Heat transfer problem solver pdf

To Mr. Osama Mahmoud of Daniya Center for publishing and printing services whose patience in editing and re — editing the manuscript of this book was the momentum that pushed me in completing successfully the present book.

I am also indebted to published texts in thermodynamics and heat and mass transfer which have been contributed to the author's thinking. The author is extremely grateful to them for constructive criticisms and valuable proposals. I express my profound gratitude to Mr. Osama Mahmoud and Mr. Ahmed Abulgasim of Daniya Center for Printing and Publishing services, Atbara, Sudan who they spent several hours in editing, re — editing and correcting the present manuscript.

Special appreciation is due to the British Council's Library for its quick response in ordering the requested bibliography, books, reviews and papers. They tend to read them as one might read a novel and fail to appreciate what is being set out in each section. The result is that the student ends his reading with a glorious feeling of knowing it all and with, in fact, no understanding of the subject whatsoever. To avoid this undesirable end a modern presentation has been adopted for this book.

The subject has been presented in the form of solution of comprehensive examples in a step by step form. The example itself should contain three major parts, the first part is concerned with the definition of terms, the second part deals with a systematic derivation of equations to terminate the problem to its final stage, the third part is pertinent to the ability and skill in solving problems in a logical manner.

This book aims to give students of engineering a thorough grounding in the subject of heat transfer. The book is comprehensive in its coverage without sacrificing the necessary theoretical details. The book is designed as a complete course test in heat transfer for degree courses in mechanical and production engineering and combined studies courses in which heat transfer and related topics are an important part of the curriculum.

Students on technician diploma and certificate courses in engineering will also find the book suitable although the content is deeper than they might require.

The entire book has been thoroughly revised and a large number of solved examples and additional unsolved problems have been added. This book contains comprehensive treatment of the subject matter in simple and direct language. The book comprises eight chapters.

All chapters are saturated with much needed text supported and by simple and self-explanatory examples. Chapter one includes general introduction to transient conduction or unsteady conduction, definition of its fundamental terms, derivation of equations and a wide spectrum of solved examples.

This chapter was supported by different solved examples. Chapter three discusses the importance of transient heat conduction in solids with finite conduction and convective resistances. At the end of this chapter a wide range of solved examples were added. These examples were solved using Heisler charts. In chapter four transient heat conduction in semi — infinite solids were introduced and explained through the solution of different examples using Gaussian error function in the form of tables and graphs.

Chapter five deals with the periodic variation of surface temperature where the periodic type of heat flow was explained in a neat and regular manner.

At the end of this chapter a wide range of solved examples was introduced. Chapter six concerns with temperature distribution in transient conduction.

In using such distribution, the one dimensional transient heat conduction problems could be solved easily as explained in examples. In chapter seven additional examples in lumped capacitance system or negligible internal resistance theory were solved in a systematic manner, so as to enable the students to understand and digest the subject properly. Chapter eight which is the last chapter of this book contains unsolved theoretical questions and further problems in lumped capacitance system.

How these problems are solved will depend on the full understanding of the previous chapters and the facilities available e. In engineering, success depends on the reliability of the results achieved, not on the method of achieving them. I would like to express my appreciation of the assistance which I have received from my colleagues in the teaching profession. I am particularly indebted to Professor Mahmoud Yassin Osman for his advice on the preparation of this textbook. When author, printer and publisher have all done their best, some errors may still remain.

For these I apologies and I will be glad to receive any correction or constructive criticism. These interactions are called work and heat. However, thermodynamics deals with the end states of the process during which an interaction occurs and provides no information concerning the nature of the interaction or the time rate at which it occurs. The objective of this textbook is to extend thermodynamic analysis through the study of transient conduction heat transfer and through the development of relations to calculate different variables of lumped capacitance theory.

In our treatment of conduction in previous studies we have gradually considered more complicated conditions.

We began with the simple case of one dimensional, steady state conduction with no internal generation, and we subsequently considered more realistic situations involving multidimensional and generation effects.

