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In one stage of an annealing process, 304 stainless steel sheet is taken from \(300 \mathrm{~K}\) to \(1250 \mathrm{~K}\) as it passes through an electrically heated oven at a speed of \(V_{s}=10 \mathrm{~mm} / \mathrm{s}\). The sheet thickness and width are \(t_{s}=8 \mathrm{~mm}\) and \(W_{s}=2 \mathrm{~m}\), respectively, while the height, width, and length of the oven are \(H_{o}=2 \mathrm{~m}\), \(W_{o}=2.4 \mathrm{~m}\), and \(L_{o}=25 \mathrm{~m}\), respectively. The top and four sides of the oven are exposed to ambient air and large surroundings, each at \(300 \mathrm{~K}\), and the corresponding surface temperature, convection coefficient, and emissivity are \(T_{s}=350 \mathrm{~K}, h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(\varepsilon_{s}=0.8\). The bottom surface of the oven is also at \(350 \mathrm{~K}\) and rests on a \(0.5\)-m-thick concrete pad whose base is at \(300 \mathrm{~K}\). Estimate the required electric power input, \(P_{\text {elec }}\), to the oven.

Short Answer

Expert verified
The required electric power input to the oven can be estimated using the following steps: 1. Calculate the oven's surface area for heat transfer: \(A_{total} = 4 \times (2 \times 2.4) + 2.4 \times 25 = 316.8 \, m^2\) 2. Calculate the heat loss due to convection: \(Q_{conv} = 10 \times 316.8 \times (350 - 300) = 158400 \, W\) 3. Calculate the heat loss due to radiation: \(Q_{rad} = 0.8 \times 5.67 \times 10^{-8} \times 316.8 \times (350^4 - 300^4) = 142439.84 \, W\) 4. Calculate the total heat loss: \(Q_{total} = 158400 + 142439.84 = 300839.84 \, W\) 5. Determine the heating capacity of the oven: \(m = 8000 \times 10 \times 10^{-3} \times 8 \times 10^{-3} \times 2 = 0.16 \, kg/s\), \(Q_{capacity} = 0.16 \times 480 \times (1250-300) = 122880 \, W\) 6. Estimate the required electric power input: \(P_{elec} = Q_{capacity} + Q_{total} = 122880 + 300839.84 = 423719.84 \, W\) Therefore, the required electric power input to the oven is approximately \(423.72 \, kW\).

Step by step solution

01

1. Calculate the oven's surface area for heat transfer

By considering the geometry of the oven, we can calculate the oven surface area for heat transfer exposed to ambient air. The sides, top, and bottom of the oven are exposed to ambient air, making up five rectangular surfaces for heat transfer. \(A_{total} = 4 \times (H_o \times W_o) + W_o \times L_o\) where \(H_o\) is the height of the oven, \(W_o\) is the width of the oven, and \(L_o\) is the length of the oven.
02

2. Calculate the heat loss due to convection

By using the convection heat transfer equation, we can determine the heat loss due to convection: \(Q_{conv} = h \times A_{total} \times (T_s - T_{air})\) where \(h\) is the convection coefficient, \(A_{total}\) is the total surface area calculated in step 1, \(T_s\) is the surface temperature of the oven, and \(T_{air}\) is the temperature of the ambient air.
03

3. Calculate the heat loss due to radiation

To determine the heat loss due to radiation, we utilize the radiation heat transfer equation: \(Q_{rad} = \varepsilon_s \times \sigma \times A_{total} \times (T_s^4 - T_{air}^4)\) where \(\varepsilon_s\) is the emissivity of the oven surface, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W/m^2 K^4\)), \(A_{total}\) is the total surface area calculated in step 1, \(T_s\) is the surface temperature of the oven, and \(T_{air}\) is the temperature of the ambient air.
04

4. Calculate the total heat loss

By adding the heat losses due to convection and radiation, we can determine the total heat loss from the oven: \(Q_{total} = Q_{conv} + Q_{rad}\)
05

5. Determine the heating capacity of the oven

To calculate the heating capacity required for the oven to heat the stainless sheet from 300K to 1250K, we can use the equation: \(Q_{capacity} = m \cdot c_p \cdot \Delta T\) where m is the mass flow rate of the steel sheet, \(c_p\) is the specific heat capacity of 304 stainless steel (approximately \(480 J / kg \cdot K\)), and \(\Delta T\) is the temperature difference between the final and initial temperatures. To find the mass flow rate (m), we can use the following: \(m = \rho \cdot V_s \cdot t_s \cdot W_s\) where \(\rho\) is the density of 304 stainless steel (approximately \(8000 kg/m^3\)), \(V_s\) is the speed of the stainless sheet, \(t_s\) is the thickness of the sheet, and \(W_s\) is the width of the sheet.
06

6. Estimate the required electric power input

Finally, knowing the heating capacity and the total heat loss, we can estimate the required electric power input to the oven: \(P_{elec} = Q_{capacity} + Q_{total}\) Calculate each term using the provided values and perform the necessary calculations to estimate the required electric power input to the oven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection Heat Loss
The transfer of heat through a fluid – such as air or a liquid – is known as convection. In the context of the annealing oven described in the exercise, convection heat loss occurs because the oven is hotter than the surrounding air. The cooler air around the oven absorbs heat from the oven surfaces, leading to a loss of thermal energy.

