The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity \$\mathrm{K}\$ and \$2\mathrm{K}\$ and thickness \$\mathrm{x}\$ and \$4\mathrm{x}\$, respectively are \$\mathrm{T}_{2}\$ and \$\mathrm{T}_{1}(\mathrm{T}_{2}>\mathrm{T}_{1})\$. The rate of heat transfer through the slab, in a steady state is \$\left ( \dfrac{A(T_{2}-T_{1})K}{X} \right )f,\$ with \$f\$ equals to

The thermal resistance of both the blocks will be\$\dfrac { x }{ KA } \$ and \$\dfrac { 4x }{ 2KA } =\dfrac { 2x }{ KA } \$ respectively. Since the two resistors are in series, (as same heat current will flow through both), equivalent thermal resistance will be\${ R }_{ eq. }=\dfrac { x }{ KA } +\dfrac { 2x }{ KA } =\dfrac { 3x }{ KA } \$Thus, heat current will be \$\dfrac { dQ }{ dt } =\dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { R }_{ eq. } } =\dfrac { 1 }{ 3 } \dfrac { KA({ T }_{ 2 }-{ T }_{ 1 }) }{ x } \$Hence, \$f=\dfrac { 1 }{ 3 } \$

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Conduction

Q: A wall has two layers \$A\$ and \$B\$, each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material of \$A\$ is twice that of \$B\$. Under thermal equilibrium, the temperature difference across the wall is \$36^{\circ}C\$. The temperature difference across the layer \$A\$ is

Consider two walls '\$A\$' and '\$B\$' of thickness '\$t\$' each, and thermal conductivity of '\$2K\$' and '\$K\$' respectively.The temperatures at the left of \$A\$ = \$T_{a}\$The temperatures at the right of \$B\$ = \$T_{b}\$The temperature at the junction = \$T\$Heat flow is a constant in steady state:\$\dot{Q} = KA\dfrac{dT}{dx} = constant\$Equating heat flow for both walls we get:\$2KA\dfrac{T-T_{a}}{t-0} = KA\dfrac{T_{b}-T}{2t -t}\$Which simplifies to: \$3T = T_{b} + 2T_{a}\$Also given that the temperature difference between the walls is \$36^{\circ}C\$\$T_{b} - T_{a} = 36\$Combining the two equation in \$T,T_{a},T_{b}\$ and eliminating \$T_{b}\$We get \$T - T_{a} = 12^{\circ}C\$

Q: The dimensional formula for coefficient of thermal conductivity is:

The for the coefficient of thermal conductivity is\$K_{th}=\dfrac{Q\Delta x}{A\Delta T t}\$Dimensional formula of \$K_{th}\$\$[K_{th}]=\dfrac{[Heat].[length]}{[Area].[Temperature].[time]}\$\$[K_{th}]=\dfrac{[ML^2T^{-2}].[L]}{[L^2].[K].[T]}\$\$[K_{th}]=[MLT^{-3}K^{-1}]\$The correct option is D.

Q: Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities \$K\$ and \$2K\$ respectively. The equivalent thermal conductivity of the slab is

The thermal resistances can be added as they are in series. But, note that the width of the combined slab is twice the width of individual slab. So, we have:\$\Rightarrow \dfrac { 2L }{ { K }_{ eq }A } =\dfrac { L }{ KA } +\dfrac { L }{ 2KA }\$\$\Rightarrow { K }_{ eq }=\dfrac { 4K }{ 3 } \$

Q: Two rods \$A\$ and \$B\$ of different materials are welded together as shown in figure. Their thermal conductivities are \$K_1\$ and \$K_2\$. The thermal conductivity of the composite rod will be :

\$R_1=\dfrac{d}{k_1A}\$\$R_2={d}{K_2A}\$\$Req=\dfrac{R_1R_2}{R_1+R_2}\$\$\dfrac{d}{keq(2A)} = \dfrac{(\dfrac{d}{k_1A})(\dfrac{d}{k_2A})}{\dfrac{d}{A}(\dfrac{1}{k_1}+\frac{1}{k_2})}\$\$\dfrac{1}{2keq}=\dfrac{\dfrac{1}{k_1k_2}}{\dfrac{k_1+k_2}{k_1k_2}}=\dfrac{1}{k_1+k_2}\$\$keq = \dfrac{k_1+k_2}{2}\$

Q: Two rods of same length and material transfer a given amount of heat in \$12 \$ seconds , when they are joined in parallel. But when they are joined in series , then they will transfer same heat in same condition in