However, we have not yet considered situations for which conditions change with time. We now recognize that many heat transfer problems are time dependent. Such unsteady, or transient problems typically arise when the boundary conditions of a system are changed. For example, if the surface temperature of a system is altered, the temperature at each point in the system will also begin to change.

The changes will continue to occur until a steady state temperature distribution is reached. Consider a hot metal billet that is removed from a furnace and exposed to a cool air stream. Energy is transferred by convection and radiation from its surface to the surroundings.

Energy transfer by conduction also occurs from the interior of the metal to the surface, and the temperature at each point in the billet decreases until a steady state condition is reached. The final properties of the metal will depend significantly on the time — temperature history that results from heat transfer. Controlling the heat transfer is one key to fabricating new materials with enhanced properties.

The nature of the procedure depends on assumptions that may be made for the process. If, for example, temperature gradients within the solid may be neglected, a comparatively simple approach, termed the lumped capacitance method or negligible internal resistance theory, may be used to determine the variation of temperature with time. Under conditions for which temperature gradients are not negligible, but heat transfer within the solid is one dimensional, exact solution to the heat equation may be used to compute the dependence of temperature on both location and time.

Such solutions are for finite and infinite solids. Also, the response of a semi — infinite solid to periodic heating conditions at its surface is explored. What happens during this lap of time may be detrimental. Again, when quenching a piece of metal, the time history of the temperature should be known i.

One of the cases to be considered is when the internal or conductive resistance of the body is small and negligible compared to the external or convective resistance. This system is also called lumped capacity or capacitance system or negligible internal resistance system because internal resistance is small, conductivity is high and the rate of heat flow by conduction is high and therefore, the variation in temperature through the body is negligible. The measure of the internal resistance is done by the Biot Bi number which is the ratio of the conductive to the convective resistance.

Definition of Lumped Capacity or Capacitance System: Is the system where the internal or conductive resistance of a body is very small or negligible compared to the external or convective resistance. Biot number Bi : is the ratio between the conductive and convective resistance. Where is the characteristic linear dimension and can be written as , is the heat transfer coefficient by convection and k is the thermal conductivity.

When , the system is assumed to be of lumped capacity. Characteristic Linear Dimensions of Different Geometries: The characteristic linear dimension of a body, Characteristic linear dimension of a plane surface, Characteristic linear dimension of a cylinder, Characteristic linear dimension of a sphere ball , Characteristic linear dimension of a cube, Where: t is the plate thickness, r is the radius of a cylinder or sphere, and a is the length side of a cube.

They are heated to a temperature of and then quenched in oil that is at a temperature of. The ball bearings have a diameter of 4cm and the convective heat transfer coefficient between the bearings and oil is. These pellets are initially at and must be cooled to maximum before entering storage vessel. This proposed to cool these pellets to the required temperature by passing them down slightly inclined channel where they are subjected to a stream of air at K. If the length of the channel is limited to 3m, calculate the maximum velocity of the pellets along the channel and the total heat transferred from one pellet.

Heat transfer from pellet surface to the air stream may be considered to be the limiting process with. Show that the lumped capacitance system analysis is applicable and find the temperature of the cylinder after 4min. What is the total heat transfer during this period? Taking the aluminum as a sphere having the same mass as that given, estimate the time required to cool the aluminum to.

Find also the total heat transferred during this period. Justify your use of the lumped capacity method of analysis. Solution: A piece of aluminum Taking aluminum as a sphere. Since , therefore the lumped capacity method of analysis is used. The response of a thermocouple is defined as the time required for the thermocouple to attain the source temperature.

It is evident from equation 4 , that the larger the quantity , the faster the exponential term will approach zero or the more rapid will be the response of the temperature measuring device. This can be accomplished either by increasing the value of "h" or by decreasing the wire diameter, density and specific heat. Hence, a very thin wire is recommended for use in thermocouples to ensure a rapid response especially when the thermocouples are employed for measuring transient temperatures.

From equation ; The quantity which has units of time is called time constant and is denoted by the symbol. In other words, temperature difference would be reduced by The time required by a thermocouple to reach its Depending upon the type of fluid used the response times for different sizes of thermocouple wires usually vary between 0. Properties of material are: This junction is initially at and inserted in a stream of hot air at.

