Question

Consider a \(2-\mathrm{m}\)-high electric hot-water heater that has a diameter of \(40 \mathrm{~cm}\) and maintains the hot water at \(60^{\circ} \mathrm{C}\). The tank is located in a small room at \(20^{\circ} \mathrm{C}\) whose walls and ceiling are at about the same temperature. The tank is placed in a 44 -cm- diameter sheet metal shell of negligible thickness, and the space between the tank and the shell is filled with foam insulation. The average temperature and emissivity of the outer surface of the shell are \(40^{\circ} \mathrm{C}\) and \(0.7\), respectively. The price of electricity is \(\$ 0.08 / \mathrm{kWh}\). Hot-water tank insulation kits large enough to wrap the entire tank are available on the market for about \(\$ 60\). If such an insulation kit is installed on this water tank by the homeowner himself, how long will it take for this additional insulation to pay for itself? Disregard any heat loss from the top and bottom surfaces, and assume the insulation reduces the heat losses by 80 percent.

Short answer

Answer: It takes approximately 12,244 hours for the additional insulation to pay for itself.

Step-by-step solution

Calculate the initial heat loss per unit time from the heater

We are given the following information: - The diameter of the hot water heater is \(40\,\text{cm}\), this corresponds to a radius \(r_1=20\,\text{cm}\). - The diameter of the shell is \(44\,\text{cm}\), this corresponds to a radius \(r_2=22\,\text{cm}\). - The outer surface temperature of the shell is \(40^{\circ}\,\text{C}\). - The emissivity of the outer surface of the shell is \(0.7\). - The ambient temperature is \(20^{\circ}\,\text{C}\). We will use the formula for heat loss due to radiation from a cylinder \(Q_{\text{loss}}\): \(Q_{\text{loss}}=2\pi r_2 L \sigma \varepsilon \left(T_{\text{shell}}^4-T_{\text{room}}^4\right)\) where: - \(L\) is the height of the water heater (\(2\,\text{m}\)) - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8}\, \text{W/m}^2\text{K}^4\)) - \(\varepsilon\) is the emissivity of the shell (0.7) - \(T_{\text{shell}}\) is the temperature of the outer surface of the shell in Kelvin, (\(273+40=313\,\text{K}\)) - \(T_{\text{room}}\) is the temperature of the room in Kelvin, (\(273+20=293\,\text{K}\))

Calculate how much this heat loss costs per unit time

To calculate the cost of heat loss per unit time, we need to divide the heat loss an hour by the energy conversion factor (3600J = 1Wh) and multiply it by the price of electricity: \(C_{\text{loss}} = \frac{Q_{\text{loss}}}{3600} \times \$0.08/\text{kWh}\)

Calculate the reduction in heat loss and cost after the insulation is installed

After the insulation is installed, we are given that the heat losses are reduced by 80%. Therefore, we have: \(C_{\text{reduced}} = C_{\text{loss}} \times 0.20\)

Determine the time it takes for the additional insulation to pay for itself

To determine how long it takes for the additional insulation to pay for itself, we need to divide the cost of the insulation kit by the difference in cost before and after installing the insulation: \(t = \frac{\$60}{C_{\text{loss}} - C_{\text{reduced}}}\) Now we can plug in the values to find the answer. \(Q_{\text{loss}}=2\pi(0.22)(2) \times 5.67\times10^{-8} \times 0.7 \left(313^4-293^4\right)= 27.85\,\text{W}\) \(C_{\text{loss}} = \frac{27.85}{3600} \times \$0.08/\text{kWh} = \$0.0062/\text{h}\) \(C_{\text{reduced}} = \$0.0062/\text{h} \times 0.20 = \$0.00124/\text{h}\) \(t = \frac{\$60}{\$0.0062/\text{h} - \$0.00124/\text{h}} \approx 12244 \,\text{hours}\) So, it will take about 12,244 hours for the additional insulation to pay for itself.