8.2 Defrost Controls, Supplemental Electric Heat & Balance Points

Defrost Controls, Supplemental Electric Heat & Balance Points

Core Principle: In heating mode, the outdoor coil operates at a saturation temperature $10^\circ\text{F}$ to $20^\circ\text{F}$ below the outdoor ambient air temperature. Whenever outdoor ambient drops below approximately $42^\circ\text{F}$ to $45^\circ\text{F}$ and relative humidity is moderate to high, the outdoor coil surface falls below $32^\circ\text{F}$ ($0^\circ\text{C}$). Atmospheric moisture condenses and freezes into frost on the aluminum fins and copper tubes, acting as thermal insulation and choking coil airflow.

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Frost Accumulation & Defrost Control Architecture

Frost Accumulation Mechanism:
Outdoor Air (35°F Dry Bulb / 32°F Dew Point)
  ├── Passes over Outdoor Coil Boiling at 18°F (-7.8°C)
  ├── Moisture Condenses on Sub-Freezing Fin Surfaces (< 32°F)
  ├── Condensate Freezes into Solid Frost / Ice Layers
  ├── Airflow CFM Drops (Static pressure across coil face spikes)
  ├── Evaporating Temperature & Suction Pressure Collapse
  └── Heat Pump Heating Capacity Drops by 40% - 70% if Not Defrosted

1. Time-Temperature Defrost Systems

Time-temperature defrost is an electro-mechanical or simple solid-state control method governed by two independent criteria:

1. Time Accumulation: An electronic timer on the defrost control board records cumulative compressor run time. Selectable pin jumpers allow field configuration for $30\text{, }60\text{, or }90\text{ minutes}$ of compressor operation (commonly set to $60\text{ or }90\text{ minutes}$ in North Carolina's Piedmont and Coastal regions, and $30\text{ or }60\text{ minutes}$ in the Western Blue Ridge Mountains where winter humidity and snowfall are elevated).
2. Coil Temperature Sensor (Defrost Thermostat / Klixon): A bi-metal disc switch or thermistor clamped to the coldest return bend of the outdoor coil (near the bottom liquid distributor). The switch closes on temperature drop at approximately $30^\circ\text{F} - 32^\circ\text{F}$ ($0^\circ\text{C}$).

• Initiation: Defrost begins only when the accumulated run timer expires AND the defrost thermostat switch is closed.
• Termination: The cycle terminates when the coil sensor warms to $55^\circ\text{F} - 70^\circ\text{F}$ ($13^\circ\text{C} - 21^\circ\text{C}$), confirming complete frost melting, or when an internal safety timer overrides the cycle after $10\text{ to }14\text{ minutes}$.

2. Demand Defrost (Microprocessor Sensor Differential)

Demand defrost eliminates unnecessary defrost cycles by initiating defrost only when frost accumulation physically restricts heat transfer:

• Temperature Differential Method ($\Delta T$ Method): The control board monitors two calibrated thermistors: Outdoor Ambient Air Temperature ($T_{\text{ambient}}$) and Outdoor Coil Temperature ($T_{\text{coil}}$). As frost accumulates, coil heat transfer degrades and evaporating temperature drops. When the differential ($T_{\text{ambient}} - T_{\text{coil}}$) exceeds a dynamically calculated threshold curve, defrost is triggered.
• Air Pressure Differential Method: An air pressure differential switch measures static pressure drop across the outdoor coil face. As ice blocks fin passages, the static pressure differential rises, closing the switch contacts to initiate defrost.

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The Complete Defrost Cycle: Step-by-Step Sequence of Operations

When defrost initiates, the heat pump orchestrates a rapid sequence of electrical and mechanical transitions to melt outdoor ice as quickly as possible (typically in 2 to 7 minutes):

+-----------------------------------------------------------------------------------------+
|                        THE DEFROST SEQUENCE OF OPERATIONS                               |
+-----------------------------------------------------------------------------------------+
| Step 1: Defrost Initiation Trigger (Timer + Sensor closed OR Demand ΔT threshold met)   |
| Step 2: 4-Way Reversing Valve Energizes/De-energizes (Shifts into COOLING MODE)         |
| Step 3: Outdoor Fan Motor De-energizes (Outdoor fan contactor opens)                    |
| Step 4: Auxiliary Electric Strip Heat Energizes (W2/D terminal powered for tempering)   |
| Step 5: High-Pressure Hot Discharge Gas Floods Outdoor Coil (Rapid ice melting)         |
| Step 6: Coil Thermistor Reaches 55°F - 70°F (Or 10-14 min safety override expires)      |
| Step 7: Reversing Valve Shifts back to HEATING MODE                                     |
| Step 8: Outdoor Fan Motor Re-energizes                                                  |
| Step 9: Auxiliary Electric Heat De-energizes (Thermostat returns to normal staging)     |
+-----------------------------------------------------------------------------------------+

