Question

Consider a \(1.5\)-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 whose average temperature is $27^{\circ} \mathrm{C}$, and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and $12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The tank is placed in another \(46-\mathrm{cm}\)-diameter sheet metal tank of negligible thickness, and the space between the two tanks is filled with foam insulation $(k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. The thermal resistances of the water tank and the outer thin sheet metal shell are very small and can be neglected. The price of electricity is \(\$ 0.08 / \mathrm{kWh}\), and the homeowner pays \(\$ 280\) a year for water heating. Determine the fraction of the hot-water energy cost of this household that is due to the heat loss from the tank. Hot-water tank insulation kits consisting of 3 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) large enough to wrap the entire tank are available in the market for about \(\$ 30\). If such an insulation is installed on this water tank by the homeowner himself, how long will it take for this additional insulation to pay for itself?

Short answer

Now, we can calculate the heat loss rate: $$ \dot{Q} = \frac{5 \cdot 1.885 (80 - 20)}{(5/10)(1.885/2.670) + (1)} $$ #tag_title#Step 2: Calculate the cost of heat loss per year and the fraction of total hot-water energy cost#tag_content#To find the cost of heat loss per year, we multiply the heat loss rate by the number of hours in a year and the cost of electricity: $$ \text{Cost per year} = \dot{Q} \cdot \text{Hours per year} \cdot \text{Cost of electricity} $$ The fraction of total hot-water energy cost can be calculated using the following formula: $$ \text{Fraction of hot-water energy cost} = \frac{\text{Cost of heat loss per year}}{\text{Total hot-water energy cost per year}} $$ #tag_title#Step 3: Determine the new heat loss rate after installing the additional insulation layer#tag_content#After installing the additional insulation layer, we need to calculate the new heat loss rate. We first need to calculate the new thermal resistance of insulation: $$ R_{new\ insulation} = R_{insulation} + \frac{L_{additional}}{k_{additional}} $$ Next, we can calculate the new heat loss rate using the same formula as in Step 1, but with the updated \(R_{new\ insulation}\). $$ \dot{Q}_{new} = \frac{h_{i} A_{i}(T_{s}- T_{w})}{(h_{i}/h_{o})(A_{i}/A_{o}) + (R_{new\ insulation})} $$ #tag_title#Step 4: Calculate the payback time for the insulation kit#tag_content#To calculate the payback time, we first need to determine the annual energy savings: $$ \text{Annual energy savings} = (\dot{Q} - \dot{Q}_{new}) \cdot \text{Hours per year} \cdot \text{Cost of electricity} $$ Next, we can find the payback time by dividing the insulation kit's initial cost by the annual energy savings: $$ \text{Payback time} = \frac{\text{Initial cost of insulation kit}}{\text{Annual energy savings}} $$ Finally, the payback time is calculated, and we can determine whether the insulation kit is worth the investment.

Step-by-step solution

Calculate the heat loss rate from the tank without insulation kit

First, we need to find the heat loss rate from the tank without an insulation kit. We can use the formula for the heat loss rate through a cylindrical wall: $$ \dot{Q} = \frac{h_{i} A_{i}(T_{s}- T_{w})}{(h_{i}/h_{o})(A_{i}/A_{o}) + (R_{insulation})} $$ Where, \(\dot{Q}\) is the heat loss rate (W) \(h_{i}\) and \(h_{o}\) are the inner and outer heat transfer coefficients (\(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(A_{i}\) and \(A_{o}\) are the inner and outer surface areas of the tank (m²) \(T_{s}\) and \(T_{w}\) are the tank surface and room temperatures (°C) \(R_{insulation}\) is the thermal resistance of insulation (m²·K/W) First, we calculate the inner and outer surface areas of the tank: $$A_{i} = 2 \pi r_{i} h = 2 \pi (0.2 \text{ m})(1.5 \text{ m})$$ $$A_{o} = 2 \pi r_{o} h = 2 \pi (0.23 \text{ m})(1.5 \text{ m})$$ Next, we find the thermal resistance of insulation: $$R_{insulation} = \frac{L}{k} = \frac{0.03}{0.03}$$