ME5320 Advanced heat transferLecturer : K. BadarinathLecture 13/8/2015 Lectures (Room 203) Monday : 0900 to 1025 ; Thursday : 1030 to 1155 Lecturer and TAs Dr. K. Badarinath (Lecturer) Teaching Assistants Y.S. Kannan (PhD student), Dilin Sharaf(3rdsemester M.Tech) Consultation hoursFriday:3:00 5:30 P.MRoom No. 39,badarinath@iith.ac.in2Course informationGrading Policy1. Attendance 10%2. Assignments 10 % 3. Critical Literature review/presentation 10%4. Mid- semester exam 35 %5. End-semester exam 35 %3Syllabus Topic wise ME3110(Basic Heat transfer course for undergraduates)Introduction Steady State heat conduction in one-dimensional systems. One dimensional unsteady state conduction; extended surface heat transfer (Fins). Convection: Basic equations, Dimensional analysis, Boundary layers; Forced convection: External and internal flows, correlations, Natural convection and Mixed convection. Design of heat exchangers: LMTD and NTU methods. Radiation heat transfer: Basic laws, Properties of surfaces, view factors, network method and enclosure analysis for gray-diffuse enclosures containing transparent media. Concepts of Mass transfer. Current trends of research in the field of heat transfer. ME5320(Advanced Heat transfer)Introduction - Review of fundamentals of heat transfer. Conduction: General heat conduction equation, Analytical solutions of two dimensional steady state heat conduction; Transient conduction. Convection: Governing equations, boundary layer equations, Forced convection over external surfaces and internal ducts; Similarity solutions. Free and Mixed convection flows, Conjugate heat transfer analysis. Radiative Heat Transfer: Thermal radiation, Emissive Power, Solid Angles, Radiative Intensity, Heat Flux, Pressure and Characteristics, Radiative transport equation4Reference booksReferences:1. A Textbook on Heat Transfer by S. P. Sukhatme, Universities Press 2. NPTEL Heat and Mass Transfer: http://nptel.ac.in/courses/112101097/3. Fundamentals of Heat and Mass Transfer, T. L. Bergman, A. S. Lavine, F. P. Incropera, D. P. DeWitt, Wiley 5Course learning objectives To understand a few different analytical approaches used in Heat transfer studies to some depth. To be able to ascertain the modes of heat transfer in a given (Real-life) problem and come up with a formulation of the equations. To learn how to do a critical literature review.Introduction to heat transfer Energy can be transferred by interaction between a system and its surroundings in the form of work and heat. While thermodynamics deals with the energy states of a system and energy transferred, heat transfer deals with the modes of that energy transfer (in form of heat). 7What and how? Heat transfer (or heat) is thermal energy in transit due to a spatial temperature difference. Modes of heat transfer 8Modes of heat transfer Conduction : When a temperature gradient exists in a stationary medium, which may be a solid or a fluid, the heat transfer that will occur across the medium is termed conduction. Convection : Heat transfer that will occur between a surface and a moving fluid when they are at different temperatures. Radiation : All surfaces at finite temperature emit energy in the form of electromagnetic radiation. If there is a temperature difference between two surfaces and in the absence (or presence) of a medium, radiation heat transfer will take place. 9Conduction10Temperature is associated with the energy of molecules a higher temperature indicates a higher energy. The mechanism behind conduction is the transfer of energy from more energetic molecules to less energetic ones by collision and plain random movement of molecules 11The rate of heat transfer in different processes can be quantified through equations.We have the Fouriers law of conduction (which for 1-D conduction is )

=

Where

is the heat flux or heat transferred per unit area (

2) is a transport property known as the thermal conductivity (

)If the temperature distribution is linear, we have

=

1 2

= k

Example 1 The wall of an industrial furnace is constructed from 0.15-m-thick fireclay brick having a thermal conductivity of =1.7

. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces. What is the rate of heat loss through a wall that is 0.5 1.2 on a side? Solution : The heat flux through the wall is given as

