/*! 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 60 A long 35-cm-diameter cylindrica... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 60

A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Short Answer

Expert verified
The temperature at the center of the shaft after 20 minutes of cooling is 356.9掳C, and the heat transfer per unit length during this time is 6.278 x 10^6 J/m.

Step by step solution

01

Determine the initial conditions and conversion of given values

Initially, note down the given values and convert the units provided into SI units where necessary. In our case, the time needs to be converted from minute to seconds. - Initial temperature of the shaft, \(T_i = 400 ^\circ C\) - Cooling chamber temperature, \(T_\infty = 150 ^\circ C\) - Thermal conductivity, \(k = 14.9 W/(m \cdot K)\) - Density, \(\rho = 7900 kg/m^3\) - Specific heat, \(c_p = 477 J/(kg \cdot K)\) - Shaft diameter, \(D = 0.35m\) - Convection heat transfer coefficient, \(h = 60 W/(m^2 \cdot K)\) - Time, \(t = 20 \times 60 = 1200s\)
02

Calculate the Fourier number

Now, let's calculate the Fourier number (\(Fo\)) which is given by the formula, \(Fo = \frac{\alpha t}{(D/2)^2}\), where \(\alpha = k/(\rho c_p)\). First, find the value of \(\alpha\): \(\alpha = \frac{14.9}{7900 \times 477} = 3.95 \times 10^{-6} m^2/s\) Now calculate the Fourier number: \(Fo = \frac{(3.95 \times 10^{-6} \times 1200)}{(0.35/2)^2} = 0.162\)
03

Calculate the Biot number

Next, calculate the Biot number (\(Bi\)) using the following formula, \(Bi = \frac{h(D/2)}{k}\) Now plug in the values to find \(Bi\): \(Bi = \frac{60(0.35/2)}{14.9} = 0.703\)
04

Analyze the lumped capacitance approach's suitability and calculate the temperature

Since \(Bi \times Fo > 0.1\), we cannot use the lumped capacitance approach and have to use the analytical one-term approximation method. With \(Bi = 0.703\) and \(Fo = 0.162\), the dimensionless temperature is found from the analytical one-term approximation as \(\theta = 0.827\). Therefore, \(T(t) = T_\infty + \theta \times (T_i - T_\infty)\) Now substitute the values to find the temperature after 20 minutes: \(T(t) = 150 + 0.827 \times (400 - 150) = 356.9 ^\circ C\)
05

