This document contains solutions to tutorial problems for a thermodynamics and heat transfer textbook. It includes 9 problems solved relating to concepts like heat transfer processes, Fourier's law, heat loss through windows, convection coefficients, and the Stefan-Boltzmann law. The problems calculate values like heat loss rates, temperature differences, convection coefficients, and maximum allowable chip powers using equations of heat transfer.
This document contains 10 solved problems related to electronics cooling. The problems provide data such as component dimensions, power dissipation, coolant conditions, and environmental temperatures. The solutions involve calculating temperature differences, maximum allowable power, heat transfer rates, and other thermal parameters using equations for one-dimensional steady-state heat conduction, Newton's law of cooling, and radiation heat transfer between surfaces. Graphs and diagrams are provided in some solutions to illustrate the analysis and results.
Heat transfer 5th ed incropera solution manualManish Kumar
This document presents 13 problems related to heat transfer through walls, slabs, windows, and other materials via one-dimensional conduction. The problems provide known parameters like material properties, dimensions, temperatures, and heat flux/rates. The goal is to use Fourier's law of conduction to find unknowns like temperature differences, thermal conductivities, thickness of insulation needed, or heat loss/flux. Assumptions include one-dimensional steady-state conduction with constant properties. Solutions show the analysis and application of Fourier's law to solve for the requested unknowns.
6th ed solution manual---fundamentals-of-heat-and-mass-transferRonald Tenesaca
This document contains 10 problems related to heat transfer by conduction. Each problem provides known information such as materials, dimensions, temperatures, and heat transfer rates. The problems then ask the reader to find unknown values like heat fluxes, surface temperatures, or thicknesses using Fourier's Law of heat conduction and assumptions of one-dimensional and steady-state heat transfer. The problems cover a variety of applications including insulation, walls, windows, refrigeration, and cooking.
This document discusses heat transfer via conduction and presents a numerical solution to the heat equation in one dimension using finite difference approximations in MATLAB. It begins by introducing the three modes of heat transfer and Fourier's law of heat conduction. It then describes the problem of subsurface temperature fluctuations over time. The solution section presents the heat equation and initial/boundary conditions. It describes using forward, backward and central finite differences to discretize the equation. The MATLAB code implements this solution over 400 depth steps and 5000 time steps to plot the temperature at various depths over time. The code converges and plots are shown of the temperature distribution and profiles at different depths.
This document provides the problem statement, solution approach, and analysis for determining the total acceleration in an axisymmetric parallel flow in a tube. The key steps are:
1) The axial velocity distribution is given and is independent of axial and angular position.
2) Total acceleration expressions in cylindrical coordinates are applied.
3) For parallel streamlines, the radial velocity is zero.
4) The axial acceleration is found to be zero, and the radial and angular accelerations are also zero. All acceleration components vanish for this parallel flow case.
The document discusses unsteady state heat conduction, which occurs when the temperature gradient across a solid changes over time. It provides examples of unsteady state conduction and examines the one-dimensional case. It introduces the governing equation and describes using charts like the Geankoplis charts to solve problems involving startup conditions for plates, cylinders, and spheres. It then provides examples of using the charts to solve problems involving cooling/heating of various objects.
The document provides data from a test on a flat-walled furnace with inner and outer brick walls of unknown conductivity. The temperatures at various points within the furnace walls are given. Using this data and the thickness of each material, the conductivity of each brick type is calculated. The percentage of heat loss saved by adding 5cm of magnesia insulation to the furnace is calculated to be 52.73%.
This document contains 10 solved problems related to electronics cooling. The problems provide data such as component dimensions, power dissipation, coolant conditions, and environmental temperatures. The solutions involve calculating temperature differences, maximum allowable power, heat transfer rates, and other thermal parameters using equations for one-dimensional steady-state heat conduction, Newton's law of cooling, and radiation heat transfer between surfaces. Graphs and diagrams are provided in some solutions to illustrate the analysis and results.
Heat transfer 5th ed incropera solution manualManish Kumar
This document presents 13 problems related to heat transfer through walls, slabs, windows, and other materials via one-dimensional conduction. The problems provide known parameters like material properties, dimensions, temperatures, and heat flux/rates. The goal is to use Fourier's law of conduction to find unknowns like temperature differences, thermal conductivities, thickness of insulation needed, or heat loss/flux. Assumptions include one-dimensional steady-state conduction with constant properties. Solutions show the analysis and application of Fourier's law to solve for the requested unknowns.
6th ed solution manual---fundamentals-of-heat-and-mass-transferRonald Tenesaca
This document contains 10 problems related to heat transfer by conduction. Each problem provides known information such as materials, dimensions, temperatures, and heat transfer rates. The problems then ask the reader to find unknown values like heat fluxes, surface temperatures, or thicknesses using Fourier's Law of heat conduction and assumptions of one-dimensional and steady-state heat transfer. The problems cover a variety of applications including insulation, walls, windows, refrigeration, and cooking.
