Question

The side surfaces of a 3-m-high cubic industrial (?) furnace burning natural gas are not insulated, and the temperature at the outer surface of this section is measured to be \(110^{\circ} \mathrm{C}\). The temperature of the furnace room, including its surfaces, is \(30^{\circ} \mathrm{C}\), and the emissivity of the outer surface of the furnace is 0.7. It is proposed that this section of the furnace wall be insulated with glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) wrapped by a reflective sheet \((\varepsilon=0.2)\) in order to reduce the heat loss by 90 percent. Assuming the outer surface temperature of the metal section still remains at about \(110^{\circ} \mathrm{C}\), determine the thickness of the insulation that needs to be used. The furnace operates continuously throughout the year and has an efficiency of 78 percent. The price of the natural gas is \(\$ 1.10 /\) therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content). If the installation of the insulation will cost \(\$ 550\) for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves.

Short answer

Answer: To answer this question, follow the steps in the solution and calculate the payback period using the given information and formulas.

Step-by-step solution

1: Calculate initial heat loss without insulation

First, let's find the total surface area of the furnace's side surfaces. This is a cubic furnace, so it has 6 sides. We're only looking for the heat loss from the side surfaces, so we'll multiply the surface area of one side by 4 (ignoring the top and bottom). The surface area of one side is \(A = (3 \mathrm{m})^2 = 9 \mathrm{m}^2\), and the total surface area for the side surfaces is \(A_{tot} = 4A = 36 \mathrm{m}^2\). Next, calculate the initial heat loss through radiation, using the formula: \(q_{rad,initial} = \varepsilon \sigma A_{tot}(T_s^4 - T_r^4)\), where \(T_s= (110 + 273.15)\ \mathrm{K}\) is the surface temperature and \(T_r=(30+273.15)\ \mathrm{K}\) is the room temperature. Plug in values to find \(q_{rad,initial}\).

2: Calculate the desired heat loss rate after insulation

We need to reduce the heat loss rate by 90%, so let's find the desired heat loss rate after insulation: \(q_{desired} = 0.1q_{rad,initial}\).

3: Calculate heat loss through radiation on the reflective sheet surface

Now we want to calculate the heat loss through radiation after the reflective sheet is installed. Use the emissivity of the sheet (\(\varepsilon=0.2\)) in the radiation formula to find the heat loss rate through the sheet, \(q_{rad,sheet}=\varepsilon\sigma A_{tot}(T_s^4 - T_r^4)\).

4: Calculate heat loss through conduction in the insulation

The remaining heat loss must be through conduction in the insulation. The formula for the heat loss rate in conduction is \(q_{cond} = k \cdot A_{tot} \frac{T_s - T_r}{\delta}\), where \(\delta\) is the thickness of the insulation. Set \(q_{cond}\) equal to the desired heat loss rate minus radiation heat loss through the sheet: \(q_{cond} = q_{desired} - q_{rad,sheet}\).

5: Calculate the insulation thickness

Now, we can rearrange the conduction formula to solve for insulation thickness \(\delta\): \(\delta = k \cdot A_{tot} \frac{T_s - T_r}{q_{cond}}\). Plug in all the values and calculate the thickness of the insulation.

6: Calculate energy loss per year without insulation

We now know the initial radiant heat loss rate, \(q_{rad,initial}\). To find the energy loss per year without insulation, we need to convert the heat loss rate into energy using the relationship \(E = qt\), where \(t = 24 \cdot 365\ \mathrm{hr}\) (one year). Calculate the energy loss per year without insulation, \(E_{initial} = q_{rad,initial}t\).

7: Calculate energy loss per year with insulation

Use the same formula, but with the desired heat loss rate to find the energy loss per year with insulation, \(E_{desired} = q_{desired}t\).

8: Calculate energy and cost savings

To find the energy savings, subtract the energy loss per year with insulation from the energy loss per year without insulation: \(\Delta E = E_{initial} - E_{desired}\). Next, calculate the cost savings from natural gas. Convert the energy into therms by dividing by 105,500 kJ/therm. Then, multiply by 78% furnace efficiency, and multiply by the gas cost \(1.10/\text{therm}\).

9: Calculate the payback period

To find how long it will take for the insulation to pay for itself, we simply divide the cost of the insulation by the cost savings: \(\text{Payback period} = \frac{\text{Installation cost}}{\text{Cost savings per year}}\). Calculate the payback period.

Key concepts

Heat Transfer

Understanding how heat moves is crucial when considering ways to improve the energy efficiency of systems like furnaces. Heat transfer occurs through three primary mechanisms: conduction, convection, and radiation. In our furnace insulation optimization scenario, we're focused on reducing heat transfer via radiation from the hot surfaces of the furnace and through conduction across the proposed insulation layer.

Radiation is the transfer of energy through electromagnetic waves and doesn't require a medium, meaning it can occur even through a vacuum. Conduction, however, is the process by which heat moves through a material, in this case, the insulation we intend to use. Conductive heat transfer is directly proportional to the temperature difference across the material and inversely proportional to the thickness of the insulating layer. By optimizing insulation, we address and minimize these heat transfer processes to reduce energy loss.

Thermal Insulation

Thermal insulation plays a pivotal role in reducing unwanted heat loss or gain by providing a barrier between areas that are significantly different in temperature. In our furnace example, adding insulation will minimize the heat escaping into the surrounding environment. The performance of insulation is measured by its thermal conductivity (\( k \)), which indicates how easily heat passes through a material.

The lower the thermal conductivity, the better the material is at insulating. Glass wool, chosen for the furnace, has a low thermal conductivity and is effective for high-temperature applications. Furthermore, covering the insulation with a reflective sheet serves to reduce radiation heat transfer because of its low emissivity, meaning it does not readily emit heat as radiation, reflecting it instead. Choosing the right materials and thickness is essential for ensuring optimal insulation and energy conservation.

Energy Efficiency

Energy efficiency refers to using less energy to perform the same task, thus eliminating energy waste. In industrial settings like a furnace operation, improving energy efficiency can result in substantial cost savings and environmental benefits. For the furnace, the suggestion to insulate would help achieve a drastic reduction in heat loss, thus requiring less natural gas to maintain the operating temperature, which ultimately means improved energy efficiency of the system.

While upfront costs for improvements can be significant, as with the insulation installation, the long-term savings from reduced energy use are typically greater. This creates a compelling financial incentive for implementing energy-efficient practices and technologies, in addition to the positive impact on sustainability and resource conservation.

Heat Loss Calculation

Calculating heat loss is a mathematical way to estimate the energy being wasted through the surfaces of a heated volume such as a furnace. Understanding and being able to calculate heat loss informs decisions on insulation requirements and efficiency improvements. The initial heat loss calculation considers factors like the temperature difference between inside and outside surfaces, the surface’s emissivity, and the total area through which the heat is lost.

In our exercise, we first calculated the heat loss without insulation and then figured out how much insulation would be needed to achieve a 90% reduction in this heat loss. The calculations involved understanding the properties of the chosen insulation, particularly its ability to conduct heat, represented by its thermal conductivity. By reducing the heat loss, we're improving the furnace's energy efficiency and can then estimate the savings in energy costs, which allows us to calculate the payback period for the investment in insulation.