Heat Pump COP Calculation¶

The Coefficient of Performance (COP) is crucial for optimization - it determines how efficiently the heat pump converts electricity into heat. This page explains the COP model used by the integration.

COP Definition¶

COP is the ratio of heat output to electrical input:

For example, a COP of 4.0 means the heat pump produces 4 kW of heat for every 1 kW of electricity consumed.

Theoretical COP Model¶

Carnot Efficiency¶

The theoretical maximum COP is given by the Carnot efficiency:

Where temperatures are in Kelvin.

Real heat pumps achieve 40-60% of Carnot efficiency due to:

• Compressor inefficiency
• Heat exchanger losses
• Refrigerant properties
• Defrost cycles

Practical COP Model¶

The integration uses a linearized empirical model that's simpler than Carnot but accurate for typical operating ranges:

Parameters¶

1. Base COP (\(\text{COP}_{base}\))¶

Heat pump COP at reference conditions:

• Outdoor temperature: 7°C (A7)
• Supply temperature: 35°C (W35)
• Typical range: 3.0 - 5.0

This value is found in heat pump datasheets under "A7/W35" rating.

2. Outdoor Temperature Coefficient (\(\alpha\))¶

How much COP improves per °C increase in outdoor temperature.

• Fixed value: 0.025
• Physical meaning: Higher outdoor temp = less work to lift heat

3. K-Factor (\(k\))¶

How much COP degrades per °C increase in supply temperature.

• Typical range: 0.015 - 0.045
• Physical meaning: Higher supply temp = more compressor work

K-factor varies by heat pump type:

4. COP Compensation Factor (\(f\))¶

Accounts for real-world system losses:

• Distribution losses in pipes
• Circulation pump consumption
• Defrost cycle energy

• Control system overhead

• Typical range: 0.80 - 0.95

• Well-designed system: 0.90 - 0.95
• Older system: 0.80 - 0.85

Complete Calculation Example¶

Scenario¶

Given:

• Base COP: 4.0
• K-factor: 0.03
• Compensation factor: 0.90
• Outdoor temperature: 5°C
• Supply temperature: 40°C

Step 1: Calculate temperature effects $$ \text{Outdoor effect} = 0.025 \times 5 = +0.125 $$

Step 2: Calculate theoretical COP $$ \text{COP}_{theoretical} = 4.0 + 0.125 - 0.15 = 3.975 $$

Step 3: Apply compensation $$ \text{COP}_{actual} = 3.975 \times 0.90 = 3.58 $$

Result: COP = 3.58

COP Variation Charts¶

Effect of Outdoor Temperature¶

At constant supply temperature (35°C):

Effect of Supply Temperature¶

At constant outdoor temperature (7°C):

Why This Model Works¶

Advantages¶

1. Simple: Only 3 parameters to configure (base COP, k-factor, compensation)
2. Fast: Linear calculation, no iterative solving
3. Accurate: Within ±5% of actual COP in typical range (-10°C to +15°C outdoor)
4. Intuitive: Parameters have clear physical meaning

Limitations¶

1. Linearization: Less accurate at extreme temperatures
2. No humidity: Doesn't account for defrost (handled by compensation factor)
3. No modulation: Assumes continuous operation (not cycling)
4. Fixed refrigerant: Doesn't account for refrigerant type

Despite these limitations, the model is sufficient for optimization because:

• Most operation occurs in the linear range
• Errors are systematic (cancel out in relative comparisons)
• Compensation factor absorbs second-order effects

Calibrating Your COP Model¶

Step 1: Find Base COP¶

Check your heat pump datasheet for A7/W35 rating. If not available:

1. Set heat pump to 35°C supply temperature
2. Wait for outdoor temperature of ~7°C
3. Measure electrical input and heat output
4. Calculate COP = heat_out / electricity_in

Step 2: Estimate K-Factor¶

Use typical values from the table, or:

1. Measure COP at two different supply temperatures (e.g., 35°C and 45°C)
2. Calculate k = (COP₁ - COP₂) / (T₂ - T₁)

Step 3: Determine Compensation Factor¶

1. Use the model to calculate theoretical COP
2. Measure actual system COP (including pumps, distribution)
3. Calculate f = actual_COP / theoretical_COP

COP in Optimization¶

The optimizer uses COP to calculate electricity consumption:

Lower supply temperature → Higher COP → Less electricity

This creates a trade-off:

• Higher offset: More heat output, lower COP, higher consumption
• Lower offset: Less heat output, higher COP, lower consumption

The dynamic programming algorithm finds the optimal balance.

Advanced: Multi-Point COP Models¶

For users with detailed heat pump data, a piecewise-linear model can be used:

def advanced_cop(outdoor_temp, supply_temp):
    """Multi-point interpolation for precise COP."""
    # Define COP at known operating points
    cop_table = {
        (-7, 35): 3.2,
        (2, 35): 3.8,
        (7, 35): 4.2,
        (7, 45): 3.4,
        (7, 55): 2.8,
    }

    # Bilinear interpolation
    return interpolate_2d(cop_table, outdoor_temp, supply_temp)

This is not currently implemented but could be added for expert users.

Related:

• Dynamic Programming - How COP affects optimization
• Optimization Strategy - Balancing COP vs price timing