Heat & Mass Transfer

For Class use only

ACHARYA N.G. RANGA AGRICULTURAL UNIVERSITY

Course No. FDEN- 221 HEAT AND MASS TRANSFER Course Title: Credits:

2 (1 + 1)

Prepared by

Er. B. SREENIVASULA REDDY Assistant Professor (Food Engineering)

College of Food Science and Technology Chinnarangapuram, Pulivendula – 516390 YSR (KADAPA) District, Andhra Pradesh

DEPARTMENT OF FOOD ENGINEERING 1 2 3 4

Course No Title Credit hours General Objective

: : : :

FDEN - 221 Heat and Mass Transfer 2 (1+1) To impart knowledge to students on different modes of heat transfer through extended surfaces, study of heat exchanges and principles of mass transfer

5

Specific Objectives a) Theory

: : By the end of the course, the students will acquire knowledge from different modes of heat transfer, extended surfaces, boiling and condensation process and principles of heat exchangers which

b) Practical

are very essential in dairy and food industries By the end of the course, the students will learn efficient design of heat exchangers on the basis of overall heat transfer coefficient and LMTD

1

A) Theory Lecture Outlines Introduction to Heat Transfer- Basic Mechanisms of Heat Transfer Conduction Heat transfer - Fourier's Law of Heat Conduction- Convection

2

Heat Transfer- Radiation Heat Transfer The basic equation that governs the transfer of heat in a solid - Thermal

3

conductivity One-dimensional steady-state conduction of heat through some simple geometries - Conduction Through a Flat Slab or Wall - Conduction Through

4

a Hollow Cylinder - Conduction Through a Hollow Sphere Conduction Heat Transfer Through A Composite Plane Wall - Conduction

5 6

Heat Flow Through A Composite Cylinder The Overall Heat-Transfer Coefficient - Critical Thickness of Insulation Heat Source Systems: One-dimensional steady state heat conduction with

7

heat generation : Heat Flow through slab / Plane Wall Steady state heat conduction with heat dissipation to environmentIntroduction to extended surfaces (FINS) of uniform area of cross section -

8

Different fin configurations - General Conduction Analysis Equation Equation of temperature distribution with different boundary conditions, Fin

9

Performance and Overall surface efficiency of FINS Principles Of Unsteady-State Heat Transfer: Derivation of Basic Equation;

Simplified Case For Systems With Negligible Internal Resistance ; Total 10

Amount of Heat Transferred ; Dimensional Analysis in Momentum Transfer Some important empirical relations used for determination of heat transfer coefficient: Nusselt’s number, Prandtl number, Reynold’s number, Grashoff

11

number Radiation - heat transfer, Radiation Properties, radiation through black and

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grey surfaces, determination of shape factors Introduction to condensing and boiling heat transfer, Condensation HeatTransfer Phenomena, Film Condensation Inside Horizontal Tubes , Boiling

13

Heat Transfer Heat exchangers- general introduction; Double-pipe heat exchanger; Shell-

14

and-tube heat exchanger; Cross-flow exchanger; fouling factors, LMTD Design problems on heat exchangers: Calculation of heat exchanger size from known temperatures, Problem on Shell-and-tube heat exchanger ,

15

Design of shell-and-tube heat exchanger Introduction To Mass Transfer: A Similarity of Mass, Heat, and Momentum

16

Transfer Processes; Fick's Law for Molecular Diffusion Molecular Diffusion In Gases : Equimolar Counter diffusion in Gases

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

B) Practical Class Outlines Determination of thermal conductivity of milk and dairy products Tutorials on heat conduction through slab, cylinder and sphere Tutorials on heat conduction through slab, cylinder and sphere Tutorials on heat conduction through slab, cylinder and sphere Tutorials on extended surfaces (FINS) Tutorials on unsteady state heat conduction Determination of specific heat of food materials Tutorials on determination of Nusselt’s number by dimensional analysis Tutorials on LMTD and NTU method of analysis of heat exchangers Study of shell and tube heat exchanger Study of plate heat exchanger Study on temperature distribution and heat transfer in HTST pasteurizer Study on temperature distribution and heat transfer in HTST pasteurizer Design problems on heat exchangers - I Design problems on heat exchangers - II Design problems on heat exchangers - III

1

References Geankoplis, C.J. 1978. Transport Processes and Unit Operations. Allyn and

2 3

Bacon Inc., Newton, Massachusetts. Holman, J. P. 1989. Heat Transfer. McGraw Hill Book Co., New Delhi. Incropera, F. P. and De Witt, D .P. 1980. Fundamentals of Heat and Mass

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Transfer. John Wiley and Sons, New York. Gupta, C. P. and Prakash, R. 1994. Engineering Heat Transfer. Nem Chand and Bros., Roorkee

LECTURE NO.1 INTRODUCTION TO HEAT TRANSFER- BASIC MECHANISMS OF HEAT TRANSFER - CONDUCTION HEAT TRANSFER - FOURIER'S LAW OF HEAT CONDUCTION- CONVECTION HEAT TRANSFER- RADIATION HEAT TRANSFER Introduction Heat transfer is the science that seeks to predict the energy transfer that may take place between material bodies as a result of a temperature difference. Thermodynamics deals with systems in equilibrium; it may be used to predict the amount of energy required to change a system from one equilibrium state to another; it may not be used to predict how fast a change will take place since the system is not in equilibrium during the process. Heat transfer supplements the first and second principles of thermodynamics by providing additional experimental rules which may be used to establish energy-transfer rates. As an example of the different kinds of problems that are treated by thermodynamics and heat transfer, consider the cooling of a hot steel bar that is placed in a pail of water. Thermodynamics may be used to predict the final equilibrium temperature of the steel bar-water combination. Thermodynamics will not tell us how long it takes to reach this equilibrium condition or what the temperature of the bar will be after a certain length of time before the equilibrium condition is attained. Heat transfer may be used to predict the temperature of both the bar and the water as a function of time. Basic Mechanisms of Heat Transfer Heat transfer may occur by anyone or more of the three basic mechanisms of heat transfer: conduction, convection, and radiation. 1. Conduction Heat transfer. In conduction, heat can be conducted through solids, liquids, and gases. The heat is conducted by the transfer of the energy of motion between adjacent molecules. In a gas the "hotter" molecules, which have greater energy and motions, impart energy to the adjacent molecules at lower energy levels. This type of transfer is present to some extent in all solids, gases, or liquids in which a temperature gradient exists. In conduction,

