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To reduce heat losses, a horizontal rectangular duct that is \(W=0.80 \mathrm{~m}\) wide and \(H=0.3 \mathrm{~m}\) high is encased in a metal radiation shield. The duct wall and shield are separated by an air gap of thickness \(t=0.06 \mathrm{~m}\). For a duct wall temperature of \(T_{d}=40^{\circ} \mathrm{C}\) and a shield temperature of \(T_{\mathrm{sh}}=20^{\circ} \mathrm{C}\), determine the convection heat loss per unit length from the duct.

Short Answer

Expert verified
Using the given dimensions and temperatures, we first calculate the Rayleigh number (Ra) and subsequently the Nusselt number (Nu). With the computed Nusselt number, we calculate the convective heat transfer coefficient (h) using the thermal conductivity of air (k) and the height of the gap (L). Finally, we compute the convection heat loss per unit length (q) by multiplying the convective heat transfer coefficient (h), the area of the air gap (A), and the temperature difference between the duct wall and shield (ΔT).

Step by step solution

01

Determine the Nusselt number

First, we will need to find the Nusselt number, which is a dimensionless number that represents the ratio of convective to conductive heat transfer. In this case, we can use the following correlation for the average Nusselt number for natural convection in a horizontal gap filled with air: \(Nu = 1.0 + 1.441 \left(\frac{Ra}{28}\right)^{1/2} \) Here, \(Nu\) is the Nusselt number, and \(Ra\) is the Rayleigh number, another dimensionless value that characterizes the driving force for natural convection. We need first to calculate the Rayleigh number.
02

Calculate the Rayleigh number

The Rayleigh number (Ra) can be calculated using the following formula: \(Ra = \frac{g \beta \abs{T_d - T_{sh}} L^3}{\alpha \nu} \) Where, \(g\) - acceleration due to gravity (\(9.81 m/s^2\)) \(\beta \) - coefficient of thermal expansion for air (\(3.67 \times 10^{-3} K^{-1}\)) \(T_d\) - duct wall temperature (\(40^{\circ} C\)) \(T_{sh}\) - shield temperature (\(20^{\circ} C\)) \(L\) - height of the gap (\(0.06 m\)) \(\alpha\) - thermal diffusivity of air (\(2.22 \times 10^{-5} m^2/s\)) \(\nu\) - kinematic viscosity of air (\(15.89 \times 10^{-6} m^2/s\)) Now, plug in these values into the Rayleigh number formula and calculate the value.
03

Calculate the Nusselt number

Now that we have the Rayleigh number, we can use the correlation provided in Step 1 to find the Nusselt number. Plug in the Rayleigh number value into the formula and calculate the Nusselt number.
04

Calculate the convective heat transfer coefficient

Once we have the Nusselt number, we can use it to find the convective heat transfer coefficient, \(h\), using the following expression: \(h = \frac{Nu \cdot k}{L} \) Where, \(Nu\) - Nusselt number calculated in Step 3 \(k\) - thermal conductivity of air (\(0.0262 W/m \cdot K\)) \(L\) - height of the gap (\(0.06 m\)) Plug in the values and calculate the convective heat transfer coefficient.
05

Determine the convection heat loss per unit length

Finally, we can calculate the convection heat loss per unit length (\(q\)) using the following formula: \(q = h A \Delta T \) Where, \(h\) - convective heat transfer coefficient calculated in Step 4 \(A\) - area of the air gap, which is the product of width and height (\(A = W \times H = 0.8 m \times 0.3 m\)) \(\Delta T\) - temperature difference between the duct wall and shield (\(T_d - T_{sh} = 40^{\circ} C - 20^{\circ} C\)) Plug in these values into the formula and calculate the convection heat loss per unit length (\(q\)). This is the final answer.

