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Heat pumps design and sizing guide

Learn how to properly size a heat pump for your building with this comprehensive guide.

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Section summary
Introduction to air cooled heat exchangers
Step by step design of an air cooled heat exchanger
Step by step example calculation of an air cooled heat exchanger
Free air cooled heat exchanger design calculator Excel

To size a heat pump for a building effectively, follow these detailed steps. Each step incorporates relevant calculations and considerations to ensure optimal performance.

Step 1: Determine Heating Requirements

Calculate the Heating Load

The first step is to estimate the heating load of the building, which is the amount of heat required to maintain a comfortable indoor temperature. This can be done using the following formula:

Heating Load (W) = Area (m2) * Power Coefficient (W/m2)

The power coefficient typically ranges from 30 to 100 W/m2, depending on factors like insulation quality and climate. For example, for a 100 m² home with moderate insulation, you might use 50 W/m²:

Heating Load = 100 m2 * 50 W/m2 = 5000 W

In focus
Choosing the appropriate power coefficient for determining the heating load of a building is crucial for effective heat pump sizing. The power coefficient varies based on several factors, including insulation quality, building type, climate zone, and specific heating requirements. Below is a detailed explanation of how to select the power coefficient along with a table summarizing typical values for different building conditions.

Factors Influencing Power Coefficient

  1. Insulation Quality : Well-insulated buildings retain heat better, requiring a lower power coefficient.
  2. Building Type : Different types of buildings (e.g., residential homes, commercial spaces) have varying thermal characteristics.
  3. Climate Zone : Colder climates generally require higher coefficients due to increased heat loss.
  4. Desired Indoor Temperature : Higher indoor temperatures will increase the heating load.
  5. Window Area and Quality : More windows or poorly insulated windows can lead to higher heat loss.
  6. Occupancy and Usage Patterns : Buildings with more occupants or frequent use may need more heating capacity.
Power Coefficient Selection Table
Building Type/Condition Power Coefficient (W/m²) Description
Well-Insulated Residential 30-50 Modern homes with good insulation and energy-efficient windows.
Average Insulation Residential 50-70 Typical homes with moderate insulation levels.
Poorly Insulated Residential 70-100 Older homes or those with significant air leaks.
Commercial Buildings (Well-Insulated) 40-60 Energy-efficient commercial buildings with good insulation.
Commercial Buildings (Average Insulation) 60-80 Standard commercial buildings without special insulation features.
Commercial Buildings (Poorly Insulated) 80-120 Older commercial structures or those with high air leakage.
Passive Houses 15-30 Extremely well-insulated buildings designed for minimal energy use.
High Ceilings +10-20 Add additional W/m² for every meter above standard ceiling height (2.4 m).

Heat Pump components illustration
(AI generated representation of a heat pump :)

Note : the use of the power coefficient as explained above allows to have a quick calculation but is not very precise. A full building study will be required in detail design.

A more accurate formula would be Q = UA∆T + infiltration losses, where:

  • U is the overall heat transfer coefficient of the building envelope (walls, windows, etc.).
  • A is the surface area.
  • ∆T is the temperature difference between inside and outside
  • This would provide a more precise estimation of heat loss.

Step 2: Consider Base Outdoor Temperature

Identify Base Outdoor Temperature

This is the lowest temperature expected during the heating season in your location. The heat pump's capacity must be sufficient to meet the heating load at this temperature.

Step 3: Calculate Required Heat Pump Capacity

Convert Heating Load to Kilowatts

Since heat pumps are often rated in kilowatts (kW), convert the heating load:

Heating Load (kW) = Heating Load (W)/1000

For our example:

Heating Load = 5000 / 1000 = 5 kW

Step 4: Adjust for Efficiency and Conditions

Consider Heat Pump Efficiency

Heat pumps have varying efficiencies based on outdoor conditions. The Coefficient of Performance (COP) indicates how effectively a heat pump operates. A COP of 3 means that for every kW of electricity consumed, the heat pump produces 3 kW of heat.