energy can also be transferred by "free" electrons, which is quite important in metallic solids. Examples of heat transfer mainly by conduction are heat transfer through walls of exchangers or a refrigerator, heat treatment of steel forgings, freezing of the ground during the winter, and so on. Fourier's Law of Heat Conduction The basic rate transfer process equation for processes such as momentum transfer, heat transfer, mass transfer and electric current is as follows: rate of a transfer process =

driving force resis tan ce

----------- (1)

This equation states what we know intuitively: that in order to transfer a property such as heat or mass, we need a driving force to overcome a resistance. The transfer of heat by conduction also follows this basic equation and is written as Fourier's law for heat conduction in fluids or solids: qx dT =− k A dx

--------------- (2)

where q x is the heat-transfer rate in the

x direction in watts (W), A is

the cross-sectional area normal to the direction of flow of heat in m2, T is temperature in K,

x is distance in m, and

k is the thermal conductivity in

W/m·K in the SI system. The quantity q x / A is called the heat flux in W/m2. The quantity dT / dx is the temperature gradient in the

x direction. The minus

sign in eq. (2) is required because if the heat flow is positive in a given direction, the temperature decreases in this direction. Fourier's law, eq.(2), can be integrated for the case of steady-state heat transfer through a flat wall of constant cross-sectional area A, where the inside temperature is T1 at point 1 and T2 at point 2, a distance of x2 −x1 m away. Rearranging eq. (2),

qx A

x2

T2

∫ dx = − k ∫ x1

T1

dT

-------(3)

Integrating, assuming that k is constant and does not vary with temperature and dropping the subscript

x on

q x for convenience,

q k = (T1 −T2 ) A x2 −x1

-------- (4)

2. Convection Heat Transfer. The transfer of heat by convection implies the transfer of heat by bulk transport and mixing of macroscopic elements of warmer portions with cooler portions of a gas or liquid. It also often refers to the energy exchange between a solid surface and a fluid. A distinction must be made between forced-convection heat transfer, where a fluid is forced to flow past a solid surface by a pump, fan, or other mechanical means, and natural or free convection, where warmer or cooler fluid next to the solid surface causes a circulation because of a density difference resulting from the temperature differences in the fluid. Examples of heat transfer by convection are loss of heat from a car radiator where the air is being circulated by a fan, cooking of foods in a vessel being stirred, cooling of a hot cup of coffee by blowing over the surface, and so on. Convective Heat-Transfer Coefficient It is well known that a hot piece of material will cool faster when air is blown or forced past the object. When the fluid outside the solid surface is in forced or natural convective motion, we express the rate of heat transfer from the solid to the fluid or vice versa, by the following equation: q =hA (Tw −T f )

-------------- (5)

where q is the heat-transfer rate in W, A is the area in m 2, Tw is the temperature of the solid surface in K, Tf is the average or bulk temperature of the fluid flowing past in K, and h is the convective heat-transfer coefficient in W/m2.K. The coefficient h is a function of the system geometry, fluid properties, flow velocity, and temperature difference. In many cases, empirical correlations are available to predict this coefficient, since it often cannot be predicted theoretically. Since we know that when a fluid flows past a surface there is a thin, almost stationary layer or film of fluid adjacent to the wall which presents most of the resistance to heat transfer, we often call the coefficient h a film coefficient. 3. Radiation Heat Transfer:- Radiation differs from heat transfer by conduction and convection in that no physical medium is needed for its

propagation. Radiation is the transfer of energy through space by means of electromagnetic waves in much the same way as electromagnetic light waves transfer light. The same laws that govern the transfer of light govern the radiant transfer of heat. Solids and liquids tend to absorb the radiation being transferred through them, so that radiation is important primarily in transfer through space or gases. The most important example of radiation is the transport of heat to the earth from the sun. Other examples are cooking of food when passed below red-hot electric heaters, heating of fluids in coils of tubing inside a combustion furnace, and so on. The rate of energy emitted by a black body is proportional to the fourth power of the absolute temperature of the body and directly proportional to its surface area. Thus qemitted =σAT

where

4

σ is the proportionality constant and is called the Stefan-

Boltzmann constant with the value of 5.669 x 10-8 W/m2.K4. The equation is called the Stefan-Boltzmann law of thermal radiation, and it applies only to blackbodies. Problem Heat Loss Through an Insulating Wall Calculate the heat loss per m2 of surface area for an insulating wall composed of 25.4 -mm-thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K. The thermal conductivity of fiber insulating board is 0.048 W/m.K Solution: The thickness x2 −x1 = 0.0254 m. Substituting into the eq. q k 0.048 = (T1 −T2 ) = (352 .7 −297 .1) A x2 −x1 0.0254

= 105.1 W/m2

LECTURE NO.2 THE BASIC EQUATION THAT GOVERNS THE TRANSFER OF HEAT IN A SOLID - THERMAL CONDUCTIVITY The basic equation that governs the transfer of heat in a solid

Fig.2.1 Elemental volume for one-dimensional heat conduction analysis Consider (Fig. 2.1) the general case where the temperature may be changing with time and heat sources may be present within the body. For the element of thickness dx the following energy balance may be made: Energy conducted in left face + heat generated within element = change in internal energy + energy conducted out right face These energy quantities are given as follows: Energy in left face = q x =−kA

∂T ∂x .

