Question

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

Short answer

Answer: The key factors taken into consideration when calculating the money lost through the roof during a 14-hour period include the conduction, radiation, and convection heat transfers, the efficiency of the furnace, the unit cost of natural gas, and the duration of time.

Step-by-step solution

Find conduction heat transfer through the slab

For this, we will use Fourier's law of heat conduction: \(q_{cond} = k * A * \frac{T_{in} - T_{out}}{d}\) Where \(k = 2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity, \(A = 15 \mathrm{~m} * 20 \mathrm{~m}\) is the area of the slab, \(T_{in} = 15 ^{\circ} \mathrm{C}\) is the inner surface temperature, \(T_{out}\) is the outer surface temperature, and \(d = 0.22 \mathrm{~m}\) is the thickness of the slab.

Find the radiation heat transfer from the slab to the sky

We will use the radiation heat transfer equation: \(q_{rad} = \epsilon * A * \sigma * (T_{out}^{4} - T_{sky}^{4})\) Where \(\epsilon = 0.9\) is the emissivity, \(\sigma = 5.67 × 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\) is the Stefan-Boltzmann constant, and \(T_{sky} = 255 \mathrm{~K}\) is the sky temperature during the night.

Find the convection heat transfer from the slab to the ambient air

Using Newton's law of cooling: \(q_{conv} = h * A * (T_{out} - T_{amb})\) Where \(h = 15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the convection heat transfer coefficient and \(T_{amb} = 10 ^{\circ} \mathrm{C}\) is the ambient temperature.

Combine the heat transfer equations and find the outer surface temperature

Add the conduction, radiation, and convection heat transfer equations together: \(q_{cond} = q_{rad} + q_{conv}\) Substitute the equations from steps 1, 2, and 3 into this equation and solve for \(T_{out}\).

Find the total heat transfer through the roof

Plug the value of \(T_{out}\) obtained in step 4 into the conduction heat transfer equation obtained in step 1 to calculate the heat transfer rate, \(q_{cond}\).

Determine the energy lost during a 14-hour period

Multiply the heat transfer rate by the duration of time: \(E_{lost} = q_{cond} * (14 \mathrm{~hours} * 3600 \mathrm{~s} / \mathrm{hour})\)

Determine the amount of natural gas energy required to compensate for the energy lost

We're given the efficiency of the furnace, which is 85 percent, so divide the energy lost by the efficiency to find the amount of natural gas energy required: \(E_{ng} = \frac{E_{lost}}{0.85}\)

Determine the cost of natural gas energy

Finally, using the unit cost of natural gas, we can calculate the money lost through the roof for a 14-hour period: \(Cost_{ng} = \frac{E_{ng}}{105,500 \mathrm{~kJ / therm}} * \$1.20 / \mathrm{therm}\) Now you have the outer surface temperature, rate of heat transfer through the roof, and the money lost through the roof during a 14-hour period.

Key concepts

Fourier's Law of Heat Conduction

At the heart of understanding how heat moves through solids like the concrete slab roof in the problem is Fourier's Law of Heat Conduction. This principle provides us with the mathematical formulation to calculate the rate of heat transfer, or heat flux, through a material. The law states that the heat flux (\(q_{cond}\)) is proportional to the negative gradient in temperature and the area through which heat is being transferred.

More simply, it can be expressed by the equation: \[q_{cond} = k \cdot A \cdot \frac{T_{in} - T_{out}}{d}\] where:

• \(k\) is the thermal conductivity of the material, indicating how well it conducts heat,
• \(A\) is the area perpendicular to the direction of heat flow,
• \(T_{in} - T_{out}\) is the temperature difference between the hot and cold sides of the material,
• and \(d\) is the thickness of the material.

In our problem, by rearranging and solving the equation for \(T_{out}\), we can determine the outside temperature of the concrete roof slab through which the heat is being transferred.

Radiation Heat Transfer

When it comes to the roof losing heat to the cold night sky, radiation heat transfer is the key mechanism at play. This mode of heat transfer doesn't require any medium and can occur across the vacuum of space, which is why the roof radiates heat to the sky even though the air temperature is different. \

Radiation Heat Transfer Equation\

The formula to calculate radiative heat transfer is: \[q_{rad} = \epsilon \cdot A \cdot \sigma \cdot (T_{out}^{4} - T_{sky}^{4})\] where:

• \(\epsilon\) is the emissivity of the material's surface,
• \(A\) is the area that is emitting or absorbing radiation,
• \(\sigma\) is the Stefan-Boltzmann constant, a fundamental constant of nature,
• and \(T_{out}^{4} - T_{sky}^{4}\) represents the difference in the fourth power of absolute temperature between the roof surface and the sky temperature.

By considering both \(T_{out}\) and \(T_{sky}\), and knowing all other variables, this part of the equation lets us calculate the energy radiating away into space.

Convection Heat Transfer

Finally, convection heat transfer is the process that describes how heat is exchanged between a surface and a fluid moving over it—in our case, the cool air brushing against the roof. This is represented in Newton's Law of Cooling which is often used to approximate the heat transfer rate due to convection.

The formula for convection heat transfer is: \[q_{conv} = h \cdot A \cdot (T_{out} - T_{amb})\] where:

• \(h\) is the convection heat transfer coefficient, which characterizes the convective heat transfer behavior between the surface and the fluid,
• \(A\) is the surface area,
• \(T_{out}\) is the surface temperature,
• and \(T_{amb}\) is the ambient temperature of the fluid.

By computing \(T_{out}\) with considerations from both radiation and convection heat loss, we can estimate how much heat is being lost from the house into the cooler air of the environment.

Merging these calculations gives us a wholesome picture of how the roof temperature affects and is affected by the surrounding environment, allowing us to determine the outer surface temperature and the rate of heat loss from the roof.