04.conduction part2

8/10/2019 04.Conduction Part2

1/41

1D, Steady State Heat Transfer with HeatGeneration

Fins and Extended Surfaces

1


2/41

Chapter 3c : One-dimensional, Steady state conduction (with

thermal energy generation) (Section 3.5Textbook)

2

3.1 Implications of energy generation Involve a local source of thermal energy due to conversion from another

form of energy in a conducting medium.

The source may be uniformlydistributed, as in the conversion fromelectrical to thermal energy

or it may be non-uniformlydistributed, as in the absorption of radiationpassing through a semi-transparent medium

Generation affects the temperature distribution in the medium and causesthe heat rate to varywith location, thereby precluding inclusion of themedium in a thermal circuit. (Cannot use electrical analogy!)

Eq. (1.11c)


3/41

Chapter 3 : One-dimensional, Steady state conduction

(without thermal generation)

3

*Recall previous case: a steady state plane wall with constant k &

no heat generation Assuming steady-state conditions and nointernal heat generation (i.e. q = 0), thenthe 1-D heat conduction equation reducesto:

For constant kandA

.

This means:Heat flux (qx) is independent of xHeat rate (qx) is independent of x

Boundary conditions: T(0)= Ts,1T(L)= Ts,2


4/41

Chapter 3 : One-dimensional, Steady state conduction (with

thermal generation)

4

3.2 A plane wall with internal heat generation

Assuming steady-state conditions and internal heatgeneration (i.e. q = 0), from the 1-D heat conductionequation:

.

*Consider a plane wall between two fluids of different temperatures

- general heat equation reduces to:

Boundary conditions: T(-L)= Ts,1T(L)= Ts,2

This means:

Heat flux (qx) is not independent of x

Heat rate (qx) is not independent of x


5/41


thermal generation)

5

, for constant kandA

2ndorder DE: Integrate twice to getT(x)

at x = -L, T(-L) = Ts,1 , at x = L, T(L) = Ts,2

for boundary conditions:

Thisgives,

and

Substituting the values

for C1and C2into eq. T(x)Tem erature distribution e uation


6/41


thermal generation)

6

Then, apply Fouriers Law to get heat transfer(BUT qxis now dependent on x)

Heat flux (W/m2):

Heat rate (W):


7/41


thermal generation)

7

What happens if both surfaces are maintained at the same temperature,

Ts,1 = Ts,2 = Ts

This case is called A case of symmetricsurface conditions or one surface wasinsulated.

Therefore, the temperature distribution eq. reducesto :

The max temperature exists at the mid plane:

Rearranging the temp distribution

*this means, at the planeof symmetry the temp

gradient is ZERO.

Eq. (3.42)

Eq. (3.43)


8/41


thermal generation)

8

Symmetric surface conditions or one surface was

insulated. Because the temp gradient at the centerline is zero,there is ZEROheat flow at that point and it behaveslike an insulated wall.

The insulated wall has the same parabolictemperature profile as half the un-insulated full wall

*recall the previous chapter: Boundaryconditions (Table 2.2)


9/41


thermal generation)

9



The insulated wall has the same parabolic temperatureprofile as half the un-insulated full wall

Why does the magnitude of the temperaturegradient increase with increasingx ?


10/41


thermal generation)

10





- due to increase in temperature difference


11/41


thermal generation)

11





How do we determine Ts?- due to increase in temperature difference


12/41


thermal generation)

12





How do we determine Ts?

How do we determine the heat rate at x = L ?

(no energy in, neglecting radiation,energy balance becomes)


13/41


thermal generation)

13





How do we determine Ts?

How do we determine the heat rate at x = L ?Using the surface energy balance, energy

generated must equal to energy outflow

(Neglecting radiation, energy balancebecomes)


14/41


thermal energy generation)

Referring to the Example 3.7 in textbook

a) Parabolic in material Ab) Zero slope at insulated

boundaryc) Linear slope in material

Bd) Slope change kB/kA=2 at

interfacee) Large gradients near the

surface


15/41



Example (3.74):

Consider a plane composite wall that is composed of three materials(materials A,B and C are arranged left to right) of thermal conductivitieskA=0.24 W/mK, kB=0.13 W/mK and kC=0.50 W/mK. The thickness of thethree sections of the wall are LA= 20mm, LB= 13mm and LC= 20mm. Acontact resistance of Rt,c=10

-2m2K/W exists at the interface between

materials A and B, as well as interface between B and C. The left face of thecomposite wall is insulated, while the right face is exposed to convectiveconditions characterised by h=10 W/m2K, T=20C. For case 1, thermalenergy is generated within material A at rate 5000 W/m3. For case 2,thermal energy is generated within material C at rate 5000 W/m3.

a) Determine the maximum temperature within the composite wall under

steady state conditions for Case 1b) Sketch the steady state temperature distribution on T-x coordinates for

Case 1

c) Find the maximum temperature within the composite wall and sketchthe steady state temperature distribution for Case 2


16/41



16

3.3 Radial systems (cylinder and sphere)


17/41



17

Temp distribution for solid cylinder:

Eq. (3.53)or

(C.23)

Temp distribution for hollow cylinder:

Eq. (C.2)


18/41



18

Temp distribution for solid sphere:

Temp distribution for spherical wall:

Eq. (C.24)

Eq. (C.3)

*A summary of temp distributions, heat fluxes & heat rates for all cases is provided inA endix C.


