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Chapter 9: Problem 143

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 \(0.85\). If the cost of natural gas is \(\$ 1.15 /\) therm and the furnace operates \(2800 \mathrm{~h}\) per year, determine the annual cost of this heat loss to the plant. Evaluate properties of air at a film temperature of \(107.5^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Short Answer

Expert verified
Answer: To find the highest allowable surface temperature of the furnace, we need to equate the heat loss through radiation and natural convection with the maximum heat loss allowed. After solving for the surface temperature, we can calculate the annual cost of heat loss using the given parameters. The highest allowable surface temperature will need to be calculated using trial and error or iterative methods, and the annual cost of heat loss for this furnace is approximately $1269.336.

Step by step solution

01

Determine the heat generated by burning natural gas

First, we will find the heat generated in the furnace. The furnace burns 48 therms/h of natural gas, and the combustion efficiency is 82%, so the heat generated can be calculated as follows: Heat generated = Combustion efficiency * Natural gas burned rate Heat generated = 0.82 * 48 therms/h Heat generated = 39.36 therms/h
02

Calculate the maximum heat loss allowed

The heat loss from the outer surfaces of the furnace should not exceed 1% of the heat generated inside. So we can calculate the maximum heat loss allowed as follows: Max heat loss allowed = Heat generated * 1% Max heat loss allowed = 39.36 therms/h * 0.01 Max heat loss allowed = 0.3936 therms/h
03

Calculate the heat loss through radiation and natural convection

We are given that the emissivity of the outer surface of the furnace is 0.85 and the temperature of the air and wall surface is 75°F. We can write the heat loss through radiation and natural convection as follows: Heat loss = Emissivity * Stefan-Boltzmann constant * Surface area * (T_furnace^4 - T_ambient^4) + h * Surface area * (T_furnace - T_ambient) where h is the convection heat transfer coefficient, T_furnace is the surface temperature of the furnace, and T_ambient is the ambient temperature.
04

Calculate the highest allowable surface temperature of the furnace

To find the highest allowable surface temperature, we will equate the heat loss with the maximum heat loss allowed: 0.3936 therms/h = 0.85 * (5.67 * 10^(-8){W/ (m^2 *K^4)}) * 2 * pi * (8/2 ft * 0.3048 m/ft) * 13 ft * 0.3048 m/ft * (T_furnace^4 - T_ambient^4) + h * 2 * pi * (8/2 ft * 0.3048 m/ft) * 13 ft * 0.3048 m/ft * (T_furnace - T_ambient) Now we can solve for the highest allowable surface temperature, T_furnace, using trial and error or iterative methods considering the properties of air at a film temperature of 107.5°F and 1 atm pressure.
05

Calculate the annual cost of heat loss

The annual cost of heat loss can be calculated as follows: Annual heat loss = Max heat loss allowed * operating hours per year * cost of natural gas Annual heat loss = 0.3936 therms/h * 2800 h * $1.15/therm Annual heat loss = $1269.336
06

Evaluate the assumption

We assumed the properties of air at a film temperature of 107.5°F and 1 atm pressure to find the highest allowable surface temperature. Since the air temperature in the room is 75°F which is close to the film temperature, this assumption is reasonable and can be considered valid for solving the problem.

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Most popular questions from this chapter

Flat-plate solar collectors are often tilted up toward the sun in order to intercept a greater amount of direct solar radiation. The tilt angle from the horizontal also affects the rate of heat loss from the collector. Consider a \(1.5-\mathrm{m}\)-high and 3-m-wide solar collector that is tilted at an angle \(\theta\) from the horizontal. The back side of the absorber is heavily insulated. The absorber plate and the glass cover, which are spaced \(2.5 \mathrm{~cm}\) from each other, are maintained at temperatures of \(80^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss from the absorber plate by natural convection for \(\theta=0^{\circ}, 30^{\circ}\), and \(90^{\circ}\).

Under what conditions does natural convection enhance forced convection, and under what conditions does it hurt forced convection?

The density of liquid water can be correlated as \(\rho(T)=1000-0.0736 T-0.00355 T^{2}\) where \(\rho\) and \(T\) are in \(\mathrm{kg} / \mathrm{m}\) and \({ }^{\circ} \mathrm{C}\), respectively. Determine the volume expansion coefficient at \(70^{\circ} \mathrm{C}\). Compare the result with the value tabulated in Table A-9.

A vertical \(0.9\)-m-high and \(1.8\)-m-wide double-pane window consists of two sheets of glass separated by a \(2.2-\mathrm{cm}\) air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be \(20^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), the rate of heat transfer through the window is (a) \(19.8 \mathrm{~W}\) (b) \(26.1 \mathrm{~W}\) (c) \(30.5 \mathrm{~W}\) (d) \(34.7 \mathrm{~W}\) (e) \(55.0 \mathrm{~W}\) (For air, use \(k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7296, v=\) \(1.562 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Also, the applicable correlation is \(\mathrm{Nu}\) \(\left.=0.42 \mathrm{Ra}^{1 / 4} \mathrm{Pr}^{0.012}(H / L)^{-0.3}\right)\) (For air, use \(k=0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

A vertical double-pane window consists of two sheets of glass separated by a \(1.5-\mathrm{cm}\) air gap at atmospheric pressure. The glass surface temperatures across the air gap are measured to be \(278 \mathrm{~K}\) and \(288 \mathrm{~K}\). If it is estimated that the heat transfer by convection through the enclosure is \(1.5\) times that by pure conduction and that the rate of heat transfer by radiation through the enclosure is about the same magnitude as the convection, the effective emissivity of the two glass surfaces is (a) \(0.47\) (b) \(0.53\) (c) \(0.61\) (d) \(0.65\) (e) \(0.72\)
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