/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 A \(4 \mathrm{~mm}\)-diameter, \... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(4 \mathrm{~mm}\)-diameter, \(25 \mathrm{~cm}\)-long aluminum alloy rod has an electric heater wound over the central \(5 \mathrm{~cm}\) length. The outside of the heater is well insulated. The two \(10 \mathrm{~cm}\)-long exposed portions of rod are cooled by an air stream at \(300 \mathrm{~K}\) giving an average convective heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the power input to the heater is \(10 \mathrm{~W}\), determine the temperature at the ends of the rod. Take \(k=190\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) for the aluminum alloy.

Short Answer

Expert verified
The temperature at the ends of the rod is approximately 379.67 K.

Step by step solution

01

Analyze the problem setup

We have a cylindrical rod that is exposed to convection at both ends, while the central portion is heated. The problem requires us to find the temperature at the ends of the rod.
02

Define governing equation

We will assume steady-state heat conduction in the rod and use the rod's thermal conductivity and convective heat transfer to find the end temperatures. We apply energy balance involving conduction and convection.
03

Apply energy balance at the end of the rod

At one end of the rod, the heat conduction equation in steady-state is given by Fourier's law: \[ q'' = k \frac{dT}{dx} \] where \( q'' \) is the heat flux, \( k \) is the thermal conductivity, and \( \frac{dT}{dx} \) is the temperature gradient.
04

Calculate heat loss through convection

The convective heat loss at each end can be expressed as:\[ q' = h A (T_s - T_{\infty}) \]where \( h = 50 \, \text{W/m}^2\text{K} \) is the heat transfer coefficient, \( A \) is the surface area for each end section of the rod, \( T_s \) is the surface temperature at the end, and \( T_{\infty} = 300 \, \text{K} \) is the ambient temperature.
05

Express energy balance using heat input and convection

The power input to the heater must equal the total heat loss at both ends:\[ q = 2 \times q' \]Substitute the convection equation into this balance:\[ 10 = 2 \times (h \cdot \pi \cdot D \cdot L (T_s - 300)) \]Here, \( D = 0.004 \) m is the diameter and \( L = 0.10 \) m is the length of the convective section.
06

Solve for the surface temperature \( T_s \)

Simplify and solve for \( T_s \):\[ 10 = 2 \times (50 \times \pi \times 0.004 \times 0.10 \times (T_s - 300)) \]\[ 10 = 0.1256 (T_s - 300) \]\[ T_s - 300 = \frac{10}{0.1256} \]Solve:\[ T_s = \frac{10}{0.1256} + 300 = 379.67 \text{K} \]
07

Conclusion

The calculated surface temperature at the ends of the rod, under the conditions given, is approximately 379.67 K.