This document discusses heat transfer via conduction and presents a numerical solution to the heat equation in one dimension using finite difference approximations in MATLAB. It begins by introducing the three modes of heat transfer and Fourier's law of heat conduction. It then describes the problem of subsurface temperature fluctuations over time. The solution section presents the heat equation and initial/boundary conditions. It describes using forward, backward and central finite differences to discretize the equation. The MATLAB code implements this solution over 400 depth steps and 5000 time steps to plot the temperature at various depths over time. The code converges and plots are shown of the temperature distribution and profiles at different depths.
This document provides the problem statement, solution approach, and analysis for determining the total acceleration in an axisymmetric parallel flow in a tube. The key steps are:
1) The axial velocity distribution is given and is independent of axial and angular position.
2) Total acceleration expressions in cylindrical coordinates are applied.
3) For parallel streamlines, the radial velocity is zero.
4) The axial acceleration is found to be zero, and the radial and angular accelerations are also zero. All acceleration components vanish for this parallel flow case.
The document discusses unsteady state heat conduction, which occurs when the temperature gradient across a solid changes over time. It provides examples of unsteady state conduction and examines the one-dimensional case. It introduces the governing equation and describes using charts like the Geankoplis charts to solve problems involving startup conditions for plates, cylinders, and spheres. It then provides examples of using the charts to solve problems involving cooling/heating of various objects.
The document provides data from a test on a flat-walled furnace with inner and outer brick walls of unknown conductivity. The temperatures at various points within the furnace walls are given. Using this data and the thickness of each material, the conductivity of each brick type is calculated. The percentage of heat loss saved by adding 5cm of magnesia insulation to the furnace is calculated to be 52.73%.
1. The document discusses 1D steady state heat conduction through two solid plates in contact. It describes how heat transfer occurs through the interface via solid contact spots and gaps, and defines thermal contact resistance.
2. Thermal contact resistance values reported in experiments typically fall between 0.000005 and 0.0005 m2·°C/W. The corresponding thermal contact conductance ranges from 2000 to 200,000 W/m2·°C.
3. The importance of considering thermal contact resistance in heat transfer problems is discussed.
The document describes a problem calculating the heat loss from a furnace with inner dimensions of 1m x 1.2m x 0.75m and 11cm thick refractory walls. It provides the inner and outer surface temperatures and conductivity and calculates the total heat loss as 881798399.8 J over 24 hours by conduction through the walls. A second problem calculates the heat loss from a duct with ceramic and insulation layers given temperatures, thicknesses, and conductivities. It uses an iterative process to calculate the interface temperature and heat loss.
This document provides an overview of heat conduction concepts and equations. It begins with an introduction to Fourier's law of heat conduction and the heat diffusion equation. It then discusses one-dimensional, steady-state heat conduction problems like conduction through plane walls, cylinders, and spheres. It also covers transient heat conduction and internal heat generation. Boundary conditions like convection, radiation, and contact resistance are presented. Finally, the document discusses extended surfaces like fins and provides the equations to calculate fin efficiency.
1) The document derives the one-dimensional heat conduction equation for plane walls, cylinders, and spheres through an energy balance on a thin volume element.
2) For all three geometries, the heat conduction equation can be written compactly as (1/r^n)∂(r^n ∂T/∂r) + ġ = ρC∂T/∂t, where n=0 for a plane wall, n=1 for a cylinder, and n=2 for a sphere.
3) The heat conduction equation is also derived for several special cases including steady-state and transient conditions with and without heat generation.
The document discusses heat transfer through a thermopane double window. It calculates that for a temperature drop of 27.8 K, the heat loss through a 0.914 m by 1.83 m window would be 179.58 W. It also presents a solution to calculating the temperature of lead shot after falling 15 seconds into a quenching oil bath held at 32.2°C.
This document summarizes research that uses a binary coded genetic algorithm to optimize the design of fin arrays on an internal combustion engine cylinder. The goal is to maximize heat transfer through either rectangular or triangular fin profiles. Mathematical models are developed to calculate the heat transfer based on parameters like fin length, thickness, spacing, and material properties. Constraints related to fin mass and dimensions are also considered. The genetic algorithm operates on a population of potential designs encoded as binary strings. Over generations, the algorithms applies reproduction, crossover and mutation to arrive at optimal fin array designs that maximize heat transfer within the given constraints. Results show that the triangular fin profile provides advantages over the rectangular profile in terms of heat transfer per unit mass.