Mathematically, convection can be expressed with Newton's law of cooling, which is given by the relation:
\[Q_{conv} = h \times A_{total} \times (T_s - T_{air})\]
where:\[h\] is the convection coefficient that quantifies the convective heat transfer rate per unit area per unit temperature difference, \[A_{total}\] is the total surface area through which convection occurs, \[T_s\] is the surface temperature, and \[T_{air}\] is the ambient air temperature.

The value of \[h\] depends on various factors such as fluid properties, fluid velocity, and surface characteristics. The steps in the provided solution involve calculating the total surface area and using that with the given convection coefficient to quantify the convection heat loss.
Radiation Heat Loss
Radiation is another form of heat transfer, occurring through electromagnetic waves, and does not require a medium. This means radiation can happen through a vacuum. All objects emit and absorb infrared radiation according to their temperature.

The amount of radiant energy emitted by a surface can be calculated using the Stefan-Boltzmann law, which is stated as:
\[Q_{rad} = \varepsilon_s \times \sigma \times A_{total} \times (T_s^4 - T_{air}^4)\]
In this equation:\[\varepsilon_s\] is the emissivity, representing how effectively a surface emits thermal radiation, \[\sigma\] is the Stefan-Boltzmann constant, \[A_{total}\] is the area from which radiation is emitted, and \[T_s^4 - T_{air}^4\] denotes the difference in the fourth power of the absolute temperatures of the surface and the ambient air. This equation encapsulates the radiative heat loss for the annealing oven.
Heating Capacity
Heating capacity refers to the amount of thermal energy needed to raise the temperature of an object. It is a crucial concept when determining the energy requirement for heating processes like the annealing of steel sheets. The thermal energy required depends on the material properties and the extent of the temperature increase.

To calculate heating capacity, we use the specific heat formula:
\[Q_{capacity} = m \cdot c_p \cdot \Delta T\]
where \[m\] is the mass of the object, \[c_p\] is the specific heat capacity of the material (the amount of energy needed to raise one kilogram of the material by one degree Celsius), and \[\Delta T\] represents the change in temperature.

For continuous processes, we consider the rate of mass flow through the system, which involves the material density and the speed at which the steel sheet passes through the oven.
Electric Power Input Estimation
The amount of electric power input needed for a heating process is estimated to ensure the equipment can supply sufficient energy to maintain the required temperature profile. It involves considering both the heating capacity needed to raise the material's temperature and the heat losses due to convection and radiation.

The required electric power input, denoted by \(P_{elec}\), is the sum of the heating capacity and the total heat loss:
\[P_{elec} = Q_{capacity} + Q_{total}\]
Where \[Q_{total}\] is the sum of convective and radiative heat losses. Estimating the electric power input correctly is vital for achieving the desired material processing, energy efficiency, and cost-effectiveness in industrial heating applications like the annealing process.

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Most popular questions from this chapter

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a temperature of \(T_{\infty}=T_{\text {sur }}=5^{\circ} \mathrm{C}\). The duct is \(15 \mathrm{~m}\) long, and its cross section is \(350 \mathrm{~mm} \times 200 \mathrm{~mm}\). (a) For an uninsulated duct whose average surface temperature is \(50^{\circ} \mathrm{C}\), estimate the rate of heat loss from the duct. The surface emissivity and convection coefficient are approximately \(0.5\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (b) If heated air enters the duct at \(58^{\circ} \mathrm{C}\) and a velocity of \(4 \mathrm{~m} / \mathrm{s}\) and the heat loss corresponds to the result of part (a), what is the outlet temperature? The density and specific heat of the air may be assumed to be \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

Following the hot vacuum forming of a paper-pulp mixture, the product, an egg carton, is transported on a conveyor for \(18 \mathrm{~s}\) toward the entrance of a gas-fired oven where it is dried to a desired final water content. Very little water evaporates during the travel time. So, to increase the productivity of the line, it is proposed that a bank of infrared radiation heaters, which provide a uniform radiant flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), be installed over the conveyor. The carton has an exposed area of \(0.0625 \mathrm{~m}^{2}\) and a mass of \(0.220 \mathrm{~kg}, 75 \%\) of which is water after the forming process. The chief engineer of your plant will approve the purchase of the heaters if they can reduce the water content by \(10 \%\) of the total mass. Would you recommend the purchase? Assume the heat of vaporization of water is \(h_{f g}=2400 \mathrm{~kJ} / \mathrm{kg}\).

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a \(1 \mathrm{~m} \times 3 \mathrm{~m}\) window? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Most of the energy we consume as food is converted to thermal energy in the process of performing all our bodily functions and is ultimately lost as heat from our bodies. Consider a person who consumes \(2100 \mathrm{kcal}\) per day (note that what are commonly referred to as food calories are actually kilocalories), of which \(2000 \mathrm{kcal}\) is converted to thermal energy. (The remaining \(100 \mathrm{kcal}\) is used to do work on the environment.) The person has a surface area of \(1.8 \mathrm{~m}^{2}\) and is dressed in a bathing suit. (a) The person is in a room at \(20^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). At this air temperature, the person is not perspiring much. Estimate the person's average skin temperature. (b) If the temperature of the environment were \(33^{\circ} \mathrm{C}\), what rate of perspiration would be needed to maintain a comfortable skin temperature of \(33^{\circ} \mathrm{C}\) ?

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