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Koya University Faculty of Engineering Chemical Engineering Department Laboratory of Heat transfer Experiment Number One Linear heat conduction Instructor: Dr.Barham & Mrs.Anfal Author Name: Aree Salah Tahir Experiment Contacted on: 20/Oct/2014 Report Submitted on: 10/Nov /2014 Group:A
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  • Koya University Faculty of Engineering

    Chemical Engineering Department

    Laboratory of Heat transfer Experiment Number One Linear heat conduction

    Instructor: Dr.Barham & Mrs.Anfal

    Author Name: Aree Salah Tahir

    Experiment Contacted on: 20/Oct/2014

    Report Submitted on: 10/Nov /2014

    Group:A

  • List of content:

    Objectives...............................................3 Introduction.. .4 Background Theory ...........5 Procedure .......6 Equipment and components used...7 Table of Reading........8 Calculation........................................9-10 Table...11-12 Discussion ......13-14 Conclusion........15 References .......16

  • Objectives:

    The aim of this experiment is to measurement linear thermal along z

    direction conductivity and to investigate and verify Fouriers Law for linear heat conduction along z direction.

    {1}

  • Introduction:

    Conduction is defined as the transfer of energy from more energetic

    particles to adjacent less energetic particles as a result of interactions

    between the particles. In solids, conduction is the combined result of

    molecular vibrations and free electron mobility. Metals typically have

    high free electron mobility, which explains why they are good heat

    conductors. Conduction can be easily understood if we imagine two

    blocks, one very hot and the other cold. If we put these blocks in contact

    with one another but insulate them from the surroundings, thermal

    energy will be transferred from the hot to the cold block, as evidenced by

    the increase in temperature of the cold block.

    This mode of heat transfer between the two solid blocks is termed

    conduction.

  • Background Theory:

    Linear Conduction of heat along a simple bar.

    If a plane wall of thickness (X) and area (A) , supports a temperature difference (T) then the heat transfer rate per unit time (Q) by conduction through the wall is found to be:

    If the material of the wall is

    homogeneous and

    has a thermal conductivity

    (k) then:

    Heat flow is positive in the direction of temperature fall. What is the

    effect of average temperature on the values of thermal conductivity for

    brass?

    The heat flow through a material can not always be evaluated at steady

    state e.g. through the wall of a furnace that is being heated or cooled. To

    calculate the heat flow under these conditions

    it is necessary to find the temperature distribution through the solid and

    how the distribution varies with time. Using the equipment set-up

    described above, it is a simple matter of monitoring the temperature

    profile variation during either a heating or cooling cycle thus

    facilitating the study of unsteady state conduction. {2}

  • Procedure:

    1-Install the insert and adjust the cooling water flow rate and the heater

    power.

    2-Switch on the unit and adjust the desired temperature drop through the

    power setting on the control and display unit.

    3-When the thermal conduction process has reached a steady state

    condition i.e. the temperature at individual measuring points are stable

    note the measurement results at the individual measuring points and the

    electrical power supplied to the heater.

  • Equipment and components used:

    1-display and control unit,

    2-measuring object,

    3-experimental set-up for radial heat conduction,

    4-experimental set-up for linear heat conduction {3}

  • Table of Reading:

    Temperature of barrel is 23.2 c

    Q T1 T2 T3 T4 T5 T6 T7 T8

    30 0 100.6 97.9 X 89.2 88.9 X 73.2

    60 0 91.2 89.6 X 86.1 85.7 X 74.8

    Q T1 T2 T3 T4 T5 T6 T7 T8

    30 0 100.6 97.9 93.6 89.2 88.9 81.1 73.2

    60 0 91.2 89.6 87.9 86.1 85.7 80.3 74.8

  • Calculation:

    At Q= 30 watt

    For No 2:

    Diameter for (brass and steel) =25mm = 0.025m

    Q= 30watt T2=100.6 T3=97.9 T=T2 T1 =2.7 K

    X=10 mm = 0.01m

    A=

    4d2 =

    4 (0.025)2 =4.910 -4

    Q=KA (X

    T) K=

    Q

    A(

    X

    T)

    K =30(0.01)

    4.9104(2.7) = 226.5

    W

    m.C

    The same way for others.

    No. X (m)

    T

    () T

    ( ) K

    (

    . )

    1 ------- ------- ------ -------

    2 0.010 373.6 2.7 226.75

    3 0.020 370.9 4.3 142.38

    4 0.030 366.6 4.4 139.14

    5 0.040 362.2 0.3 2040.8

    6 0.050 361.9 7.8 78.49

    7 0.060 354.1 7.9 77.49

    8 0.070 346.2 ------ -------

  • At Q= 60 watt

    For No 2:

    Diameter for brass and steel = 25mm = 0.025m

    Q= 60 watt T2=91.2 T3=89.6 T=1.6 K

    X=10 mm = 0.01m

    A=

    4d2 =

    4 (0.025)2 =4.910 -4

    Q=KA (X

    T) K=

    Q

    A(

    X

    T)

    K =360(0.01)

    4.9104(1.6) =

    KW

    m.K

    The same way for others.

    No. X (m)

    T

    ( ) T

    ( ) K

    (

    . )

    1 ----- ----- ------- ------

    2 0.010 364.2 1.6 756.3

    3 0.020 362.6 1.7 720.3

    4 0.030 360.9 1.8 680.2

    5 0.040 359.1 0.4 3061.2

    6 0.050 358.7 5.4 226.7

    7 0.060 353.3 5.5 222.6

    8 0.070 347.8 ----- ------

  • Table:

    2.7

    4.3 4.4

    0.3

    7.8 7.9

    0

    1

    2

    3

    4

    5

    6

    7

    8

    9

    0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

    Tem

    per

    atu

    re (

    K)

    distance(m)

    Plot between Temperature & distanceQ=30

    1.6 1.71.8

    0.4

    5.4 5.5

    0

    1

    2

    3

    4

    5

    6

    0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

    Tem

    per

    atu

    re (

    K)

    distance(m)

    Plot between Temperature & distanceQ=60

  • Discussion: 1-

    Here are the factors that affect the rate of conduction

    A) Temperature difference

    B) Cross-sectional area

    C) Length (distance heat must travel)

    D) Time

    2-

    The thermal conductivity of an object is dependent on its composition and

    dimensions(cross-sectional area and length).

    for two connected objects of the same dimension connected to hot and

    cold reservoirs, the higher the temperature drop, the lower the thermal

    conductivity

    3-

    In contact point will make error because when we join the peace of

    material will make a space that cause to heat losses.

  • 4-

    To measure the temperature distribution for steady-state conduction of

    energy through a composite plane wall and determine the Overall Heat

    Transfer Coefficient for the flow of heat through a combination of

    different

    materials in series

    5-

    By increasing area and T thermal conductivity decrease:

    But by increasing X, thermal conductivity increase

    6-

    Why we neglect the first rate of temp. ?

    Because of the distance is zero and read temp.as minus.

    7-

    What is the advantage of cooling water?

    To make the difference between the temp.

  • Conclusion:

    In this experiment we proved that K is inversely proportional with T, and we have many errors in our experiment that made the result not clear.

  • References:

    1-

    www.me.uprm.edu/o_meza/.../Heat%20Transfer%20

    Experiment-1.doc

    2-

    http://www.d.umn.edu/~dlong/exhtcond.pdf

    3-

    www.gunt.de/static/s3684_1.php?p1=0&p2=&pN=se

    arch;Volltext;linear%20heat%20conduction