This document discusses heat transfer through conduction and various methods for solving the heat equation using finite difference approximations. It introduces the heat equation in Cartesian, cylindrical and spherical coordinates. It discusses boundary conditions and describes setting up a nodal network to discretize the domain. It then presents the finite difference form of the heat equation and describes different cases for nodal finite difference equations, including for interior nodes, nodes at corners or surfaces with convection, and nodes at surfaces with uniform heat flux. It discusses solving the finite difference equations using matrix inversion, Gauss-Seidel iteration, and provides examples.
The document provides information about calculating heat loss from a furnace and exhaust duct. It gives dimensions, material properties, and temperature values. It then shows calculations to determine the total heat loss over 24 hours from the furnace, which is 801.65 MJ. For the exhaust duct, it uses a trial-and-error method to calculate the interface temperature and heat loss.
Conductive Heat Transfer Laboratory Experimentdp93
This document outlines an experiment to measure the thermal conductivity of various materials. The objectives were to measure thermal conductivity for different materials and analyze the effect of plexiglas thickness. Materials tested included stainless steel, plywood, and plexiglas plates of varying thicknesses. Fourier's law of heat conduction was used to calculate thermal conductivity from temperature and heat flux measurements. Results showed metals have higher conductivity than plexiglas or wood. Thickness was found to not impact conductivity as expected, likely due to air gaps and external currents affecting measurements. Recommendations include improving apparatus sealing and shielding from air currents.
The document discusses the one-dimensional heat conduction equation. It begins by introducing heat transfer concepts like temperature, heat transfer, and Fourier's Law of heat conduction. It then derives the general heat conduction equation for one-dimensional conduction in a plane wall, long cylinder, and sphere. The equation can be simplified for special cases like steady-state and transient problems without heat generation.
1-D Steady State Heat Transfer With Heat GenerationMihir Patel
1) The document describes heat transfer through a sphere with uniform heat generation under steady-state conditions.
2) The temperature distribution is derived to be parabolic, with the maximum temperature at the center of the sphere.
3) The maximum temperature and temperature distribution are expressed in terms of the heat generation rate, thermal conductivity, radius of the sphere, and temperature at the outer surface.
This document provides examples of temperature conversions between Celsius, Fahrenheit, and Kelvin scales. It also includes examples of calculating thermal expansion of materials when temperature changes. Key points covered include:
- Converting between Celsius, Fahrenheit, and Kelvin scales using defined formulas.
- Calculating changes in length, area, and volume of various materials like metals and liquids as their temperatures change, using thermal expansion coefficients.
- Examples involve temperatures ranges from -1960C to 4500F and materials like copper, steel, glass, and liquids. Thermal expansion is used to calculate changes in dimensions or if materials will overflow containers.
The document summarizes solutions to an in-class exercise on heat transfer problems. It includes:
1) Calculation of heat loss through a window glass using conduction, finding higher loss for thinner glass.
2) Use of the convection equation to find heat transfer from hot air to a plate.
3) Calculation of heat loss from a person by radiation to room walls at different temperatures.
4) Use of conduction, convection, radiation equations to determine the effective thermal conductivity of a wall from its inner and outer temperatures and heat transfer rates.
The rate of heat flow through both rods is the same.
dT/dx is constant along each rod.
dT/dx along D = dT/dx along E
(100-80)/0.2 = (T-80)/0.4
T = 120 °C
Therefore, the temperature at the free end of rod E is 120 °C.
20
The document discusses heat conduction equations in various coordinate systems and geometries:
1) It presents Fourier's law of heat conduction and derives the one-dimensional heat conduction equation.
2) It then extends the derivation to multidimensional systems, deriving equations for cylindrical and spherical geometries with variable or constant conductivity.
3) Finally, it presents the general three-dimensional heat conduction equation applicable to problems involving transient or steady heat transfer in solids with arbitrary geometries and material properties.
Solution Manual for Heat Convection second edition by Latif M. Jijiphysicsbook
Solution Manual for Heat Convection
https://unihelp.xyz/solution-manual-for-heat-convection-by-latif-jiji/
****
Solution Manual for Heat Conduction
https://unihelp.xyz/solution-manual-heat-conduction-latif-jiji/
Solution Manual for Heat Convection second edition by Latif M. Jiji
The document provides information about the heat loss calculation of a furnace with inner dimensions of 1m x 1.2m x 0.75m and wall thickness of 11 cm made of refractory material. The inner and outer surface temperatures are 200°C and 38°C respectively. The total heat loss by conduction in 24 hours is calculated to be 801.65 MJ.
The document discusses calculating the temperature at the boundary between a refractory brick layer and insulating material layer in a plane wall. Given information includes the thicknesses of each layer, thermal conductivities, and temperatures on both surfaces. The calculated temperature at the boundary is 867.75°C. Another example calculates the time to cool a glass sheet submerged in water from an initial to average final temperature. The calculated time is 130.21 hours.