19/41



19

Problem 3.92:

A long cylindrical rod of diameter 200 mm with thermal conductivity of0.5 W/mK experiences uniform volumetric heat generation of 24,000W/m3. The rod is encapsulated by a circular sleeve having an outerdiameter of 400 mm and a thermal conductivity of 4 W/mK. The outersurface of the sleeve is exposed to cross flow of air at 27C with a

convection coefficient of 25 W/m2

K.a) Find the temperature at the interface between the rod and sleeve and

on the outer surface.

b) What is the temperature at the center of the rod ?


20/41



20

Problem 3.95:

Radioactive wastes (krw=20 W/mK) are stored in a spherical stainless steel(kss=15 W/mK) container of inner and outer radii equal to ri=0.5 m andro=0.6 m. Heat is generated volumetrically within the wastes at a uniformrate of 105 W/m3, and the outer surfaces of the container is exposed to awater flow for which h=1000 W/m2K and T=20C

a) Evaluate the steady-state outer surface temperature, Ts,o

b) Evaluate the steady-state inner surface temperature, Ts,i

c) Obtain an expression for the temperature distribution, T(r), in theradioactive wastes. Express your result in term of ri, Ts,i, krw and q.Evaluate the temperature at r = 0

.


21/41

Chapter 3d : Heat transfer from extended surface

21

3.1 Introduction

Extended surface (also known as fins) is commonly used todepict an important special case involving combination ofconduction-convection system.

Consider a strut that connects two walls at differenttemperatures and across which there is fluid f low

(Section 3.6 Textbook)


22/41


22

3.1 Introduction

Extended surface (also known as fins) is commonly used todepict an important special case involving combination ofconduction-convection system.

Why its important ?

Basically, there are 2 ways of increasing heat transfer

i) Increase fluid velocity to reduce temperature (manylimitation)

ii) Increase surface area

*Particularly beneficial when h

is small i.e. gas and natural

convection

The most frequent application to enhance heat transferbetween a solid joining and an adjoining fluid


23/41


23

Applications ?


24/41


24

Typical fin configurations (after simplification)

Straight fins of (a) uniform;(b) non-uniform cross sections;(c) annular fin;(d) pin fin of non-uniform cross section

Figure 3.14


25/41


25

3.2 A general conduction analysis for an extended surfaces

Applying the conservation of energy

Using,

Then, the heat equation becomes:

General form of the energy equation for an extendedsurface

Eq. (3.61)


26/41


26

3.3 The Fin Equation

Assuming 1-D case, steady state conduction in an extendedsurface, constant k, uniform cross sectional area, negligiblegeneration and radiation.

Cross section area, Acis constantand fin surface area, As= Px, this

mean dAc/dx = 0 and dAs/dx = P

General equation becomes:

Eq. (3.62)


27/41


27

To simplify the equation, we define an excess temperature ( thereduced temperature) as:

The previous equation becomes:

where,

* malso known as fin parameter

Eq. (3.63)

Eq. (3.65)

P is the fin perimeter


28/41


28

By referring to Table 3.4 : at

different case of heat transferanalysis

Temperature distribution, /b


29/41


29


30/41

30


31/41


31

Example (3.120):

A brass rod 100 mm long and 5 mm in diameter extendshorizontally from a casting at 200C. The rod is in an airenvironment with T= 20C and h = 30 W/m

2K. What is thetemperature of the rod at 25, 50 and 100 mm from the castingbody ?


32/41


32

3.4 Fin performance parameters (single fin case)

Fin efficiency max. potential heat transfer rate

Fin effectiveness ratio of heat transfer with and without fin

Fin resistance

Eq. (3.81)

Eq. (3.83) and (3.92)

Eq. (3.86)

Expressions for f are provided in Table 3.5 for common geometries, for example atriangular fin: - Surface area of the fin

- Profile area


33/41

33


34/41

34

(cont.)


35/41

35


36/41

36


37/41


37

Example (3.123):

A straight fin fabricated from 2024 aluminium alloy (k = 185 W/mK) has abase thickness of t = 3 mm and a length of L = 15 mm. Its basetemperature is Tb= 100 C, and it is exposed to a fluid for which T=20C and h = 50 W/m2K. For the foregoing conditions and a fin of unitwidth, compare the fin heat rate, efficiency and volume for

i) Rectangular profileii) Triangular profile

iii) Parabolic profile


38/41


38

3.5 Fin arrays

Representative arrays of

a) Rectangular fins

b) Annular fins

Eq. (3.99)

Eq. (3.100)

Eq. (3.102)

Eq. (3.103)


39/41


39

Previous equations are for fins that are produced by machining or

casting which as an integral part of the wall ( as in Fig. 3.20 & Fig3.21a)

Eq. (3.102)

Eq. (3.103)


40/41


40

However, some fins are manufactured separately and attached by

a metallurgical or adhesive joint or press fit. Such cases need toconsider contact resistance (as in Fig 3.21b)

Eq.(3.105a)Eq.(3.105b)

Eq. (3.104)


41/41

04.conduction part2

Documents