Energy generated within element = q A dx Change in internal energy = ρcA

∂T dx ∂τ

∂T Energy out right face = q x +dx =−kA ∂x

x +dx

 ∂T ∂  ∂T   =− A k + k dx   ∂x ∂x  ∂x  

(In the derivations, the expression for the derivative at x + dx has been written in the form of a Taylor-Series expression with only the first two terms of the series employed for the development.) where .

3 q = energy generated per unit volume, W /m

c = specific heat of material, J /kg .°C ρ = density, kg/m3 Combining the relations above gives −kA

 ∂T ∂  ∂T   . ∂T ∂T + k + q A dx = ρcA dx − A k  dx  ∂x ∂τ  ∂x ∂x  ∂x  

or . ∂  ∂T  ∂T k dx + q =ρc ∂x  ∂x  ∂τ

This is the one-dimensional steady state heat-conduction equation with heat source.

Figure 2.2 Elemental volume for three-dimensional heat-conduction analysis: (a) cartesian coordinates; (b) cylindrical coordinates; (c) spherical coordinates.

The general three-dimensional heat-conduction equation is ∂  ∂T  ∂  ∂T  ∂  ∂T  . ∂T k  k + + ∂z k ∂z + q =ρc ∂τ ∂x  ∂x  ∂y  ∂ y    

For constant thermal conductivity equation can be written as .

∂2T ∂2T ∂2T q 1 ∂T + + 2 + = ∂x 2 ∂x 2 ∂x k α ∂τ where the quantity α =k / ρc is called the thermal diffusivity of the material. The larger the value of high value of

α , the faster heat will diffuse through the material. A

α could result either from a high value of thermal conductivity,

which would indicate a rapid energy-transfer rate, or from a low value of the thermal heat capacity ρc . A low value of the heat capacity would mean that less of the energy moving through the material would be absorbed and used to raise the temperature of the material; thus more energy would be available for further transfer.

α has the units of m2/s.

The three-dimensional heat-conduction equation with heat generation for cylindrical or spherical co-ordinates is Cylindrical coordinates: .

∂2T 1 ∂T 1 ∂2T ∂2T q 1 ∂T + + + 2 + = 2 2 2 ∂r r ∂r r ∂φ ∂z k α ∂τ

Spherical coordinates:

.

1 ∂2 1 ∂  ∂T  1 ∂2T q 1 ∂T ( ) rT + sin θ + + =   r ∂r 2 r 2 sin θ ∂θ  ∂θ  r 2 sin 2 θ ∂φ2 k α ∂τ

The reduced form of the general equations for several cases of practical interest. Steady-state one-dimensional heat flow (no heat generation): d 2T =0 dx 2

Steady-state one-dimensional heat flow in cylindrical coordinates (No heat generation): d 2T 1 dT + =0 dr 2 r dr

Steady-state one-dimensional heat flow with heat sources: .

d 2T q + =0 dx 2 k

Thermal conductivity Thermal conductivity, k, is the property of a material that indicates its ability to conduct heat. Thermal conductivity is measured in W/m·K. The thermal conductivity predicts the rate of energy loss (in watts, W) through a piece of material. Thermal conductivity, k, also defined as the quantity of heat Q that flows per unit time through a food of unit thickness and unit area having unit temperature difference between faces.

The reciprocal of thermal conductivity is thermal resistivity. In general, the thermal conductivity is strongly temperature-dependent. Thermal energy may be conducted in solids by two modes: lattice vibration and transport by free electrons. In good electrical conductors a rather large number of free electrons move about in the lattice structure of the material. Just as these electrons may transport electric charge, they may also carry thermal energy from a high-temperature region to a low-temperature region, as in the case of gases. In fact, these electrons are frequently referred to as the electron gas. Energy may also be transmitted as vibrational energy in the lattice structure of the material. In general, however, this latter mode of energy transfer is not as large as the electron transport, and for this reason good electrical conductors are almost always good heat conductors, viz., copper, aluminum, and silver, and electrical insulators are usually good heat insulators. A notable exception is diamond, which is an electrical insulator, but which can have a thermal conductivity five times as high as silver or copper. It is this fact that enables a jeweler to distinguish between genuine diamonds and fake stones. A small instrument is available that measures the response of the stones to a thermal heat pulse. A true diamond will exhibit a far more rapid response than the non genuine stone.

LECTURE NO.3 ONE-DIMENSIONAL STEADY-STATE CONDUCTION OF HEAT THROUGH SOME SIMPLE GEOMETRIES - CONDUCTION THROUGH A FLAT SLAB OR WALL - CONDUCTION THROUGH A HOLLOW CYLINDER CONDUCTION THROUGH A HOLLOW SPHERE CONDUCTION HEAT TRANSFER One-dimensional steady-state conduction of heat through some simple geometries

Fig. 3.1 Heat conduction in a flat wall: (a) geometry of wall, (b) temperature plot. Conduction Through a Flat Slab or Wall Consider a flat slab or wall (Fig. 3.1) where the cross-sectional area A and k in are constant, q

k

The eq. A =x −x (T1 −T2 ) can be rewrite as 2 1 q k = (T1 −T2 ) A ∆x

where ∆x =x2 −x1 . The above indicates that if T is substituted for T2 and

x for

x2 , the

temperature varies linearly with distance, as shown in Fig.3.1(b). If the thermal conductivity is not constant but varies linearly with temperature, then substituting k =a =bT into the above equation and integrating,

q = A

T1 +T2 k 2 (T1 −T2 ) = m (T1 −T2 ) ∆x ∆x

a +b

where k m =a +b

T1 +T2 2

This means that the mean value of k (i.e., km) to use in q k = (T1 −T2 ) is the value of k evaluated at the linear average of T1 and A ∆x

T2. The rate of a transfer process equals the driving force over the resistance and the equation

q k = (T1 −T2 ) can be rewritten in that form A ∆x

as: q=

T1 −T2 T1 −T2 driving force = = ∆x / kA R resis tan ce

where R =∆x / kA A and is the resistance in K/W.