The document describes a problem calculating the heat loss from a furnace with inner dimensions of 1m x 1.2m x 0.75m and 11cm thick refractory walls. It provides the inner and outer surface temperatures and conductivity and calculates the total heat loss as 881798399.8 J over 24 hours by conduction through the walls. A second problem calculates the heat loss from a duct with ceramic and insulation layers given temperatures, thicknesses, and conductivities. It uses an iterative process to calculate the interface temperature and heat loss.
1. The document discusses 1D steady state heat conduction through two solid plates in contact. It describes how heat transfer occurs through the interface via solid contact spots and gaps, and defines thermal contact resistance.
2. Thermal contact resistance values reported in experiments typically fall between 0.000005 and 0.0005 m2·°C/W. The corresponding thermal contact conductance ranges from 2000 to 200,000 W/m2·°C.
3. The importance of considering thermal contact resistance in heat transfer problems is discussed.
The document describes a problem calculating the heat loss from a furnace with inner dimensions of 1m x 1.2m x 0.75m and 11cm thick refractory walls. It provides the inner and outer surface temperatures and conductivity and calculates the total heat loss as 881798399.8 J over 24 hours by conduction through the walls. A second problem calculates the heat loss from a duct with ceramic and insulation layers given temperatures, thicknesses, and conductivities. It uses an iterative process to calculate the interface temperature and heat loss.
This document provides an overview of heat conduction concepts and equations. It begins with an introduction to Fourier's law of heat conduction and the heat diffusion equation. It then discusses one-dimensional, steady-state heat conduction problems like conduction through plane walls, cylinders, and spheres. It also covers transient heat conduction and internal heat generation. Boundary conditions like convection, radiation, and contact resistance are presented. Finally, the document discusses extended surfaces like fins and provides the equations to calculate fin efficiency.
1) The document derives the one-dimensional heat conduction equation for plane walls, cylinders, and spheres through an energy balance on a thin volume element.
2) For all three geometries, the heat conduction equation can be written compactly as (1/r^n)∂(r^n ∂T/∂r) + ġ = ρC∂T/∂t, where n=0 for a plane wall, n=1 for a cylinder, and n=2 for a sphere.
3) The heat conduction equation is also derived for several special cases including steady-state and transient conditions with and without heat generation.
The document discusses heat transfer through a thermopane double window. It calculates that for a temperature drop of 27.8 K, the heat loss through a 0.914 m by 1.83 m window would be 179.58 W. It also presents a solution to calculating the temperature of lead shot after falling 15 seconds into a quenching oil bath held at 32.2°C.
This document summarizes research that uses a binary coded genetic algorithm to optimize the design of fin arrays on an internal combustion engine cylinder. The goal is to maximize heat transfer through either rectangular or triangular fin profiles. Mathematical models are developed to calculate the heat transfer based on parameters like fin length, thickness, spacing, and material properties. Constraints related to fin mass and dimensions are also considered. The genetic algorithm operates on a population of potential designs encoded as binary strings. Over generations, the algorithms applies reproduction, crossover and mutation to arrive at optimal fin array designs that maximize heat transfer within the given constraints. Results show that the triangular fin profile provides advantages over the rectangular profile in terms of heat transfer per unit mass.
This document discusses heat transfer through conduction and various methods for solving the heat equation using finite difference approximations. It introduces the heat equation in Cartesian, cylindrical and spherical coordinates. It discusses boundary conditions and describes setting up a nodal network to discretize the domain. It then presents the finite difference form of the heat equation and describes different cases for nodal finite difference equations, including for interior nodes, nodes at corners or surfaces with convection, and nodes at surfaces with uniform heat flux. It discusses solving the finite difference equations using matrix inversion, Gauss-Seidel iteration, and provides examples.
The document provides information about calculating heat loss from a furnace and exhaust duct. It gives dimensions, material properties, and temperature values. It then shows calculations to determine the total heat loss over 24 hours from the furnace, which is 801.65 MJ. For the exhaust duct, it uses a trial-and-error method to calculate the interface temperature and heat loss.
Conductive Heat Transfer Laboratory Experimentdp93
This document outlines an experiment to measure the thermal conductivity of various materials. The objectives were to measure thermal conductivity for different materials and analyze the effect of plexiglas thickness. Materials tested included stainless steel, plywood, and plexiglas plates of varying thicknesses. Fourier's law of heat conduction was used to calculate thermal conductivity from temperature and heat flux measurements. Results showed metals have higher conductivity than plexiglas or wood. Thickness was found to not impact conductivity as expected, likely due to air gaps and external currents affecting measurements. Recommendations include improving apparatus sealing and shielding from air currents.
The document discusses the one-dimensional heat conduction equation. It begins by introducing heat transfer concepts like temperature, heat transfer, and Fourier's Law of heat conduction. It then derives the general heat conduction equation for one-dimensional conduction in a plane wall, long cylinder, and sphere. The equation can be simplified for special cases like steady-state and transient problems without heat generation.
