Question

During a plant visit, it was observed that a \(1.5\)-m-high and \(1-\mathrm{m}\)-wide section of the vertical front section of a natural gas furnace wall was too hot to touch. The temperature measurements on the surface revealed that the average temperature of the exposed hot surface was \(110^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(25^{\circ} \mathrm{C}\). The surface appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(25^{\circ} \mathrm{C}\) also, determine the rate of heat loss from this furnace. The furnace has an efficiency of 79 percent, and the plant pays \(\$ 1.20\) per therm of natural gas. If the plant operates \(10 \mathrm{~h}\) a day, 310 days a year, and thus \(3100 \mathrm{~h}\) a year, determine the annual cost of the heat loss from this vertical hot surface on the front section of the furnace wall.

Short answer

Answer: The annual cost of heat loss from the vertical hot surface on the front section of the furnace wall is $359.56.

Step-by-step solution

Heat transfer by radiation

To calculate the heat transfer by radiation, we will use the Stefan-Boltzmann law: \(q_{rad} = A\epsilon\sigma(T_s^4 - T_{sur}^4)\) where \(A\) is the area of the surface, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8} \mathrm{W/m^2K^4}\)), \(T_s\) is the surface temperature, and \(T_{sur}\) is the surrounding surface temperature. Given the dimensions of the surface, we can find its area: \(A = (1.5 \mathrm{m})(1 \mathrm{m}) = 1.5 \mathrm{m^2}\) Next, we convert the surface and surrounding temperatures to Kelvin: \(T_s = 110^{\circ}\mathrm{C} + 273.15 = 383.15 \mathrm{K}\) \(T_{sur} = 25^{\circ}\mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Now we can calculate the heat transfer by radiation: \(q_{rad} = (1.5 \mathrm{m^2})(0.7)(5.67\times10^{-8} \mathrm{W/m^2K^4})(383.15^4 - 298.15^4 \mathrm{K^4}) = 963.86 \mathrm{W}\).

Heat transfer by convection

We will use the following formula for heat transfer by convection: \(q_{conv} = hA(T_s - T_\infty)\) where \(h\) is the heat transfer coefficient, \(A\) is the area, \(T_s\) is the surface temperature, and \(T_\infty\) is the air temperature. For this problem, we can assume a common value for the heat transfer coefficient (natural convection): \(h = 10 \mathrm{W/m^2K}\) Now we can calculate the heat transfer by convection: \(q_{conv} = (10 \mathrm{W/m^2K})(1.5 \mathrm{m^2})(383.15 \mathrm{K} - 298.15 \mathrm{K}) = 1275 \mathrm{W}\).

Total heat loss

Now we can find the total heat loss by adding the heat losses due to radiation and convection: \(q_{total} = q_{rad} + q_{conv} = 963.86 \mathrm{W} + 1275 \mathrm{W} = 2238.86 \mathrm{W}\)

Annual energy loss in therms

First, let's find the total operation time in seconds and the energy loss in joules: \(t = 3100 \mathrm{h}\times3600 \mathrm{s/h} = 11160000 \mathrm{s}\) \(E_{loss} = q_{total}\times t = (2238.86 \mathrm{W})(11160000 \mathrm{s}) = 2.498\times10^{10} \mathrm{J}\) Now we can find the energy loss in therms: \(1 \mathrm{therm} = 1.055\times10^8 \mathrm{J}\) \(E_{loss} = \frac{2.498\times10^{10} \mathrm{J}}{1.055\times10^8 \mathrm{J/therm}} = 236.8\mathrm{~therms}\)

Annual cost of heat loss

Finally, we can calculate the annual cost of the heat loss using the given cost per therm and the furnace efficiency: \(C_{annual} = \frac{236.8\mathrm{~therms}}{0.79}\times\$1.20\mathrm{/therm} = \$359.56\) Therefore, the annual cost of the heat loss from this vertical hot surface on the front section of the furnace wall is \(\$359.56\).