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Key Concepts

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

Nusselt Number
The Nusselt number is a dimensionless number used to describe heat transfer in different situations. It is particularly helpful in distinguishing between different types of heat transfer: conductive and convective.
In simpler terms, you can think of the Nusselt number as a way to measure how effectively heat is moving through fluids like air or water. If we talk about conduction, we're focusing on heat moving through a solid or still fluid. But if convection is involved, this heat is being transported by fluid movement.
  • Formula: The Nusselt number is calculated using the formula: \(Nu = 1.0 + 1.441 \left(\frac{Ra}{28}\right)^{1/2} \).
  • Application: The formula helps identify how much the fluid motion (convection) is contributing to overall heat transfer.
By knowing the Nusselt number, we can decide the right methods or materials to enhance heat transfer efficiency. In the context of the exercise, calculating the Nusselt number allows us to find the convection heat transfer coefficient, which is critical for determining heat loss.
Rayleigh Number
Think of the Rayleigh number like a tool that helps predict how heat transfer will behave in the air or fluids due to natural convection. Simply put, it's all about the natural currents that develop when there is a temperature difference.
  • Formula: The Rayleigh number is calculated using: \[Ra = \frac{g \beta \abs{T_d - T_{sh}} L^3}{\alpha u} \]
  • Parameters: Here's a breakdown of what every term means:
    • \(g\): Acceleration due to gravity - it's always pulling downwards
    • \(\beta\): Thermal expansion coefficient - tells us how much a material expands with temperature
    • \(T_d\) and \(T_{sh}\): Temperatures of the duct wall and shield respectively
    • \(L\): Height of the air gap - determines how far the heat must travel
    • \(\alpha\) and \(u\): Thermal diffusivity and kinematic viscosity - both relate to the fluid's resistance to deform and conduct heat
Understanding the Rayleigh number is crucial as it helps estimate the flow conditions. High Rayleigh numbers imply strong natural convection currents. In the given problem, calculating the Rayleigh number helps us later use it to find the Nusselt number.
Heat Loss Calculation
Calculating heat loss is an important step in understanding how well or poorly insulated a system like a duct is. It's about figuring out how much warmth escapes from a surface into the surrounding air, which, in this case, consists of both conduction and convection.
The process can be broken into a series of simple stages.
  • Convection Heat Transfer Coefficient (\(h\)): First, once we have the Nusselt number from the prior section, we find \(h\) using the formula: \(h = \frac{Nu \cdot k}{L} \), where \(k\) is the thermal conductivity of air.
  • Heat Loss per Unit Length (\(q\)): Then, the convection heat loss is calculated using: \(q = h A \Delta T \), where \(A\) is the area of air gap, determined by multiplying the width and height of the gap, and \(\Delta T\) symbolizes the temperature difference.
This calculation gives us an idea of how much energy is lost, and from there, we can plan ways to reduce this loss through better insulation or design changes. It’s all about optimizing for energy efficiency, which is crucial for cost savings and energy conservation.

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Most popular questions from this chapter

A solid object is to be cooled by submerging it in a quiescent fluid, and the associated free convection coefficient is given by \(\bar{h}=C \Delta T^{1 / 4}\), where \(C\) is a constant and \(\Delta T=T-T_{\infty}\). (a) Using the results of Section \(5.3 .3\), obtain an expression for the time required for the object to cool from an initial temperature \(T_{i}\) to a final temperature \(T_{f}\). (b) Consider a highly polished, \(150-\mathrm{mm}\) square aluminum alloy (2024) plate of \(5-\mathrm{mm}\) thickness, initially at \(225^{\circ} \mathrm{C}\), and suspended in ambient air at \(25^{\circ} \mathrm{C}\). Using the appropriate approximate correlation from Problem 9.11, determine the time required for the plate to reach \(80^{\circ} \mathrm{C}\). (c) Plot the temperature-time history obtained from part (b) and compare with the results from a lumped capacitance analysis using a constant free convection coefficient, \(\bar{h}_{o}\). Evaluate \(\bar{h}_{o}\) from an appropriate correlation based on an average surface temperature of \(\bar{T}=\left(T_{i}+T_{f}\right) / 2\).

A household oven door of \(0.5-\mathrm{m}\) height and \(0.7-\mathrm{m}\) width reaches an average surface temperature of \(32^{\circ} \mathrm{C}\) during operation. Estimate the heat loss to the room with ambient air at \(22^{\circ} \mathrm{C}\). If the door has an emissivity of \(1.0\) and the surroundings are also at \(22^{\circ} \mathrm{C}\), comment on the heat loss by free convection relative to that by radiation.

Hot air flows from a furnace through a \(0.15\)-m-diameter, thin-walled steel duct with a velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through the crawlspace of a house, and its uninsulated exterior surface is exposed to quiescent air and surroundings at \(0^{\circ} \mathrm{C}\). (a) At a location in the duct for which the mean air temperature is \(70^{\circ} \mathrm{C}\), determine the heat loss per unit duct length and the duct wall temperature. The duct outer surface has an emissivity of \(0.5\). (b) If the duct is wrapped with a \(25-\mathrm{mm}\)-thick layer of \(85 \%\) magnesia insulation \((k=0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) having a surface emissivity of \(\varepsilon=0.60\), what are the duct wall temperature, the outer surface temperature, and the heat loss per unit length?

As discussed in Section 5.2, the lumped capacitance approximation may be applied if \(B_{i}<0.1\), and, when implemented in a conservative fashion for a long cylinder, the characteristic length is the cylinder radius. After its extrusion, a long glass rod of diameter \(D=15 \mathrm{~mm}\) is suspended horizontally in a room and cooled from its initial temperature by natural convection and radiation. At what rod temperatures may the lumped capacitance approximation be applied? The temperature of the quiescent air is the same as that of the surroundings, \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\), and the glass emissivity is \(\varepsilon=0.94\).

Free convection occurs between concentric spheres. The inner sphere is of diameter \(D_{i}=50 \mathrm{~mm}\) and temperature \(T_{i}=50^{\circ} \mathrm{C}\), while the outer sphere is maintained at \(T_{o}=20^{\circ} \mathrm{C}\). Air is in the gap between the spheres. What outer sphere diameter is required so that the convection heat transfer from the inner sphere is the same as if it were placed in a large, quiescent environment with air at \(T_{\infty}=20^{\circ} \mathrm{C}\) ?

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