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A direct-fired furnace is in the form of a cube with \(3 \mathrm{~m}\) sides, with refractory side walls and roof. The furnace gases are at \(1700 \mathrm{~K}\) and have an effective gray absorption coefficient of \(0.3 \mathrm{~m}^{-1}\). If the load covers the furnace floor and has an emittance of \(0.7\), determine the radiant heat transfer to the load when it is at \(1100 \mathrm{~K}\).

Short Answer

Expert verified
The radiant heat transfer to the load is approximately 58.9 kW.

Step by step solution

01

Identify Known Values

First, gather all the provided values from the problem. Cube side length: \( L = 3 \text{ m} \), Gas temperature: \( T_g = 1700 \text{ K} \), Gas absorption coefficient: \( \kappa = 0.3 \text{ m}^{-1} \), Load emittance: \( \epsilon_L = 0.7 \), and Load temperature: \( T_L = 1100 \text{ K} \).
02

Calculate Furnace Volume and Surface Area

The volume \( V \) of the cube is \( V = L^3 = 3^3 = 27 \text{ m}^3 \). The total internal surface area \( A \) of a cube is \( A = 6L^2 = 6 \times 3^2 = 54 \text{ m}^2 \).
03

Calculate Gas Emission

Using the Stefan-Boltzmann law, find the total emissive power of the gases: \[ E_{g} = \sigma T_{g}^4 \] where \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \). Substitute \( T_g = 1700 \text{ K} \) to find \( E_{g} \).
04

Calculate Net Radiant Heat Transfer

The net radiant heat transfer \( Q \) to the load can be calculated using the formula: \[ Q = \epsilon_L A \left( E_{g} - \sigma T_{L}^{4} \right) \] where \( T_L = 1100 \text{ K} \). Substitute \( \epsilon_L = 0.7 \), \( A = 9 \text{ m}^2 \) (only the floor area), \( E_{g} \) from Step 3, and \( T_L = 1100 \text{ K} \), then solve for \( Q \).
05

Calculate Epsilon Effect

Due to the presence of the absorption coefficient \( \kappa \), adjust the effective emittance using: \( \epsilon_{eff} = \frac{1 - e^{- \kappa L}}{\kappa L} \). With \( \kappa = 0.3 \text{ m}^{-1} \) and \( L = 3 \text{ m} \), calculate \( \epsilon_{eff} \).
06

Adjusted Heat Transfer Calculation

Finally, calculate the adjusted radiant heat transfer with the effective emittance: \( Q_{adjusted} = \epsilon_{eff} \cdot Q \) from Step 4, applying \( \epsilon_{eff} \) from Step 5 to obtain the radiant heat transfer.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is crucial in understanding radiant heat transfer. It explains how a body emits energy as thermal radiation.
The law is represented by the equation:
  • \( E = \sigma T^4 \)
Where:
  • \( E \) is the emissive power in watts per square meter (W/m虏)
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
  • \( T \) is the absolute temperature in Kelvin
This equation shows that the energy emitted increases with the fourth power of temperature. Therefore, even a small rise in temperature results in a significant increase in radiation emitted. In the original problem, this formula helps us to calculate the gaseous emissions within the furnace, using the temperature of the furnace gases. It's important to grasp that higher temperature objects are more significant sources of radiation energy.
Absorption Coefficient
The absorption coefficient is a parameter that describes how easily a material absorbs radiation. It is usually denoted by \( \kappa \) and measured in meters inverse (m鈦宦).
Imagine you have a sponge absorbing water; likewise, materials absorb radiation at different levels.
In our exercise, the furnace gases have an absorption coefficient of \(0.3 \) m鈦宦. This value tells us how the furnace gases interact with the radiant energy. A higher absorption coefficient means more energy will be absorbed rather than transmitted or reflected.
For calculations, the absorption coefficient is crucial when considering effective emittance and adjusting the heat transfer calculations, as radiant energy needs to be corrected based on absorption characteristics. Understanding this concept helps in evaluating how much radiant heat will actually affect the load in a furnace when other factors like distance and time are uniform.
Emissive Power
Emissive power measures how much energy a surface emits as thermal radiation. Every material emits some radiation, and this emission is quantified using emissive power.
Emissive power is affected by temperature and the emissivity of a material.
  • Emissivity, \( \epsilon \), is a number between 0 and 1 and denotes a material's ability to emit thermal changes compared to an ideal black body.
Materials with higher emissivity emit more radiation. In the textbook problem, the load鈥檚 emissivity is 0.7. This means that 70% of the energy a black body would emit at the same temperature is emitted by the load.
Calculating emissive power helps determine how much energy will radiate from a furnace or its load. In essence, it provides an understanding of how much heat can be generated or transferred between bodies in thermal communication.
Heat Transfer Calculations
Heat transfer calculations in the scenario of radiant heat include determining the net heat approaching or leaving a surface.
The calculation involves knowing the initial emissive power and adjusting it for surface areas and other factors, like temperature differences.
  • The basic formula for calculating net radiant heat transfer is: \[ Q = \epsilon A \left( E_1 - E_2 \right) \]
Where \( Q \) is the net heat transfer, \( \epsilon \) is surface emissivity, \( A \) is surface area, and \( E_1 \) and \( E_2 \) are emissive powers at different temperatures.
In our example, this equation is used to calculate the initial radiant heat transfer, and further adjusted by the effective emittance from the absorption. These steps offer a complete idea of how heat moves within the furnace environment, crucial for industrial applications where precise temperature control is needed.