=

1 2

Substituting, we get

=

1 2

= 1.7

1400 11500.15Km = 2833 Wm2The rate of heat loss through the wall is =

A = 2833 0.5 1.2 = 1700 W12Convection Convection heat transfer is comprised of two mechanisms superimposed on each other Energy transfer due to random molecular motion (diffusion) Energy transfer due to the bulk or macroscopic motion of the fluid. (advection) When a fluid moves as a bulk over a surface and there is a temperature gradient, the bulk motion contribution to the heat transfer is referred to as advection 13 In fluid flow over a surface, there is a development of a velocity distribution from =0 at the surface to = far away. This is the velocity boundary layer. If the temperatures of fluid and surface are different, there is also the development of a thermal boundary layer with =

at the surface and = in the fluid far away. If

> convective heat transfer will take place from the surface to the fluid. 14 The contribution of random molecular motion (diffusion) to convection dominates close to the surface where the fluid velocity is low. The contribution due to the bulk motion comes from the fact that the boundary layer grows as the flow progresses in the x direction. Heat conducted into the boundary layer is swept downstream and transferred to the bulk fluid outside the boundary layer. Convective heat transfer can be classified into Forced convection Natural convection Mixed convection In addition heat can also be transferred via convection when there is a phase change phenomenon, boiling or condensation 1516Convection equation Regardless of the form of convection, the equation for this mode of heat transfer can be written as

= (

)Where is the convective heat flux (

2) is the convective heat transfer coefficient The above equation is known as the Newtons law of cooling. Heat transfer is considered positive if it is from the surface and negative if it is to the surface. Any study of convection ultimately reduces to the means by which h can be determined and its determination.17Radiation 18G incident radiation heat flux ; E emitted radiation heat fluxRadiation Thermal radiation is energy emitted by matter that is at a non-zero temperature. Emission can occur from all of solids, liquids and gases although we focus here (for analysis) on radiation by solids. Radiation emitted by a surface originates from the thermal energy of matter bounded by the surface. The rate at which energy is released per unit area is termed as the Emissive power . There is an upper limit to the emissive power which is given by the Stefan-Boltzmann law

=

4Where is the Stefan-Boltzmann constant ( = 5.67 108

2

)

is the absolute temperature in Kelvin19 A body which emits radiation according to Stefan-Boltzmann law is called an ideal blackbody. A real body emits radiation which is less than a blackbody and the heat flux is given by =

4Where is a radiative property of the surface called as emissivity. 0 1 Radiation may also be incident upon a surface by its surroundings. We designate the rate at which all such radiation is incident on a surface as the irradiation A portion (or all) of the incident thermal energy may be absorbed by the surface increasing its energy. This is quantified as

= Where is the absorptivity of the material. 20Radiation equation In many engineering problems, liquids can be considered opaque and gases can be considered transparent to radiation heat transfer. Solids can be opaque or semi-transparent. A special case (slide 18) that occurs frequently involves radiation exchange between a small surface at temperature

and a much larger isothermal surface (

) . We approximate the irradiation as emission from a black-body at

which gives =

4Further, if we assume the surface to have = we get

=

=

G =

4

4= (

4

4)21 In certain applications, it is convenient to express

=

Where

= (

+

)(

2 +

2) Now, considering the figure in slide 18, the net heat transfer can be expressed as =

+

=

+(

4

4)22Example 1 An uninsulated steam pipe passes through a room in which the air and walls are at 25. The outside diameter of the pipe is 70 mm, and its surface temperature and emissivity are 200and 0.8 respectively. 1. What are the surface emissive power and irradiation?2. If the coefficient associated with the free convection heat transfer from the surface to the air is 15

2

, what is the rate of heat loss from the surface per unit length of the pipe. Answers1. = 2270 /2, = 447

22. = 998 /23Energy equation A useful tool for analyzing heat transfer problems is the first law of thermodynamics energy equation. The general statement of the first law is

= 24 The total energy can be classified as mechanical energy (kinetic + potential +flow energy), thermal energy, and others such as chemical, electrical etc. For the purpose of heat transfer analysis, we focus on only the sum of the thermal and mechanical energies. If other sources are present, we consider their effect only in terms of change in these two energies. E.g: Heat release due to reaction, or due to resistive heating etc. In such a case, the energy equation can be rewritten as

=

+

Subscripts st- stored energy, g- energy generated.

