Question

In a plant that manufactures canned aerosol paints, the cans are temperature- tested in water baths at \(60^{\circ} \mathrm{C}\) before they are shipped to ensure that they withstand temperatures up to \(55^{\circ} \mathrm{C}\) during transportation and shelving (as shown in Fig. P9-51). The cans, moving on a conveyor, enter the open hot water bath, which is \(0.5 \mathrm{~m}\) deep, $1 \mathrm{~m}\( wide, and \)3.5 \mathrm{~m}$ long, and they move slowly in the hot water toward the other end. Some of the cans fail the test and explode in the water bath. The water container is made of sheet metal, and the entire container is at about the same temperature as the hot water. The emissivity of the outer surface of the container is \(0.7\). If the temperature of the surrounding air and surfaces is \(20^{\circ} \mathrm{C}\), determine the rate of heat loss from the four side surfaces of the container (disregard the top surface, which is open). The water is heated electrically by resistance heaters, and the cost of electricity is \(\$ 0.085 / \mathrm{kWh}\). If the plant operates $24 \mathrm{~h}\( a day 365 days a year and thus \)8760 \mathrm{~h}$ a year, determine the annual cost of the heat losses from the container for this facility.

Short answer

Answer: The annual cost of heat losses from the container for this facility is approximately $8230.69 per year.

Step-by-step solution

Evaluating the container's surface area

To determine the heat loss from the container, we first need to find its surface area. The container is 0.5 m deep, 1 m wide, and 3.5 m long. We only need to find the surface area of the four side surfaces as we disregard the top surface. Surface Area = (2 x depth x length) + (2 x depth x width) Surface Area = (2 x 0.5 x 3.5) + (2 x 0.5 x 1) = 3.5 + 1 = 4.5 m²

Applying Stefan-Boltzmann law

Now, we apply the Stefan-Boltzmann law to calculate the rate of heat loss from the container due to radiation. The Stefan-Boltzmann law is given by: $$ q_{loss} = ε \cdot σ \cdot A \cdot (T_{1}^4 - T_{2}^4) $$ where q_loss is the rate of heat loss, ε is the emissivity, σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)), A is the surface area, T1 is the container's temperature, and T2 is the surrounding temperature. The temperatures must be in Kelvin. First, convert the given temperatures to Kelvin: T1 = 60°C = (60+273.15) K = 333.15 K T2 = 20°C = (20+273.15) K = 293.15 K Now we can plug in the values: $$ q_{loss} = 0.7 \cdot (5.67 \times 10^{-8}) \cdot 4.5 \cdot (333.15^4 - 293.15^4) $$

Calculating the rate of heat loss

Calculate the value of q_loss by evaluating the expression for the rate of heat loss: $$ q_{loss} = 0.7 \cdot (5.67 \times 10^{-8}) \cdot 4.5 \cdot (333.15^4 - 293.15^4) \approx 1096.32 \mathrm{W} $$ The rate of heat loss from the four side surfaces of the container is 1096.32 W.

Calculating the annual cost of heat losses

Now that we have the rate of heat loss, we can calculate the annual cost of the heat losses using the cost of electricity and plant operating hours. $$Cost_{annual} = \frac{q_{loss} \cdot operating \ hours \cdot cost \ per \ kWh}{1000 W/kW} $$ Operating hours = 24 h/day x 365 days/year = 8760 h/year Cost per kWh = \$ 0.085 / kWh Plug in the values: $$Cost_{annual} = \frac{1096.32 \mathrm{W} \cdot 8760 \mathrm{h/year} \cdot \$ 0.085 / \mathrm{kWh}}{1000 \mathrm{W/kW}} $$

Finding the annual cost

Calculate the annual cost of heat losses: $$Cost_{annual} = \frac{1096.32 \mathrm{W} \cdot 8760 \mathrm{h/year} \cdot \$ 0.085 / \mathrm{kWh}}{1000 \mathrm{W/kW}} \approx \$ 8230.69 / \mathrm{year}$$ The annual cost of heat losses from the container for this facility is approximately $8230.69 per year.