When heating or cooling homes, we have different technologies to choose from. Sometimes, the costs of a new system outweigh the benefits, even if a new system is more efficient. We should know the basics of these calculations so that we can make the decisions ourselves, rather than potentially being mislead or making a mistake with a costly new system.
1.
Assume a home needs 20 BTUs per hour per square foot (BTUs are a
measurement of energy needed for heating/cooling). If a home is 1,500
square feet, how many BTUs per hour are needed to heat the home?
We can multiply these numbers so that the units cross cancel to leave us with just BTUs.
\[\frac{20 \text{ BTUs per hour}}{1 \text{ ft}^2} \cdot 1,500 \text{ ft}^2 =\]
\[\frac{20 \text{ BTUs per hour}}{1 \cancel{\text{ ft}^2}} \cdot 1,500 \cancel{\text{ ft}^2} =\]
\[30,000 \text{ BTUs per hour}\]
2.
A coefficient of performance is used to determine the efficiency
of a system. The coefficient of performance for a heating system, \(C\), is calculated by dividing the amount
of heat produced, \(Q\), by the amount
of work needed to produce that heat, \(W\).
\[C = \frac{Q}{W}\]
The higher the coefficient of performance of a system, the more efficient the system is.
A heat pump provides 30,000 BTUs per hour of heat. The heat pump requires 8,500 BTUs per hour of electricity. What is the heat pump’s coefficient of performance?
We substitute \(Q = 30,000\) and \(W = 8,500\) into the above equation.
\[C = \frac{Q}{W}\]
\[C = \frac{30,000}{8,500}\]
\[C \approx 3.53\]
We can use this coefficient of performance for a calculation below.
3.
An oil furnace consumes 2.5 gallons of heating oil per hour and
produces 175,000 BTUs of heat per hour. Assume one gallon of heating oil
is 140,000 BTUs. What is the coefficient of performance of the oil
furnace?
To use the coefficient of performance equation, we have to first calculate the BTUs of heat produced per hour. We are told we need 2.5 gallons per hour and that we get 140,000 BTUs per gallon.
\[\frac{2.5 \text{ gallons}}{1 \text{ hour}} \cdot \frac{140,000 \text{ BTUs}}{1 \text{ gallon}} =\]
\[\frac{2.5 \cancel{\text{ gallons}}}{1 \text{ hour}} \cdot \frac{140,000 \text{ BTUs}}{1 \cancel{\text{ gallon}}} =\]
\[\frac{350,000 \text{ BTUs}}{1 \text{ hour}}\]
Now we can substitute \(W = 350,000\) and \(Q = 175,000\) into the formula.
\[C = \frac{Q}{W}\]
\[C = \frac{175,000}{350,000}\]
\[C = 0.5\]
The coefficient of performance of the oil furance is much lower than the coefficient of performance of the heat pump. That means that the heat pump is more efficient in producing heat with the energy that is put into its system.
4.
A home is 1,800 square feet. The home needs 50,000 BTUs per year
per square foot. The electric heating system has a coefficient of
performance of 1. The heat pump system has a coefficient of performance
of 3.5. A heat pump costs $8,000 to install. An electric heating system
costs $4,000 to install. The cost of electricity is $0.16 per kilowatt
hour. 1 kilowatt hour is equal to 3,412 BTUs.
Eventually, assuming the systems don’t break down and cost extra money to repair, the heat pump will save money. How many years will it take before the heat pump is saving this homeowner money?
There are many pieces of information in this problem. When that occurs, it can help to focus on the goal of the problem. We should create equations for the cost of both systems. The equations should be based on the key variable, time, \(t\), in years.
The cost of the electric system can be called \(C_1\). It has a starting cost of $4,000. We need to calculate the energy cost per year.
If the home is 1,800 square feet, and if the home needs 50,000 BTUs per year per square foot, then the home needs
\[1,800 \text{ ft}^2 \cdot 50,000 \frac{\text{BTUs}}{\text{ft}^2} \text{ per year}=\]
\[1,800 \cancel{\text{ ft}^2} \cdot 50,000 \frac{\text{BTUs}}{\cancel{\text{ft}^2}} \text{ per year} =\]
\[90,000,000 \text{ BTUs per year}\]
Our electric heating costs are given to us in kWh (kilowatt hours). We have to convert BTUs to kWh using the conversion factor 1 kilowatt hour is equal to 3,412 BTUs.
