We chose a geothermal system to heat and cool our home since the location of our home is in Iowa City. Deciding on which system to use was basically done by the process of elimination. The judicious choice was made to rule out both solar and wind-powered systems because of the available natural resources (or the lack of). Iowa City, Iowa doesn’t have enough wind to justify the instillation of a wind turbine, nor does it receive enough energy from the sun on a constant basis for a solar heating system to seem very practical.
Advantages of using a geothermal system to heat and cool a house include high energy efficiency, simplicity, low maintenance costs, no outdoor equipment, environmentally “green”, and low life-cycle cost. The ground will always be a source for heat since it’s constantly heated by the shifting layers of the earth and by solar energy. Once the initial cost is invested into the project some of the money may be received back through government incentives or company incentives. Finding a qualified contractor is the toughest part of installing the system, and with the type of drilling that needs done machinery does not come cheap.
Our design is centered around choosing the proper heat pump. When we initially set up our design equations we left everything in terms of the heat rates that we would need at each stage. Finding the heat we would need to keep the house warm on the coldest days will help us size the pump. Using other equations (flow rate, power rating, COP, entering water temperature, etc.) we will leave variables as unknowns that we will need to size for our loop. Examples of unknowns include the length of pipe for the ground loop, the size of the pump for the ground water, the size of the blower fan required for the heating coils, and the length of overall ductwork we will need.
To model the system it is easiest to split the system into three separate loops: the ground loop, the heat pump loop, and the circulating air inside the house.
Advantages of using a geothermal system to heat and cool a house include high energy efficiency, simplicity, low maintenance costs, no outdoor equipment, environmentally “green”, and low life-cycle cost. The ground will always be a source for heat since it’s constantly heated by the shifting layers of the earth and by solar energy. Once the initial cost is invested into the project some of the money may be received back through government incentives or company incentives. Finding a qualified contractor is the toughest part of installing the system, and with the type of drilling that needs done machinery does not come cheap.
Our design is centered around choosing the proper heat pump. When we initially set up our design equations we left everything in terms of the heat rates that we would need at each stage. Finding the heat we would need to keep the house warm on the coldest days will help us size the pump. Using other equations (flow rate, power rating, COP, entering water temperature, etc.) we will leave variables as unknowns that we will need to size for our loop. Examples of unknowns include the length of pipe for the ground loop, the size of the pump for the ground water, the size of the blower fan required for the heating coils, and the length of overall ductwork we will need.
To model the system it is easiest to split the system into three separate loops: the ground loop, the heat pump loop, and the circulating air inside the house.
PART 1: Ground Loop
The closed loop consists of the piping and water (or water-antifreeze mixture), and acts as a heat exchanger with the earth. We found that in the Winter the ground doesn’t get much below 50 oF between and 10 and 20 feet down, and in the summer the temperature is fairly consistent at 60 oF. We chose to use high-density polyethylene (HDPE) for all underground piping, which is in accordance with ASTM standards because of the high strength and stiffness properties. Since HDPE has a much lower roughness than any other piping we could have chosen we will neglect this variable and assume the pipe is smooth. We will use HDPE that has a 1-inch inner diameter SDR11 with an outer diameter of 1.5 inches. From the geothermal specifications chart we found that for 1” SDR11 piping, we will need 4.7 gallons of brine per 100 feet of pipe installed in the ground loop. Since Iowa typically has temperatures that drop below freezing in the winter we want to add methanol to our water to form a brine solution. This brine is 25 percent methanol and 75 percent water, which does change the thermal properties of the ground loop, but we will assume that flow properties of water can be used for modeling. We selected brine because of its low freezing point, good heat transfer abilities, low viscosity, low corrosiveness, and it’s fairly cheap. Using the thermal conduction equation, heat transfer effectiveness and NTU equations, the basic heat transfer equation, and setting up equations for the minimum and maximum flow rates that we could use with the heat pump, we were able to find an optimal length of tubing that we would need to install underground. This optimal length was found with the intent of getting the best heat transfer from the ground while keeping the size of our water pump small enough to avoid unnecessary high capital costs.
