Question

A composite slab is prepared by pasting three slabs of thickness ${L}_{1}$, ${L}_{2}$, ${L}_{3}$ and thermal conductivity ${K}_{1}$, ${K}_{2}$, ${K}_{3}$. The slab has equal cross-sectional areas. Find the equivalent thermal conductivity.
A.$\dfrac { { K }_{ 1 }+{ K }_{ 2 }+{ K }_{ 3 } }{ { L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 } }$
B.$\dfrac { { K }_{ 1 }{ K }_{ 2 }{ K }_{ 3 }({ L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 }) }{ { K }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ L }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ K }_{ 1 }{ L }_{ 2 }{ K }_{ 3 } }$
C.$\dfrac { { K }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ L }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ K }_{ 1 }{ L }_{ 2 }{ K }_{ 3 } }{ { K }_{ 1 }+{ K }_{ 2 }+{ K }_{ 3 } }$
D.$\dfrac { { L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 } }{ { K }_{ 1 }+{ K }_{ 2 }+{ K }_{ 3 } }$

Answer 1

Hint: Thermal Conductivity denoted by K is the property of the material indicating its ability to conduct heat. Use the formula for Fourier’s law of heat conduction showing the relation between slabs thickness, thermal conductivity and cross-sectional areas. Substitute the values in that formula and calculate equivalent thermal conductivity.

Complete answer:
Effective Resistance is given by,
$R={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ …(1)
From Fourier’s law of heat conduction, we know,
${ R }_{ N }=\dfrac { { L }_{ N } }{ { K }_{ N }A }$
Where, K is the materials conductivity
              L is the plane thickness
             A is the plane area
Substituting the values in the equation. (1) we get,
$\dfrac { { L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 } }{ { K }_{ eq }A } =\dfrac { { L }_{ 1 } }{ { K }_{ 1 }A } +\dfrac { { L }_{ 2 } }{ { K }_{ 2 }A } +\dfrac { { L }_{ 3 } }{ { K }_{ 3 }A }$
Take A common from both the sides and cancel them.
$\Rightarrow \dfrac { { L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 } }{ { K }_{ eq } } =\dfrac { { L }_{ 1 } }{ { K }_{ 1 } } +\dfrac { { L }_{ 2 } }{ { K }_{ 2 } } +\dfrac { { L }_{ 3 } }{ { K }_{ 3 } }$
$\Rightarrow \dfrac { { L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 } }{ { K }_{ eq } } =\dfrac { { K }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ L }_{ 1 }{ K }_{ 2 }{ L }_{ 3 }+{ K }_{ 1 }{ L }_{ 2 }{ K }_{ 3 } }{ { K }_{ 1 }{ K }_{ 2 }{ K }_{ 3 } }$
$\Rightarrow { K }_{ eq }= \dfrac { { K }_{ 1 }{ K }_{ 2 }{ K }_{ 3 }({ L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 }) }{ { K }_{ 3 }{ K }_{ 2 }{ L }_{ 1 }+{ K }_{ 1 }{ K }_{ 3 }{ L }_{ 2 }+{ K }_{ 1 }{ K }_{ 2 }{ L }_{ 3 } }$
Thus, the equivalent thermal conductivity is $\dfrac { { K }_{ 1 }{ K }_{ 2 }{ K }_{ 3 }({ L }_{ 1 }+{ L }_{ 2 }+{ L }_{ 3 }) }{ { K }_{ 3 }{ K }_{ 2 }{ L }_{ 1 }+{ K }_{ 1 }{ K }_{ 3 }{ L }_{ 2 }+{ K }_{ 1 }{ K }_{ 2 }{ L }_{ 3 } }$

Hence, the correct answer is option B.

Note:
Thermal conductivity is reciprocal of thermal resistivity. Heat transfer in materials of low thermal conductivity occurs at a lower rate as compared to the materials of high thermal conductivity. Thermal conductivity is an intrinsic property of the materials. It does not depend on the dimensions of the material. It depends on the temperature, density and moisture content of the material.