1-D Steady State Heat Transfer With Heat GenerationMihir Patel
1) The document describes heat transfer through a sphere with uniform heat generation under steady-state conditions.
2) The temperature distribution is derived to be parabolic, with the maximum temperature at the center of the sphere.
3) The maximum temperature and temperature distribution are expressed in terms of the heat generation rate, thermal conductivity, radius of the sphere, and temperature at the outer surface.
This document provides examples of temperature conversions between Celsius, Fahrenheit, and Kelvin scales. It also includes examples of calculating thermal expansion of materials when temperature changes. Key points covered include:
- Converting between Celsius, Fahrenheit, and Kelvin scales using defined formulas.
- Calculating changes in length, area, and volume of various materials like metals and liquids as their temperatures change, using thermal expansion coefficients.
- Examples involve temperatures ranges from -1960C to 4500F and materials like copper, steel, glass, and liquids. Thermal expansion is used to calculate changes in dimensions or if materials will overflow containers.
The document summarizes solutions to an in-class exercise on heat transfer problems. It includes:
1) Calculation of heat loss through a window glass using conduction, finding higher loss for thinner glass.
2) Use of the convection equation to find heat transfer from hot air to a plate.
3) Calculation of heat loss from a person by radiation to room walls at different temperatures.
4) Use of conduction, convection, radiation equations to determine the effective thermal conductivity of a wall from its inner and outer temperatures and heat transfer rates.
The rate of heat flow through both rods is the same.
dT/dx is constant along each rod.
dT/dx along D = dT/dx along E
(100-80)/0.2 = (T-80)/0.4
T = 120 °C
Therefore, the temperature at the free end of rod E is 120 °C.
20
The document discusses heat conduction equations in various coordinate systems and geometries:
1) It presents Fourier's law of heat conduction and derives the one-dimensional heat conduction equation.
2) It then extends the derivation to multidimensional systems, deriving equations for cylindrical and spherical geometries with variable or constant conductivity.
3) Finally, it presents the general three-dimensional heat conduction equation applicable to problems involving transient or steady heat transfer in solids with arbitrary geometries and material properties.
Solution Manual for Heat Convection second edition by Latif M. Jijiphysicsbook
Solution Manual for Heat Convection
https://unihelp.xyz/solution-manual-for-heat-convection-by-latif-jiji/
****
Solution Manual for Heat Conduction
https://unihelp.xyz/solution-manual-heat-conduction-latif-jiji/
Solution Manual for Heat Convection second edition by Latif M. Jiji
The document provides information about the heat loss calculation of a furnace with inner dimensions of 1m x 1.2m x 0.75m and wall thickness of 11 cm made of refractory material. The inner and outer surface temperatures are 200°C and 38°C respectively. The total heat loss by conduction in 24 hours is calculated to be 801.65 MJ.
The document discusses calculating the temperature at the boundary between a refractory brick layer and insulating material layer in a plane wall. Given information includes the thicknesses of each layer, thermal conductivities, and temperatures on both surfaces. The calculated temperature at the boundary is 867.75°C. Another example calculates the time to cool a glass sheet submerged in water from an initial to average final temperature. The calculated time is 130.21 hours.
The document describes a problem calculating the heat loss from a furnace with inner dimensions of 1m x 1.2m x 0.75m and 11cm thick refractory walls. It provides the inner and outer surface temperatures and conductivity and calculates the total heat loss as 881798399.8 J over 24 hours by conduction through the walls. A second problem calculates the heat loss from a duct with ceramic and insulation layers given temperatures, thicknesses, and conductivities. It uses an iterative process to calculate the interface temperature and heat loss.
Fundamentals of heat transfer lecture notesYuri Melliza
This document discusses various modes of heat transfer including conduction, convection, and radiation. It provides equations to calculate heat transfer via these different modes. For conduction, Fourier's law and its relation to electrical resistance is explained. For convection, Newton's law of cooling and relationships between heat transfer coefficients, temperature differences and surface areas are given. Stefan-Boltzmann law is described for radiation heat transfer between black and gray bodies. Methods to relate convection and radiation heat transfer are also presented. Several examples are provided to demonstrate calculations of heat transfer via different modes.
This document provides an introduction to heat transfer and the different modes of heat transfer including conduction, convection, radiation, boiling, and condensation. It defines key terms like thermal conductivity and heat flux. Examples of one-dimensional heat conduction through plane walls, cylindrical walls, and multilayer walls are presented. The document also provides sample problems and their solutions for determining heat transfer rates through composite walls and insulated cylinders.
1) Heat transfer by conduction occurs due to random molecular motion within a material. The rate of heat transfer by conduction is proportional to the temperature gradient and the thermal conductivity of the material.