Adjust your required capacity based on the COP:

Adjusted Capacity (kW) = Heating Load (kW)/COP

If your heat pump has a COP of 3:

Adjusted Capacity = 5/3 ~ 1.67 kW

In Focus
The Coefficient of Performance (COP) of a heat pump is a crucial measure of its efficiency, defined as the ratio of useful heating or cooling provided to the energy input required. A higher COP indicates a more efficient heat pump, meaning it can deliver more heating or cooling for each unit of energy consumed.

How COP is Calculated

The formula for calculating COP is :

COP = Q/W

With :
- Q = Useful heat output (in kW)
- W = Electrical energy input (in kW)

Example Calculation

For instance, if a heat pump provides a heating output of 36,000 BTUs and consumes 3,000 watts of electricity, the calculation would be as follows:

1. Convert watts to BTUs:
 3,000 times 3.412 = 10,236  BTUs

2. Calculate COP:
  COP = 36,000/10,236 ~ 3.5
 
This means that for every unit of electrical energy consumed, the heat pump provides approximately 3.5 units of heating output.

Typical COP Values Based on Technology

The COP can vary significantly based on the type of heat pump technology and operating conditions. Below is a table summarizing typical COP estimates for different types of heat pumps:

Heat Pump Technology Typical COP Range Description
Air Source Heat Pumps 2.5 - 4.0 Efficiency decreases in colder temperatures.
Ground Source Heat Pumps 3.0 - 5.0 More stable performance due to consistent ground temperatures.
Water Source Heat Pumps 3.0 - 6.0 High efficiency when using water bodies as heat sources.
Hybrid Heat Pumps 2.5 - 4.5 Combines air and ground sources for improved efficiency.
Ductless Mini-Split Systems 3.0 - 5.0 High efficiency in targeted heating/cooling zones.

Factors Affecting COP

  1. Temperature Difference : The greater the temperature difference between the heat source and sink, the lower the COP.
  2. Operating Conditions : The efficiency can vary based on whether the system is operating at full load or partial load.
  3. Environmental Conditions : Humidity and air quality can influence performance.

Note : The Adjusted Capacity formula uses a simple ratio, Adjusted Capacity (kW) = Heating Load (kW)/COP, to size the heat pump. While this works for rough estimates, heat pumps lose efficiency (lower COP) in colder temperatures. The heat pump must be sized for peak load at the lowest expected temperatures.

In detail design, ensure you account for the heat pump's performance curve at different temperatures. Many heat pumps operate at reduced capacity in extreme cold.

Step 5: Calculate Flow Rate for Hydronic Systems

If using a hydronic system, calculate the flow rate required using the formula:

Q = m * c * dT

With :
- Q = Heat transfer (kW)
- m = Mass flow rate (kg/s)
- c = Specific heat capacity of water (4.2 kJ/kgK)
- dT = Temperature difference (K)

Rearranging gives:

m = Q / (c*dT)

Assuming a dT of 5 K for optimal performance:

For a heat output of 5 kW:

m = 5000/(4.2 * 5) = 238.1 kg/h

In focus
A hydronic system is a heating and cooling system that utilizes water or steam as a heat transfer medium. This system circulates heated or cooled fluid through a network of pipes to provide temperature control in residential or commercial spaces.

How Hydronic Systems Work

1. Heating Source : The process begins with a boiler, which heats water to a desired temperature (typically between 140°F and 180°F). This can be achieved using various energy sources such as natural gas, electricity, or even renewable energy like solar power.

2. Circulation : Once heated, the water is pumped through insulated pipes to radiators, underfloor heating systems, or other heat emitters within the building.

3. Heat Distribution : The heat is transferred from the water to the surrounding air and surfaces through:
   - Radiant Heating : Directly warming surfaces (like floors) that then radiate heat into the room.
   - Convective Heating : Heating air that rises from radiators or other emitters.