The closed loop consists of the piping and water (or water-antifreeze mixture), and acts as a heat exchanger with the earth. We found that in the Winter the ground doesn’t get much below 50 oF between and 10 and 20 feet down, and in the summer the temperature is fairly consistent at 60 oF. We chose to use high-density polyethylene (HDPE) for all underground piping, which is in accordance with ASTM standards because of the high strength and stiffness properties. Since HDPE has a much lower roughness than any other piping we could have chosen we will neglect this variable and assume the pipe is smooth. We will use HDPE that has a 1-inch inner diameter SDR11 with an outer diameter of 1.5 inches. From the geothermal specifications chart we found that for 1” SDR11 piping, we will need 4.7 gallons of brine per 100 feet of pipe installed in the ground loop. Since Iowa typically has temperatures that drop below freezing in the winter we want to add methanol to our water to form a brine solution. This brine is 25 percent methanol and 75 percent water, which does change the thermal properties of the ground loop, but we will assume that flow properties of water can be used for modeling. We selected brine because of its low freezing point, good heat transfer abilities, low viscosity, low corrosiveness, and it’s fairly cheap. Using the thermal conduction equation, heat transfer effectiveness and NTU equations, the basic heat transfer equation, and setting up equations for the minimum and maximum flow rates that we could use with the heat pump, we were able to find an optimal length of tubing that we would need to install underground. This optimal length was found with the intent of getting the best heat transfer from the ground while keeping the size of our water pump small enough to avoid unnecessary high capital costs.
When we solved our 7 equations and 7 unknowns we found that the length of pipe would be 1307 feet. With this length of pipe we found the effectiveness of heat transfer from the ground to the ground loop to be 0.871. To have this much pipe in the ground requires some sort of looping (drilling 1307/2 feet in one direction and then coming back in the opposite generally occupies too much space). The figure to the right shows how we have designed our ground loop. The loop goes down to 12 feet and then goes into the ground horizontally after a 90 degree elbow. The pipe goes 150 feet, and then is attached to a return bend where another 150 feet is bored into the ground. This keeps going until the 9th and 10th lines are installed, with the final line at a depth of 17 feet. These lines are only 118 feet, with another 90 degree elbow on the end of the return (bottom) line installed. There are a total of 9 return bends and 2 90 degree elbows. The dynamic loss coefficients are 1.29 and 0.75. Using these number we were able to come up with a total head loss that the pump would need to overcome.
Hf = 67.1 feet Hz = 17 feet Hd = 2.25 feet Hp = 86.35 feet
We were able to size a pump for the ground loop using this head loss. This pump curve shows the numbers for the pump performance.
The pump is an Orenco BEP10DD pump worth about 1100 dollars. Although this is a 10 gpm pump, it has a flow control so that a smaller flow rate can be achieved. To find the appropriate flow rate we combined several equations including the mass flow rates and heat transfer equations. After we found that we needed 8 gallons per minute, the pump to the left was our choice because of the total head loss of 95 feet.
For the ground loop, one key unknown is the rate at which we will extract heat. This variable will be used later in the simulation figure out the heat rates at each stage of heat exchange.
Jumping ahead, after we sized our heat pump and found the heat transfer rates for the evaporator and the ground loop, we can find a COP for the ground loop. Multiplying the evaporator heat rate by our effectiveness we find that the rate we get heat from the ground loop to be Qground = 24,911 BTU/hr. Next we can find the work put into the loop from the water pump to be Wp = 3,689 BTU/hr.
PART 2: Heat Pump Loop
Designing the heat pump is the most vital part of this system. For the design of this portion was focused on the heat rates at the condenser and evaporator, the temperatures at the different stages, and the mass flow rate of the refrigerant. The refrigerant we selected is R404a because of the very low ozone depletion potential (0) and because it has desirable heat transfer characteristics. R404a (44% HFC-125, 52% HFC-143a, and 4% HFC-134a) is a fairly new refrigerant and its popularity continues to grow. As the refrigerant flows through the copper lines in the heat pump, the pressure and temperature changes help the refrigerant “carry” heat through the cycle. The heat is transferred from the ground loop to the evaporator, which is where the low pressure liquid R404a turns to a gas. After the compressor the gas is now high pressure and so the temperature increases. As the R404a goes through the condenser, heat is transferred by a blower fan to the interior of the house. The refrigerant then comes out of the condenser as a high pressure liquid before returning to a low pressure liquid after the expansion valve.