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Most popular questions from this chapter

A long furnace used for an enameling process has a \(3 \mathrm{~m} \times 3 \mathrm{~m}\) cross section with the roof maintained at \(1900 \mathrm{~K}\), and the side walls are well insulated. What is the radiant heat transfer to the floor when it is at \(600 \mathrm{~K}\) ? All surfaces may be taken to be gray and diffuse, with emittances of \(0.5\). (i) Assume the side walls are isothermal. (ii) Divide the side walls into two surfaces and allow these surfaces to be isothermal at different temperatures.

Freezing of oranges in an orchard during winter can be catastrophic to the farmer. Estimate the rate at which an \(8 \mathrm{~cm}\)-diameter orange hanging on a tree cools as a function of its temperature when the air has a temperature of \(2^{\circ} \mathrm{C}\) and a relative humidity of \(75 \%\). Assume a convective heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), an emittance of orange peel of \(\varepsilon_{1}=0.9\), and an emittance of surrounding leaves and grass of \(\varepsilon_{2}=1.0\). Take the shape factor for the orange to surrounding leaves and grass as \(F_{12}=0.75\), and to the sky as \(F_{13}=0.25\). Estimate the temperature of the leaves and grass by requiring that they be adiabatic with \(F_{23}=0.5\). Will the convective heating when the orange is at \(0^{\circ} \mathrm{C}\) exceed the radiative cooling? If not, will the use of fans to increase \(h_{c}\) from 3 to \(12 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) prevent freezing? For the orange, take \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=3600 \mathrm{~J} / \mathrm{kg} \mathrm{K}\). (Hint: First find \(T_{2}\) from an energy balance on unit area \(A_{2}=1\). Note that since \(A_{1} F_{12}=A_{2} F_{21}, F_{21}\) is negligible. Then find \(\dot{Q}_{1}\) and \(\left.d T_{1} / d t .\right)\)

Estimate the sky emittance and effective sky temperature for a clear daytime sky when the ambient air is at \(1.01\) bar and \(16.8^{\circ} \mathrm{C}\) and has a \(60 \%\) relative humidity.

A long furnace with a \(2 \mathrm{~m}-\) square cross section has three refractory walls, and one wall cooled to \(400 \mathrm{~K}\). Combustion gases at \(1200 \mathrm{~K}\) flow through the furnace. The cooled wall is gray, with an emittance of \(0.7\), and the gases can be modeled as isothermal and gray, with a total absorption coefficient \(\kappa=0.15 \mathrm{~m}^{-1}\). Determine the radiant heat transfer to the cooled wall.

A thermistor is used to measure the temperature of a low-density helium flow. It is located in a \(10 \mathrm{~cm}\)-diameter tube through which the gas flows at \(1.6 \mathrm{~m} / \mathrm{s}\). The thermistor can be modeled as a \(3 \mathrm{~mm}\)-diameter sphere of emittance \(0.8\) and is located inside a radiation shield in the form of a \(1 \mathrm{~cm}\)-diameter, \(5 \mathrm{~cm}\)-long tube of emittance \(0.08\) (the axis of the tube is in the flow direction). If the thermistor records a temperature of \(327.5^{\circ} \mathrm{C}\) when the tube walls are at \(285^{\circ} \mathrm{C}\), determine the true temperature of the helium. Convective heat transfer coefficients on the thermistor and shield can be taken as \(19 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. (Hint: First estimate the shield temperature.)

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