Fig. 3.2 Heat conduction in a cylinder Conduction Through a Hollow Cylinder In many instances in the process industries, heat is being transferred through the walls of a thick-walled cylinder, such as a pipe that may or may not be insulated. Consider the hollow cylinder in Fig.3.2 with an inside radius of r1, where the temperature is T1, an outside radius of r2 having a temperature of T2, and a length of L m. Heat is flowing radially from the inside surface to the outside. Rewriting Fourier's law, with distance dr instead of dx, qx dT =− k A dr

The cross-sectional area normal to the heat flow is A =2πrL

Substituting A value and rearranging, and integrating,

q

r2

T2 dr = − k ∫T1 dT 2πL ∫ r r1

q =k

2πL (T1 −T2 ) ln( r2 / r1 )

Conduction Through a Hollow Sphere Heat conduction through a hollow sphere is another case of onedimensional conduction. Using Fourier's law for constant thermal conductivity with distance dr, where r is the radius of the sphere, qx dT =− k A dr The cross-sectional area normal to the heat flow is A =2πr 2

Substituting A value and rearranging, and integrating,

q

r2

dr

4π ∫ r r1

q=

2

T2

= − k ∫ dT T1

4πk (T1 −T2 ) T1 −T2 = 1 / r1 −1 / r2 (1 / r1 −1 / r2 ) / 4πk

It can easily be shown that the temperature varies hyperbolically with the radius.

LECTURE NO.4 CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE WALL - CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE WALL

Fig. 4.1 Heat flow through multilayer wall Consider the heat flow through composite wall made of several materials of different thermal conductivities and thicknesses. An example is a wall of a cold storage, constructed of different layers of materials of different insulating properties. All materials are arranged in series in the direction of heat transfer, as shown in the above Figure. The thickness of the walls are x1, x2, and x3 and the thermal conductivites of the walls are K1, K2, and K3, respectively. The temperatures at the contact surfaces are T2, T3, and T4. From Fourier’s Law, q A

=− K A

dT dx

This may be written as ∆T =T2 −T1 =− T1 −T2 =

x1 .q K1 A

T2 −T3 =

x2 .q K2 A

T4 −T3 =

x3 .q K3 A

q . ∆x KA

Total temperature difference, ∆T =∆T1 +∆T2 +∆T3  x x3  x T1 −T4 = q.  1 + 2 +   K1 A K 2 A K 31 A 

where,

T1 −T4 =

thermal potential responsible for

heat flow. The

 x1 x x3  + 2 +   is known as the total thermal resistance of the  K1 A K 2 A K 31 A 

composite ass. It is similar to the electrical resistance in series. The thermal circuit for multilayer rectangular system is shown in the following figure.

Fig 4.2 Electrical analog of one dimensional heat transfer through composite wall q=

T1 −T4 R1 + R2 + R3

CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER

Fig.4.3 One-dimensional heat flow through multiple cylindrical sections and electrical analog Consider a long cylinder of inside radius ri, outside radius ro, and length L, such as the one shown in above Figure 4.3. We expose this cylinder to a

temperature differential Ti - To and determine what the heat flow will be. For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction, so that the only space coordinate needed to specify the system is r. Again, Fourier's law is used by inserting the proper area relation. The area for heat flow in the cylindrical system is A =2πrL

so that Fourier's law is written qr =−k Ar

dT dr

qr =− 2π k rL

or dT dr

with the boundary conditions T = Ti at r = ri, T = To at r = ro The solution to equation qr =−k Ar

dT is dr

q =k

2πL (T1 −T2 ) ln( r2 / r1 )

and the thermal resistance in this case is Rth =

ln( ro / ri ) 2πkL

The thermal-resistance concept may be used for multiple-layer cylindrical walls just as it was used for plane walls. For the three-layer system shown in Figure the solution is

q=

2π L (T1 −T4 ) ln( r2 / r1 ) ln( r3 / r2 ) ln( r4 / r2 ) + + Ka KB KB

The thermal circuit is also shown in Figure.

LECTURE NO.5 THE OVERALL HEAT-TRANSFER COEFFICIENT - CRITICAL THICKNESS OF INSULATION 5.1 THE OVERALL HEAT-TRANSFER COEFFICIENT

Fig. 5.1 Heat flow with convective boundaries in plane wall

Fig. 5.2 Heat flow with convective boundaries in plane wall- Electrical Analog In many practical situations the surface temperatures (or boundary conditions at the surface) are not known, but there is a fluid on both sides of the solid surfaces. Consider the plane wall in the Figure 5.1, with a hot fluid at temperature T1 on the inside surface and a cold fluid at T4 on the outside surface. The convective coefficient on the outside is ho W/m2.K and hi on the inside. The heat transfer is expressed by q =hi A (T1 −T2 ) =

KA A (T2 −T3 ) =ho A (T3 −T4 ) ∆x A

The heat-transfer process may be represented by the resistance network in electrical analog Figure, and the overall heat transfer is calculated

as the ratio of the overall temperature difference to the sum of the thermal resistances:

∴q =

T1 −T4 T −T4 = 1 1 ∆x A 1 ∑R + + hi A K A A ho A

The overall heat transfer by combined conduction and convection is frequently expressed in terms of an overall heat-transfer coefficient U, defined by the relation q =U A ∆Toverall

where ∆Toverall

=T1 −T4 and U is

U=

1 ∆x A 1 + + W/m2.K hi K A ho

The overall heat-transfer coefficient is also related to the R value as U=

1 Rvalue

A more important application is heat transfer from a fluid outside a cylinder, through a metal wall, to a fluid inside the tube, as often occurs in heat exchangers.