,

represents the energy advected in and out of the volume. Alternatively, in terms of rate, the equation is

=

+

2526The energy equation for a control volume as shown above is + +12

2 +

+ +12

2 +

+

= 0Under steady state conditions, and specific conditions, the equation above reduces to the steady-flow thermal energy equation given as =

(

) The conditions under which the steady flow thermal energy equation is valid are as follows1. An ideal gas with negligible kinetic and potential energy changes and negligible work (other than flow work).Here, =

(

), represents the net change in enthalpy + between the outlet and the inlet. 2. An incompressible liquid with negligible kinetic and potential energy changes, and negligible work including flow work. The flow work can be taken as negligible if the pressure variation is not too great (usually true for incompressible liquids). 3. An ideal gas with negligible viscous dissipation and negligible pressure variation.4. An incompressible liquid with negligible viscous dissipation. Viscous dissipation converts the frictional losses (due to viscosity) to thermal energy. It becomes significant only for high-speed or high viscosity flows. 27Surface energy equation In many situations, we only need to consider the change of energy at the surface of a body. An appropriate form of the energy equation then is

= 0This equation holds at the surface and is valid both for steady state and transient conditions.It is also independent of the energy generation that may be occurring inside the medium. 28Surface energy balance 29In a situation like above, the surface energy balance becomes

= 0Applying conservation laws general guidelines1. Define an appropriate control volumewith relevant control surfaces shown (and sketch a schematic).2. Identify the appropriate time basis (whether rate form of the equation or the energy change over a time period)3. Identify the relevant energy processes and the direction of transfer (conduction, convection, radiation etc.)4. Write the appropriate conservation equation, simplify and solve. 30Problem solving methodology1. After reading the problem, concisely state what is known about the problem. 2. State briefly and concisely what needs to be found. 3. Draw a schematic of the physical system (with appropriate control volume sketched) and the processes indicated.4. List all the simplifying assumptions you need to make.5. Compile all the property data needed to solve the problem.6. Begin your analysis by identifying appropriate conservation laws, and then introduce the rate equations as needed. Develop the analysis as completely as possible before substituting numerical values. Perform the calculations substituting the values. 7. * Comment on the results : This includes cross-checking your assumptions, and any other insights you get. 31Example 2 A long conducting rod of diameter D and electrical resistance per unit length

is initially in thermal equilibrium with the ambient air and its surroundings. This equilibrium is disturbed when an electric current is passed through the rod. Develop an equation that can be used to compute the variation of the rod temperature with time during the passage of the current. You may take the convective heat transfer coefficient for air as and the surrounding temperature to be

.[Assume properties for the rod as , , , = ( )]32 Solution Known data : Properties of the rod (Diameter, D, resistance per unit length, , , ) andthe heat transfer coefficient of air, To evaluate : An expression for the change of temperature with time in the rod under resistive heating. Assumptions1. At any time t, the temperature of the rod is uniform.2. Radiation exchange between the rod and the surroundings is between a small surface and a large enclosure. A relevant control volume for the problem is 33The first law of thermodynamics gives

=

+

We have in this case

= 2

= 0

= + 4

4

= ()

=

=

24

Substituting we can get the expression for

as

= 2

+ 4

4

24 34Example 3 Large Proton exchange membrane fuel cells, such as those used in automotive applications often require internal cooling using pure liquid water to maintain their temperature at a desired level. In cold climates, the cooling water must be drained from the fuel cell to an adjoining container when the automobile is turned off so that harmful freezing does not occur within the fuel cell. Consider a mass M of ice that was frozen while the automobile was not being operated. The ice is at a fusion temperature (

=0) and is enclosed in a cubical container of side . The container wall is of thickness and thermal conductivity . If the outer surface of the wall is heated to a temperature 1 >

to melt the ice, obtain an expression for the time needed to melt the entire mass of ice, and in turn deliver cooling water to, and energize the fuel cell.35Example 4Humans are able to control their heat production rate, and heat loss rate to maintaina nearly constant core temperature of

= 37under a wide range of environmental conditions. This process is called thermoregulation. From the perspective of calculating heat transfer between a human body and its surroundings, we focus on a layer of skin and fat, with its outer surface exposed to the environment, and its inner surface at a temperature slightly less than the core temperature

=35 = 308 .Consider a person with a skin/fat layer of thickness = 3 mmand effective thermal conductivity = 0.3 / . The person has a surface area = 1.8 2and is dressed in a bathing suit. The emissivity of the skin is = 0.95.1. When the person is in still air at = 297 , what is the skin surface temperature and rate of heat loss to the environment? Convection heat transfer coefficient to air is = 2 /2.2. Repeat problem 1 with still water at = 297 , here =200 /2.36Example 537