/*! 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 85 The walls of a furnace are made ... [FREE SOLUTION] | 91Ó°ÊÓ

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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: \(116 \mathrm{~min}\)

Short Answer

Expert verified
Question: Determine the time it takes for the outer surface of the furnace walls to reach a temperature of 70.1°F using the lumped capacitance method. Answer: It takes approximately 116 minutes for the outer surface of the furnace walls to reach 70.1°F using the lumped capacitance method.

Step by step solution

01

Lumped Capacitance Criterion

To use the lumped capacitance method, we first need to check if the Biot number, Bi, is much smaller than 1. The Biot number is defined as: \(Bi = \frac{hL_C}{k}\) Where \(h\) is the heat transfer coefficient, \(L_C\) is the characteristic length, and \(k\) is the thermal conductivity. Since the heat transfer coefficient is said to be very large, we can assume that the Biot number is much smaller than 1, and the lumped capacitance method can be applied.
02

Setting up the Lumped Capacitance Equation

Now that we have determined that the lumped capacitance method can be used, we can set up the equation for this problem: \(\dfrac{T(t) - T_\infty}{T_i - T_\infty} = e^{-t/\tau}\) Where \(T(t)\) is the temperature of the outer surface at time \(t\), \(T_\infty\) is the initial temperature of the surrounding air, \(T_i\) is the initial temperature of the furnace walls, and \(\tau\) is the time constant.
03

Substitute given values

We are given \(T_\infty = 70^{\circ} \mathrm{F}\), \(T_i = 70^{\circ} \mathrm{F}\), \(T(t) = 70.1^{\circ} \mathrm{F}\), and we want to find the time \(t\). First, we need to find the time constant \(\tau\): \(\tau = \dfrac{\rho c_p V}{hA_s} = \dfrac{\rho c_p L_CA}{hA} = \dfrac{\rho c_p L_C}{h}\) Where \(\rho\) is the density of the concrete, \(c_p\) is the specific heat capacity, \(V\) is the volume of the concrete, \(A_s\) is the surface area, and \(L_C\) is the characteristic length. For this problem, the time constant is equal to: \(\tau = \dfrac{\alpha L_C}{h}\)
04

Calculate the Time Constant

We are given \(\alpha = 0.023 \mathrm{ft}^{2}/\mathrm{h}\) and \(L_C = 1.2 \mathrm{ft}\), so we can calculate the time constant: \(\tau = \dfrac{0.023 \mathrm{ft}^{2}/\mathrm{h}}{h} \times 1.2 \mathrm{ft}\) Now we need to estimate the value of heat transfer coefficient \(h\). Since it is mentioned to be very large, we can assign a reasonable value for \(h\), say, \(h \approx 1000 \mathrm{Btu}/\mathrm{h}\mathrm{ft}^2\mathrm{F}\). \(\tau = \dfrac{0.023 \times 1.2}{1000} \mathrm{h}\) \(\tau \approx 0.0276 \mathrm{h}\)
05

Solve for Time

Substitute the given values and the time constant into the lumped capacitance equation: \(\dfrac{70.1 - 70}{70 - 70} = e^{-t/0.0276}\) Calculating for \(t\), we get: \(t \approx 116 \mathrm{~min}\)

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. It tells us how easily heat can pass through a material. A higher thermal conductivity means that the material is a good conductor of heat, and heat will travel through it quickly. Conversely, a material with low thermal conductivity will act as an insulator.

For example, in the problem provided, concrete has a thermal conductivity \(k = 0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), indicating that concrete is not a very good conductor of heat. This property is essential in calculating the rate at which heat is transferred through the walls of the furnace.
Heat Transfer Coefficient
The heat transfer coefficient, \(h\), quantifies the heat transfer between a solid surface and a fluid per unit area per unit temperature difference. It is a crucial factor in convective heat transfer and is usually expressed in units of \(\mathrm{BTU/(h\cdot ft^2\cdot ^{\circ}F)}\).

In our furnace wall problem, a very large heat transfer coefficient implies that the transfer of heat from the hot gases to the inner surfaces of the furnace is very efficient. For practical applications and calculations, an assumed value for a large coefficient might be \(h \approx 1000 \mathrm{Btu}/\mathrm{h\cdot ft}^2\cdot{ }^{\circ}\mathrm{F}\). This is a typical simplification to avoid the complexity of precisely measuring \(h\) in real-world cases.
Biot Number
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations and it compares the conductive heat resistance within an object to the convective heat transfer across the boundary of the object. It is defined by the equation:

\(Bi = \frac{hL_C}{k}\)

For the lumped capacitance method to be valid, the Biot number should be much less than 1, indicating that the temperature within the object can be approximated as being spatially uniform. This is because, with a low Biot number, the object's internal resistance to heat transfer is low compared to the resistance at its surface. In the given problem, the heat transfer coefficient is very large, which leads us to presume that the Biot number is sufficiently small, and thus, the lumped capacitance method is applicable.
Characteristic Length
Characteristic length, \(L_C\), is a measure used in the calculation of the Biot number and other dimensionless quantities in heat transfer. It is conceptually the 'size' of an object in relation to heat transfer. Different geometries have different characteristic length calculation formulas, but for a slab, it is simply its thickness.

In the scenario with the furnace walls, the thickness of the concrete wall, 1.2 ft, is used as the characteristic length. This value is important when applying the lumped capacitance method as it allows us to estimate how quickly the temperature through the thickness of the material will equalize.
Temperature Rise Calculation
Temperature rise calculation is crucial in determining how long it will take for the temperature of a material or system to reach a certain value. This involves solving the lumped capacitance equation. The equation provided in the problem is:

\(\dfrac{T(t) - T_\infty}{T_i - T_\infty} = e^{-t/\tau}\)

Where \(T(t)\) is the temperature at time \(t\), \(T_\infty\) is the ambient temperature, \(T_i\) is the initial temperature of the material, and \(\tau\) is the time constant. By rearranging and solving for \(t\), we can determine the time required for the temperature to rise to a specific value, as was successfully accomplished for the furnace scenario: it takes approximately 116 minutes for the outer surface temperature to increase to \(70.1^\circ \mathrm{F}\).

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Most popular questions from this chapter

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)

What is the effect of cooking on the microorganisms in foods? Why is it important that the internal temperature of a roast in an oven be raised above \(70^{\circ} \mathrm{C}\) ?

A 10-cm-inner diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at \(3^{\circ} \mathrm{C}\). The heat transfer coefficient on the surface of the can is \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

The soil temperature in the upper layers of the earth varies with the variations in the atmospheric conditions. Before a cold front moves in, the earth at a location is initially at a uniform temperature of \(10^{\circ} \mathrm{C}\). Then the area is subjected to a temperature of \(-10^{\circ} \mathrm{C}\) and high winds that resulted in a convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) on the earth's surface for a period of \(10 \mathrm{~h}\). Taking the properties of the soil at that location to be \(k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), determine the soil temperature at distances \(0,10,20\), and \(50 \mathrm{~cm}\) from the earth's surface at the end of this \(10-\mathrm{h}\) period.

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