Figure 5.3 Resistance analogy for hollow cylinder with convection boundaries.

Note that the area for convection is not the same for both fluids in this case, these areas depend on the inside tube diameter and wall thickness. In this case the overall heat transfer would be expressed by

q=

T1 −T4 1 ln ( ro / ri ) 1 + + hi Ai 2πKL ho Ao

The terms Ai and Ao represent the inside and outside surface areas of the inner tube. The overall heat-transfer coefficient may be based on either the inside or the outside area of the tube. Accordingly,

Ui =

1 1 Ai ln ( ro / ri ) Ai 1 + + hi 2πKL Ao ho

Uo =

1 Ao 1 A ln ( ro / ri ) 1 + o + Ai hi 2πKL ho

The general notion, for either the plane wall or cylindrical coordinate system, is that UA =

1 1 = ∑Rth Rth ,Overall

5.2 CRITICAL THICKNESS OF INSULATION

Fig. 5.4 Critical radius for insulation of cylinder or pipe Consider a layer of insulation is installed around the outside of a cylinder whose radius r1, is fixed and with a length L. The cylinder has a high thermal conductivity and the inner temperature T1 at point r1 outside the cylinder is fixed. An example is the case where the cylinder is a metal pipe

with saturated steam inside. The outer surface of the insulation at T 2 is exposed to an environment at To where convective heat transfer occurs. It is not obvious if adding more insulation with a thermal conductivity of k will decrease the heat-transfer rate. At steady state the heat-transfer rate q through the cylinder and the insulation equals the rate of convection from the surface: q =ho A (T2 −To )

q=

-------------- (1)

T1 −T4 1 ln ( ro / ri ) 1 --------- (2) + + hi Ai 2πKL ho Ao

As insulation is added, the outside area, which is A =2πr2 L , increases, but T2 decreases. However, it is not apparent whether q increases or decreases. To determine this, an equation similar to Eq. (2) with the resistance of the insulation represented by Eq.(3) is written using the two resistances:  r2  ln  r   r2 − r1 R= =  1 K Alm 2πKL Alm = log mean area

q=

2πL ( T1 − To ) ln ( r2 / r1 ) 1 + K r2 ho

As the outside radius, r2, increases, then in the denominator, the first term increases but the second term decreases. Thus, there must be a critical radius, rc, that will allow maximum rate of heat transfer, q. To determine the effect of the thickness of insulation on q, we take the derivative of q with respect to r2, equate this result to zero, and obtain the following for maximum heat flow. The maximization condition is

 1 1 − 2πL(T1 −To ) − 2 r K r h dq 2 o  2 = 2 dr2  ln ( r2 / r1 ) 1  +   K r 2 h0  

   

T1,T0, K, L, ro, ri are constant terms.  1 1   r K − r 2h 2 o  2

   =0 

Therefore, 1 1 = 2 r2 K r2 ho

When outside radius becomes equal to critical radius, or r2 = rc, we get ( r2 ) cr =

k ho

Where (r2)cr is the value of the critical radius when the heat transfer rate is a maximum. Hence, if the outer radius r2 is less than the critical value, adding more insulation will actually increase the heat- transfer rate q. Also, if the outer radius is greater than the critical, adding more insulation will decrease the heat transfer rate. Using values of K and ho typically encountered, the critical radius is only a few mm. As a result, adding insulation on small electrical wires could increase the heat loss. Adding insulation to large pipes decreases the heat transfer rate. When no insulation is provided then for a metal pipe with an outside radius of r2, qbare =2πr2 L ho (T2 −To )

The rate of heat transfer from an insulated pipe, where the annular insulating shell has an inside radius of r2 and an outer radius of r3,

qinsulated =

2πr3 L ho ( T2 − To ) r h r 1 + 3 o ln 3 K r2

Then, qinsulated qbare

 r3  1  = r h r2 1 + 3 o ln r3  K r2 

     

LECTURE NO.6 HEAT SOURCE SYSTEMS: ONE-DIMENSIONAL STEADY STATE HEAT CONDUCTION WITH HEAT GENERATION: HEAT FLOW THROUGH SLAB / PLANE WALL

HEAT-SOURCE SYSTEMS A number of interesting applications of the principles of heat transfer are concerned with systems in which heat may be generated internally. Nuclear reactors are one example; electrical conductors and chemically reacting systems are others.

Fig.6.1 Sketch illustrating one- dimensional conduction problem with heat generation. One-dimensional steady state heat conduction with heat generation: Heat Flow through slab / Plane Wall Consider the plane wall with uniformly distributed heat sources shown in the above Figure. The thickness of the wall in the x direction is 2L, and it is assumed that, the dimensions in the other directions are sufficiently large that the heat flow may be considered as one-dimensional. The heat generated per •

unit volume is q , and we assume that the thermal conductivity does not vary with temperature. (This situation might be produced in a practical situation by passing a current through an electrically conducting material) The differential equation which governs the heat flow is

.

d 2T q + =0 dx 2 k .

d 2T q =− 2 dx k Integrating twice with respect to x results in .

dT q =− x +c1 dx k .

q 2 -------------- (1) T =− x +c1 x +c2 2k For the boundary conditions we specify the temperatures on either side of the wall, i.e., T = Tw at x =±L

Since the temperature must be the same on each side of the wall, c 1 must be zero. •

dT =0, q =0 ⇒c1 =0 The temperature at the mid plane (x = 0) is denoted by To and from Equation (1) At mid plane = x=0 and T= T0 To = c2 The temperature distribution equation (1) becomes .

q 2 T =− x +T0 2k .

q 2 --------------- (2) T −T0 =− x 2k

Assumed T=Tw at x= L .

q 2 ----------------- (3) Tw −T0 =− L 2k 2

( 2) T −T0 x ⇒ =  (3) Tw −T0  L 

------------ (4)

An expression for the midplane temperature To may be obtained through an energy balance. At steady-state conditions the total heat generated must equal the heat lost at the faces. Thus  dT  2−KA  dx  x =L 

 • =q . A. 2 L  

where A is the cross-sectional area of the plate. The temperature gradient at the wall is obtained by differentiating Equation (4): 2

T −T0 x =  Tw −T0  L 

2

x  T −T0 =(Tw −T0 )  L 

dT =(Tw −T0 )

2x . dx L2

dT  2 =(Tw −To ). dx  L x =L

at x= L

− K.