2) Fourier's law of heat conduction describes conduction in Cartesian, cylindrical, and spherical coordinate systems. It relates the heat flux to the temperature gradient through the thermal conductivity.
3) The heat equation can be derived by applying the law of conservation of energy combined with Fourier's law. It describes the distribution of temperature as a function of time and space within a body undergoing transient or steady-state heat conduction.
1) An electronic chip attached to an air-cooled heat sink with 11 fins is analyzed to determine the maximum power the chip can dissipate without exceeding 85°C.
2) The thermal resistance network is analyzed using the fin equation and equations for thermal resistance of each component.
3) The maximum allowable power is found to be 31.8W based on the geometry, material properties, and convection coefficients given.
4) Without the heat sink, the maximum power would be much lower at only 2.60W due to the higher thermal resistance of natural convection alone. The heat sink significantly increases the chip's cooling ability.
The document discusses two problems involving heat transfer through insulation layers. Problem 1 calculates the heat loss per square meter for a furnace wall that is 23 cm thick, with inner and outer surface temperatures of 315°C and 38°C respectively. Problem 2 calculates the reduction in heat transfer when better insulating material is placed either on the inside or outside of two layers of pipe insulation each 0.0245 m thick. It is found that heat transfer reduces by 35.98% when the better insulator is near the pipe.
The document provides details on two problems involving heat exchanger design and optimization. Problem 11 involves designing a condenser to condense 2000 kg/hr of vapor at 132°C using cooling water. The optimal solutions are a cooling water flow rate of 2284.58 kg/hr and an exit temperature of 106.54°C. Problem 17 requires determining the optimal exit temperature and optimal tube length but no other details are provided.
This lecture discusses steady state heat transfer through composite slabs. It provides examples of calculating heat transfer rate, thermal resistances, and intermediate temperatures in multi-layer slabs. The examples solve for overall heat transfer coefficient and surface temperatures in furnace walls, double pane windows, and oven walls consisting of multiple materials and layers.
James Clerk Maxwell's equations represent the fundamentals of electricity and magnetism in an elegant and concise form. The document discusses various units used to measure magnetic flux, such as the Maxwell and Weber. It then examines Maxwell's modifications to Ampere's law by including the concept of displacement current to account for changing electric fields producing magnetic fields. As an example, the document calculates the magnetic field produced near a parallel plate capacitor due to the changing electric field between its plates.
The document contains two engineering problems involving heat transfer calculations:
1) Calculating heat loss from a spherical furnace wall over 24 hours, and the temperature at a radius of 1.1m.
2) Modeling transient heat conduction in a slab, constructing a temperature profile table up to 4000 seconds using finite differences.
The problems provide heat transfer parameters and geometry, and require calculating heat loss, temperature, or constructing a temperature profile table.
The document contains two engineering problems involving heat transfer calculations:
1) Calculating heat loss from a spherical furnace wall over 24 hours, finding the temperature at a radius of 1.1m is 543.64°C.
2) Modeling transient heat transfer in a slab, constructing a temperature profile table up to 4000 seconds using finite differences, with the temperature at 1500s being 150.8°C.
The document contains multiple engineering problems involving heat transfer calculations. Problem 7 calculates the temperature at the boundary between a refractory brick layer and insulating material layer based on given temperatures, thicknesses, and thermal conductivities. Problem 15 calculates the time required for a glass sheet to cool from an initial to average temperature when placed in running water. Problem Geankoplis 4.1-2 calculates thermal conductivity given initial temperatures, heat flux, and material thickness. Problem Geankoplis 5.3-7 calculates the internal temperature of a steel rod after 1 hour of cooling in an oil bath using given properties and a Heisler chart.
The document contains multiple engineering problems involving heat transfer calculations. Problem 7 calculates the temperature at the boundary between a refractory brick layer and insulating material layer based on given temperatures, thicknesses, and thermal conductivities. Problem 15 calculates the time required for a glass sheet to cool from an initial to average temperature in a water stream. Problem Geankoplis 4.1-2 calculates thermal conductivity using given temperatures, heat flux, and material thickness. Problem Geankoplis 5.3-7 calculates the internal temperature of a steel rod after 1 hour of cooling in an oil bath using given properties and conditions.
This document provides answers to questions about electric current and resistance from Chapter 27. It includes calculations of current, resistance, resistivity, and other electrical properties. Key points addressed include:
- How geometry and temperature affect resistance
- Models for electrical conduction and how resistance changes with temperature
- Properties of superconductors and electrical power calculations
- Questions cover topics like current density, resistivity, Ohm's law, and resistances of different materials.
1. The document contains 11 electrostatics problems involving calculating electric field intensity, charge, capacitance, and energy for various charged objects like spheres, plates, and cylinders in different materials and configurations.
2. The problems apply equations for electric field intensity outside charged objects, charge density on surfaces, capacitance of parallel plates, and energy density in dielectrics to solve for unknown values.
