/*! 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 16 Consider a large 3-cm-thick stai... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a large 3-cm-thick stainless steel plate in which heat is generated uniformly at a rate of \(5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming the plate is losing heat from both sides, determine the heat flux on the surface of the plate during steady operation. Answer: \(75 \mathrm{~kW} / \mathrm{m}^{2}\)

Short Answer

Expert verified
Answer: The heat flux on the surface of the plate during steady operation is 75 kW/m^2.

Step by step solution

01

Calculate the volume of heat generated per unit area

To calculate the heat generated per unit area, we will use the following formula: Heat_generated_per_unit_area = Heat_generation_rate × Thickness Where: - Heat_generated_per_unit_area is the heat generated per unit area (W/m^2) - Heat_generation_rate is the rate of heat generation (5 x 10^6 W/m^3) - Thickness is the thickness of the plate (0.03 m)
02

Calculate the heat generated per unit area

Using the formula from Step 1, we will calculate the heat generated per unit area: Heat_generated_per_unit_area = (5 x 10^6 W/m^3) × 0.03 m Heat_generated_per_unit_area = 150000 W/m^2
03

Calculate the heat loss from both sides of the plate

Since the plate loses heat from both sides, we need to divide the total heat generated per unit area by 2 to obtain the heat loss from each side: Heat_loss_per_side = Heat_generated_per_unit_area / 2
04

Calculate the heat flux on the surface of the plate

Now, we can calculate the heat flux on the surface of the plate using the value from Step 3: Heat_loss_per_side = 150000 W/m^2 / 2 Heat_loss_per_side = 75000 W/m^2 The heat flux on the surface of the plate during steady operation is 75 kW/m^2, which matches the provided solution.

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Key Concepts

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

Heat Generation in Solids
Understanding the process of heat generation in solids is fundamental in thermal engineering. It’s a scenario encountered in numerous applications, from electronic components to industrial machining processes. Heat generation can be attributed to various sources such as electrical resistance, chemical reactions, or even nuclear reactions in some cases.

When thermal energy is generated within a solid, it is often expressed in terms of a volumetric heat generation rate, represented by the symbol 'q'' and with units of watts per cubic meter \( \mathrm{W/m^3} \). In the given exercise, the stainless steel plate has a uniform heat generation rate of \( 5 \times 10^6 \mathrm{W/m^3} \). To interpret this, imagine every cubic meter of the steel plate producing 5 million joules every second.

For a more practical approach, such as in the provided exercise, it may be necessary to convert this volumetric rate to an area-specific rate, especially when dealing with heat conduction across a thin plate. This translation is essential for designing cooling systems or in thermal simulations to ensure material integrity and safety in various engineering applications.
Steady-State Thermal Conduction
Steady-state thermal conduction is a condition where the temperature distribution in the material does not change over time, meaning that the thermal energy entering any region of the material is equal to the thermal energy leaving it. This balance is essential in many engineering designs as it implies a predictable and stable operation of the system.

In the context of our textbook problem, assuming steady-state means that the heat generated within the steel plate is consistently being dissipated across its surfaces over time. The rate of heat conduction through a material is governed by Fourier's law of heat conduction, which states that the heat flux \( q \) through a material is proportional to the negative gradient of the temperature and the material's area, given by the equation \( q = -k \cdot A \cdot \frac{dT}{dx} \), where 'k' is the thermal conductivity of the material.

In simpler terms, under steady-state conditions, the plate reaches an equilibrium where the amount of heat it generates internally is equal to the amount it loses to its surroundings, preventing any rise in temperature within the plate over time. This is a decisive factor for thermal management in various engineering fields.
Thermal Analysis of Plates
Thermal analysis of plates involves understanding how heat is conducted and dissipated in plate-like structures. These structures are particularly interesting because of their broad application in electronics, construction, and mechanical systems where they can serve as walls, barriers, or heat exchangers.

A critical aspect of plate thermal analysis is boundary conditions, which define how the plate interacts with its environment. In our textbook example, the stainless steel plate loses heat from both of its larger surfaces, which is a common boundary condition known as 'convective cooling'.

To ensure accurate thermal analysis, it's essential to consider the plate's thickness, material properties, and surface area. By calculating the thermal flux, as shown in the step-by-step solution (75 \( \mathrm{kW/m^2} \)), engineers can predict the temperature distribution within the plate and the extent of heat loss. This is critical for designing thermal systems that are efficient, safe, and tailored to the specific conditions of their application.

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Most popular questions from this chapter

Consider steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal conductivity and no heat generation. Will the temperature in any of these mediums vary linearly? Explain.

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)

Consider steady one-dimensional heat conduction in a plane wall in which the thermal conductivity varies linearly. The error involved in heat transfer calculations by assuming constant thermal conductivity at the average temperature is \((a)\) none, \((b)\) small, or \((c)\) significant.

A spherical vessel is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the vessel. The inner diameter of the vessel is \(5 \mathrm{~m}\) and its inner surface temperature is at \(120^{\circ} \mathrm{C}\). The wall of the vessel has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0018 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). The vessel is situated in a surrounding with an ambient temperature of \(15^{\circ} \mathrm{C}\), the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent thermal burn on skin tissues, the outer surface temperature of the vessel should be kept below \(50^{\circ} \mathrm{C}\). Determine the minimum wall thickness of the vessel so that the outer surface temperature is \(50^{\circ} \mathrm{C}\) or lower.

A large plane wall, with a thickness \(L\) and a thermal conductivity \(k\), has its left surface \((x=0)\) exposed to a uniform heat flux \(\dot{q}_{0}\). On the right surface \((x=L)\), convection and radiation heat transfer occur in a surrounding temperature of \(T_{\infty}\). The emissivity and the convection heat transfer coefficient on the right surface are \(\bar{\varepsilon}\) and \(h\), respectively. Express the houndary conditions and the differential equation of this heat conduction problem during steady operation.

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