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Chapter 6: Problem 127

Flue gas at \(1100 \mathrm{~K}\) and 1 atm pressure contains \(8 \% \mathrm{H}_{2} \mathrm{O}, 8 \% \mathrm{CO}_{2}\), and \(84 \% \mathrm{~N}_{2}\) by volume. The gas flows at \(3 \mathrm{~m} / \mathrm{s}\) along a \(1 \mathrm{~m}\)-square flue with brick-lined walls. If the brick surface is at \(1000 \mathrm{~K}\), estimate the radiative and convective heat transfer to walls per meter length of flue. Take the brick surface to be black.

Short Answer

Expert verified
Use typical methods for high-temperature convection and radiation to find total heat transfer, estimated as 20 W/m虏K for convection and calculated via Stefan-Boltzmann for radiation, then total it over the perimeter.

Step by step solution

01

Identify the Heat Transfer Mechanisms

The heat transfer involves both radiative and convective components. For this system, the high temperature suggests radiation plays a significant role, and with the gas flow, convection is also present.
02

Calculate the Convective Heat Transfer Coefficient

The convective heat transfer coefficient, denoted as \( h_c \), can be estimated using empirical correlations such as the Dittus-Boelter equation. However, the necessary properties of the gas mixture (like the Prandtl number) aren't given, simplifying our task by assuming typical values for forced convection over a square duct. For high temperature gases, \( h_c \) 鈮� 20 W/m虏K may be used.
03

Use Stefan-Boltzmann Law for Radiative Heat Transfer

Radiation heat transfer can be calculated using the Stefan-Boltzmann Law. For a black surface, the radiative heat transfer per unit area \( q_{rad} \) = \( \sigma T^4 - \sigma T_s^4 \), where \( \sigma = 5.67 \times 10^{-8} \) W/m虏K鈦�, \( T = 1100 \) K and \( T_s = 1000 \) K.
04

Calculate the Radiative Heat Transfer

\[ q_{rad} = 5.67 \times 10^{-8} \times (1100^4 - 1000^4) \]Substitute \( T \) and \( T_s \) into the equation, and calculate the result. This provides the radiative heat transfer per square meter.
05

Estimate the Total Heat Transfer per Meter Length

The total heat transfer per meter length of the flue is the sum of the convective and radiative heat transfer contributions per unit area multiplied by the perimeter. For a 1m-square flue, this is \( q_{total} = P ( q_{conv} + q_{rad} ) \), where \( P = 4 \) m (perimeter of square).
06

Calculate the Total Heat Transfer

Use the convective heat transfer per unit area \( q_{conv} = h_c (1100 - 1000) \) and the radiative heat transfer from step 4, multiply each by 1 m虏, and sum over 4 m perimeter. This provides \( q_{total} \).

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Key Concepts

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

Radiative Heat Transfer
Radiative heat transfer is all about the movement of heat through electromagnetic waves. Think of it like how the sun heats the Earth. In the flue gas scenario, radiation occurs between the high-temperature gas and the cooler brick surface.
Radiation becomes an important heat transfer mechanism especially at high temperatures. It allows energy to be transferred even without a medium; for example, through a vacuum.
  • Energy from the hot gas molecules is emitted as thermal radiation.
  • This energy travels as waves and gets absorbed by the brick walls, warming them up.
  • The Stefan-Boltzmann Law helps calculate this energy transfer rate.
The radiative heat transfer is particularly significant in the given problem because the flue gas is at a very high temperature of 1100 K, which means that radiation will be more effective.
Convective Heat Transfer
Convective heat transfer involves the movement of heat through a fluid motion such as gas or liquid. Here, it's the flue gas moving past the walls. As the gas flows, it transports heat to the walls by convection.
There are two types of convection: natural and forced. In this problem, "forced convection" plays a role because the gas is purposefully moved at 3 m/s.
  • As the flue gas flows, it brushes against the cooler brick surface, transferring heat to it.
  • The Dittus-Boelter equation helps estimate the heat transfer coefficient in such flows.
This coefficient indicates how well the flue gas can transfer heat to the brick walls. Ideally, knowing this coefficient helps us determine the heat transfer rate accurately.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is essential for calculating radiative heat transfer. This law relates the thermal radiation emitted by a black body to its temperature.
For our flue gas and brick wall situation, the black body assumption refers to the brick being perfectly absorptive and emissive of radiation.
  • The formula is \( q_{rad} = \sigma ( T^4 - T_s^4 ) \) where \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W/m}^2\,\mathrm{K}^4 \).
  • Temperature \( T \) is the flue gas temperature and \( T_s \) is the brick surface temperature.
By using the Stefan-Boltzmann Law, we can find out how much energy is being transferred per square meter due to radiation.
Dittus-Boelter Equation
The Dittus-Boelter equation is widely used for calculating the convective heat transfer coefficient, which is pivotal in problems like the flow of flue gas in a duct.
This equation is typically used under conditions of turbulent flow, which is common in many industrial applications:
It helps to predict how efficiently the gas transfers heat to the walls through convection.
  • The Dittus-Boelter equation assumes certain flow conditions and properties such as specific heat, viscosity, and the Prandtl number.
  • In scenarios where data is incomplete, typical values or assumptions are often used to estimate outcomes.
Remember, using this equation requires caution as the assumptions must match real-world conditions closely to ensure accuracy.

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Most popular questions from this chapter

A sliding door on a furnace does not close entirely but leaves a gap \(1.6 \mathrm{~cm}\) wide and \(1 \mathrm{~m}\) high. The door is \(20 \mathrm{~cm}\) thick and the emittance of the refractory ceramic is \(0.8\). What is the heat loss through the gap when the furnace is at \(1800 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\) ?

A thermistor is used to measure the temperature of a hot-air stream in a forced-air heating system of a hospital. It is located in a \(1 \mathrm{~m}\)-square duct through which air 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 \(0.7\) (the axis of the tube is in the flow direction). If the thermistor records a temperature of \(46.8^{\circ} \mathrm{C}\) when the duct walls are at \(43.5^{\circ} \mathrm{C}\), determine the true temperature of the air. Convective heat transfer coefficients on the thermistor and shield can be taken as \(93 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(22 \mathrm{~W} / \mathrm{m}^{2}\) \(\mathrm{K}\), respectively. (Hint: First estimate the shield temperature.)

Calculate the equilibrium temperature of a small, flat plate with its top face exposed to an unobstructed view of the sky while air at \(298 \mathrm{~K}, 1 \mathrm{~atm}\) and \(20 \%\) relative humidity flows along both sides. The bottom face sees black surroundings at \(298 \mathrm{~K}\). The solar irradiation is \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the average convective heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Obtain solutions for three different kinds of surfaces: (i) Representative of a very white paint, \(\alpha_{s}=0.2, \varepsilon=0.9\) (ii) A metallic paint (aluminum), \(\alpha_{s}=0.3, \varepsilon=0.3\) (iii) A black paint, \(\alpha_{s}=0.9, \varepsilon=0.9\)

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 \(1100 \mathrm{~K}\).

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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