Question

Consider an industrial furnace that resembles a 13-ft-long horizontal cylindrical enclosure \(8 \mathrm{ft}\) in diameter whose end surfaces are well insulated. The furnace burns natural gas at a rate of 48 therms/h. The combustion efficiency of the furnace is 82 percent (i.e., 18 percent of the chemical energy of the fuel is lost through the flue gases as a result of incomplete combustion and the flue gases leaving the furnace at high temperature). If the heat loss from the outer surfaces of the furnace by natural convection and radiation is not to exceed 1 percent of the heat generated inside, determine the highest allowable surface temperature of the furnace. Assume the air and wall surface temperature of the room to be \(75^{\circ} \mathrm{F}\), and take the emissivity of the outer surface of the furnace to be \(0.85\).

Short answer

Given an industrial furnace with dimensions, heat generation, and combustion efficiency, find the highest allowable surface temperature. Using the given data and formulas, we calculate the heat generation rate, heat loss limit, and the outer surface area of the furnace. With this information, we then calculate the heat transfer coefficient by natural convection. Finally, we determine the highest allowable surface temperature using the heat loss limit, surface area, and heat transfer coefficient. The highest allowable surface temperature of the furnace is approximately \(551.54^{\circ}\mathrm{F}\).

Step-by-step solution

Determine the heat generation rate inside the furnace.

The furnace burns 48 therms/h of natural gas, and its combustion efficiency is 82 percent. So, 18 percent of the energy is lost due to incomplete combustion and flue gases leaving the furnace at a high temperature. Thus, we can determine the heat generation rate as follows: \(q_\text{gen} = (0.82 \times 48 \, \text{therms/h}) \times 100000 \, \text{Btu/therm}\) \(q_\text{gen} \approx 3936000 \, \text{Btu/h}\)

Calculate the heat loss limit from the outer surface.

We are given that the heat loss by natural convection and radiation from the outer surface must not exceed 1 percent of the heat generated inside. So, we can determine the heat loss limit: \(q_\text{loss} = 0.01 \times q_\text{gen}\) \(q_\text{loss} \approx 39360 \, \text{Btu/h}\)

Determine the outer surface area of the furnace.

The furnace has a cylindrical shape with a diameter of 8 ft and length of 13 ft. We can calculate the surface area as follows: \(A_\text{surface} = \pi D L\) \(A_\text{surface} = \pi (8)(13) \approx 327.17 \, \text{ft}^2\)

Calculate the heat transfer coefficient by natural convection.

We are given that the air and wall surface temperature of the room are \(75^{\circ} \mathrm{F}\). We'll assume an average heat transfer coefficient for natural convection: \(h \approx 1.5 \, \text{Btu/(h.ft².°F)}.\)

Calculate the highest allowable surface temperature.

Now, we can calculate the highest allowable surface temperature (\(T_\text{surface}\)) of the furnace using the heat loss limit, surface area, and heat transfer coefficient: \(q_\text{loss} = h \cdot A_\text{surface} \cdot (T_\text{surface} - T_\text{room})\) \(39360 = 1.5 \cdot 327.17 \cdot (T_\text{surface} - 75)\) \(T_\text{surface} = \frac{39360}{(1.5 \cdot 327.17)} + 75 \approx 551.54^{\circ}\mathrm{F}\) Therefore, the highest allowable surface temperature of the furnace is approximately \(551.54^{\circ}\mathrm{F}\).