Jumping ahead, after we sized our heat pump and found the heat transfer rates for the evaporator and the ground loop, we can find a COP for the ground loop. Multiplying the evaporator heat rate by our effectiveness we find that the rate we get heat from the ground loop to be Qground = 24,911 BTU/hr. Next we can find the work put into the loop from the water pump to be Wp = 3,689 BTU/hr.
PART 2: Heat Pump Loop
Designing the heat pump is the most vital part of this system. For the design of this portion was focused on the heat rates at the condenser and evaporator, the temperatures at the different stages, and the mass flow rate of the refrigerant. The refrigerant we selected is R404a because of the very low ozone depletion potential (0) and because it has desirable heat transfer characteristics. R404a (44% HFC-125, 52% HFC-143a, and 4% HFC-134a) is a fairly new refrigerant and its popularity continues to grow. As the refrigerant flows through the copper lines in the heat pump, the pressure and temperature changes help the refrigerant “carry” heat through the cycle. The heat is transferred from the ground loop to the evaporator, which is where the low pressure liquid R404a turns to a gas. After the compressor the gas is now high pressure and so the temperature increases. As the R404a goes through the condenser, heat is transferred by a blower fan to the interior of the house. The refrigerant then comes out of the condenser as a high pressure liquid before returning to a low pressure liquid after the expansion valve.
In the summer months, there is a reversal valve (the expansion valve) that changes the direction of flow for the R404a. However, the refrigerant is always pumping into the compressor. The diagrams below show the details of how this operation is performed.
After finding the flow rate for the ground loop and the required CFM we were able to size a heat pump from the Enertech Global XT Series of heat pumps. We selected the Model 036C 3.0 ton heat pump. This pump has an 8 GPM rating and has an air flow rating of 1350 CFM. Since it is a 3.0 ton pump, the amount of heat it is rated for is roughly 36,000 BTU/hr. From our heat transfer equation for the ground loop we found that the brine enters the evaporator at a temperature very near the temperature of the ground, 50 oF. Using this pump with our numbers we were then able to get the heat rates at the condenser and compressor for this heat pump.
Qev = 28,600 BTU/hr Qcd = 36,800 BTU/hr
So for the heat pump cycle we can easily find a COP of 4.48. The energy difference between the two stages comes from the power input to the compressor, Wc = 8200 BTU/h
PART 3: Circulating Air
When the R404a goes through the condenser, the airflow driven by a blower fan cools down the refrigerant and “carries away” heat from the heat pump. When we set up the equations for this loop we had a number of equations involving the mass flow rate of the air, the temperature differences, and the overall length of ducting required for our house. We assumed the air temperature of the house, 70oF, is the inlet temperature at the blower fan.
The heat transfer from the refrigerant to the air in the house is accomplished via mixed radiation and convection. We included both equations for our simulation. We will assume that the condenser/radiator has a heat transfer effectiveness of 0.9. This means that the heat transfer process from the refrigerant in the coils to the air is 90 percent effective.
The size of the ducting we will use is 2 ft x 1 ft. We will use 24-gauge aluminum sheet metal, which has a thickness of 0.0239 in. The ducting begins at the surface of the enclosure for the coils and will have an opening facing downwards. The blower fan will push air upward across the coils and the heat rate from the condenser will be distributed throughout the house by the ducting. The ducting will go up to the ceiling and then across 30 feet of the room. At the end of the ducting will be two grilles, one dumping heat into each room of the house. That means the hydraulic diameter of the ducting is 1.33 ft. We were able to set up equations to find our ducting size by using the external static pressure of the blower fan we selected. This pressure was equal of 0.75 inches of water so we were able to divide by 400 ft, a value we obtained from the ACCA manual D (residential duct systems guide). Once we obtain our friction rate in inches per 100 feet, selecting the size of the ducting was straight forward.