\[90,000,000 \text{ BTUs per year} \cdot \frac{1 \text{ kWh}}{3,412 \text{ BTUs}} =\]
\[90,000,000 \cancel{ \text{ BTUs}} \text{ per year} \cdot \frac{1 \text{ kWh}}{3,412 \cancel{\text{ BTUs}}} =\]
\[\frac{90,000,000 \text{ kWh}}{3,412} \text{ per year} =\]
\[26,377.49 \text{ kwH per year}\]
We are told that the cost of electricity is $0.16 per kilowatt hour. The cost for this system per year is
\[26,377.49 \text{ kwH per year} \cdot \frac{\$0.16}{1 \text{ kWh}} =\]
\[26,377.49 \cancel{\text{ kwH}}\text{ per year} \cdot \frac{\$0.16}{1 \cancel{\text{ kWh}}} \approx\]
\[\$4,220.39 \text{ per year}\]
The equation for the cost of the electric heating system is
\[C_1 = \$4,220.39t + 4,000\]
We need to make a similar calculation for the cost of the heat pump system. We can call the cost \(C_2\).
The home still needs 90,000,000 \(\text{ BTUs of heat per year}\), which we calculated above. The heat pump is said to have a coefficient of performance of 3.5. That means it produces 3.5 BTUs of heat for every 1 BTU it is supplied by the power company. We can convert the 90,000,000 BTUs of heat per year to the amount of BTUs spent by this heating system.
\[90,000,000 \text{ BTUs of heat per year} \cdot \frac{1 \text{ BTUs of energy from power company }}{3.5 \text{ BTUs of heat}} =\]
\[90,000,000 \cancel{\text{ BTUs of heat}} \text{ per year} \cdot \frac{1 \text{ BTUs of energy from power company }}{3.5 \cancel{\text{ BTUs of heat}}} =\]
\[\frac{90,000,000 \text{ BTUs from power company}}{3.5} \text{ per year} \approx\]
\[25,714,286 \text{ BTUs from power company per year}\]
The power company charges in kWh, so we again have to convert these BTUs to kWh. 1 kilowatt hour is equal to 3,412 BTUs.
\[25,714,286 \text{ BTUs per year} \cdot \frac{1 \text{ kWh}}{3,412 \text{ BTUs}} =\]
\[25,714,286 \cancel{\text{ BTUs}} \text{ per year} \cdot \frac{1 \text{ kWh}}{3,412 \cancel{\text{ BTUs}}} =\]
\[7,536.43 \text{ kWh per year}\]
We are told that the cost of electricity is $0.16 per kilowatt hour. The cost for this system per year is
\[7,536.43 \text{ kWh per year} \cdot \frac{\$0.16}{1 \text{ kWh}} =\]
\[7,536.43 \cancel{\text{ kWh}} \text{ per year} \cdot \frac{\$0.16}{1 \cancel{\text{ kWh}}} \approx\]
\[\$1,205.83 \text{ per year}\]
The equation for the cost of the heat pump heating system is
\[C_2 = \$1,205.83t + 8,000\]
Compare the equations for \(C_1\) and \(C_2\). They are called linear equations. The slope is the number next to the variable. The slope of \(C_2\) is smaller. That means the cost of the heat pump is not rising as fast per year as the cost of the electric system. To find out when the heat pump has become cheaper, we can set the two equations to each other. This will be when the two costs equal one another. After that time, the heat pump will be cheaper because of its smaller slope.
\[C_1 = 4,220.39t + 4,000\]
\[C_2 = 1,205.83t + 8,000\]
Setting the right side of the first equation equal to the right side of the second equation, we have
\[4,220.39t + 4,000 = 1,205.83t + 8,000\]
Subtract $1,205.83t from both sides.
\[4,220.39t - 1,205.83t + 4,000 = 1,205.83t - 1,205.83t + 8,000\]
\[3,014.56t + 4,000 = 8,000\]
Subtract 4,000 from both sides.
\[3,014.56t + 4,000 - 4,000 = 8,000 - 4,000\]
\[3,014.56t =4,000\]
Divide both sides by 3,014.56.
\[\frac{3,014.56t}{3,014.56} =\frac{4,000}{3,014.56}\]
\[t = 1.32\]
Assuming the systems don’t break down, this heat pump system would be saving the customer money within a year and a half!
1. Houses can sometimes be made more efficient for heating or cooling without changing the heating/cooling technology itself. If the amount of work needed to produce heating/cooling, \(W\), stays the same, but the amount of heat/cooling produced, \(Q\), can change, what is the relationship between the coefficient of performance \(C\) and \(Q\)? Is it linear/direct? Indirect? Exponential? Quadratic? etc.
2. What is the relationship between the total cost of a heating system and the efficiency of a system? Is it linear/direct? Indirect? Exponential? Quadratic? etc.