Critical Operating Details During Defrost

• Outdoor Fan Shutdown: Turning off the outdoor fan stops cold outdoor air from blowing across the coil, allowing the hot discharge gas ($140^\circ\text{F} - 180^\circ\text{F}$) to rapidly warm the coil to $60^\circ\text{F}+$, melting the frost within minutes.
• Indoor Cold Blow Prevention (Tempering Heat): Because the system is operating in cooling mode during defrost, the indoor coil becomes the evaporator, absorbing heat from the indoor air. To prevent discharging chilling $45^\circ\text{F}$ air into the occupied living space, the defrost control board energizes the "W2" / "D" (Defrost) terminal, staging on indoor electric resistance heat banks to temper the airstream to a comfortable $75^\circ\text{F} - 85^\circ\text{F}$.

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Supplemental vs. Emergency Electric Heat & Electrical Staging

Electric resistance strip heaters installed in the indoor air handler serve two distinct functions:

Electric Resistance Heating Operations:
├── 1. Supplemental Heat (Staged with Compressor)
│   ├── Operates concurrently with heat pump when ambient is below Thermal Balance Point
│   ├── Energized by 2nd-stage thermostat (W2) or outdoor ambient staging thermostats (OT)
│   └── Sized to satisfy the deficit between building heat loss and heat pump capacity
└── 2. Emergency Heat (Compressor Locked Out)
    ├── Manually switched at thermostat when compressor mechanically fails or trips on safety
    ├── 100% of building design heating load is carried exclusively by electric strip heaters
    └── Must be sized to handle 100% of ACCA Manual J heating load at 99% design dry bulb

1. Strip Heater Sizing & Kilowatt ($kW$) Calculations

Electric strip heater capacity is rated in kilowatts ($\text{kW}$):

$\text{Heat Output (BTU/hr)} = \text{kW} \times 3,412.14$

$\text{Current Draw (Amperes)} = \frac{\text{kW} \times 1,000}{\text{Operating Voltage } (V)}$

Electric Heater Kilowatt to BTU/hr Matrix (at 240V, 1-Phase)

[!NOTE] Under NEC Article 424 and NC Electrical Code, electric resistance space heating is classified as a continuous load. Branch circuit conductors and overcurrent protective devices must be sized at a minimum of $125%$ of total heater amperage plus blower motor load: $\text{MCA} = (I_{\text{heaters}} \times 1.25) + (I_{\text{motor}} \times 1.25)$

2. Staging Controls & Outdoor Thermostats (OT)

To prevent massive electrical demand spikes and unnecessary high electric utility bills, supplemental heat banks must be staged in increments of $5\text{ kW}$ or $10\text{ kW}$ using:

• Microprocessor Electronic Thermostat Staging: Uses intelligent proportional-integral (PI) algorithms that delay 2nd-stage (W2) auxiliary heat unless the indoor temperature droop exceeds $1.5^\circ\text{F} - 2.0^\circ\text{F}$ or recovery time exceeds a pre-set threshold.
• Outdoor Thermostats (OT-1, OT-2): Adjustable bimetal or thermistor switches mounted in the outdoor unit wired in series with the W2 heat staging relays. An outdoor thermostat locks out strip heat banks until the outdoor ambient temperature drops below the predetermined thermal balance point (e.g., locking out Stage 2 electric heat above $32^\circ\text{F}$).

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Thermal Balance Point vs. Economic Balance Point

Proper heat pump application engineering requires analyzing two distinct balance points on a capacity-versus-heat-loss chart.

