Physics > Heat Transfer > 1.0 Conduction
Heat Transfer
1.0 Conduction
2.0 Convection
3.0 Radiation
4.0 Kirchhoff's Law
5.0 Wein's displacement law
6.0 Stefan-Boltzman law
7.0 Newton's laws of cooling
1.1 Series connection of slabs
If two slabs are connected in series as shown in figure.
Let the thermal resistance of the slabs be ${{R_1}}$ and ${{R_2}}$.
They are maintained at temperature ${{T_1}}$ and ${{T_2}}$ where ( ${{T_1}}$>${{T_2}}$ ).
After some time they attain steady state, so during this process the heat crossing per unit cross-sectional area at any cross-section in both the slabs are equal. ( ${i_1} = {i_2} = i$ ).$$i = \frac{{\Delta Q}}{{\Delta t}} = \frac{{{T_1} - T}}{{{R_1}}} \Rightarrow i{R_1} = {T_1} - T$$and$$i = \frac{{\Delta Q}}{{\Delta t}} = \frac{{{T} - {T-2}}}{{{R_2}}} \Rightarrow i{R_2} = {T} -{T_2}$$ By adding the two equations$$\begin{equation} \begin{aligned} {T_1} - {T_2} = i\left( {{R_1} + {R_2}} \right) \\ \Rightarrow i = \frac{{{T_1} - {T_2}}}{{\left( {{R_1} + {R_2}} \right)}} \\\end{aligned} \end{equation} $$By the above equation, we can write$${R_{equivalent}} = \left( {{R_1} + {R_2}} \right)$$We can extend this $${R_{equivalent}} = {R_1} + {R_2} + {R_3} + .......$$
Question 4. An electric is used in a room of total wall area $137\,{m^2}$ to maintain a temperature of ${20^o}C$ inside it, when the outside temperature is ${-10^o}C$. The walls have three different types of materials. The innermost layer is of wood of thickness 2.5 $cm$, the middle is of cement of thickness $1.0\ cm$ and the outermost layer is of thickness $25.0\ cm$. Find the power of electric heater( ${K_{W}} = 0.125\,W{m^{ - 1}}^o{C^{ - 1}},{K_{C}} = 1.5\,W{m^{ - 1}}^o{C^{ - 1}},{K_{B}} = 1.0\,W{m^{ - 1}}^o{C^{ - 1}}$ ).
Solution: $$\begin{equation} \begin{aligned} {R_W} = \frac{1}{{{K_W}}}\frac{x}{A} \\ = \frac{1}{{0.125\,W{m^{ - 1}}^o{C^{ - 1}}}}\frac{{2.5 \times {{10}^{ - 2}}\,m}}{{137\,{m^2}}} \\ = \frac{{0.20}}{{137}}{}^oC\,{W^{ - 1}} \\ {R_C} = \frac{1}{{{K_C}}}\frac{x}{A} \\ = \frac{1}{{1.5\,W{m^{ - 1}}^o{C^{ - 1}}}}\frac{{1.0 \times {{10}^{ - 2}}\,m}}{{137\,{m^2}}} \\ = \frac{{0.0067}}{{137}}{}^oC\,{W^{ - 1}} \\ {R_B} = \frac{1}{{{K_B}}}\frac{x}{A} \\ = \frac{1}{{1.0\,W{m^{ - 1}}^o{C^{ - 1}}}}\frac{{25.0 \times {{10}^{ - 2}}\,m}}{{137\,{m^2}}} \\ = \frac{{0.25}}{{137}}{}^oC\,{W^{ - 1}} \\\end{aligned} \end{equation} $$As the layers are in series connection,$$\begin{equation} \begin{aligned} R = {R_W} + {R_C} + {R_B} \\ = \frac{{0.20 + 0.0067 + 0.25}}{{137}}{}^oC\,{W^{ - 1}} \\ = 3.33 \times {10^{ - 3}}\,{}^oC\,{W^{ - 1}} \\ i = \frac{{{T_1} - {T_2}}}{R} \\ = \frac{{20{}^oC - \left( { - 10{}^oC} \right)}}{{3.33 \times {{10}^{ - 3}}\,{}^oC\,{W^{ - 1}}}} = 9000\,W \\\end{aligned} \end{equation} $$The heater must supply $9000\ W$ to compensate the outflow of heat.
Question 5. Two plates each of area $A$, thickness ${L_1}$ and ${L_2}$, thermal conductivity ${K_1}$ and ${K_2}$ are joined to form a single plate of thickness ( ${L_1}$+${L_2}$ ). If the temperature of free surfaces are ${T_1}$ and ${T_2}$. Find
(a) Rate of flow of heat
(b) Temperature of the interface
(c) Equivalent thermal conductivity
Solution: (a) If the thermal resistance of two plates are ${R_1}$ and ${R_2}$, then as plates are in series
$$\begin{equation} \begin{aligned} {R_S} = {R_1} + {R_2} = \frac{{{L_1}}}{{A{K_1}}} + \frac{{{L_2}}}{{A{K_2}}}\quad as\,R = \frac{L}{{KA}} \\ \frac{{dQ}}{{dt}} = \frac{{\Delta T}}{R} = \frac{{\left( {{T_1} - {T_2}} \right)}}{{\left( {{R_1} + {R_2}} \right)}} = \frac{{A\left( {{T_1} - {T_2}} \right)}}{{\left[ {\frac{{{L_1}}}{{{K_1}}} + \frac{{{L_2}}}{{{K_2}}}} \right]}} \\\end{aligned} \end{equation} $$
(b) Let $T$ be the interface temperature, as they are in series rate of flow of heat remains same.
$$\begin{equation} \begin{aligned} \frac{{\left( {{T_1} - {T_2}} \right)}}{{\left( {{R_1} + {R_2}} \right)}} = \frac{{\left( {{T_1} - T} \right)}}{{\left( {{R_1}} \right)}} \\ T = \frac{{{T_1}{R_2} + {T_2}{R_1}}}{{\left( {{R_1} + {R_2}} \right)}} \\ T = \frac{{\left[ {{T_1}\frac{{{L_1}}}{{{K_1}}} + {T_2}\frac{{{L_2}}}{{{K_2}}}} \right]}}{{\left[ {\frac{{{L_1}}}{{{K_1}}} + \frac{{{L_2}}}{{{K_2}}}} \right]}}\quad as\,R = \frac{L}{{KA}} \\\end{aligned} \end{equation} $$
(c) If $K$ is the equivalent constant of composite slab,
$$\begin{equation} \begin{aligned} {R_S} = {R_1} + {R_2}\quad or\quad \frac{{{L_1} + {L_2}}}{{A{K_{eq}}}} = {R_1} + {R_2} \\ {K_{eq}} = \frac{{{L_1} + {L_2}}}{{A\left( {{R_1} + {R_2}} \right)}} = \frac{{{L_1} + {L_2}}}{{\left[ {\frac{{{L_1}}}{{{K_1}}} + \frac{{{L_2}}}{{{K_2}}}} \right]}}\quad as\,R = \frac{L}{{KA}} \\\end{aligned} \end{equation} $$