3. Dielectric constants are provided for different materials like water, acetone, and bakelite, and are incorporated into the calculations.
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
This presentation was provided by Steph Pollock of The American Psychological Association’s Journals Program, and Damita Snow, of The American Society of Civil Engineers (ASCE), for the initial session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session One: 'Setting Expectations: a DEIA Primer,' was held June 6, 2024.
Executive Directors Chat Leveraging AI for Diversity, Equity, and InclusionTechSoup
Let’s explore the intersection of technology and equity in the final session of our DEI series. Discover how AI tools, like ChatGPT, can be used to support and enhance your nonprofit's DEI initiatives. Participants will gain insights into practical AI applications and get tips for leveraging technology to advance their DEI goals.
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
Physiology and chemistry of skin and pigmentation, hairs, scalp, lips and nail, Cleansing cream, Lotions, Face powders, Face packs, Lipsticks, Bath products, soaps and baby product,
Preparation and standardization of the following : Tonic, Bleaches, Dentifrices and Mouth washes & Tooth Pastes, Cosmetics for Nails.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...PECB
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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1. MECH3400 Thermodynamics and Heat Transfer
Solutions for tutorial problems (Tutorial 1)
Textbook: Incropera and Dewitt (5th edition)
(by Bo Feng, 2006)
PART II
1. Q1.73, page 48. Identification of heat transfer processes.
2.
3.
4.
5. 2. Q1.5, page 34. The inner and outer surface temperatures of a glass window 5 mm thick are 15
and 5 °C. What is the heat loss through a window that is 1m by 3 m on a side? The thermal
conductivity of glass is 1.4 W/mK.
Solution:
Assume the process is at steady state and the direction of x is from the inner surface to the
outer surface. The heat loss may be obtained from Fourier’s law:
ΔT (5 − 15) K
q = −kA = −1.4W / mK × (1m × 3m ) = 8.4kW
L 5 × 10 −3 m
The value is positive, indicating the direction of heat flow is from the inner surface to the
outer surface.
3. Q1.6, page 34. A glass window of width W=1m and height H=2m is 5 mm thick and has a
thermal conductivity of kg=1.4 W/mK. If the inner and outer surface temperatures of the glass
are 15 and -20 °C, respectively, on a cold winter day, what is the rate of heat loss through the
glass? To reduce heat loss through windows, it is customary to use a double pane construction
in which adjoining panes are separated by an air space. If the spacing is 10 mm and the glass
surfaces in contact with the air have temperatures of 10 °C and -15 °C, what is the rate of heat
loss from a 1m * 2m window? The thermal conductivity of air is ka=0.024 W/mK.
Solution:
Assume steady state for the first case and the direction of x is from inner to outer surface. The
heat loss is
ΔT (−20 − 15) K
q = −kA = −1.4W / mK × (1m × 2m ) = 19.6kW
L 5 × 10 −3 m
The value is positive, indicating the direction of heat flow is from the inner surface to the
outer surface.
In the second case with double pane construction, the heat loss is through three layers, the
inner window panel, the air separation layer and the outer panel. The heat loss is determined
by the layer with the least heat conduction. Assume the air between the two windows is
motionless, or convection is negligible. The heat loss through the air layer is
ΔT (−15 − 10) K
q a = −k a A = −0.024W / mK × (1m × 2m) = 120W
L 10 × 10 −3 m
ΔT (10 − 15) K
q g ,in = −k g A = −1.4W / mK × (1m × 2m ) = 2.8kW
L 5 × 10 −3 m
ΔT (−20 − (−15)) K
q g ,out = −k g A = −1.4W / mK × (1m × 2m ) = 2.8kW
L 5 × 10 −3 m
Since the heat loss from the air layer is much less than that from the glass, the heat loss from
the window is limited by the heat loss from the air layer, i.e. is 120W. It also indicates that at
this moment steady state has not established.
Comments: We may find out the surface temperatures of the window panels at steady state
(the innermost surface has a temperature of Ti=15 °C and the outermost surface has a
temperature of To=-20 °C). At steady state, we have (Ts1 is the temperature of the inner panel
surface in contact with air and Ts2 is the temperature of the outer pane surface in contact with
air)
6. Ts1 − Ti To − Ts 2 To − Ti
q= = =
R g1 Rg 2 RT
L g1 5 × 10 −3 m
in which R g1 = R g 2 = = = 1.79 × 10 −3 K / W
k g A 1.4W / mK (1m × 2m )
La 10 × 10 −3 m
RT = R g1 + Ra + R g 2 = 2 R g1 + = 2 × 1.79 × 10 −3 K / W + = 0.21K / W
ka A 0.024W / mK (1m × 2m )
Therefore,
−3
(To − Ti ) = 15 + 1.79 × 10
R g1
Ts1 = Ti +
K /W
(− 20 − 15) = 14.7C
RT 0.21K / W
and Ts 2 = −19.7C .
