lab 1-linear conduction

8/12/2019 Lab 1-Linear Conduction

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1.0 OBJECTIVES

1.1 To investigate the thermal conductivity and thermal contact resistance of

different types of material.

1.2 To study on the different method of insulation of the system.

1.3 To observe unsteady conduction of heat

1.4 To understand the use of the Fourier Rate Equation in determining rate of

heat flow through solid materials for one-dimensional steady flow of heat.


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2.0 INTRODUCTION TO LINEAR CONDUCTION HEAT TRANSFER

When a temperature gradient exists in either a solid or stationary fluid medium, the heat

transfer which takes place is known as conduction. When neighbouring molecules in a fluid

collide, energy is transferred from the more energetic to the less energetic molecules. Because

higher temperatures are associated with higher molecular energies, conduction must occur in

the direction of decreasing temperature.

Fig 2.1 : Rate of heat flow in a rod

2.1 Concepts

The temperature distribution in the wall can be determined by solving the heat

equation with the proper boundary conditions. For steady state conditions with no

distributed source or sink within the wall, the form of the heat equation is:

Fig 2.2 : Linear heat flow of material

0

dx

dTk

dx

dt


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2.2 Principles

For one-dimensional, steady-state conduction in a plane wall with no heat generation, the

heat flux is a constant, independent of x. If the thermal conductivity of the wall material is

assumed to be constant, the equation may be integrated twice to obtain the general solution:

For one-dimensional, steady-state conduction in a plane wall with no heat generation and

constant thermal conductivity, the temperature varies linearly with x:

2.3 Thermal Resistance

Thermal resistance for conduction in a plane wall is given as:

2.4 Contact Resistance

Although neglected until now, it is important to recognize that, in composite systems, the

temperature drop across the interface between materials may be appreciable. This temperature

change is attributed to what is known as the thermal contact resistance.

Where qx= qx/A

21)( CxCxT

BAt

tx TTL

Ak

dx

dTAkq

Ak

L

q

TTR

tx

BAcondt

,

"

"

,

x

BAct

q

TTR


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2.5 Linear Heat Conduction

Fig. 2.3: Layout for Linear Heat Conduction Test Specimen

Fig. 2.4: Module for Linear Heat Conduction Test Specimen

The rate of linear conduction heat transfer for this system (Fig. 3.1):

Where,

k - Thermal conductivity

A - Cylindrical area of specimen

L - Heat traveling distance

TA- Temperature near heater

TB- Temperature further heater

BAt

tx TTL

Ak

dx

dTAkq


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3.0 COMPONENTANDEQUIPMENT

Fig. 3.1:AssemblyDiagramofLinear & Radial Heat ConductionApparatus

No. Item No. Item

1. Linear heat conduction module 9. ON / OFF switch

2. Cold water inlet & outlet port 10. 220VAC fuse

3. Thermocouple ports 11. Linear/Radial module selector switch

4. Temperature selector switch 12. Module clip

5. Temperature meter 13. Radial heat conduction module

6. Heater supply 14. Thermocouples

7. Power meter 15. Thermocouple ports

8. Power controller

***SafetyInst ruct ion :

Theequipmentsyouareusingarepotentiallydangerous.Youarestrictlyrequiredtofollow the

proceduresoutlinedbelow.Donotmakeanyunnecessaryactionswhicharenotstatedinthe

procedure. Ifnot,anaccidentmayoccur. Incaseofdoubt,contact theTechnicianorTraining

Engineerimmediate.

2

1

3

4 5 6

8

7

9

10

15

1

1

13

1


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4.0 PROCEDURE

4.1 Heat Conductionon brass usingLinear Module

No. Procedure

1. Connect the cold water supply. Allow the cooling

water to flow continuously throughout the experiment

(Turn until 180)

2. Do not connect the extra 30 mm length of brass or

stainless steel in the middle of the Apparatus.

3. Apply some heat transfer compound on the surface ofthe thermocouple to further improve the experiment

performance

4. Place the 6 thermocouples into the holes located

directly above the linear heat conduction apparatus

according to the numberings attached

Figure


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5. Tighten the set screws to keep the thermocouples in

place. (do not over tighten)

6. Connect the thermocouple to the thermocouple ports

according to the number.

7. Connect the heater cable between the control panel

and make sure the clips properly locked

8. Record the initial temperature values for each

measuring point by switching the selector switch.

Read the temperature value from the temperature

meter.

9. Switch ON the selector switch to Linearand set the

power of the heater to 20 W by turning the power

regulator.


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10. Wait for 25 to 30 minutes until the temperature

achieved stable at every measuring points.

11. Observe and record down the respective final

temperature values at every point.

12. Turn OFF the ON/OFF switch after finish the

experiment.

*Note:

Re-adjust the thermocouple if the experimental result is not ideal. Make sure the

thermocouples are inserted into the holes touched on the brass or stainless steel.

Ensure the heater selector On/Off switch is set at the correct experiment setting (linear or

radial) prior to carry out the experiment.

For linear conduction heat transfer, the clips must be properly locked to allow the test

specimens stay in contact during the experiment for effective results.