Both rods of same length, material and axis sectional area so their thermal conductivities \$(k)\$ would be be samealso, thermal resistance\$P_m =\dfrac {L}{KA}\ \Rightarrow \dfrac {R_1}{R_2}=\dfrac {K_2}{K_1}\$ [\$K\to\$ conductivity in series / parallel]also thermal transfer\$\dfrac {\Delta Q}{\Delta t}=\dfrac {\Delta T}{R} \$ [\$\Delta Q=\$ Heat transformed \$\Delta T: Temperature difference]That mass for same \$\Delta Q\$ & \$\Delta T\$\$\Delta \ t \alpha R\ t (2)\$ [\$\Delta t=\$ time taken]From \$(1)\$ & \$(2)\$ gives\$\dfrac {\Delta t_1}{\Delta t_2}=\dfrac {R_1}{R_2}=\dfrac {K_2}{K_1}\$ given \$\Delta t_1=125\$In parallel connection \$K_1=2K\$ But in series \$K_2=K/2\$ so:-\$\dfrac {\Delta t_1}{\Delta t_2}=\dfrac {K/2}{2K}=\dfrac {1}{4}\Rightarrow \Delta t_2=4\Delta t_1\$\$=4\times 12=48

Q: The coefficient of thermal conductivity of copper is \$9\$ times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?

heat current along the rod will be same\$\dfrac{dq}{dt}=\dfrac{9kA(100-T)}{18}=\dfrac{kA(T-0)}{6}\$\$T=75^oC\$

Q: A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities \$K_1\$ and \$K_2\$. The equivalent thermal conductivity of the slab is :

At thermal equilibrium\$ H = H_{1} + H_{2} \$Given that two slab are in parallel and have same dimensions. \$ \dfrac{K\times A\times (T_{2} - T_{1})}{L} = \dfrac{K_{1}\times \dfrac{A}{2}\times (T_{2} - T_{1})}{L} + \dfrac{K_{2}\times \dfrac{A}{2}\times (T_{2} - T_{1})}{L} \$,where \$K\$ is the effective thermal conductivity, \$T_{2}-T_{1}\$ is the temperature difference between the ends of the slab, \$A\$ is area of the slab and \$L\$ is the thickness of the each layer of the slabSo, \$ \dfrac{K\times A}{L} = \dfrac{K_{1}\times \dfrac{A}{2}}{L} + \dfrac{K_{2}\times \dfrac{A}{2}}{L} \$or, \$ K = \dfrac{K_{1} +K_{2}}{2}\$

Q: Define coefficient of thermal conductivity.

Coefficient of thermal conductivity of a material is defined as the quantity of heat conducted per second through a unit area of a slab of unit thickness when the temperature difference between its ends is 1K.

Q: Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio \$1 : 2\$ and their lengths are in the ratio \$2 : 1\$. If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be :

Diameters are in ratio \$1:2\$, so areas are in ratio \$1:4\$Lengths are in ratio \$2:1\$Hence, \$A/L\$ ratio is \$1:8\$Since other factors are constant, ratio of heat flow is \$1:8\$

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Important Questions

A wall has two layers $A$ and $B$ , each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material of $A$ is twice that of $B$ . Under thermal equilibrium, the temperature difference across the wall is $3 6^{\circ} C$ . The temperature difference across the layer $A$ is

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The dimensional formula for coefficient of thermal conductivity is:

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Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities $K$ and $2 K$ respectively. The equivalent thermal conductivity of the slab is

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Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be :

Define coefficient of thermal conductivity.

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Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio $1 : 2$ and their lengths are in the ratio $2 : 1$ . If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be :

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The dimensional formula for coefficient of thermal conductivity is:

Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K1​ and K2​. The thermal conductivity of the composite rod will be :

Two rods of same length and material transfer a given amount of heat in 12 seconds , when they are joined in parallel. But when they are joined in series , then they will transfer same heat in same condition in

The coefficient of thermal conductivity of copper is 9 times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities K1​ and K2​. The equivalent thermal conductivity of the slab is :

Define coefficient of thermal conductivity.

Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio 1:2 and their lengths are in the ratio 2:1. If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be :

Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities $K$ and $2 K$ respectively. The equivalent thermal conductivity of the slab is

Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be :

Two rods of same length and material transfer a given amount of heat in $12$ seconds , when they are joined in parallel. But when they are joined in series , then they will transfer same heat in same condition in

The coefficient of thermal conductivity of copper is $9$ times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities $K_{1}$ and $K_{2}$ . The equivalent thermal conductivity of the slab is :

Heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio $1 : 2$ and their lengths are in the ratio $2 : 1$ . If the temperature difference between their ends is same, then the ratio of amounts of heat conducted through them per unit time will be :