Since,

• dT =q L dx

•

dT qL =− dx K •

(Tw −T0 ) 2 =− qL L K •

qL 2 Tw −T0 =− 2K

•

qL 2 T0 = +Tw 2K

----- (5)

LECTURE NO.7 STEADY STATE HEAT CONDUCTION WITH HEAT DISSIPATION TO ENVIRONMENT- INTRODUCTION TO EXTENDED SURFACES (FINS) OF UNIFORM AREA OF CROSS SECTION - DIFFERENT FIN CONFIGURATIONS

-

GENERAL

CONDUCTION

ANALYSIS

EQUATION Steady State heat conduction with heat dissipation to environment: Introduction to extended surfaces (FINS) of Uniform area of cross section The term extended surface is commonly used to depict an important special case involving heat transfer by conduction within a solid and heat transfer by convection (and/or radiation) from the boundaries of the solid. Heat transfer from the boundaries of a solid to be in the same direction as heat transfer by conduction in the solid. In contrast, for an extended surface, the direction of heat transfer from the boundaries is perpendicular to the principal direction of heat transfer in the solid.

Fig. 7.1 Combined conduction and convection in a structural element. Consider a strut that connects two walls at different temperatures and across which there is fluid flow (Figure 7.1). With T1>T2, temperature gradients in the x-direction sustain heat transfer by conduction in the strut. However, with T1>T2> T∞there is concurrent heat transfer by convection to the fluid, causing qx, and hence the magnitude of the temperature gradient,

dT / dx

, to

decrease with increasing x. An extended surface is used specifically to enhance heat transfer between a solid and adjoining fluid. Such an extended surface is termed a fin.

The heat transfer rate may be increased by increasing the surface area across which the convection occurs. This is done by employing fins that extend from the wall into the surrounding fluid. The thermal conductivity of the fin material can have a strong effect on the temperature distribution along the fin and therefore influences the degree to which the heat transfer rate is enhanced. Ideally, the fin material should have a large thermal conductivity to minimize temperature variations from its base to its tip. In the limit of infinite thermal conductivity, the entire fin would be at the temperature of the base surface, thereby providing the maximum possible heat transfer enhancement.

Fig. 7.2 Use of fins to enhance heat transfer from a plane wall (a) Bare surface (b) Finned surface. Examples of fin applications are easy to find. Consider the arrangement for cooling engine heads on motorcycles and lawn mowers or for cooling electric power transformers. Consider also the tubes with attached fins used to promote heat exchange between air and the working fluid of an air conditioner. Two common finned tube arrangements are shown in Figure 7.3

Fig.7.3 Schematic of typical finned-tube heat exchangers.

Different fin configurations Different fin configurations are illustrated in Figure 7.4. A straight fin is any extended surface that is attached to a plane wall. It may be of uniform cross-sectional area, or its cross-sectional area may vary with the distance x from the wall. An annular fin is one that is circumferentially attached to a cylinder, and its cross section varies with radius from the wall of the cylinder. The foregoing fin types have rectangular cross sections, whose area may be expressed as a product of the fin thickness t and the width w for straight fins or the circumference 2πr for annular fins. In contrast a pin fin, or spine, is an extended surface of circular cross section. Pin fins may also be of uniform or non-uniform cross section. In any application, selection of a particular fin configuration may depend on space, weight, manufacturing, and cost considerations, as well as on the extent to which the fins reduce the surface convection coefficient and increase the pressure drop associated with flow over the fins.

FIGURE 7.4 Fin configurations. (a) Straight fin of uniform cross section. (b) Straight fin of non-uniform cross section. (c) Annular fin. (d) Pin fin / spine. General Conduction Analysis Equation As engineers we are primarily interested in knowing the extent to which particular extended surfaces or fin arrangements could improve heat transfer from a surface to the surrounding fluid. To determine the heat transfer rate associated with a fin, we must first obtain the temperature distribution along the fin.

Fig.7.5 Energy balance for an extended surface A general form of the energy equation for an extended surface is as follows: d 2T  1 dAc  dT  1 h dAs    + −   (T −T∞ ) =0 dx 2   Ac dx  dx  Ac k dx 

Its

solution

for

appropriate

boundary

conditions

provides

the

temperature distribution, which may be used with Fourier’s equation ( q x =− kA

dT ) to calculate the conduction rate at any x. dx

LECTURE NO.8 EQUATION OF TEMPERATURE DISTRIBUTION WITH DIFFERENT BOUNDARY CONDITIONS, FIN PERFORMANCE AND OVERALL SURFACE EFFICIENCY OF FINS Fins of Uniform Cross-Sectional Area To solve general form of fin energy equation it is necessary to be more specific about the geometry. We begin with the simplest case of straight rectangular and pin fins of uniform cross section (Figure 8.1). Each fin is attached to a base surface of temperature T(O) = Tb and extends into a fluid of temperature T∞.

