91Ó°ÊÓ

The products of combustion from a burner are rouled to an industrial application through a thin-walled metllic duct of diameter \(D_{4}=1 \mathrm{~m}\) and length \(L=100 \mathrm{~m}\). The \(\mathrm{ar}\) enters the duct at atmospheric pressure and a mean tant perature and velocity of \(T_{m i}=1600 \mathrm{~K}\) and \(u_{m j}=10 \mathrm{mk}\). respectively. It must exit the duct at a temperature fiat is no less than \(T_{m e}=1400 \mathrm{~K}\). What is the minimum thick ness of an alumina-silica insulation ( \(k_{\text {iun }}=0.125 \mathrm{~W} / \mathrm{m}\). \(\mathrm{K}\) ) needed to meet the outlet requirement under wors case conditions for which the duct is exposed to ambient air iu \(T_{\mathrm{m}}=250 \mathrm{~K}\) and a cross-fow velocity of \(V=15 \mathrm{~m} / \mathrm{s}\) ? The properties of the gas may be approximated as those of aic, and as a first estimate, the effect of the insulation thickness on the convection coefficient and thermal resistanoe associated with the cross flow may be neglecled.

Step by step solution

01

Identify Known Values

First, let's gather all the given values from the problem: - Diameter of the duct: \( D_4 = 1 \) m- Length of the duct: \( L = 100 \) m- Initial mean temperature: \( T_{mi} = 1600 \) K- Exit temperature: \( T_{me} = 1400 \) K- Mean velocity: \( u_{mj} = 10 \) m/s- Insulation thermal conductivity: \( k_{ins} = 0.125 \) W/(m·K)- Ambient temperature: \( T_{amb} = 250 \) K- Cross-flow velocity: \( V = 15 \) m/sWe need to calculate the minimum thickness of insulation.

02

Setup Energy Balance Equation

To determine the minimum thickness of insulation, we must ensure that the energy lost by the flowing gases in the duct does not drop the exit temperature below \( T_{me} = 1400 \) K. We will use an energy balance approach:\[ q = \dot{m} \cdot c_p \cdot (T_{mi} - T_{me})\]where \( \dot{m} \) is the mass flow rate and \( c_p \) is the specific heat capacity of air (approx 1005 J/(kg·K) at constant pressure).

03

Calculate Mass Flow Rate

The mass flow rate \( \dot{m} \) can be calculated using the formula:\[ \dot{m} = \rho \cdot A \cdot u_{mj}\]Where:- \( \rho \) is the density of air at \( T_{mi} \) which can be approximated using ideal gas law: \( \rho = \frac{P}{R \cdot T} \).- \( A \) is the cross-sectional area of the duct: \( A = \frac{\pi D_4^2}{4} \).Assume air as an ideal gas with \( P = 101325 \) Pa and \( R = 287 \) J/(kg·K).

04

Calculate Heat Loss Through Conduction in Insulation

The heat loss through the insulation is given by Fourier's law:\[ q = \frac{2 \pi L k_{ins} (T_{me} - T_{amb})}{\ln(\frac{r_{outer}}{r_{inner}})}\]Since the problem neglects the effect on the convection coefficients, we only look at the conduction through the insulation.We will solve for \( r_{outer} \) which is \( r_{inner} + t \) (where \( t \) is the thickness of insulation).

05

Equalize Energy Loss and Required Heat Transfer

By setting the energy coming out equal to the allowable heat transfer, our equations become:\[ \dot{m} \cdot c_p \cdot (T_{mi} - T_{me}) = \frac{2 \pi L k_{ins} (T_{me} - T_{amb})}{\ln(\frac{D_4/2 + t}{D_4/2})}\]This is a complex equation in terms of \( t \), and typically requires iterative solutions or the use of computational tools to find \( t \).

06

Solve for Thickness (t)

Using numerical methods (such as the bisection method, Newton's method, or computational software), solve for \( t \). Substitute known values in the equation and iterate until you find a value for \( t \) that balances the equation.An approximate solution should be calculated to find the required insulation thickness, making sure the exit temperature condition and energy balance are satisfied.

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Key Concepts

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

Energy Balance

In the study of heat transfer, an energy balance is crucial for ensuring that the amount of energy entering a system matches the energy leaving it. This balance helps prevent undesired temperature changes. For the given exercise, we want to maintain the temperature of gases exiting the duct at a minimum of 1400 K.

• The concept of energy balance involves calculating the energy flowing into the duct through combustion and how much energy is lost during its travel.
• The main idea is based on the equation: \[ q = \dot{m} \cdot c_p \cdot (T_{mi} - T_{me}) \]
• \(\dot{m}\) is the mass flow rate, and \(c_p\) is the specific heat capacity of the gas.
• This allows us to calculate how much heat must be conserved or how much is lost through the walls of the duct.

Performing this energy balance accurately enables the determination of the minimum insulation thickness required to meet the exit temperature specifications without unnecessary loss.

Thermal Insulation

Thermal insulation is a material's ability to reduce heat flow. Its role, especially in industrial applications like this one, is critical in managing temperatures and energy efficiency.

• In our exercise, we're interested in calculating the thickness needed for aluminum-silica insulation to prevent excessive heat loss.
• The insulation's thermal conductivity, \(k_{\text{ins}} = 0.125\) W/(m·K), is given, indicating how much heat the material will transmit per unit area per degree of temperature difference.
• The thickness \(t\) of insulation acts as a barrier between hot gases and the cooler surroundings.

The purpose of the insulation is to minimize heat conduction from the inner surface of the duct to the outer surface exposed to ambient conditions. Good insulation ensures the temperature of the gases at the duct exit remains above the required limit.

Heat Conduction

Heat conduction through a material is a fundamental concept of heat transfer and is described by Fourier's law. It is the process of heat energy transfer in a solid or stationary fluid.

• For the duct, Fourier’s law is expressed as \[ q = \frac{2 \pi L k_{ins} (T_{me} - T_{amb})}{\ln\left(\frac{r_{outer}}{r_{inner}}\right)} \]
• This equation considers a cylindrical coordinate system suitable for the duct's geometry.
• \(r_{outer}\) and \(r_{inner}\) are the outer and inner radii of the duct respectively, and \(L\) is its length.
• The natural logarithm term accounts for the curved surface area of the duct wall, influencing heat loss.

Solving this equation for heat conducted will determine the required insulation thickness to maintain adequate heat within the system.

Fluid Dynamics

Fluid dynamics in this context refers to the study of gases flowing through the duct. It affects both the heat transfer processes and energy balances.

• The exercise considers gases modeled as air flowing with a mean velocity \(u_{mj} = 10\) m/s through the duct.
• Fluid properties, like density \(\rho\), can be estimated using the ideal gas law \(\rho = \frac{P}{R \cdot T}\).
• The velocity and density are crucial for determining the mass flow rate \(\dot{m} = \rho \cdot A \cdot u_{mj}\).

Because the exercise assumes the cross-flow velocity \(V = 15\) m/s and neglects the insulation thickness's effect on convection, these simplifications help to focus on conduction as the primary heat transfer mode. Fluid dynamics shapes the behavior of the gases, influencing energy conservation and heat transfer efficiency.