Find the following: i Time constant of the thermocouple. Assuming the heat transfer coefficient in air , find the temperature attained by the junction 20 seconds after removal from hot air. Consider cylindrical thermometer bulb to consist of mercury only for which and.

Calculate the time required for the temperature change to reach half its final value. Initially the junction and air are at a temperature of. The air temperature suddenly changes to and is maintained at. Also, determine the thermal time constant and the temperature indicated by the thermocouple at that instant. The temperature distribution is given by: Thus, the thermocouple requires The actual time requirement will, however, be greater because of radiation from the probe and conduction along the thermocouple lead wires.

Or Or At 9. Let us assume that the wall, initially, is at uniform temperature To and both the surfaces are suddenly exposed to and maintained at the ambient i. The governing differential equation is: The boundary conditions are: i At ii At iii At The conduction heat transfer equals convective heat transfer at the wall surface.

The dimensionless parameter 2 3 is replaced by 2 3 in case of cylinders and spheres. For the equation 3. In the Figs. These charts provide the temperature history of the solid at its mid — plane and the temperatures at other locations are worked out by multiplying the mid — plane temperature by correction factors read from charts given in figs.

The following relationship is used: [ ] [ ] The values Biot number and Fourier number , as used in Heisler charts, are evaluated on the basis of a characteristic parameter which is the semi — thickness in the case of plates and the surface radius in case of cylinders and spheres. When both conduction and convection resistances are almost of equal importance the Heister charts are extensively used to determine the temperature distribution.

Solution: Given: Temperature at the mid — plane centerline of the plate : The characteristic length, Fourier number, Biot number, At , the internal temperature gradients are not small, therefore, internal resistance cannot be neglected.

Thus, the plate cannot be considered as a lumped system. Further, as the , Heisler charts can be used to find the solution of the problem. Corresponding to the following parametric values, from Heisler charts Fig.

It is held initially at a uniform temperature of. When the missile enters the denser layers of the atmosphere at a very high velocity the effective temperature of air surrounding the nose region attains ; the surface convective heat transfer coefficient is estimated.

If the maximum metal temperature is not to exceed , determine: i Maximum permissible time in these surroundings. Solution: Given: i Maximum permissible time, Characteristic length, Biot number, As , therefore, lumped analysis cannot be applied in this case. Further, as , Heisler charts can be used to obtain the solution of the problem.

Corresponding to and outside surface of nose section, from Fig. The surface coefficient of heat transferred between the bar surface and the coolant is. Determine: i The time taken by the shaft center to reach. Also, calculate the temperature gradient at outside surface at the same instant of time.

Solution: Given: i The time taken by the shaft center to reach , Characteristic length, Biot number, As , therefore, lumped analysis cannot be applied in this case. The parametric values for the cylindrical bar are: At the center of the bar, Corresponding to the above values, from the chart for an infinite cylinder Fig.

Determine the temperature at the center of the apple after a period of 2 hours. Solution: Given: The characteristic length, Biot number, , Since , a lumped capacitance approach is appropriate.

The parametric values for the spherical apple are: [ ] [ ] [ ] Corresponding to the above values, from the chart for a sphere Fig. Heisler position — correction factor chart for temperature history in plate Fig. If an infinite solid is split in the middle by a plane, each half is known as semi — infinite solid.

In a semi — infinite body, at any instant of time, there is always a point where the effect of heating or cooling at one of its boundaries is not felt at all. At the point the temperature remains unaltered. The transient temperature change in a plane of infinitely thick wall is similar to that of a semi — infinite body until enough time has passed for the surface temperature effect to penetrate through it.

As shown in Fig. The entire body is initially at uniform temperature including the surface at. The surface temperature at is suddenly raised to for all times greater than. The governing equation is: Fig. Table shows a few representative values of Suitable values of error functions may be obtained from Fig. The temperature at the center of cylinder or sphere of radius R, under similar conditions of heating or cooling, is given as follows: 0 1 For the cylindrical and spherical surfaces the values of function 0 1 can be obtained from Fig.

