Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of \(150 \mathrm{~m}^{2}\) and an average height of \(3 \mathrm{~m}\) at \(1000 \mathrm{~m}\) elevation, where the standard atmospheric pressure is \(89.6 \mathrm{kPa}\). The house is maintained at a temperature of \(22^{\circ} \mathrm{C}\), and the infiltration losses are estimated to amount to \(0.7 \mathrm{ACH}\). Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is \(5^{\circ} \mathrm{C}\). Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $$\$ 0.082 / \mathrm{kWh}$$

Step by step solution

01

Determine the volume of air changed in the house

First, we need to determine the total volume of the house. We can do this by multiplying the floor area by the average height: Total Volume = Floor Area × Average Height Total Volume = \(150\mathrm{~m}^{2} × 3\mathrm{~m} = 450\mathrm{~m}^{3}\) Since there is \(0.7\mathrm{ACH}\) (air changes per hour), for every hour, \(0.7 × 450\mathrm{~m}^{3}\) of air is replaced. In a day, it is replaced by: Volume of air replaced per day = \(0.7 × 450\mathrm{~m}^{3} × 24\mathrm{~h}\)

02

Calculate the mass of the infiltrated air

Next, we will use the ideal gas law to calculate the mass of the air replaced by the infiltration (\(m\)). The air is assumed to be ideal, so we can use the ideal gas law: \(PV = nRT\) Where: P = Pressure (\(89.6 \mathrm{kPa}\)) V = Volume of air replaced per day (We will use this value in Step 1) n = Number of moles of air R = Gas Constant (\(8.314 \mathrm{J/(mol \cdot K)}\)) T = Absolute atmospheric temperature outside (\(5^{\circ} \mathrm{C} = 278\mathrm{K}\)) As we need to find the mass, we can use the formula: Mass = \(M × n\) Where: M = Molar mass of air (\(28.97 \mathrm{kg/kmol}\)) By combining the ideal gas law and mass formula, we will get: \(m = \dfrac{P \cdot V \cdot M}{R \cdot T}\)

03

Calculate the energy loss due to infiltration

Now that we have the mass of the air replaced by the infiltration, we can calculate the energy loss due to infiltration. We will use the following formula for calculating energy loss (\(Q\)): \(Q = m × c_p × \Delta T\) Where: \(m\) = Mass of air (calculated in Step 2) \(c_p\) = Specific heat capacity of air at constant pressure (\(1005\,\mathrm{J/(kg\cdot K)}\)) \(\Delta T\) = Difference in temperature between indoor and outdoor (Inlet temperature \(5^{\circ} \mathrm{C}\) and room temperature \(22^{\circ} \mathrm{C}\)) \(\Delta T = 22^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 17\,\mathrm{K}\)

04

Find the cost of the energy loss for one day

We have calculated the energy loss due to infiltration, so now we need to find the cost of this energy loss on the given day. To do this, we can use the following formula: Cost = \(\dfrac{Q}{3600} × unit\,cost\) Where: \(Q\) = Energy loss (calculated in Step 3) \(unit\,cost\) = Cost per kWh (\(0.082 \/\mathrm{USD/kWh}\)) Now, we just need to plug in the values from Steps 1 to 3 into the equations and solve for the energy loss and its associated cost.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Law

The ideal gas law is a fundamental principle that plays a crucial role in understanding how gases behave under various conditions. It relates the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas through the equation:
\(PV = nRT\)
In this formula, R represents the ideal gas constant. To solve problems involving changes in the properties of air due to infiltration or other processes, the ideal gas law comes in handy. For instance, in the context of our exercise, we use the ideal gas law to calculate the mass of infiltrated air by rearranging the formula to solve for 'n' (the number of moles), then multiplying by the molar mass (M) of air to find the mass. This step is crucial because it allows us to determine how much air (by mass) is entering the house due to infiltration. With the mass of the air known, we can move on to calculate the energy required to heat this air to room temperature.

Air Changes per Hour (ACH)

Air Changes per Hour (ACH) is a measure used to express the rate at which the air within a space is replaced with air from outside. Specifically, it tells us how many times the total volume of air inside a room or building is exchanged with outdoor air in one hour. An ACH value provides insight into the airtightness and ventilation effectiveness of a space.
In our exercise, an ACH of 0.7 means that 70% of the house's air volume is exchanged with outdoor air every hour. To calculate the total volume of air exchanged in a day, we simply multiply the house's total air volume by the ACH value and then by 24 hours. This figure is the cornerstone for our infiltration energy loss calculations, as it represents the total amount of cold air we need to warm up to the indoor temperature over the duration of one day.

Specific Heat Capacity

The specific heat capacity (denoted as \(c_p\)) of a substance is the amount of heat required to raise the temperature of one kilogram of the substance by one degree Celsius (or Kelvin, as the increment is the same for both scales). This property is critical when calculating the energy needed to heat or cool materials. In the context of our exercise, the specific heat capacity of air informs us how much energy is required to increase the temperature of the infiltrated air to equal that of the room's air.
The equation \(Q = m \times c_p \times \Delta T\) uses the specific heat capacity (\(c_p\)), the mass of the air (\(m\)), and the temperature difference (\(\Delta T\)) to calculate the total energy lost due to infiltration. By knowing the specific heat capacity, we can understand and quantify the energy implications of heating the incoming cold air to maintain comfortable indoor conditions during winter.

Energy Cost Calculation

Once the energy loss due to infiltration has been calculated, the next step is to determine its cost implication, which essentially is the energy cost calculation. This process entails converting the total energy lost (in joules) to kilowatt-hours (kWh) since electricity is commonly billed in kWh. The conversion factor is 1 kWh = \(3,600,000\) joules.
We use the calculated total energy loss, divide it by 3,600,000 to get the kWh, and then multiply by the cost per kWh to find the total cost of the energy loss for the day. This step is essential for homeowners or facility managers who need to budget for heating costs or for those looking to improve energy efficiency and reduce expenses. Understanding how to perform energy cost calculations allows individuals to make informed decisions about the cost-effectiveness of potential upgrades to insulation or heating systems.