4. Return Cycle : After releasing its heat, the cooled water returns to the boiler to be reheated, completing a closed-loop system.

Advantages of Hydronic Systems

- Energy Efficiency : Hydronic systems are often more energy-efficient than traditional forced-air systems because they can maintain a more consistent temperature and reduce energy waste.
- Improved Air Quality : Unlike forced-air systems that can circulate dust and allergens, hydronic systems do not rely on air movement, leading to better indoor air quality.
- Comfort : They provide even heating without cold spots and maintain warmth longer after the system shuts off due to the thermal mass of heated surfaces.

Typical Components of a Hydronic System

- Boiler : Heats the water.
- Pump : Circulates the hot water through the system.
- Pipes : Transport hot water to and from heating elements.
- Radiators/Underfloor Heating : Disperse heat into living spaces.
- Thermostats/Controls :Allow for temperature regulation.

Hydronic systems and forced-air heating systems are two distinct methods for providing heat in buildings, each with its own advantages and disadvantages. Here’s a comparison of their efficiency based on various factors.

Efficiency Comparison

1. Heat Transfer Medium
- Hydronic Systems : Use water as the heat transfer medium, which is more efficient at conducting heat than air. This allows for better heat retention and less energy loss during distribution    .
- Forced-Air Systems : Rely on air, which is less effective at transferring heat. This can lead to increased energy consumption as the system works harder to maintain desired temperatures   .

2. Energy Loss
- Hydronic Systems : Operate in a closed-loop system, minimizing energy loss. Insulated pipes help retain heat, and when water cools, it is recycled back to the boiler for reheating    .
- Forced-Air Systems : Often suffer from significant energy loss due to duct leaks and poor insulation. Hot air can escape through gaps in ducts or around vents, leading to uneven heating and higher operational costs     .

3. Temperature Consistency
- Hydronic Systems : Provide more consistent and even heating throughout a space, reducing cold spots. The radiant heat emitted from surfaces warms the room uniformly     .
- Forced-Air Systems : Can create drafts and uneven temperatures, as hot air tends to rise and cool air settles near the floor. This can lead to discomfort in areas further from the vents    .

4. Operational Costs
- Hydronic Systems : Generally have lower operational costs due to their efficiency. They can utilize renewable energy sources, such as solar or geothermal systems, further reducing expenses   .
- Forced-Air Systems : Tend to have higher operational costs due to energy losses and the need for constant cycling of the furnace or heat pump    .

5. Maintenance
- Hydronic Systems : Require less maintenance over time since they do not have filters that need frequent changing or ducts that must be cleaned regularly    .
- Forced-Air Systems : Require regular maintenance of filters and ducts to ensure efficient operation and good air quality      .

Overall, hydronic systems tend to be more efficient than forced-air heating systems due to their superior heat transfer capabilities, reduced energy loss, consistent temperature distribution, and lower operational costs. While initial installation costs may be higher for hydronic systems, the long-term savings and benefits often make them a more attractive option for heating in residential and commercial applications.

Step 6: Finalize Sizing and Selection

Select Appropriate Heat Pump Model
Based on calculated requirements, choose a heat pump model that meets or slightly exceeds your adjusted capacity needs while considering local climate conditions and building specifics.

Ensure that:
- The selected unit can handle the peak heating load.
- The system is designed for proper airflow and water flow rates.

Bonus : Comparing gas fired heater and heat pump operating costs

When comparing the installation of a heat pump and a traditional gas-fired heater, several key differences emerge, including how each system operates, their efficiency, costs, and maintenance requirements. Below is a detailed comparison followed by a step-by-step guide to calculating the expected energy savings with a heat pump.

Comparison of Heat Pump and Gas-Fired Heater

1. Operating Mechanism
- Heat Pump : Transfers heat from the outside air (even in cold conditions) to heat the interior of a building. It can also reverse this process for cooling in warmer months.
- Gas-Fired Heater : Burns natural gas to generate heat, which is then distributed throughout the home via ductwork.

