In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is T(in) = 20 degrees celcius and the temperature outside is T(out) = 0 degrees celcius. The walls have fiberglass insulation, which dominates the heat conduction properties of the wall So we can consider the wall to have a thermal conductivity of k(wall) = 0.048 W/m/K
. We will take the thickness of the walls and ceiling to be L(wall) = 12 cm. Assume that the house is a cube of length L = 9.0 m on a side. Assume that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air. Ignore heat loss through the ground.
a)What is H, the total rate of energy loss due to heat conduction for this house?
b)Let us assume that the winter consists of 150 days in which the outside temperature is 0*C on average. This will give the typical number of "heating degree days" observed in a winter in Vancouver. Given that heating oil has an energy content of 35 MJ per litre when burned, how much oil will be needed to supply the heat lost by conduction from this house over a winter? Assume that the heating system is 75% efficient.
To calculate the total rate of energy loss due to heat conduction (H), we can use the formula for heat conduction:
H = (k * A * ΔT) / L
Where:
k is the thermal conductivity of the wall (0.048 W/m/K),
A is the surface area of the house (6 * L^2),
ΔT is the temperature difference between inside and outside (20 - 0 = 20°C), and
L is the thickness of the walls (12 cm or 0.12 m).
Let's calculate the values and then substitute them into the equation:
A = 6 * (9.0 m)^2 = 486 m^2
H = (0.048 W/m/K) * (486 m^2) * (20°C) / (0.12 m)
H = 800.4 W
So, the total rate of energy loss due to heat conduction for this house is approximately 800.4 watts (W) or 0.8 kilowatts (kW).
Moving on to the next part of the question, to calculate the amount of oil needed to supply the heat lost by conduction over the winter, we have the following information:
- Winter consists of 150 days with an average outside temperature of 0°C, giving us 150 heating degree days.
- Heating oil has an energy content of 35 MJ per liter when burned.
- The heating system is 75% efficient.
To calculate the energy needed to supply the heat loss, we can use the formula:
Energy = H * t
Where:
H is the total rate of energy loss due to heat conduction (0.8 kW),
t is the duration of the winter in hours (150 days * 24 hours/day = 3600 hours).
Energy = (0.8 kW) * (3600 hours) = 2880 kWh
Since the heating system is 75% efficient, we need to divide the energy by 0.75 to account for the efficiency:
Energy_needed = 2880 kWh / 0.75 = 3840 kWh
Now, to get the amount of oil needed, we can convert the energy from kilowatt-hours (kWh) to megajoules (MJ) using the conversion factor of 1 kWh = 3.6 MJ:
Energy_needed_MJ = 3840 kWh * 3.6 MJ/kWh = 13824 MJ
Finally, we can calculate the amount of oil needed by dividing the energy needed by the energy content of heating oil:
Amount_of_oil = Energy_needed_MJ / 35 MJ/L = 395.83 L
Therefore, approximately 395.8 liters of oil will be needed to supply the heat lost by conduction from this house over the winter.