A \(2-\mathrm{m} \times 1.5-\mathrm{m}\) section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the outer surface of this section is measured to be \(80^{\circ} \mathrm{C}\). The temperature of the furnace room is \(30^{\circ} \mathrm{C}\), and the combined convection and radiation heat transfer coefficient at the surface of the outer furnace is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). It is proposed to insulate this section of the furnace wall with glass wool insulation \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) 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 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 $$\$ 250$$ for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves.

The initial heat loss rate can be determined using the combined convection and radiation heat transfer coefficient and the temperature difference between the outer surface of the furnace and the room. The heat loss rate per unit area is given by: \(q''=h(T_1 - T_2)\) Where \(q''\) is the heat loss rate per unit area, \(h\) is the combined convection and radiation heat transfer coefficient (\(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), \(T_1\) is the temperature of the outer surface of the furnace (\(80^{\circ} \mathrm{C}\)), and \(T_2\) is the room temperature (\(30^{\circ} \mathrm{C}\)). \(q'' = 10(80 - 30) = 500 \mathrm{~W} / \mathrm{m}^{2}\) Next, we need to find the heat loss rate, \(Q\), by multiplying \(q''\) by the area (\(A\)) of the section: \(Q = q''A\) \(A = 2\mathrm{-m} \times 1.5\mathrm{-m} = 3 \mathrm{~m}^{2}\) \(Q = 500 \mathrm{~W} / \mathrm{m}^{2} \times 3 \mathrm{~m}^{2} = 1500 \mathrm{~W}\)

The proposed insulation should reduce the heat loss rate by 90%, so the new heat loss rate would be: \(Q_{new} = 0.10 \times Q = 0.10 \times 1500 \mathrm{~W} = 150 \mathrm{~W}\) Now we need to determine the thickness of the insulation that would achieve the desired heat loss reduction. We can use the conduction equation for this purpose: \(q'' = -k\frac{\Delta T}{\Delta x}\) Since the outer surface temperature remains at about \(110^{\circ} \mathrm{C}\): \(\Delta T = 110^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 80^{\circ} \mathrm{C}\) We also know the new heat loss rate per unit area and the thermal conductivity of the glass wool insulation (\(k = 0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)): \(q'_{new}' = \frac{Q_{new}}{A} = \frac{150 \mathrm{~W}}{3 \mathrm{~m}^{2}} = 50 \mathrm{~W} / \mathrm{m}^{2}\) Now, we can find the insulation thickness (\(\Delta x\)): \(\Delta x = -k\frac{\Delta T}{q'_{new}'} = -0.038 \frac{80\mathrm{~K}}{50 \mathrm{~W} / \mathrm{m}^{2}} = 0.0608 \mathrm{~m}\) The required insulation thickness is 0.0608 m or 60.8 mm.

First, we need to calculate the energy savings due to the insulation. The energy savings can be calculated by finding the difference between the initial and new heat losses and then converting the difference into energy: \(E_{savings} = (Q - Q_{new}) \times t\) Since the furnace operates continuously, we can consider the savings per year: \(t = 1 \mathrm{~year} = 365 \times 24 \mathrm{~hours} = 8760 \mathrm{~hours}\) \(E_{savings} = (1500 \mathrm{~W} - 150 \mathrm{~W}) \times 8760 \mathrm{~hours} = 11826000 \mathrm{~Wh} = 11826 \mathrm{kWh}\) The efficiency of the furnace is 78%, so we need to divide the energy savings by the efficiency to find how much natural gas is saved: \(E_{gas} = \frac{E_{savings}}{0.78} = 15161 \mathrm{kWh}\) Now, we need to convert the energy saved in natural gas to therms: \(E_{therms} = \frac{E_{gas} \times 1000 \mathrm{kJ}}{105500 \mathrm{~kJ/therm}} = 143.5 \mathrm{~therms}\) Now we can calculate the cost savings per year by multiplying the energy saved in therms by the cost per therm: \(Savings = E_{therms} \times \$1.10 = \$157.85 \text{ per year}\) Finally, we can calculate the payback period for the insulation investment by dividing the cost of installation by the annual cost savings: \(Payback Period = \frac{\$250}{\$157.85 \text{ per year}} = 1.58 \text{ years}\) The insulation will pay for itself in approximately 1.58 years.