Calculate the heat transfer per unit length of the shaft

To compute the heat transfer per unit length of the shaft, we use the total energy equation \(q_L = \rho \cdot V \cdot c_p \cdot (T_i - T_\infty) = m \cdot c_p \cdot (T_i - T_\infty)\) Where \(m\) is the mass of the steel per unit length given by: \(m = \rho \cdot V = \rho \cdot \pi (\frac{D}{2})^2\) per unit length Now use the values provided to find the mass per unit length of the steel: \(m = 7900 \cdot \pi (\frac{0.35}{2})^2 = 3221.86 kg/m\) Now substitute the values into the total energy equation and find \(q_L\): \(q_L = 3221.86 \cdot 477 \cdot (400 - 356.9) = 6.278 \times 10^6 J/m\) The temperature at the center of the shaft after 20 minutes of cooling is \(356.9 ^\circ C\), and the heat transfer per unit length during this time is \(6.278 \times 10^6 J/m\).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. It's crucial in determining how quickly a material can transfer heat from one part to another. For stainless steel 304, the thermal conductivity is given as \(k = 14.9\, \text{W/m} \cdot \text{K}\). This relatively moderate figure indicates that stainless steel does not transfer heat as effectively as metals like copper, but better than most insulators.
Understanding thermal conductivity helps in calculating heat transfer rates through the material. When designing or analyzing processes involving heat transfer, knowing the thermal conductivity can lead to more efficient and effective designs.
For our cylindrical shaft cooling problem, the thermal conductivity is used to calculate the thermal diffusivity \(\alpha\) which assists in evaluating the material's response to heat over time.
Biot Number
The Biot number (\(Bi\)) is a dimensionless number that compares the internal resistance to heat conduction within an object to the external resistance to heat convection away from its surface. It's calculated using the formula \(Bi = \frac{hL_c}{k}\), where \(L_c\) is the characteristic length, in our case, \(\frac{D}{2}\) due to the cylindrical shape of the shaft.
If \(Bi \ll 1\), it suggests that heat conduction within the object is much faster than heat transfer away from its surface, justifying the use of simpler analytical models like the lumped capacitance method.
  • In this context, a \(Bi\) value of 0.703 indicates that the internal and external resistances are comparable, thus requiring the more detailed analytical one-term approximation method be used to solve the problem.
  • This ensures proper accuracy when solving for the temperature and other characteristics during cooling.
Fourier Number
The Fourier Number (\(Fo\)) is another dimensionless number used in heat transfer calculations. It represents the ratio of heat conduction rate to the rate of thermal energy storage in a material. The Fourier number is calculated by \(Fo = \frac{\alpha t}{L^2}\).
A high \(Fo\) means that heat conduction occurs everywhere quickly relative to the time scale of changing temperature. For the shaft, \(Fo = 0.162\) indicating moderate heat conduction over the cooling period.
This number is essential in transient heat conduction analysis, helping to determine how temperature gradients evolve over time.
In our scenario, since \(Bi \times Fo > 0.1\), we cannot simplify the problem using lumping methods, and thus require a more detailed model.
Convection Heat Transfer
Convection heat transfer refers to the transfer of heat between a solid surface and a moving fluid at different temperatures. It's quantified by the convection heat transfer coefficient \(h\).
The value \(h = 60\, \text{W/m}^2 \cdot \text{K}\) characterizes the efficiency with which heat is carried away from the shaft surface to the cooling chamber's air. Higher \(h\) values lead to faster heat dissipation.
In analyzing the cooling process, \(h\) helps determine the rate at which the shaft temperature falls once it is placed in the cooler environment.
  • This coefficient, alongside thermal conductivity, plays a pivotal role in calculating the Biot number, which influences our choice of heat transfer model.
  • A strong understanding of convection is necessary to accurately predict temperature changes and heat loss.
Stainless Steel Properties
Stainless steel, particularly grade 304 in this context, offers a unique blend of properties that are important for thermal analysis. It possesses moderate thermal conductivity, high density \(\rho = 7900\, \text{kg/m}^3\), and substantial specific heat capacity \(c_p = 477\, \text{J/kg} \cdot \text{K}\).
These properties together define how quickly and how much heat the shaft can absorb or release during cooling. The high specific heat indicates that it can absorb a lot of heat before experiencing a temperature change.
Its moderate thermal conductivity ensures efficient heat spread while still maintaining structural integrity even at high temperatures like the initial \(400^\circ \text{C}\).
When considering heat transfer applications, recognizing these stainless steel properties is essential for designing efficient systems and predicting behavior under different thermal conditions.

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Most popular questions from this chapter

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

What is lumped system analysis? When is it applicable?

Large steel plates \(1.0\)-cm in thickness are quenched from \(600^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by submerging them in an oil reservoir held at \(30^{\circ} \mathrm{C}\). The average heat transfer coefficient for both faces of steel plates is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Average steel properties are \(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Calculate the quench time for steel plates.

Consider a hot semi-infinite solid at an initial temperature of \(T_{i}\) that is exposed to convection to a cooler medium at a constant temperature of \(T_{\infty}\), with a heat transfer coefficient of \(h\). Explain how you can determine the total amount of heat transfer from the solid up to a specified time \(t_{o}\).

A hot dog can be considered to be a \(12-\mathrm{cm}-\mathrm{long}\) cylinder whose diameter is \(2 \mathrm{~cm}\) and whose properties are \(\rho=980 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.9 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.76 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(2 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). A hot dog initially at \(5^{\circ} \mathrm{C}\) is dropped into boiling water at \(100^{\circ} \mathrm{C}\). The heat transfer coefficient at the surface of the hot dog is estimated to be \(600 \mathrm{~W} / \mathrm{m}^{2}\). K. If the hot dog is considered cooked when its center temperature reaches \(80^{\circ} \mathrm{C}\), determine how long it will take to cook it in the boiling water.
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