2. Energy Efficiency
- Heat Pump : Typically has a Coefficient of Performance (COP) of 2.5 to 4.0, meaning it can produce 2.5 to 4 times more energy in heating than it consumes in electricity  . This makes heat pumps up to three times more efficient than gas furnaces under moderate temperature conditions.
- Gas-Fired Heater : High-efficiency models can achieve up to 95% Annual Fuel Utilization Efficiency (AFUE), meaning they convert 95% of the fuel into usable heat   .

3. Installation Costs
- Heat Pump : Generally has higher upfront costs (around $6,000 or more) due to installation complexity and the need for electrical connections   . However, it may not require ductwork if using a ductless mini-split system.
- Gas-Fired Heater : Lower initial costs (starting around $2,795) but may require additional expenses for gas line installation and ventilation systems    .

4. Operating Costs
- Heat Pump : Lower operating costs over time due to higher efficiency, particularly in moderate climates where electricity rates are favorable    .
- Gas-Fired Heater : While natural gas is often cheaper than electricity, the overall operating costs can be higher due to lower efficiency compared to heat pumps, especially in milder climates     .

5. Maintenance
- Heat Pump : Requires less frequent maintenance (typically once a year) and has fewer components that can fail      .
- Gas-Fired Heater : Requires regular maintenance to ensure safe operation and efficiency, including annual inspections and cleaning.

6. Lifespan
- Heat Pump : Generally lasts about 10-15 years due to year-round usage.
- Gas-Fired Heater : Can last longer (15-20 years) as it operates primarily during colder months    .

Step-by-Step Calculation of Energy Savings with a Heat Pump

To calculate the expected energy savings when switching from a gas-fired heater to a heat pump, follow these steps:

Step 1: Determine Current Heating Costs

1. Calculate Current Energy Consumption :
   - Find out how much natural gas your current heater uses annually (in therms or BTUs).
   - Example: If your gas heater uses 1,200 therms annually:
     - Convert therms to BTUs:

     100,000  BTUs * 1,200 therms = 120,000,000 BTUs

2. Calculate Cost of Gas :
   - Multiply the annual consumption by the cost per therm.
   - Example: If natural gas costs $1 per therm:

     Annual Cost = 1,200 therms * $1 per therm = $1,200

 Step 2: Estimate Heat Pump Energy Consumption

1. Determine Required Heating Load :
   - From previous calculations, assume the heating load is still 75,000 BTU/hr.

2. Calculate Annual Heating Requirement :
   - Multiply by the number of hours used annually.
   - Example: Assuming you need heating for about 1,500 hours per year:
 
     Total Heating Requirement = 75,000 BTU/hr * 1,500 hours = 112,500,000 BTUs
    
3. Estimate Heat Pump Efficiency :
   - Use an average COP of 3 for calculations.
   - Calculate energy input required:

     Energy Input = Total Heating Requirement/COP = 112,500,000/3 = 37,500,000 BTUs
 
4. Convert BTUs to kWh:
   - Since electricity is usually measured in kWh:

   1 kWh = 3,412  BTUs implies 37,500,000/3,412 ~ 10,973 kWh

5. Calculate Cost of Electricity :
   - Multiply by your local electricity rate (e.g., $0.13 per kWh):

   Annual Cost = 10,973 * $0.13 = $1,426

Step 3: Calculate Energy Savings

Finally, compare annual costs:

Energy Savings = Cost of Gas Heater - Cost of Heat Pump
Energy Savings = $1,200 - $1,426 = -$226

In this example scenario with these specific rates and usage patterns, switching from a gas furnace to a heat pump would actually lead to higher costs; however, this could vary significantly based on local energy prices and specific heating needs. There is also the environmental aspect to consider, if electricity in the region is generated from nuclear or renewable sources, the electricity will be "clean" thus switching to the heat pump will have a positive environmental impact.

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