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Mechanical Properties Of Fluids

Question
CBSEENPH11019800

Discuss Searle’s method to find the value of thermal conductivity.

Solution
Searle's method:
i) Searle's apparatus consists of a rod XY whose thermal conductivity is to be determined.
ii) One end X is enclosed in a steam chamber and steam is continuously passed through it. The second end Y is embedded in a cooling pipe C. A constant flow of water runs through pipe C.
iii) Two thermometers T1 and T2 are fitted in the holes drilled in the rod at P and Q separated by distance d.
iv) Two other thermometers T3 and T4 are used to measure the temperature of incoming and outgoing water through pipe C.
v) The rod is thoroughly insulated to stop any heat losses by radiation.
vi) Heat is conducted through the rod and when steady state is reached, the thermometers T1 T2, T3 and T4 record constant temperature.
vi) The mass of water flowing through a pipe C is collected in a given interval of time and weighed.
 

Now, it has been observed that at a steady state, the quantity of heat flowing through every section is same.
i.e., the rate of flow of heat at point P, Q and Y is same.

Quantity of heat that flows from section P to Q in one second is given by, 


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where,
 θ1 and θ2 be the reading of thermometers T1 and T2 respectively.
Amount of heat gained by water in one second at end Y is,
              dH over dt equals ms fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight l end fraction space space space space space space space space space space space... left parenthesis 2 right parenthesis
where,
m is the mass of water collected in time t,
s be the specific heat and
straight theta3 and θ4 be the reading of thermometers T3 and T4. 
Now, equating (1) and (2), we get
space space fraction numerator K A over denominator straight d end fraction left parenthesis straight theta subscript 1 minus straight theta subscript 2 right parenthesis equals m s fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight l end fraction
space
rightwards double arrow space space space space space space space space space space space space space space space space space space straight K equals fraction numerator m s d over denominator A t end fraction open parentheses fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight theta subscript 1 minus straight theta subscript 2 end fraction close parentheses