When we sized the heat pump, one variable that the pump depended on was a CFM rating that the heat pump operated with. The only way to provide the amount of heat required to the house was to use a CFM value of at least 1294. With this in mind we decided to choose a blower fan that was capable of delivering 1350 CFM. We decided to go with an ECM (electronically commutated motor) blower instead of a conventional blower because of the fact that they require less than half the amount of electricity to operate. These fans increase the RPMs and are able to achieve a higher CFM than other fans (with the same amount provided energy). The fan cost approximately 460 dollars to buy. This fan may have a 0.4 kW reduction in demand charges, so the payback on this piece is evident. The Series FPB ½ hp ECM Blower by Carrier operates at just over 200 Watts for our selected air flow rate of 1350 CFM.
Wf = 682 BTU/hr
Multiplying the heat rate at the condenser by the effectiveness of heat transfer gives 33,120 BTU/hr. Then we can subtract the work from the pump to find the amount of heat delivered to our house to be:
Qheat = 32,438 BTU/hr
The overall COP for our geothermal heating system is then 2.58.
When a similar analysis was done for the cooling process, we found that Qcool = 42,389 BTU/hr
Qev = 28,600 BTU/hr Qcd = 36,800 BTU/hr
So for the heat pump cycle we can easily find a COP of 4.48. The energy difference between the two stages comes from the power input to the compressor, Wc = 8200 BTU/h
PART 3: Circulating Air
When the R404a goes through the condenser, the airflow driven by a blower fan cools down the refrigerant and “carries away” heat from the heat pump. When we set up the equations for this loop we had a number of equations involving the mass flow rate of the air, the temperature differences, and the overall length of ducting required for our house. We assumed the air temperature of the house, 70oF, is the inlet temperature at the blower fan.
The heat transfer from the refrigerant to the air in the house is accomplished via mixed radiation and convection. We included both equations for our simulation. We will assume that the condenser/radiator has a heat transfer effectiveness of 0.9. This means that the heat transfer process from the refrigerant in the coils to the air is 90 percent effective.
The size of the ducting we will use is 2 ft x 1 ft. We will use 24-gauge aluminum sheet metal, which has a thickness of 0.0239 in. The ducting begins at the surface of the enclosure for the coils and will have an opening facing downwards. The blower fan will push air upward across the coils and the heat rate from the condenser will be distributed throughout the house by the ducting. The ducting will go up to the ceiling and then across 30 feet of the room. At the end of the ducting will be two grilles, one dumping heat into each room of the house. That means the hydraulic diameter of the ducting is 1.33 ft. We were able to set up equations to find our ducting size by using the external static pressure of the blower fan we selected. This pressure was equal of 0.75 inches of water so we were able to divide by 400 ft, a value we obtained from the ACCA manual D (residential duct systems guide). Once we obtain our friction rate in inches per 100 feet, selecting the size of the ducting was straight forward.
When we sized the heat pump, one variable that the pump depended on was a CFM rating that the heat pump operated with. The only way to provide the amount of heat required to the house was to use a CFM value of at least 1294. With this in mind we decided to choose a blower fan that was capable of delivering 1350 CFM. We decided to go with an ECM (electronically commutated motor) blower instead of a conventional blower because of the fact that they require less than half the amount of electricity to operate. These fans increase the RPMs and are able to achieve a higher CFM than other fans (with the same amount provided energy). The fan cost approximately 460 dollars to buy. This fan may have a 0.4 kW reduction in demand charges, so the payback on this piece is evident. The Series FPB ½ hp ECM Blower by Carrier operates at just over 200 Watts for our selected air flow rate of 1350 CFM.
Wf = 682 BTU/hr
Multiplying the heat rate at the condenser by the effectiveness of heat transfer gives 33,120 BTU/hr. Then we can subtract the work from the pump to find the amount of heat delivered to our house to be:
Qheat = 32,438 BTU/hr
The overall COP for our geothermal heating system is then 2.58.