Fig. 8.1 Straight fins of uniform cross section (a) Rectangular fin (b) Pin fin For the prescribed fins, Ac is a constant and As = Px, where As is the surface area measured from the base to x and P is the fin perimeter. Accordingly, with dAc/dx = 0 and dAs/dx = P, general form of the energy equation for an extended surface reduces to d 2T hP − (T −T∞) =0 2 dx kA c

--------(1)

To simplify the form of this equation, we transform the dependent variable by defining an excess temperature θ as

θ( x ) =T ( x ) −T∞

-------------(2)

where, since T∞is a constant, dθ / dx = dT / dx . . Substituting Equation (2) into Equation (3), we then obtain

d 2θ −m 2θ =0 2 dx

--------(3)

where m2 =

hP kA c

-------------(4)

Equation (3) is a linear, homogeneous, second-order differential equation with constant coefficients. Its general solution is of the form

θ( x) =C1 e mx + C2e −mx ---------------- (5) By substitution it may readily be verified that Equation (5) is indeed a solution to Equation (3). To evaluate the constants C1 and C2 of Equation (5), it is necessary to specify appropriate boundary conditions. One such condition may be specified in terms of the temperature at the base of the fin (x = 0)

θ(0) =Tb −T∞=θb ----------------- (6) The second condition, specified at the fin tip (x = L), may correspond to one of four different physical situations. The first condition, case A, considers convection heat transfer from the fin tip. Applying an energy balance to a control surface about this tip (Figure 8.2), we obtain hA c [T ( L ) −T∞] =−kA

dT dx

x =L

or hθ( L ) =−kA

dT dx

x =L

--------------- (7)

Figure 8.2 Conduction and convection in a fin of uniform cross section.

That is, the rate at which energy is transferred to the fluid by convection from the tip must equal the rate at which energy reaches the tip by conduction through the fin. Substituting Equation (5) into Equations (6) and (7), we obtain, respectively, θb = C1 + C2 ----------- (8)

And h(C1e mL + C2e −mL ) = km (C2e −mL − C1 e mL )

Solving for C1, and C2, it may be shown, after some manipulation, that θ cosh m( L − x) + ( h / mk ) sinh m( L − x) = ------- (9) θb cosh mL + (h / mk ) sinh mL

The form of this temperature distribution is shown schematically in Figure 8.2. Note that the magnitude of the temperature gradient decreases with increasing x. This trend is a consequence of the reduction in the q x ( x ) with

conduction heat transfer

increasing x due to continuous

convection losses from the fin surface. We are particularly interested in the amount of heat transferred from the entire fin. From Figure 8.2 it is evident that the fin heat transfer rate q f may be evaluated in two alternative ways, both of which involve use of the temperature distribution. The simpler procedure, and the one that we will use, involves applying Fourier's law at the fin base. That is, q f =qb =−kA c

dT dx

=−kA c x =0

dθ dx x =0

----- (10)

Hence, knowing the temperature distribution, θ( x) ,

qf

may be

evaluated, giving q f = hPkA c θb

sinh mL +( h / mk ) cosh mL cosh mL +( h / mk ) sinh mL

------- (11)

However, conservation of energy dictates that the rate at which heat is transferred by convection from the fin must equal the rate at which it is conducted through the base of the fin. Accordingly, the alternative formulation for q f is q f =∫ h[T ( x ) −T∞] dA s Af

q f =∫ hθ( x ) dA s Af

-------------- (12)

where A f Af is the total, including the tip, fin surface area. Substitution of Equation (9) into Equation (12) would yield Equation (11): The second tip condition, case B, corresponds to the assumption that the convective heat loss from the fin tip is negligible, in which case the tip may be treated as adiabatic and dθ =0 dx x =L

-------- (13)

Substituting from Equation (5) and dividing by m, we then obtain C1 e mL − C2 e −mL =0

Using this expression with Equation (8) to solve for C1 and C2 and substituting the results into Equation (5), we obtain

θ cosh m( L − x ) = θb cosh mL

--------- (14)

Using this temperature distribution with Equation (10), the fin heat transfer rate is then q f = hPkA cθb tanh mL

----- (15)

In the same manner, we can obtain the fin temperature distribution and, heat transfer rate for case C, where the temperature is prescribed at the fin tip. That is, the second boundary condition is θ( L) =θL , and the resulting expressions are of the form

θ (θL / θb ) sinh mx + sinh m( L − x) = ------- (16) θb sinh mL q f = hPkA cθb

cosh mL −θL / θb sinh mL

The very long fin, case D, is an interesting extension of these results. In particular, as L → ∞,θL → 0 and it is easily verified that

θ =e −mx θb q f = hPkA cθb

Fin Performance Fins are used to increase the heat transfer from a surface by increasing the effective surface area. However, the fin itself represents a conduction resistance to heat transfer from the original surface. For this

reason, there is no assurance that the heat transfer rate will be increased through the use of fins. An assessment of this matter may be made by evaluating the fin effectiveness εf . It is defined as the ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin. Therefore

εf =

qf hA c , bθb

where Ac , b is the fin cross-sectional area at the base. In any rational design the value of εf should be as large as possible, and in general, the use of fins may rarely be justified unless ε f ≥ 2 . Fin effectiveness is enhanced by the choice of a material of high thermal conductivity. Aluminum alloys and copper come to mind. However, although copper is superior from the stand point of thermal conductivity. aluminum alloys are the more common choice because of additional benefits related to lower cost and weight. Fin effectiveness is also enhanced by increasing the ratio of the perimeter to the cross-sectional area. For this reason, the use of thin, but closely spaced fins, is preferred. Another measure of fin thermal performance is provided by the fin efficiencyηf . The maximum driving potential for convection is the temperature difference between the base (x = 0) and the fluid, θb =Tb −T∞. Hence the maximum rate at which a fin could dissipate energy is the rate that would exist if the entire fin surface were at the base temperature. However, since any fin is characterized by a finite conduction resistance, a temperature gradient must exist along the fin and the above condition is an idealization. A logical definition of fin efficiency is therefore

ηf =

qf qmax

=

qf hA f θb

where Af is the surface area of the fin. Overall Surface Efficiency In contrast to the fin efficiency ηf , which characterizes the performance of a single fin, the overall surface efficiency ηo characterizes an array of fins and the base surface to which they are attached. Representative arrays are shown in Figure 8.3, where S designates the fin pitch. In each case the overall efficiency is defined as

ηf =

qf qmax

=

qt hA tθb

where qt is the total heat rate from the surface area At associated with both the fins and the exposed portion of the base (often termed the prime surface). If there are N fins in the array, each of surface area Af, and the area of the prime surface is designated as Ab, the total surface area is At =NA

f

+Ab

The maximum possible heat rate would result if the entire fin surface, as well as the exposed base, were maintained at Tb .