Capacity / Heat Loss (BTU/hr)
     ^
  60k|                                      / Building Heat Loss Line
     |                                     /  (Q_loss = U · A · ΔT)
  50k|                                    /
     |                                   / 
  40k|----------------------------------X <--- THERMAL BALANCE POINT (~32°F)
     |                                 / \
  30k|   Heat Pump Heating Capacity   /   \ Supplemental Heat Required
     |   Curve (Q_capacity)          /     \ (Deficit = Q_loss - Q_capacity)
  20k|==============================/       \
     +-----------------------------+---------+---------+---------> Outdoor Temp (°F)
     0°F                          32°F      47°F      65°F

1. The Thermal Balance Point

The Thermal Balance Point is the outdoor ambient temperature at which the building's calculated heat loss rate exactly equals the heat pump's maximum heating output capacity:

$Q_{\text{building loss}}(T) = Q_{\text{heat pump capacity}}(T)$

• Above the Thermal Balance Point ($T > T_{\text{balance}}$): The heat pump supplies $100%$ of the home's heating requirement cycling on its normal first-stage thermostat (Y1). No auxiliary heat is needed.
• Below the Thermal Balance Point ($T < T_{\text{balance}}$): The heat pump cannot satisfy the total building load alone. The indoor temperature droops, energizing the second-stage auxiliary heat (W2) to run concurrently with the heat pump to make up the thermal deficit: $Q_{\text{supplemental required}} = Q_{\text{building loss}} - Q_{\text{heat pump capacity}}$

2. The Economic Balance Point (Dual-Fuel / Hybrid Systems)

In a Dual-Fuel (Hybrid) System, an electric heat pump is paired with a fossil fuel furnace (natural gas, LP propane, or fuel oil) instead of electric strip heat. The Economic Balance Point is the outdoor ambient temperature at which the cost per delivered $\text{BTU}$ of heat pump operation exactly equals the cost per delivered $\text{BTU}$ of fossil fuel furnace operation.

Cost per 100,000 BTU of Delivered Heat Equations

1. Heat Pump Cost per 100,000 BTU: $\text{Cost}_{\text{HP}} = \frac{100,000\text{ BTU}}{3,412.14\text{ BTU/kWh} \times \text{COP}} \times \text{Electric Rate (\$/kWh)}$

2. Natural Gas Furnace Cost per 100,000 BTU ($1\text{ Therm} = 100,000\text{ BTU}$): $\text{Cost}_{\text{Gas}} = \frac{\text{Gas Rate (\$/Therm)}}{\text{AFUE Rating}}$

3. LP Propane Furnace Cost per 100,000 BTU ($1\text{ Gallon LP} = 91,500\text{ BTU}$): $\text{Cost}_{\text{LP}} = \frac{100,000\text{ BTU}}{91,500\text{ BTU/gal} \times \text{AFUE}} \times \text{LP Rate (\$/Gal)}$

• Control Strategy: At outdoor temperatures above the Economic Balance Point, the heat pump runs. At outdoor temperatures below the Economic Balance Point, the heat pump compressor is locked out and the fossil fuel furnace fires exclusively.

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Step-by-Step Worked Technical Examples

Example 1: Calculating the Thermal Balance Point

Problem: A Raleigh, NC home has a calculated ACCA Manual J design heat loss of $45,000\text{ BTU/hr}$ at a winter design temperature of $20^\circ\text{F}$ based on a $70^\circ\text{F}$ indoor setpoint (Design $\Delta T = 70 - 20 = 50^\circ\text{F}$). The building heat loss factor is: $q_{\text{loss rate}} = \frac{45,000\text{ BTU/hr}}{50^\circ\text{F}} = 900\text{ BTU/hr per }^\circ\text{F}$

The installed 3-ton heat pump has the following manufacturer certified heating capacities:

• At $47^\circ\text{F}$ outdoor ambient: $36,000\text{ BTU/hr}$
• At $17^\circ\text{F}$ outdoor ambient: $21,000\text{ BTU/hr}$

Determine: (1) the mathematical equation for the heat pump capacity curve, (2) the exact Thermal Balance Point temperature ($T_{\text{balance}}$), and (3) the supplemental strip heat required at $20^\circ\text{F}$ design outdoor temperature.