Take note that the maximum working temperature for this apparatus is 90C. Working higher

than 90C would not provide better experimental results.

The water cooling supply must be running when conducting the experiment.

Do not regulate the water cooling supply flow rate when conducting the experiment.


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5.0 RESULTS

5.1 Temperature results for different measuring point

Table 5.1: Linear Conduction for Brass (25.40mm)

Fig. 5.1: Measuring point in Radial module test specimen

Measuring Point PositionDistance from

Heater (mm)

Initial Temperature

(C)

Final Temperature

(C)

1 Nearest to heater 15 27.0 39.0

2 25 27.0 39.3

3 35 27.0 39.0

4 45 27.0 37.8

5 55 27.0 36.8

6 Furthest to heater 65 27.0 36.0

Measuring

point 1 to 6


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6.0 DISCUSSION

6.1 Based on the formula following, do discuss on it,

Linear:

The most efficient method of heat transfer is conduction. This mode of heat

transfer occurs when there is a temperature gradient across a body. In this case, the

energy is transferred from a high temperature region to low temperature region due to

random molecular motion (diffusion). Conduction occurs similarly in liquids and gases.

Regions with greater molecular kinetic energy will pass their thermal energy to regions

with less molecular energy through direct molecular collisions. In metals, a significant

portion of the transported thermal energy is also carried by conduction-band electrons.

Different materials have varying abilities to conduct heat. Materials that conduct heat

poorly (wood, styrofoam) are often called insulators. However, materials that conduct

heat well (metals, glass, some plastics) have no special name.

The simplest conduction heat transfer can be described as one-dimensional

heat flow as shown in the following figure.The rate of heat flow from one side of an

object to the other, or between objects that touch, depends on the cross-sectional area

of flow, the conductivity of the material and the temperature difference between the two

surfaces or objects.

BAt

tx TTL

Ak

dx

dTAkq

"

"

,

x

BAct

q

TTR


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Fig. 6.1: Heat flow in conductor

Mathematically, it can be expressed as :

where qis the heat transfer rate in watts (W), kis the thermal conductivity of the material

(W/m.K),Ais the cross sectional area of heat path, and is the temperature gradient

in the direction of the flow (K/m).

The above equation is known as Fouriers law of heat conduction. Therefore, the

heat transfer rate by conduction through the object in the above figure can be expressed

as:

Where L is the conductor thickness (or length), DT is the temperature difference

between one side and the other (for example, DT= T1T2is the temperature difference

between side 1 and side 2).The quantity (DT/L) in equation above is called the temperature gradient: it tells

how many 0C or K the temperature changes per unit of distance moved along the path

of heat flow. The quantity L/kA is called the thermal resistance


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Thermal resistance has SI units of kelvins per watt (K/W). Notice from Equation

above that the thermal resistance depends on the nature of the material (thermalconductivity kand geometry of the body d/A). We realize from the above equations, we

realize the heat transfer rate as a flow, and the combination of thermal conductivity,

thickness of material and area as a resistance to this flow.

Considering the temperature as a potential function of the heat flow, the Fourier

law can be written as

If we define the resistance as the ratio of potential to the corresponding transfer

rate, the thermal resistance for conduction can be expressed as

It is clear from the above equation that decreasing the thickness or increasing the

cross-sectional area or thermal conductivity of an object will decrease its thermal

resistance and increase its heat transfer rate.


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6.2 What is the relationship between the rate and thermal conductivity?

Based on the above equation, we can see an obvious relation between the rate

and thermal conductivity. Increasing any of it would increase another. This means,

increasing the thermal conductivity will increase the rate of heat transfer.

Mathematically, it can be expressed as:

where q is the heat transfer rate in watts (W), k is the thermal conductivity of the material

(W/m.K), A is the cross sectional area of heat path, and is the temperature gradient ,

dT/dx in the direction of the flow (K/m).

The above equation is known as Fouriers law of heat conduction. Therefore, the

heat transfer rate by conduction through the object in the above figure can be expressed

as:

Where L is the conductor thickness (or length), DT is the temperature difference

between one side and for example, DT = T1T2 is the temperature difference between

side 1 and side 2.

BAt

tx TT

L

Ak

dx

dTAkq


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6.3 What factors may concern on the thermal conductivity. Discuss it

based on:

- Transfer distance, transfer rate , transfer area and material type.

We knew that different materials have different thermal conductivity. The

dimension of thermal conductivity is M 1 L1T3t 1. These variables are (M)mass,

(L)length, (T)time, and (t)temperature. In SI units, thermal conductivity is measured in

watts per meter kelvin (Wm1K1). In Imperial units, thermal conductivity is measured

in BTU/(hrftF)

This shows that, the thermal conductivity is increase as the transfer rate, transfer

distance and transfer area is increases. This means, having a high value of these criteria

would increase thermal conductivity of the materials.

As based on material types for thermal conductivity, the heat transfer occurs at a

higher rate across materials of high thermal conductivity than across materials of low

thermal conductivity. Correspondingly materials of high thermal conductivity are widely

used in heat sink applications and materials of low thermal conductivity are used as

thermal insulation. Thermal conductivity of materials is temperature dependent .One of

example of materials of high thermal conductivity is copper whilst for the low thermal

conductivity is wood.


