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

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Advanced Knowledge of Conduction

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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: 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: 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: 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: 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: 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: 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: A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio \$1 : 4\$. If the free face of brass is at \$100^{\circ}C\$ and that of copper at \$0^{\circ}C\$, the temperature of interface is

As we know that conductive heat flow is given as \${Q}=\dfrac{KA{\Delta}T}{L}\$and it is given that relation between thermal conductivity are as follow,\${K}_{brass}={4K}_{copper}\$so,let an intermediate temperature \${T}\$ will line at inside the slab.therefore,we equate the rate of heat flow from copper surface and brass surface,as follow\${K}_{brass}({100-T})={K}_{copper}({T-0})\$\${4}({100-T})={T-0}\$\${T}={80}^{0C}\$

Conduction

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Conduction using Kinetic Model

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Thermal Conductivity

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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

Medium

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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

Medium

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The dimensional formula for coefficient of thermal conductivity is:

Medium

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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 :

Medium

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

Hard

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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?

Medium

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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 :

Medium

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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

Medium

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Define coefficient of thermal conductivity.

Medium

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A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio $1 : 4$ . If the free face of brass is at $10 0^{\circ} C$ and that of copper at $0^{\circ} C$ , the temperature of interface is

Medium

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Revise with Concepts

Advanced Knowledge of Conduction

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Conduction using Kinetic Model

Thermal Conductivity

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 dimensional formula for coefficient of thermal conductivity 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 :

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 :

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

Define coefficient of thermal conductivity.

A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1:4. If the free face of brass is at 100∘C and that of copper at 0∘C, the temperature of interface is

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 :

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 :

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

A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio $1 : 4$ . If the free face of brass is at $10 0^{\circ} C$ and that of copper at $0^{\circ} C$ , the temperature of interface is