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Key Concepts

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

Convection
Convection refers to the transfer of heat between a surface and a fluid moving over it. This fluid could be a gas or liquid, such as the air stream in this exercise.
The heat transfer coefficient, denoted as \( h \), plays a crucial role in determining the rate of heat convected away from the object's surface. In the case of the aluminum alloy rod, the air stream cools the two exposed portions of the rod, creating a temperature gradient.
To calculate the convection heat transfer, the equation \( q' = h A (T_s - T_\infty) \) is used, where:
  • \( h \) = the convective heat transfer coefficient
  • \( A \) = the surface area
  • \( T_s \) = the surface temperature that needs to be found
  • \( T_\infty \) = the temperature of the air stream
Understanding convection is essential for solving problems where objects are subject to environmental cooling or heating.
Thermal Conductivity
Thermal conductivity, symbolized as \( k \), describes a material's ability to conduct heat. It is a property intrinsic to materials and varies from one material to another. In this exercise, the thermal conductivity of the aluminum alloy rod is essential to determine how efficiently heat transfers along the rod's length. The value given is \( k = 190 \, \text{W/mK} \).
The higher the thermal conductivity, the better the material is at conducting heat. So, knowing the thermal conductivity allows us to apply equations like Fourier's Law to solve for temperature gradients.
Materials with high thermal conductivity, like the aluminum alloy used in this exercise, are good conductors, helping effectively distribute heat from the heater along the rod.
Heat Conduction
Heat conduction is the process by which heat energy is transmitted through collisions between particles of the material. In this context, the heat conduction in the rod is analyzed using Fourier's Law, which describes the flow of heat through materials. According to Fourier's Law, the heat flux \( q'' \) in the direction of the length of the rod can be given by:
  • \( q'' = k \frac{dT}{dx} \)
Here:
  • \( q'' \) = heat flux
  • \( k \) = thermal conductivity
  • \( \frac{dT}{dx} \) = temperature gradient along the rod
For steady-state conditions, this equation helps in determining how heat moves from the heated center to the ends of the rod, as well as predicting temperature distribution across the material.
Energy Balance
Energy balance is a principle based on the conservation of energy, applied in this exercise to relate the power input with the losses due to convection at both ends of the rod. This involves ensuring that the energy introduced by the heater matches the energy leaving the rod through its ends.
In our problem, we set the heater power input equal to twice the heat loss through each end by convection:
  • \( q = 2 \times q' \)
The equation demonstrates that all of the electrical energy supplied to the heater is eventually lost through convection, thus aiding in calculating the surface temperature. This is crucial in scenarios where precise temperature control is required.
Fourier's Law
Fourier's Law is a foundational principle of heat conduction, allowing us to solve for the temperature gradient within a material. It mathematically expresses the heat transfer rate within the material, which is vital for heat conduction analysis.
The law is formulated as:
  • \( q'' = -k \frac{dT}{dx} \)
Here:
  • \( q'' \) = heat flux
  • \( k \) = thermal conductivity
  • \( \frac{dT}{dx} \) = temperature gradient
In the aluminum rod scenario, Fourier's Law helps us predict how temperature varies along the length of the rod, essential for determining the surface temperature at the rod's ends.
This law is a pivotal tool for engineers in designing systems involving heat conduction to ensure efficiency and safety.

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Most popular questions from this chapter

A thick-walled cylindrical tube has inner and outer diameters of \(2 \mathrm{~cm}\) and \(5 \mathrm{~cm}\), respectively. The tube is evacuated and contains a high-temperature radiation source along its axis giving a net radiant heat flux into the inner surface of the tube of \(10^{5} \mathrm{~W} / \mathrm{m}^{2}\). The outer surface of the tube is convectively cooled by a coolant at \(300 \mathrm{~K}\) with a convective heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the conductivity of the tube material is \(2.2 \mathrm{~W} / \mathrm{m} \mathrm{K}\), determine the temperature distribution \(T(r)\) in the tube wall. Also determine the inner- surface temperature.

Saturated steam at \(200^{\circ} \mathrm{C}\) flows through an AISI 1010 steel tube with an outer diameter of \(10 \mathrm{~cm}\) and a \(4 \mathrm{~mm}\) wall thickness. It is proposed to add a \(5 \mathrm{~cm}\)-thick layer of \(85 \%\) magnesia insulation. Compare the heat loss from the insulated tube to that from the bare tube when the ambient air temperature is \(20^{\circ} \mathrm{C}\). Take outside heat transfer coefficients of 6 and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) for the bare and insulated tubes, respectively.

A \(5 \mathrm{~mm} \times 2 \mathrm{~mm} \times 1 \mathrm{~mm}\)-thick semiconductor laser is mounted on a \(1 \mathrm{~cm}\) cube copper heat sink and enclosed in a Dewar flask. The laser dissipates \(2 \mathrm{~W}\), and a cryogenic refrigeration system maintains the copper block at a nearly uniform temperature of \(90 \mathrm{~K}\). Estimate the top surface temperature of the laser chip for the following models of the dissipation process: (i) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer underneath the top surface of the laser. (ii) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer at the midplane of the chip. (iii) The energy is dissipated uniformly through the chip. Take \(k=170 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the chip, and neglect parasitic heat gains from the Dewar flask.

A hollow cylinder, of inner and outer diameters 3 and \(5 \mathrm{~cm}\), respectively, has an inner surface temperature of \(400 \mathrm{~K}\). The outer surface temperature is \(326 \mathrm{~K}\) when exposed to fluid at \(300 \mathrm{~K}\) with an outside heat transfer coefficient of \(27 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). What is the thermal conductivity of the cylinder?

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