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6.4 How thermal conductivity and thermal contact resistance concerns to any

insulating designs.

Fig. 6.2 : Thermal conductivity and the interface resistance form part ofthe thermal interface resistance of a thermal interface material.

Thermal contact resistance occurs due to the voids created by surface

roughness effects, defects and misalignment of the interface. The voids present in the

interface are filled with air. Heat transfer is therefore due to conduction across the actual

contact area and to conduction (or natural convection) and radiation across the gaps.

If the contact area is small, as it is for rough surfaces, the major contribution to theresistance is made by the gaps. To decrease the thermal contact resistance, the

surface roughness can be decreased while the interface pressure is increased.

However, these improving methods are not always practical or possible for

electronic equipment. Thermal interface materials (TIM) are a common way to overcome

these limitations. Properly applied thermal interface materials displace the air that is

present in the gaps between the two objects with a material that has a much-higher

thermal conductivity. Air has a thermal conductivity of 0.022 W/mK while TIMs have

conductivities of 0.3 W/mKand higher.

When selecting a TIM, care must be taken with the values supplied by the

manufacturer. Most manufacturers give a value for the thermal conductivity of a material.

However, the thermal conductivity does not take into account the interface resistances.

Therefore, if a TIM has a high thermal conductivity, it does not necessarily mean that the

interface resistance will be low.


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Selection of a TIM is based on three parameters: the interface gap which the TIM

must fill, the contact pressure, and the electrical resistivity of the TIM. The contact

pressure is the pressure applied to the interface between the two materials. The

selection does not include the cost of the material. Electrical resistivity may, or may not,

be important, depending upon electrical design details.

6.5 Gives other insulating material and method.

Other insulator material that the material having low thermal conductivity such

wool. If the home is under construction, the easiest and cleanest type of insulation to

install is in the form of a blanket, also known as batts or rolls. Installation of blanket

insulation is quick and easy. Generally the batts conform to industry standards so that

the width matches the space between uprights in a wall. Some custom cutting will be

required to go around pipes, wires, and secondary uprights or cross members.

However, you'll find that most of the work can be done by just measuring the

height of the wall and cutting the insulation. Then, using an industrial stapler, simply

apply the insulation to the wall. Blanket insulation can be manufactured from a variety of

materials: slag wool, rock wool, fiberglass, cotton, and cellulose. If we put any of this

material in between high thermal conductivity materials, the rate of heat will be

decreased.


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7.0 CONCLUSION

From this lab session, we have demonstrate how the equation can be used to relate the

temperature difference, heat flow and distance in solid material of constant cross sectional area

and thermal conductivity. Increased of heat flow produces an increased temperature gradient.

We also can see for one-dimensional, steady-state conduction in a plane wall with no heat

generation, the heat flux is a constant, independent of x. we have learnt how to investigate the

thermal conductivity and thermal contact resistance of different types of material by using

formula. At the same time, we have study on the different method of insulation of the system

with that we knew materials that have low thermal conductivity, such as brick, cork, glass,

granite, limestone, wool, paper, rubber and sand stone. Any of these material is applied as

insulator put in between a high thermal conductivity materials, the rate of heat transfer will be

less down . Then, for the application of this experiment this concept can be apply in a design of

a heat sink. A heat sink is designed to increase the surface area in contact with the cooling

medium surrounding it, such as the air. Approach air velocity, choice of material, fin or other

protrusion design and surface treatment are some of the factors which affect the thermal

performance of a heat sink. Heat sink attachment methods and thermal interface materials also

affect the eventual die temperature of the integrated circuit. So, this experiment is a good

exposure to student to learn and know the concept of heat transfer in conductivity mechanism

more better. As conclusion, we can see that the objectives of the experiment have achieved

and been completed.
http://en.wikipedia.org/wiki/Die_(integrated_circuit)http://en.wikipedia.org/wiki/Die_(integrated_circuit)


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8.0 REFERENCES

8.1 Website

[1] http://en.wikipedia.org/wiki/Heat_transfer

[2] http://www.taftan.com/thermodynamics/FOURIER.HTM

[3] http://en.wikipedia.org/wiki/Heat_sink#Heat_transfer_principle

[4] http://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html

8.2 Books

[1] Heat and Mass Transfer-Fundamental and Applications.

By Yunus A. Cengel and Afshin J. Ghajar, Mc Graw Hill (2011).

[2] Fundamentals of Heat and Mass Transfer

By M. Thirumaleshwar, Pearson Education India (2006)
http://en.wikipedia.org/wiki/Heat_transferhttp://www.taftan.com/thermodynamics/FOURIER.HTMhttp://en.wikipedia.org/wiki/Heat_sink#Heat_transfer_principlehttp://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.htmlhttp://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.htmlhttp://en.wikipedia.org/wiki/Heat_sink#Heat_transfer_principlehttp://www.taftan.com/thermodynamics/FOURIER.HTMhttp://en.wikipedia.org/wiki/Heat_transfer

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