When a similar analysis was done for the cooling process, we found that Qcool = 42,389 BTU/hr
Payback Period:
Electricity needed at any time = Power of compressor + Power of water pump + Power of blower
Fraction of time that power is actually on = 65% = 0.65
Electricity needed in winter = (12,571 BTU/h)(24 h/day)(30 days/month)(4 months)(0.65) = 2.35 x 107 BTU = 6,687 kwh
Electricity needed in summer = (43,920 BTU/h)(24h/day)(30 days/month)(3 months)(0.65) = 18,050 kwh
Capital costs for geothermal system after federal incentive = Ground loop pump + Heat pump + Blower + Piping and ductwork material + Installation + Excavation/trenching/drilling – Federal Incentive
Federal incentive = (.30)(Heat pump + ground loop pump + Installation)
Ground loop pump = $1100, Heat pump = $5000, Blower = $460, Material costs = $1500, Installation = $4000, Excavation/trenching/drilling = $1000
Capital cost for geothermal after federal incentive = $10,030
Capital cost for using natural gas = Natural gas furnace + AC unit +Materials and Installation
Natural gas furnace = $3500, AC unit = $3000, Materials and Installation = $500
Capital cost for using natural gas = $7,000
Geothermal capital cost – natural gas capital cost = $3,030
Cost of electricity for geothermal = $.045/kwh, cost of electricity for natural gas = $.044/kwh
Efficiency of furnace = 80% = 0.80
Yearly heating operation costs:
Yearly cooling operation costs:
Yearly savings from geothermal = (367.80-300.90) + (992.75-812.25) = $247.40
Paypack period = (Geothermal capital cost – natural gas capital cost) / (yearly savings from geothermal) = 12.2 years
Fraction of time that power is actually on = 65% = 0.65
Electricity needed in winter = (12,571 BTU/h)(24 h/day)(30 days/month)(4 months)(0.65) = 2.35 x 107 BTU = 6,687 kwh
Electricity needed in summer = (43,920 BTU/h)(24h/day)(30 days/month)(3 months)(0.65) = 18,050 kwh
Capital costs for geothermal system after federal incentive = Ground loop pump + Heat pump + Blower + Piping and ductwork material + Installation + Excavation/trenching/drilling – Federal Incentive
Federal incentive = (.30)(Heat pump + ground loop pump + Installation)
Ground loop pump = $1100, Heat pump = $5000, Blower = $460, Material costs = $1500, Installation = $4000, Excavation/trenching/drilling = $1000
Capital cost for geothermal after federal incentive = $10,030
Capital cost for using natural gas = Natural gas furnace + AC unit +Materials and Installation
Natural gas furnace = $3500, AC unit = $3000, Materials and Installation = $500
Capital cost for using natural gas = $7,000
Geothermal capital cost – natural gas capital cost = $3,030
Cost of electricity for geothermal = $.045/kwh, cost of electricity for natural gas = $.044/kwh
Efficiency of furnace = 80% = 0.80
Yearly heating operation costs:
- Geothermal = (Electricity needed in winter)(.045) = $300.90
- Gas = (Electricity needed in winter/Efficiency of furnace)(.044) = $367.80
Yearly cooling operation costs:
- Geothermal = (Electricity needed in summer)(.045) = $812.25
- Gas = (Electricity needed in winter/efficiency of furnace)(.044) = $992.75
Yearly savings from geothermal = (367.80-300.90) + (992.75-812.25) = $247.40
Paypack period = (Geothermal capital cost – natural gas capital cost) / (yearly savings from geothermal) = 12.2 years
LEED Certification:
LEED for Homes is a rating system that promotes the design and construction of green homes. The rating system measures the overall performance of a home in 8 categories: Innovation and Design Process, Location and Linkages, Sustainable Sites, Water Efficiency, Energy & Atmosphere, Materials and Resources, Indoor Environmental Air Quality, and Awareness and Education.
The level of performance in LEED for Homes is indicated by 4 performance tiers, as shown below. Note that this is the LEED rating system for the unadjusted homes.
