Skip to main content

Worked Example: HVAC Electrical System Sizing for Dual-Mode Heating/Cooling — The Texas Winter Storm Uri Disaster

Complete HVAC electrical sizing calculation for a commercial building with heat pump and auxiliary resistance heating. Covers NEC 440 motor circuits, NEC 424 resistance heating, summer vs winter demand comparison, feeder sizing, and disconnect requirements — and why the 2021 Texas storm proved that designing only for summer cooling loads is a fatal oversight.

NEC/NFPA 7020 min readUpdated February 24, 2026
Share:

The Incident: When a Summer-Designed Grid Met a Winter Emergency

In February 2021, Winter Storm Uri struck Texas with unprecedented cold. Temperatures plummeted to −19°C (−2°F) in Dallas — a city where the design winter temperature is typically 19°F (−7°C). Over five days, 246 people died and damages exceeded $195 billion, making it one of the costliest natural disasters in US history. The ERCOT electrical grid collapsed because electricity demand reached 76.8 GW, far exceeding the winter design capacity of 67.5 GW.

The demand surge was overwhelmingly driven by electric heating. Texas relies heavily on heat pumps for space heating — systems that are efficient in mild conditions but switch to auxiliary resistance heating when the outdoor temperature drops below their balance point (typically 0–5°C). A heat pump running on its compressor draws perhaps 5–8 kW; that same unit on auxiliary strip heating draws 15–25 kW. Across millions of homes and commercial buildings, this switchover tripled the electrical demand overnight. Residential circuits overloaded as occupants plugged in portable space heaters on circuits originally designed for summer cooling loads.

The core engineering lesson is this: HVAC electrical systems must be designed for the governing load case, which may be winter heating rather than summer cooling. In climates with significant heating loads, NEC 440 (motor-driven HVAC) and NEC 424 (fixed electric space heating) prescribe different sizing rules that must be compared to determine which case governs. This worked example demonstrates that comparison for a commercial building HVAC plant.

Scenario: Commercial Building HVAC Electrical System

Size the electrical system for a commercial building HVAC plant operating in both cooling and heating modes.

ParameterValue
Building3-storey office building, 5,000 m² (54,000 ft²)
Supply voltage480/277 V three-phase, 60 Hz
Rooftop units (RTUs)4 × 75 kW (25 TR) heat pump units with auxiliary resistance heating
RTU compressor motor25 kW each, FLA 35.2 A at 480 V
RTU condenser fan motor3.7 kW each, FLA 5.5 A at 480 V
RTU auxiliary heater60 kW resistance strip, 72.2 A at 480 V per unit
AHU supply fans2 × 30 kW, FLA 40.1 A at 480 V each
Chilled water pumps2 × 7.5 kW, FLA 11.0 A at 480 V each
Primary standardNEC/NFPA 70:2023, ASHRAE 90.1-2022

Step 1: Calculate Cooling Mode Electrical Load

In cooling mode, each RTU runs its compressor motor, condenser fan, and supply fan (evaporator fan is part of the AHU). Per NEC 440.6, motor loads are based on nameplate full-load amperes (FLA).

Per RTU (cooling mode):

IRTU,cool = Icompressor + Icondenser fan = 35.2 + 5.5 = 40.7 A — (Eq. 1)

Total cooling mode load (all HVAC):

EquipmentQuantityFLA Each (A)Total FLA (A)
RTU compressor + condenser fan440.7162.8
AHU supply fans240.180.2
Chilled water pumps211.022.0

Itotal,cool = 162.8 + 80.2 + 22.0 = 265.0 A — (Eq. 2)

Total cooling mode demand: 265.0 A at 480 V three-phase = √3 × 480 × 265.0 = 220.3 kVA

Step 2: Calculate Heating Mode Electrical Load

In heating mode, the heat pump compressor runs in reverse cycle down to a balance point. Below the balance point, auxiliary resistance heating activates. In extreme cold (as during Storm Uri), the compressor may shut down entirely and the entire heating load falls on the resistance strips.

Per RTU (heating mode, extreme cold — compressor off):

IRTU,heat = Iauxiliary heater = 60,000 / (√3 × 480) = 72.2 A — (Eq. 3)

Per RTU (heating mode, mild cold — compressor + auxiliary):

IRTU,dual = Icompressor + Iauxiliary heater = 35.2 + 72.2 = 107.4 A

Critical insight: The dual-mode scenario (compressor + auxiliary running simultaneously) produces the highest current — higher than either mode alone. This occurs in the temperature range just below the balance point where the heat pump still runs but cannot meet the full heating demand. Per NEC 440.35 Exception, the HVAC equipment nameplate must list the maximum possible current for all operating modes.

Total heating mode load (worst case: dual-mode, all RTUs):

EquipmentQuantityFLA Each (A)Total FLA (A)
RTU compressor + aux heater4107.4429.6
AHU supply fans240.180.2
Chilled water pumps0 (off in heating)0

Itotal,heat = 429.6 + 80.2 = 509.8 A — (Eq. 4)

Total heating mode demand: √3 × 480 × 509.8 = 423.7 kVA

The heating load is 1.92× the cooling load. This is the fundamental lesson of Storm Uri — winter heating demand can nearly double summer cooling demand when resistance heating engages.