The results indicate that the air layer between the two panels serves well as an insulation layer.
The major temperature drop occurs in this layer.
4. Q1.9, page 35. What is the thickness required of a masonry wall having thermal conductivity
0.75 W/mK if the heat rate is to be 80 % of the heat rate through a composite structural wall
having a thermal conductivity of 0.25 W/mK and a thickness of 100 mm? Both walls are
subjected to the same surface temperature difference.
Solution:
Assume steady state, we have the heat rates for two walls
ΔT
q1 = −k1 A
L1
ΔT
q 2 = −k 2 A
L2
Compare these two equations, we have
q1 k1 L2
=
q 2 k 2 L1
for the two walls having the same temperature difference and the same heat transfer area.
Therefore
k1 q 2 0.75 1
L1 = L2 = 100mm = 375mm
k 2 q1 0.25 0.8
5. Q1.11, page 35. A square silicon chip (k=150 W/mK) is of width w=5 mm on a side and of
thickness t=1 mm. The chip is mounted in a surstrate such that its side and back surfaces are
insulated, while the front surface is exposed to a coolant. If 4 W are being dissipated in
circuits mounted to the back surface of the chip, what is the steady-state temperature
7. difference between back and front surfaces?
Solution:
Using the Fourier’s law, assuming the direction of x is from bottom to top
ΔT ΔT
q = −kA = −kw 2
L t
therefore
qt 4W × 1 × 10 −3 m
ΔT = − =− = −1.07 K
kw 2 (
150W / mK 5 × 10 −3 )
2
The minus sign here indicates that the temperature from the bottom to the top is decreasing.
6. Q1.14, page 35. Air at 40 °C flows over a long, 25-mm diameter cylinder with an embedded
electrical heater. In a series of tests, measurements were made of the power per unit length, P’,
required to maintain the cylinder surface temperature at 300 °C for different freestream
velocities V of the air. The results are as follows:
Air velocity, V(m/s) 1 2 4 8 12
Power, P’ (W/m) 450 658 983 1507 1963
(a) Determine the convection coefficient for each velocity, and display your results
graphically.
(b) Assuming the dependence of the convection coefficient on the velocity to be of the form
h = CV n , determine the parameters C and n from the results of part (a).
Solution:
(a) Assume the heat flow is positive if it is from the surface to the fluid. From Newton’s law
of cooling
q = hA(Ts − T∞ )
Therefore
q q P'
h= = =
A(Ts − T∞ ) (πDL )(Ts − T∞ ) πD(Ts − T∞ )
in which P’=q/L is the heat loss per unit length. The above equation may be used to
calculate the value of h at each velocity, and the results are shown in the table and the
figure below.
Air velocity, V(m/s) 1 2 4 8 12
Power, P’ (W/m) 450 658 983 1507 1963
Convection coefficient, h W/m2K 22.04 32.22 48.14 73.80 96.13
8. 100
convection cofficient (W/m K)
2
80
60
40
20
0
0 2 4 6 8 10 12 14
velocity (m/s)
(b) The values of C and n can be determined by fitting the experimental data. C=21.02.
n=0.61. The solid line in the above figure is the results predicted using the fitted values of
C and n.
7. Q1.18, page 36. A square isothermal chip is of width w=5 mm on a side and is mounted in a
substrate such that its side and back surfaces are well insulated, while the front surface is
exposed to the flow of a coolant at T∞=15°C. From reliability considerations, the chip
temperature must not exceed T=85°C. If the coolant is air and the corresponding convection
coefficient is h=200W/m2K, what is the maximum allowable chip power? If the coolant is a
dielectric liquid for which h=3000 W/m2K, what is the maximum allowable power?
Solution:
Assume the heat flux is positive from the chip to the air, using Newton’s law:
q = hA(Ts − T∞ )
When the convection coefficient is h=200 W/m2K, the maximum allowable chip power is
reached when the chip temperature is the highest.
(
q = hA(Ts − T∞ ) = 200W / m 2 K × 5 × 10 −3 m 2 (85 − 15)K = 0.35W)
2
When the convection coefficient is h=3000 W/m2K, the maximum allowable chip power is
reached when the chip temperature is the highest.
(
q = hA(Ts − T∞ ) = 3000W / m 2 K × 5 × 10 −3 m 2 (85 − 15)K = 5.25W )
2
Comments: Increasing the convection coefficient is an effective way of increasing heat
9. transfer. In this example, if a liquid is used the allowable power of the chip is increased
significantly. However practically it is difficult to do so and it is easier to increase the heat
transfer area, i.e. using fins.
8. Q1.30, page 38. Stefan-Boltzmann law.
Solution: