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A furnace is in the form of a cube with \(3 \mathrm{~m}\) sides. The roof is heated to \(1100^{\circ} \mathrm{C}\), and the work floor is at \(500^{\circ} \mathrm{C}\). The walls are insulated. The emittance of the roof is \(0.85\). Calculate the radiant heat flux into the floor as a function of its emittance for \(0.2<\varepsilon<1.0\).

Short Answer

Expert verified
The radiant heat flux into the floor depends on its emittance and is given by \( q_{net} = \frac{(0.85 \times 5.67 \times 10^{-8} \times (1373.15)^4 - \varepsilon_{floor} \times 5.67 \times 10^{-8} \times (773.15)^4)}{\frac{1}{\varepsilon_{floor}} + \frac{1}{0.85} - 1} \).

Step by step solution

01

Understanding the Problem

We need to calculate the radiative heat flux from the roof to the work floor of the furnace, which is a cube. We know the temperature of the roof and the floor, both given in degrees Celsius, and we need to include the emittance of both the roof and floor in our calculation.
02

Convert Temperatures to Kelvin

The first step in our calculations involves converting the temperatures from Celsius to Kelvin. The conversion is done by adding 273.15 to the Celsius temperature. Hence, the roof temperature is 1373.15 K (1100 + 273.15) and the floor temperature is 773.15 K (500 + 273.15).
03

Use the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated per unit area is proportional to the fourth power of temperature. For the roof, the power radiated per unit area can be given as \( q = \varepsilon \sigma T^4 \), where \( \varepsilon \) is the emittance, \( \sigma = 5.67 \times 10^{-8} \) W/m虏K鈦 is the Stefan-Boltzmann constant, and \( T \) is the temperature in Kelvin.
04

Calculate Heat Flux from Roof

Using the Stefan-Boltzmann law for the roof, we have \( q_r = 0.85 \times 5.67 \times 10^{-8} \times (1373.15)^4 \) W/m虏. This gives the heat flux from the heated roof.
05

Calculate Net Radiative Heat Flux to Floor

The net heat flux to the work floor is determined by the difference in radiation incoming from the roof and the radiation emitted by the floor: \[ q_{net} = \frac{(q_r - q_f)}{\frac{1}{\varepsilon_{floor}} + \frac{1}{\varepsilon_{roof}} - 1} \]where \( q_f = \varepsilon_{floor} \sigma (773.15)^4 \), and \( \varepsilon_{floor} \) is the variable emittance of the floor.
06

Formulate as a Function of Floor Emittance

Express the radiative heat flux as a continuous function of the floor emittance \( \varepsilon_{floor} \):\[ q_{net} = \frac{0.85 \times 5.67 \times 10^{-8} \times (1373.15)^4 - \varepsilon_{floor} \times 5.67 \times 10^{-8} \times (773.15)^4}{\frac{1}{\varepsilon_{floor}} + \frac{1}{0.85} - 1} \]
07

Evaluate for Specified Emittance Range

Calculate \( q_{net} \) using the formula derived in Step 6 for emittance values ranging between 0.2 and 1.0.

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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 a principal concept in radiative heat transfer. It describes how much energy an object emits as heat, based on its temperature. To put it simply, the law states that an object's energy radiated per unit area is proportional to the fourth power of its absolute temperature. This is mathematically expressed as:
  • \( q = \varepsilon \sigma T^4 \),
  • where \( q \) represents the energy emitted,
  • \( \varepsilon \) is the emittance of the object,
  • \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \,\mathrm{W/m^2K^4}) \),
  • and \( T \) is the temperature in Kelvin.
This law is essential because it helps in calculating how much heat the roof of a furnace emits towards the floor. In our example, the roof has an emittance of 0.85, a figure which denotes how efficiently it radiates heat compared to a perfect black body. This factor, combined with the roof鈥檚 absolute temperature, allows us to use the Stefan-Boltzmann Law to find the heat flux produced by the furnace roof.
emittance
Emittance, often denoted by \( \varepsilon \), is a measure of an object鈥檚 ability to emit energy as thermal radiation. This value ranges between 0 and 1, with
  • 0 indicating a perfect reflector (no emission) and 1 indicating a perfect emitter (or black body)
  • An emittance of 0.85, as with our furnace roof, shows that it emits 85% as much energy as a perfect black body would at the same temperature.
The concept of emittance is vital in radiative heat transfer calculations. It affects how much energy an object emits compared to its potential maximum at any given temperature.In scenarios such as an industrial furnace, understanding the emittance of various surfaces鈥攍ike the heated roof and the work floor鈥攈elps calculate the net heat flow. Essentially, the emittance of a material determines how much it will absorb and emit radiation, crucial for optimizing heating processes and ensuring efficient energy usage in industrial applications.
Kelvin temperature scale
The Kelvin temperature scale is fundamental in scientific calculations, particularly when dealing with thermodynamics and heat transfer. This scale starts at absolute zero, the point where no thermal energy remains in a substance.When working with calculations involving the Stefan-Boltzmann Law, temperatures must be converted to Kelvin. This is because temperature values in Kelvin represent absolute thermodynamic temperatures, ensuring more accurate results in equations.To convert a Celsius temperature to Kelvin, simply add 273.15. For example:
  • The roof temperature in our exercise, initially at \( 1100^{\circ} \mathrm{C} \), becomes \( 1373.15 \mathrm{K} \).
  • The floor temperature of \( 500^{\circ} \mathrm{C} \) converts to \( 773.15 \mathrm{K} \).
Using Kelvin in heat transfer calculations ensures consistency and precision, enabling proper application of the Stefan-Boltzmann Law for accurate heat flux predictions.
furnace heat calculation
Furnace heat calculations involve determining how much energy is transferred between the furnace roof and the floor, critical for understanding and optimizing the system's energy efficiency.First, we calculate the radiative heat emitted by the roof using the Stefan-Boltzmann Law framework: \[ q_r = 0.85 \times 5.67 \times 10^{-8} \times (1373.15)^4 \mathrm{W/m^2} \] This sets the baseline heat flux value from the roof.Next is calculating the net radiative flux to the floor. This involves subtracting the floor's emitted radiation from the roof鈥檚, with adjustments based on emittance factors. The formula considers
  • the floor's temperature,
  • its variable emittance, \( \varepsilon_{floor} \),
resulting in: \[q_{net} = \frac{0.85 \times 5.67 \times 10^{-8} \times (1373.15)^4 - \varepsilon_{floor} \times 5.67 \times 10^{-8} \times (773.15)^4}{\frac{1}{\varepsilon_{floor}} + \frac{1}{0.85} - 1} \] Finally, evaluating this formula for different floor emittance values between 0.2 and 1.0 provides a clear understanding of radiative heat flow possibilities, essential for adjusting operational parameters for optimal performance.

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Most popular questions from this chapter

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 solar heater for a swimming pool consists of an unglazed copper plate \(0.5 \mathrm{~mm}\) thick with a selective coating (solar absorptance \(=0.92\), total hemispherical emittance \(=0.15\) ). On the back side of the plate are attached \(12 \mathrm{~mm}-\mathrm{O} . \mathrm{D} ., 1\) mm-wall-thickness copper tubes that are manifolded at the inlet and outlet to form a parallel tube matrix. The tubes have a centerline spacing of \(20 \mathrm{~cm}\) and make perfect thermal contact with the plate. Water at \(25^{\circ} \mathrm{C}\) is constantly recirculated from the pool at a flow rate per tube of \(0.5\) gallons/minute. On a given day the air temperature is \(20^{\circ} \mathrm{C}\), the air speed averages \(4 \mathrm{~m} / \mathrm{s}\) across a \(3 \mathrm{~m}\)-wide array of solar collectors, the solar irradiation is \(790 \mathrm{~W} / \mathrm{m}^{2}\), and the long wavelength radiation from the sky is about \(340 \mathrm{~W} / \mathrm{m}^{2}\). Determine the tube length required to supply water back to the pool at \(29^{\circ} \mathrm{C}\). Assume that the heat loss due to conduction through the back insulation is \(5 \%\) of the solar irradiation. Take \(\operatorname{Re}_{t r}=10^{5}\). (Hint: Appropriate assumptions will allow an analytical solution to be obtained.)

A multilayer insulation (MLI) to be used on a spacecraft consists of \(25 \mu \mathrm{m}\) Kapton external foil sheets ( \(\alpha_{s}=0.35, \varepsilon=0.60\) ), aluminum-coated on the inside ( \(\varepsilon=0.03\) ), sandwiching three aluminized Mylar sheets ( \(\varepsilon=0.03\) ). The layers are separated by Dacron mesh to prevent contact of the sheets. If a MLI shield is faced to the Sun at \(1 \mathrm{AU}\) distance and the inner surface is maintained at \(300 \mathrm{~K}\), determine the heat flux through the MLI and the outer surface temperature.

Calculate the blackbody emissive power \(E_{b}\) for surfaces at \(300 \mathrm{~K}, 1000 \mathrm{~K}, 2000 \mathrm{~K}\), and \(5000 \mathrm{~K}\). Also estimate the peak wavelength \(\lambda_{\max }\) for each surface using: (i) \(\lambda_{\max } T=2898 \mu \mathrm{m} \mathrm{K}\) (from Eq. \(6.6\) and Fig. \(6.3 a\) ) (ii) \(\lambda_{\max } T=5100 \mu \mathrm{m} \mathrm{K}\) (from Fig. 6.3b)

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.

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