ceiling (2,250) + window conduction (2,520) + solar gain (45,000) + occupant sensible (1,000) + equipment (3,412) + lighting (17,060) = 73,492 BTU/hr sensible. Solar gain through glazing is clearly the dominant term in this example — a common result for houses with substantial window area, which is exactly why window orientation and shading matter so much to real-world cooling load, more than most people intuitively expect.

Step 5 — Latent Load

Occupant latent: 4 × 200 = 800 BTU/hr. Using a simplified estimate that adds roughly 10% of sensible load as additional latent load from infiltration/ventilation moisture: 73,492 × 0.10 ≈ 7,349 BTU/hr. Total latent ≈ 800 + 7,349 = 8,149 BTU/hr.

Step 6 — Total Load and Equipment Sizing

Total load = sensible + latent = 73,492 + 8,149 = 81,641 BTU/hr. Converting to tons: 81,641 ÷ 12,000 ≈ 6.8 tons. In practice, a designer would round to the nearest standard equipment size — likely a 6.5 or 7-ton system, or reconsider whether splitting the house into multiple zones makes more sense at this load level, particularly given how much of the load is driven by solar gain, which better shading or window selection could meaningfully reduce.

What This Example Deliberately Simplifies

This worked example uses average/simplified factors (a single solar gain factor across all windows regardless of orientation, a flat percentage add-on for infiltration latent load) to keep the walkthrough legible. A full Manual J calculation for permit submission would apply orientation-specific solar gain factors per window, actual infiltration rate calculations based on the building's air-tightness, and room-by-room resolution rather than one whole-house total — always use a complete Manual J calculation, not this simplified example, for any real permit or equipment-purchase decision.

Topics covered

Manual J worked examplecooling load example calculationHVAC load calculation example2500 sq ft house cooling loadresidential cooling load example