Error integral for cylinders and spheres Penetration Depth and Penetration Time: Penetration depth refers to the location of a point where the temperature change is within 1 percent of the change in the surface temperature. This corresponds to , from the table for Gaussian error integral. Determine the length of time required for the temperature to reach at a depth of.

The ingot may be approximated as a flat plate. Determine the maximum depth at which the pipes be laid if the surrounding soil temperature is to remain above without water getting frozen. Assume the soil as semi — infinite solid.

For wet soil take Solution: Given: maximum depth The temperature, at critical depth, will just reach after 9. It is suddenly exposed on one side to a fluid which causes the surface temperature to increase to and remain at. Determine: i The maximum time that the slab be treated as a semi — infinite body; ii The temperature at the center of the slab 1.

Determine: i The temperature at a point 80mm from the surface after 8 hours. The sphere is suspended in a slow-moving air stream at 0 C. Radiation can be neglected. Since copper is highly conductive, temperature gradients in the sphere will smooth out rapidly, and its temperature can be taken as uniform throughout the cooling process i. Write the instantaneous energy balance between the sphere and the surrounding air. Solve this equation and plot the resulting temperatures as a function of time between 40 C and 0 C.

K Properties of copper, Table A. Plotting the net-entropy increase: Equation 1. The diameter at the top is 15 cm and at the bottom is 7. The lower surface is maintained at 6 C and the top at 40 C. The outer surface is insulated. Assume one dimensional heat transfer and calculate the rate of heat transfer in watts from top to bottom. To do this, note that the heat transfer, Q, must be the same at every cross section. A hot water heater contains kg of water at 75 C in a 20 C room.

Its surface area is 1. Notice that this problem will be greatly simplified if the temperature drop in the steel casing and the temperature drop in the convective boundary layers are negligible. Can you make such assumptions? What is the temperature at the left-hand wall shown in Fig.

Both walls are thin, very large in extent, highly conducting, and thermally black. Develop S. Solution: 1. K4 The conversion factor for English units is: 0. The view factor F The view factor is dimensionless, so there is no need for conversion factor. K The conversion factor for English units is: 0. Three infinite, parallel, black, opaque plates, transfer heat by radiation,as shown in Fig.

Find T2. Four infinite, parallel black, opaque plates transfer heat by radiation, as shown in Fig. Find T2 and T3. Two large, black, horizontal plates are spaced a distance L from one another.

A gas separates them. The gas is stationary because it is warm on top and cold on the bottom. Further suppose that you wish to operate in such a way that the conduction and radiation heat fluxes are identical.

Identify the operating point on your curve and report the value of Th that you must maintain. A blackened copper sphere 2 cm in diameter and uniformly at C is introduced into an evacuated black chamber that us maintained at 20 C. As part of space experiment, a small instrumentation packaged is released from a space vehicle. It can be approximated as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 30 C and it contains a pressurized hydrogen component that will condense and malfunction at 30 K.

If we take the surrounding space to be at 0 K, how long may we expect the implementation package to function properly? Is it legitimate to use the lumped-capacity method in solving the problem? Hint: See the directions for Problem 1. Solution: Properties of aluminum, Table A.

K 30 C From Prob. Consider heat calculation through the wall as shown in Fig. Calculate q and the temperature of the right-hand side of the wall. Throughout Chapter 1 we have assumed that the steady temperature distribution in a plane uniform wall is linear.

To prove this, simplify the heat diffusion equation to the form appropriate for steady flow. The temperatures are T1 and T2 on either side of the wall, and its thickness is L. Develop an expression for q. Of what might it be made? Figure 1. If h is K, estimate when the beverage will be at 15 C. Ignore thermal radiation. State all of your other assumptions. One large, black wall at 27 C faces another whose surface is C.

The gap between the two walls is evacuated. If the second wall is 0. K, what is its temperature on the back side? Assume steady state. In this case, h is not independent of temperature. Plot Tsphere as a function of t. Verify the lumped- capacity assumption. A 3-cm diameter, black spherical heater is kept at C. It radiates through an evacuated space to a surrounding spherical shell of Nichrome V.

The shell has a 9 cm inside diameter and is 0.