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Lees disc

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here u can determine the ocnductivity of bad conductor by using the lee's disc method

here u can determine the ocnductivity of bad conductor by using the lee's disc method

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  • 1. Determination of thermal conductivity of a bad conductor using Lee’s disc method
    Trisha Banerjee @ 2010
  • 2. Whole setup of the experiment:-
    Mercury thermometers
    Steam chamber
    Disc ( between the material is sandwitched)
    Trisha Banerjee @ 2010
  • 3. The coefficient of thermal conductivity of bad conductor’s ( ebonite , rubber, glass etc) is determined by taking the specimen in the form of a flat disc.
    the experimental disc D is sandwitched between a metallic heavy bottomed cylindrical steam chamber S and a brass disc B . The whole assembly is suspended from a heavy retort stand . The lower part of the chamber S and the disc B are varnished and carry mercury thermometers T1 and T2 respectively inserted in the holes drilled in them.
    Trisha Banerjee @ 2010
  • 4. Trisha Banerjee @ 2010
  • 5. In order to perform the experiment , steam is passed through the chamber S and the temperatures shown by the thermometer T1 and T2 are recorded turn by turn until they become steady. In this steady state , the arte of heat conducted through the experimental disc D is equal to the rate of heat radiated from the exposed surface of the disc B.
    let q1 and q2 be the steady temperature shown by the thermometers T1 and T2 respectively . Since both S and B are very good conductors of heat q1 and q2 may be taken sufficiently as the temperature of the upper and lower disc D. therefore the rate of heat conducted through the disc D is KA(q1-q2)/d
    where ,K is the coefficient of thermal conductivity of the material of the disc D, A is the area of the cross section , and d is the thickness of the disc D.
    now if the m be the mass of the disc B and s is the specific heat , then the rate of heat radiated from B is ms(dq1/dt)at q=q2 .
    where (dq/dt)at q=q2 is the rate of cooling of the disc B at the temperature q2.
    In the steady state , the rate of heat conducted through D equals trhe rate of heat radiated from B.therefore lastly we have
    K= (msd)/A(q1-q2) (dq/dt) at q=q2.
    Trisha Banerjee @ 2010
  • 6. Procedure:-
    a) measure the thickness d of the specimen . Determine the mass m and the radius of the lower disc . sandwitch the specimen between the steam chamber and the lower disc . Insert the thermometer T1 and the T2 into the holes provided for them.
    b) Suspend the assembly from threads and allow the steam to pass through the steam chamber . Wait till steady state , the thermometers do not register any change in its temeperatureq1 and q2.
    c) take out the lower disc and suspend it above .Heat it till its temperature rises abou10 degrees above q2 and then, put the specimen over its top surface. Allow to cool the disc till its temperature fall 10 degree below q2 and during the cooling process , note down the temperature every minute. Draw cooling curve by plotting time on X- axis and temperature on q on Y- axis . Draw a tangent at a point corresponding to the temperature q2.
    Trisha Banerjee @ 2010
  • 7. Note:-
    result will came in Joule per second per meter per degree Celsius.
    the formula is valid for thin specimen because in the derivation the heat loss from its curved surface has been neglected.
    Trisha Banerjee @ 2010