Step 3: Size RTU Branch Circuits (NEC 440 + 424)

Each RTU has both motor and resistance heating loads. Per NEC 440.33, the branch circuit for a combination load must be sized for the sum of the motor load (per NEC 440) and the heater load (per NEC 424).

Motor component per NEC 440.6:

Imotor,branch = 125% × largest motor FLA + sum of other motors — (Eq. 5)

Imotor,branch = 1.25 × 35.2 + 5.5 = 44.0 + 5.5 = 49.5 A

Heater component per NEC 424.3(A):

Fixed electric space heating is a continuous load. Per NEC 424.3(A), branch circuit conductors for resistance heaters must be rated at not less than 125% of the total heater load:

Iheater,branch = 125% × 72.2 = 90.3 A — (Eq. 6)

Combined RTU branch circuit:

IRTU,branch = Imotor,branch + Iheater,branch = 49.5 + 90.3 = 139.8 A — (Eq. 7)

Select conductor from NEC Table 310.16 (75°C column, copper, THWN-2): 1/0 AWG rated 150 A. ✓ 150 ≥ 139.8 A

Step 4: Size RTU Branch Circuit Protection

Per NEC 440.22, the branch circuit short-circuit and ground-fault protective device for the motor component must not exceed 175% of the largest motor FLA (for inverse-time circuit breaker).

IOCPD,motor = 175% × 35.2 = 61.6 A (round up to next standard: 70 A) — (Eq. 8)

Per NEC 424.19, the heater branch circuit OCPD must not exceed the heater nameplate rating. For the combined load:

IOCPD,combined = IOCPD,motor + Iheater = 70 + 72.2 = 142.2 A

Select next standard circuit breaker size: 150 A MCCB per NEC 240.6(A).

Verify conductor protection: 150 A OCPD protecting 1/0 AWG (150 A ampacity) — ✓ PASS

Step 5: Determine Governing Case — Summer vs Winter

Compare the two load scenarios to determine which governs the feeder and service sizing:

ParameterCooling Mode (Summer)Heating Mode (Winter)
RTU current (each)40.7 A107.4 A
RTU total (4 units)162.8 A429.6 A
AHU fans80.2 A80.2 A
Chilled water pumps22.0 A0 A
Total HVAC load265.0 A509.8 A
Apparent power220.3 kVA423.7 kVA

The winter heating mode governs at 509.8 A — nearly double the summer cooling load.

Per NEC 220.60: Where it is unlikely that two dissimilar loads will be in use simultaneously, it is permissible to use only the larger of the two loads. Since cooling and heating are non-coincident (the building is either cooling or heating, not both), we use the larger heating load of 509.8 A for feeder sizing. The cooling load of 265.0 A does not add to the heating load.

Step 6: Size HVAC Feeder

Per NEC 440.6 and 430.24, the feeder must be sized for 125% of the largest motor plus the sum of all other loads:

Largest motor: AHU supply fan, 40.1 A (larger than RTU compressor at 35.2 A).

Ifeeder = 125% × 40.1 + (80.2 − 40.1) + 429.6 — (Eq. 9)

Ifeeder = 50.1 + 40.1 + 429.6

Ifeeder = 519.8 A

Note: The resistance heating components are already at 125% from the branch circuit calculation (NEC 424.3 applies at every level — branch, feeder, and service).

Select feeder conductor from NEC Table 310.16 (75°C column, copper, THWN-2): 1000 kcmil rated 545 A. Alternatively, two parallel sets of 3/0 AWG (200 A each × 2 = 400 A) would be insufficient; use two parallel sets of 250 kcmil (255 A × 2 = 510 A) — marginally short. Select two parallel sets of 300 kcmil (285 A × 2 = 570 A).

Feeder: 2 sets of 300 kcmil THWN-2 copper per phase (570 A capacity)

Step 7: Calculate Voltage Drop on Longest HVAC Branch

The longest RTU branch circuit runs 60 m (200 ft) from the HVAC panel to the rooftop unit. For 1/0 AWG copper at 480 V three-phase:

From NEC Chapter 9, Table 9: 1/0 AWG copper in PVC conduit, R = 0.122 Ω/1000 ft, X = 0.044 Ω/1000 ft.

