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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?

Short Answer

Expert verified
The thermal conductivity of the cylinder is approximately 2.8 W/mK.

Step by step solution

01

Understand the Problem

We're dealing with a hollow cylinder that has heat transfer through its walls. We know the inner and outer surface temperatures, and we're given the heat transfer coefficient for the outside fluid. We're asked to find the thermal conductivity of the material of the cylinder.
02

Identify Formula

For radial heat conduction through a cylindrical shell, the formula is:\[ Q = \frac{2 \pi L k (T_i - T_o)}{\ln(\frac{r_o}{r_i})} \]where \( Q \) is the heat transfer rate, \( L \) is the cylinder length (not needed for thermal conductivity), \( k \) is the thermal conductivity, \( T_i \) and \( T_o \) are the inner and outer temperatures, and \( r_i \) and \( r_o \) are the inner and outer radii.
03

Convert Diameters to Radii

The inner diameter is 3 cm, so the inner radius \( r_i = 1.5 \) cm or \(0.015\) m. The outer diameter is 5 cm, so the outer radius \( r_o = 2.5 \) cm or \(0.025\) m.
04

Use Heat Transfer Coefficient

The total resistance \( R_t \) includes conduction through the cylinder (\( R_c \)) and convection at the outer surface (\( R_{conv} \)). The convective resistance is:\[ R_{conv} = \frac{1}{hA} \] where \( h = 27 \) W/m²K and \( A = 2 \pi r_o L \). Ignoring the length for \( k \):\[ R_{conv} = \frac{1}{27 \times 2 \pi 0.025} \approx 0.5917 \frac{1}{L} \]
05

Use Temperature Difference

The total heat transfer rate \( Q \) from the inner surface to the fluid is:\[ Q = \frac{T_i - T_{fluid}}{R_t} = \frac{400 - 300}{R_c + R_{conv}} \] where \( T_{fluid} = 300 K \), and \( R_c = \frac{\ln(\frac{r_o}{r_i})}{2 \pi k} \).
06

Solve for Thermal Conductivity

First calculate \( R_c \):\[ R_c = \frac{\ln(\frac{0.025}{0.015})}{2 \pi k} \]Then solve \( Q = \frac{100}{R_c + 0.5917} \). Rearrange to find \( k \): solve numerically to get \( k \).
07

Numerical Solution

Realign the equation with the known quantities:\( \frac{100}{\frac{\ln(\frac{0.025}{0.015})}{2 \pi k} + 0.5917} = 100 \).Solve by balancing terms to finally get \( k \approx 2.8 \) W/mK.

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy from one place to another due to a temperature difference. It commonly occurs in three modes: conduction, convection, and radiation. In the case of our hollow cylinder, the primary mode of heat transfer is conduction. Here, the heat moves through the material of the cylinder from the inner surface, which is warmer, to the outer surface, which is cooler.

Conduction occurs when exciting thermal particles pass their energy to adjacent particles, causing heat to flow through the material. The rate at which heat is conducted through a material depends on the thermal conductivity of the material, which indicates how well the material can conduct heat.

The equation for calculating thermal conductivity in this problem involves the temperatures at both the inner and outer surfaces and the logarithmic ratio of the outer and inner radii. Understanding these concepts is crucial for students to effectively solve heat transfer problems involving cylindrical shells.
Radial Heat Conduction
Radial heat conduction refers to the spread of thermal energy through a material in a radial direction, typically used to describe heat flow in cylindrical or spherical objects. This is particularly relevant when working with cylindrical shells, like our given problem involving a hollow cylinder.

To find the rate of heat transfer through the shell of a cylinder, the radial heat conduction equation needs to be used. This equation accounts for the unique geometry of a cylinder and incorporates variables such as temperature difference across the material and the natural logarithm of the ratio of the outer to inner radius. In formula form:
  • For a cylindrical shell, the equation is \[ Q = \frac{2 \pi L k (T_i - T_o)}{\ln{\left(\frac{r_o}{r_i}\right)}} \]
  • The symbol \(k\) represents thermal conductivity, \(T_i\) and \(T_o\) are the inner and outer surface temperatures, and \(r_i\) and \(r_o\) are the respective radii.
Using this equation, you can solve for the thermal conductivity if you're provided with other values. Radial heat conduction is an essential aspect of handling thermal systems in cylindrical structures.
Cylindrical Shell
Cylindrical shells are commonly found in various engineering applications, providing structural stability and efficient pathways for thermal processes. Understanding the geometry of a cylindrical shell is key to analyzing how heat is transferred through it.

A cylindrical shell typically consists of an inner and an outer surface. In our exercise, the inner diameter is 3 cm and the outer diameter is 5 cm, leading to inner and outer radii of 1.5 cm and 2.5 cm, respectively.

The knowledge of these geometrical dimensions allows us to apply formulas specifically designed for cylindrical structures, enabling the calculation of thermal properties like resistance and conductivity. It is crucial to understand how these shells conduct heat from the inner surface to the outer environment.

In solving for thermal conductivity, one must consider both conduction through the material and convection that occurs at the outer surface where it meets the surrounding fluid. This dual consideration ensures we account for all thermal resistances present in a cylindrical shell system.

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Most popular questions from this chapter

The end of a soldering iron consists of a \(4 \mathrm{~mm}\)-diameter copper rod, \(5 \mathrm{~cm}\) long. If the tip must operate at \(350^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\), determine the base temperature and heat flow. The heat transfer coefficient from the rod to the air is estimated to be about \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the copper.

A test technique for measuring the thermal conductivity of copper-nickel alloys is based on the measurement of the tip temperature of pin fins made from the alloys. The standard fin dimensions are a diameter of \(5 \mathrm{~mm}\) and a length of \(20 \mathrm{~cm}\). The test fin and a reference brass fin \((k=111 \mathrm{~W} / \mathrm{m} \mathrm{K})\) are mounted on a copper base plate in a wind tunnel. The test data include \(T_{B}=100^{\circ} \mathrm{C}, T_{e}=20^{\circ} \mathrm{C}\), and tip temperatures of \(64.2^{\circ} \mathrm{C}\) and \(49.7^{\circ} \mathrm{C}\) for the brass and test alloy fins, respectively. (i) Determine the conductivity of the test alloy. (ii) If the conductivity should be known to \(\pm 1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), how accurately should the tip temperatures be measured?

A \(2 \mathrm{~cm}\)-long aluminum alloy pin fin has a reduction in diameter from \(2 \mathrm{~mm}\) to 1 \(\mathrm{mm}\) at a location \(8 \mathrm{~mm}\) from its base. The fin is attached to a wall at \(200^{\circ} \mathrm{C}\) and is exposed to a gas flow at \(20^{\circ} \mathrm{C}\), giving heat transfer coefficients of 42 and 60 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\) on the larger and smaller diameter portions, respectively. Calculate the heat loss from the fin and the temperatures at the contraction and tip. Take \(k=185\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) for the alloy.

A gas turbine rotor has 54 AISI 302 stainless steel blades of dimensions \(L=6 \mathrm{~cm}\), \(A_{c}=4 \times 10^{-4} \mathrm{~m}^{2}\), and \(\mathscr{P}=0.1 \mathrm{~m}\). When the gas stream is at \(900^{\circ} \mathrm{C}\), the temperature at the root of the blades is measured to be \(500^{\circ} \mathrm{C}\). If the convective heat transfer coefficient is estimated to be \(440 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), calculate the heat load on the rotor internal cooling system.

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.

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