Solution:

1. Linear Heat Pump Capacity Equation ($Q_{\text{HP}} = m \cdot T + b$): $\text{Slope } (m) = \frac{36,000 - 21,000}{47 - 17} = \frac{15,000}{30} = 500\text{ BTU/hr per }^\circ\text{F}$ $Q_{\text{HP}}(T) = 21,000 + 500 \times (T - 17) = 500 \cdot T + 12,500$

2. Building Heat Loss Equation: $Q_{\text{loss}}(T) = 900 \times (70 - T) = 63,000 - 900 \cdot T$

3. Set Equations Equal to Solve for $T_{\text{balance}}$: $500 \cdot T + 12,500 = 63,000 - 900 \cdot T$ $1,400 \cdot T = 50,500$ $T_{\text{balance}} = \frac{50,500}{1,400} = \mathbf{36.07^\circ\text{F}} \approx 36^\circ\text{F}$

4. Supplemental Heat Required at $20^\circ\text{F}$ Design Temperature: $Q_{\text{loss}}(20^\circ\text{F}) = 900 \times (70 - 20) = 45,000\text{ BTU/hr}$ $Q_{\text{HP}}(20^\circ\text{F}) = 500 \times (20) + 12,500 = 22,500\text{ BTU/hr}$ $Q_{\text{supplemental}} = 45,000 - 22,500 = \mathbf{22,500\text{ BTU/hr}}$ $\text{Heater kW Required} = \frac{22,500\text{ BTU/hr}}{3,412.14\text{ BTU/kWh}} = \mathbf{6.59\text{ kW}} \rightarrow \text{Install } 10\text{ kW Heater Bank}$

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Example 2: Calculating the Economic Balance Point for a Dual-Fuel System

Problem: A dual-fuel heat pump system in Greensboro, NC is paired with a 96% AFUE natural gas furnace. Natural gas costs $1.40 per therm, and electricity costs $0.12 per kWh. Calculate the minimum heat pump $COP$ required for the heat pump to be more economical to run than the gas furnace, and determine the Economic Balance Point from manufacturer data.

Solution:

1. Calculate Cost per 100,000 BTU for the 96% Gas Furnace: $\text{Cost}_{\text{Gas}} = \frac{1.40}{0.96} = \mathbf{1.458\text{ per 100,000 BTU}}$

2. Set Heat Pump Cost Equal to Gas Cost to Find Parity COP ($COP_{\text{parity}}$): $\text{Cost}_{\text{HP}} = \frac{100,000}{3,412.14 \times COP} \times 0.12 = \frac{3.5168}{COP} = 1.458$ $COP_{\text{parity}} = \frac{3.5168}{1.458} = \mathbf{2.41}$

Economic Rule: Whenever the heat pump's operating $COP$ is greater than $2.41$ (which typically occurs at outdoor ambient temperatures above $30^\circ\text{F}$ for modern units), the heat pump is cheaper to operate. Below $30^\circ\text{F}$ where $COP < 2.41$, the control system switches off the heat pump and fires the 96% natural gas furnace exclusively.

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Example 3: Electric Auxiliary Heater Ampacity & Circuit Sizing

Problem: An air handler is installed with a $15\text{ kW}$ ($14.4\text{ kW}$ actual at $240\text{V}$) electric supplemental heater package and a blower motor rated at $3.5\text{ Full Load Amps (FLA)}$ on a $240\text{V}$, single-phase line. Calculate: (1) total heater element amperage, (2) the Minimum Circuit Ampacity (MCA), and (3) the Maximum Overcurrent Protection (MOCP) / circuit breaker size.

Solution:

1. Heater Element Full Load Amperage: $I_{\text{heaters}} = \frac{14,400\text{ Watts}}{240\text{ Volts}} = \mathbf{60.0\text{ A}}$

2. Minimum Circuit Ampacity (MCA) per NEC Article 424 (125% continuous load): $\text{MCA} = (I_{\text{heaters}} \times 1.25) + (I_{\text{motor}} \times 1.25)$ $\text{MCA} = (60.0\text{ A} \times 1.25) + (3.5\text{ A} \times 1.25) = 75.0\text{ A} + 4.38\text{ A} = \mathbf{79.38\text{ A}}$

3. Circuit Breaker / MOCP Sizing:

   • Most $15\text{ kW}$ heater kits are internally divided into two sub-circuits:
     • Circuit 1 (10 kW heater + motor): $(40.0 \times 1.25) + (3.5 \times 1.25) = 54.38\text{ A MCA} \rightarrow \mathbf{60\text{ Amp Breaker}}$
     • Circuit 2 (5 kW heater only): $20.0 \times 1.25 = 25.0\text{ A MCA} \rightarrow \mathbf{30\text{ Amp Breaker}}$
   • If supplied by a single feeder: Next standard fuse/breaker size above $79.38\text{ A}$ not exceeding manufacturer rating is $90\text{ A}$ or $100\text{ A}$ with minimum #3 AWG Copper (or #1 AWG Aluminum) conductors rated for $75^\circ\text{C}$.