ΔV = √3 × I × L × (R cosφ + X sinφ) / 1000 — (Eq. 10)

At maximum load (dual mode), I = 107.4 A, L = 200 ft, power factor = 0.95 (blended motor + resistive):

ΔV = 1.732 × 107.4 × 200 × (0.122 × 0.95 + 0.044 × 0.312) / 1000

ΔV = 1.732 × 107.4 × 200 × (0.1159 + 0.0137) / 1000

ΔV = 37,203 × 0.1296 / 1000

ΔV = 4.82 V

ΔV% = 4.82 / 480 × 100 = 1.00%

The NEC recommendation for branch circuits is 3% maximum, with 5% total (branch + feeder). At 1.00%, voltage drop is well within limits. ✓ PASS

Step 8: Size HVAC Disconnect Switches

Per NEC 440.12, each HVAC unit disconnect must have an ampere rating at least 115% of the nameplate rated-load current:

RTU disconnect (combined motor + heater):

Idisconnect = 115% × IRTU,max = 1.15 × 107.4 = 123.5 A — (Eq. 11)

Select standard disconnect: 150 A fusible disconnect switch per NEC 440.12(A)(1).

The disconnect must be within sight of the equipment and readily accessible per NEC 440.14. For rooftop units, this typically means a weather-rated NEMA 3R disconnect mounted on the roof near the unit.

AHU disconnect:

Idisconnect = 115% × 40.1 = 46.1 A → 60 A disconnect

Chilled water pump disconnect:

Idisconnect = 115% × 11.0 = 12.7 A → 30 A disconnect

Result Summary

ComponentCooling ModeHeating ModeSized For
RTU branch circuit40.7 A (motor only)107.4 A (motor + aux)1/0 AWG, 150 A MCCB
AHU branch circuit40.1 A40.1 A8 AWG, 50 A CB
CW pump branch circuit11.0 A0 A12 AWG, 20 A CB
Total HVAC load265.0 A509.8 AHeating governs
HVAC feeder2 × 300 kcmil/phase (570 A)
Voltage drop (longest branch)1.00%✓ PASS (< 3%)

The governing load case is winter heating mode at 509.8 A — 92% larger than summer cooling at 265.0 A. Every component from branch circuits to the service entrance must be sized for the heating scenario. Designing only for the summer cooling load, as was common practice in Texas before Storm Uri, results in an electrical system that is approximately half the required capacity for extreme winter conditions.

What Would Have Prevented This?

The Texas grid failure during Storm Uri was a systemic crisis, but at the building level, proper HVAC electrical sizing could have prevented individual circuit overloads and building-level electrical failures:

  • Always calculate both cooling and heating electrical loads and design for the governing caseNEC 220.60 explicitly requires this comparison; in climates with both significant cooling and heating loads, never assume that summer cooling governs
  • Account for auxiliary resistance heating in heat pump systems — a heat pump with a 25 kW compressor and 60 kW auxiliary heater can draw up to 85 kW when both operate simultaneously; the electrical system must be sized for this combined condition, not just the compressor alone
  • Apply NEC 424.3 continuous load rules to resistance heating — all conductors and overcurrent devices supplying fixed resistance heating must be rated at 125% of the heater current; this is a continuous load rule that catches many designers unaware
  • Consider demand diversity carefully for large buildings — while NEC 220.51 allows demand factors for non-coincident loads, during extreme weather events ALL heating units operate simultaneously at maximum output, eliminating the diversity that designers may have counted on
  • Evaluate gas heating as an alternative to reduce electrical demand — a building with gas-fired heating places no demand on the electrical system during winter; dual-fuel heat pumps that switch to gas below the balance point can dramatically reduce peak electrical demand and grid strain

Try the HVAC Electrical Calculator

Put this methodology into practice. Calculate results with full standard clause references — free, no sign-up required.

Or embed this calculator on your site
Calculate HVAC Electrical

Frequently Asked Questions

Heat pumps in cooling mode use a compressor to move heat (coefficient of performance 3–4, meaning 1 kW of electricity moves 3–4 kW of heat). In extreme cold, when the heat pump switches to auxiliary resistance heating, the COP drops to 1.0 — every kW of heat requires 1 kW of electricity. A building needing 200 kW of cooling in summer might use only 60 kW of electricity (COP 3.3), but the same building needing 200 kW of heating in extreme winter uses 200 kW of electricity on resistance heat. This is why Texas saw a near-doubling of electrical demand during Storm Uri.
NEC Article 440 covers motor-driven air conditioning and refrigeration equipment — compressors, fans, and pumps. It addresses motor starting current, branch circuit sizing based on nameplate FLA, and overcurrent protection with motor-specific rules (e.g., 175% maximum for circuit breakers). NEC Article 424 covers fixed electric space heating — resistance strip heaters, baseboard heaters, duct heaters. It treats the load as continuous (125% sizing rule) with standard overcurrent protection rules. For a heat pump with auxiliary resistance heating, both articles apply simultaneously to different components of the same unit.
NEC 220.51 allows demand factors for non-coincident loads, and NEC 220.60 allows omitting the smaller of two non-coincident loads (like cooling vs heating). However, within each mode (cooling or heating), all units may operate simultaneously during peak conditions. During Storm Uri, every heat pump in Texas was running at maximum capacity simultaneously — there was zero diversity. For HVAC feeders serving critical loads, using a demand factor of 1.0 (no diversity) for the governing mode is the safe and correct approach.

Related Resources