Figure 8.3 Representative fin arrays. (a) Rectangular fins. (b) Annular fins.

LECTURE NO.9 PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF BASIC

EQUATION;

SIMPLIFIED

CASE

FOR

SYSTEMS

WITH

NEGLIGIBLE INTERNAL RESISTANCE; TOTAL AMOUNT OF HEAT TRANSFERRED ; DIMENSIONAL ANALYSIS IN MOMENTUM TRANSFER PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF BASIC EQUATION Introduction In steady state heat-transfer systems the temperature at any given point and the heat flux were always constant over time. In unsteady state or transient processes the temperature at any given point in the system changes with time. Before steady-state conditions can be reached in a process, some time must elapse after the heat-transfer process is initiated to allow the unsteady-state conditions to disappear. Unsteady-state heat transfer is important because of the large number of heating and cooling problems occurring industrially. In metallurgical processes it is necessary to predict cooling and heating rates for various geometries of metals in order to predict the time required to reach certain temperatures. In food processing, for example, the canning industry, perishable canned foods are heated by immersion in steam baths or chilled by immersion in cold water. In the paper industry wood logs are immersed in steam baths before processing. In most of these processes the material is suddenly immersed in a fluid of higher or lower temperature.

Fig. 9.1 Unsteady State conduction in one direction

Derivation of Unsteady-State Conduction Equation To derive the equation for unsteady-state conduction in one direction in a solid, refer to Fig. 9.1. Heat is being conducted in the x direction in the ∆ x, ∆ y, ∆ z

in size. For conduction in the x direction, write q x =−kA

The term

∂T ∂x

------- (1)

∂T means the partial or derivative of T with respect to x, ∂x

with the other variables, y, z, and time t, being held constant. Next, making a heat balance on the cube, we can write rate of heat input + rate of generation

= rate of heat output + rate of heat accumulati on

---------- (2) The rate of heat input to the cube is ∂T rate of heat input = q x / x =−k ( ∆y ∆z ) ∂x ∂T Also, rate of heat output = q x / x +∆x =−k (∆y ∆z ) ∂x

----------(3)

x

------- (4)

x +∆ x

The rate of accumulation of heat in the volume ∆x, ∆y , ∆z in time ∂t is rate of heat accumulation = (∆x, ∆y , ∆z ) ρc p

∂T ∂t

------(5)

The rate of heat generation in volume ∆x, ∆y , ∆z is .

rate of heat generation = ( ∆x ∆y ∆z ) q ------(6) Substituting Eqs. (3)-(6) into (2) and dividing by ∆x, ∆y , ∆z  ∂T −k  ∂x  q+

.

− x

∆x

∂T ∂x

   ∂T x +∆x  =ρc p ∂t

------- (7)

Letting ∆x approach zero, we have the second partial of T with respect to x or ∂2T / ∂x 2 on the left side. Then, rearranging, .

.

∂T k ∂2T q ∂2T q = + = α + 2 2 ∂t ρc p ∂x ρc p ∂x ρc p

--------(8)

where

α is

k / ρc p

k , ρ, and c p

, thermal diffusivity. This derivation assumes constant

. In SI units,

α = m2/s, T = K, t = s, k = W/m.K, ρ = kg/m3,

.

q=

W/m3, and c p = J/kg.K. For conduction in three dimensions, a similar derivation gives  ∂2T ∂2T ∂2T ∂T =α   ∂x 2 + ∂y 2 + ∂z 2 ∂t 

.

 q   + ρc --------- (9)  p

In many cases, unsteady-state heat conduction is occurring but the rate of heat generation is zero. Then Eqs. (8) and (9) become ∂T ∂2T =α ∂t ∂x 2  ∂2T ∂T ∂2T ∂2T =α  + + 2 2  ∂x ∂t ∂y 2 ∂z 

------(10)    

----- (11)

Equations (10) and (11) relate the temperature T with position x, y, and z and time t. SIMPLIFIED CASE FOR SYSTEMS WITH NEGLIGIBLE INTERNAL RESISTANCE Basic Equation Consider a solid which has a very high thermal conductivity or very low internal conductive resistance compared to the external surface resistance, where convection occurs from the external fluid to the surface of the solid. Since the internal resistance is very small, the temperature within the solid is essentially uniform at any given time. An example would be a small, hot cube of steel at To K at time t = 0, suddenly immersed in a large bath of cold water at T∞which is held constant with time. Assume that the heat transfer coefficient h in W/m2 .K is constant with time. Making a heat balance on the solid object for a small time interval of time dt s, the heat transfer from the bath to the object must equal the change in internal energy of the object: hA (T∞ −T ) dt =c p ρV dT

where A is the surface area of the object in m2, T the average temperature of the object at time t in s, ρ the density of the object in kg/m 3,

and V the volume in m3. Rearranging the equation and integrating between the limits of T = To when t = 0 and T = T when t = t, T =T

∫

T =T0

dT hA = T∞ −T c p ρV  

hA

t =t

∫

t =0

dt

 

− t  c ρV  T −T∞ =e  p  T0 −T∞

This equation describes the time-temperature history of the solid object. The term c p ρV is often called the lumped thermal capacitance of the system. This type of analysis is often called the lumped capacity method or Newtonian heating or cooling method. Equation for Different Geometries In using the above equation the surface / volume ratio of the object must be known. The basic assumption of negligible internal resistance was made in the derivation. This assumption is reasonably accurate when N Bi = where

hx 1