LEED for Homes is a rating system that promotes the design and construction of green homes. The rating system measures the overall performance of a home in 8 categories: Innovation and Design Process, Location and Linkages, Sustainable Sites, Water Efficiency, Energy & Atmosphere, Materials and Resources, Indoor Environmental Air Quality, and Awareness and Education.
The level of performance in LEED for Homes is indicated by 4 performance tiers, as shown below. Note that this is the LEED rating system for the unadjusted homes.
To find the adjustment of these point ranges of our particular home, we used the following figure.
Based on the figure, we were able to implement a -7 point adjustment to all of the tiers in the previous table from our 2 bedroom, 1000 square foot home. The end goal for the home is to have LEED gold certification. After the adjustments, our home needs to have in between 68 and points out of a maximum of 136.
The methodology in which we received our final point total was going through each of the 8 performance measures listed previously, checking off what prerequisite and attributes that are feasible for our home and the specific point value for that attribute, then adding the point values of each of the 8 sub-components.
1. Innovation and Design Process (Maximum points = 11)
- Conduct a preliminary LEED for Homes meeting with the participation of the provider and the project team
- Integrate project team according to specifications (1)
- At least one person of the project team is credentialed for LEED for Homes as determined by the USGBC (1)
- Conduct at least one full-day integrated design workshop with the project team (1)
- Green rater verified durability management and durability planning (3)
- Points from ID = 5
2. Location and Linkages (Max Points = 10)
- Site selection: our home was not located on any of the restrictions specified (2)
- Infill: 75% the perimeter of our home borders previously developed land (2)
- Existing infrastructure: Home is located within ½ mile of existing water service lines and sewer service lines
- Home located within a ½ mile of transit services that offer 60 or more transit rides per weekday
- Points from LL = 7
3. Sustainable Sites (Max Points = 22, Min Points = 5)
- Erosion control during construction
- Minimize disturbed area of the site according to the specifications (1)
- No invasive plants
- Met requirements for basic landscaping design (2)
- Limit conventional turf to 25% (2)
- Percentage of buildable lot: 85% (2)
- Permanent erosion control (1)
- Management and runoff from the ground
- Completed necessary pest control measures (2)
- Build home so 1/7-acre is buildable lot (2)
- Points from SS = 14
4. Water Efficiency (Max Points = 15, Min Points = 3)
- Municipal Recycled Water System (3)
- Home has higher efficiency fixtures and fittings (2)
- Home has some very high efficiency fixtures and fittings (2)
- Points from WE = 7
5. Energy & Atmosphere (Max Points = 38)
- Basic insulation
- Greatly reduced envelope leakage (2)
- Greatly reduced distribution losses (2)
- High-efficiency HVAC (2)
- Efficient hot water distribution (3)
- Energy Star approved lights
- Improved lighting (1.5)
- High-efficiency appliances and water-efficient clothes washer (1)
- Renewable energy system (6)
- Refrigerant charge test
- Appropriate refrigerant (1)
- Points from EA = 18.5
6. Materials and Resources (Max Points = 16, Min Points = 2)
- Detailed framing documents (1)
- Detailed cut list and lumber number (1)
- Used precut framing packages (1)
- Used environmentally preferable products (3)
- Construction waste reduction (2)
- Points from MR = 8
7. Indoor Environmental Air Quality (Max Points = 21, Min Points = 6)
- Home has no fireplace (2)
- Basic outdoor ventilation
- Third-party performance testing of ventilation (1)
- Return air flow is adequate (1)
- Third-party performance test of return air flow (2)
- Room by room temperature controls
- Good filters
- Indoor contaminant control during construction (1)
- Installed a central vacuum system with exhaust to outdoors (1)
- Preoccupancy flush (1)
- Radon-resistant construction in high-risk areas
- No garage (3)
- Points from IE = 13
8. Awareness and Education (Max Points = 3)
- Enhanced training (1)
- Public awareness according to specifications (1)
- Points from AE = 2
Total LEED Points: 74.5 – Gold Certification.
From this checklist, we did not need to design our geothermal system any differently. However, we would have to spend some additional money on items like high-efficiency lights and appliances to achieve this Gold certification.
