Heat Transfer
1.0 Conduction
1.0 Conduction
Heat can be transferred from one place to another place in three ways:
- Conduction
- Convection
- Radiation
Conduction
It is the process by which the heat is transferred from one place to another place by the vibration of the molecules about a fixed point and heat is transferred by collision.
Let us consider a slab with two ends at different temperatures. Gradually, the temperature of the hot end decreases and the that of other end increases.
Here the heat is transferred from one end to the other by the vibration of the molecules about a fixed point and the mean position of the molecule does not change.
So, the heat is transferred without transferring the material molecules.
Consider a slab of uniform cross-section $A$ and length $x$.
Let one face is maintained at ${T_1}$ and the other face is maintained at ${T_2}$. The heat flows from the hotter end to the other end at lower temperature.
After some time a steady state is attained, during this the temperature at any point in the slab remains unchanged and heat flowing per unit time through any cross section remains same.
If $\Delta Q$ amount of heat passes through a cross-section in $\Delta t$ time, then $\frac{{\Delta Q}}{{\Delta t}}$ remains constant at any cross-section in the steady state.
$\frac{{\Delta Q}}{{\Delta t}}$ is called heat current.
In steady state it is proportional to,
- area $A$,
- temperature difference $\Delta T$( ${T_2} - {T_1}$ ),
- $\frac{1}{x}$.
We can write the equation as $$\frac{{\Delta Q}}{{\Delta t}} = K\frac{{A\left( {{T_2} - {T_1}} \right)}}{x}$$ where $K$ is called thermal conductivity of the material.
The equation can also be written as, $$\frac{{\Delta Q}}{{\Delta t}} = - KA\frac{{dT}}{{dx}}$$where the negative sign shows that the temperature gradient decreases along the direction of heat flow.
$\frac{{dT}}{{dx}}$ is called the temperture gradient.
The quantity $\frac{x}{{KA}}$ is called thermal resistance ( $R$ ). So, we can write $$\begin{equation} \begin{aligned} \frac{{\Delta Q}}{{\Delta t}} = \frac{{{T_2} - T}}{R} \\ i = \frac{{{T_2} - {T_1}}}{R} \\\end{aligned} \end{equation} $$
Question 1. One face of a cube of length $5\ cm$ is maintained at ${100^o}C$ and the other end at ${0^o}C$. All the surfaces are insulated. Find the amount of heat flowing per second through the slab. ${{K_{slab}} = 385\,W\,{m^{ - 1}}^o{C^{ - 1}}}$.
Solution: Heat transfer takes from hot end to the cool end. The area of cross section perpendicular to the flow of heat is $$A = {\left( {5\,cm} \right)^2}$$ The amount of heat flowing per second is $$\begin{equation} \begin{aligned} \frac{{\Delta Q}}{{\Delta t}} = KA\frac{{{T_1} - {T_{_2}}}}{x} \\ = \left( {385\,W\,{m^{ - 1}}^o{C^{ - 1}}} \right) \times {\left( {5\,cm} \right)^2} \times \frac{{\left( {{{100}^o}C - {0^o}C} \right)}}{{0.1\,m}} \\ = 962.5\,W \\\end{aligned} \end{equation} $$
Question 2. One end of a copper rod of length $1\ m$ and area of cross section $4.0 \times {10^{ - 4}}\,{m^2}$ is maintained at ${100^o}C$ and the other end at ${0^o}C$. Find the mass of ice melted in $1$ hour, neglecting the loss of heat from the surroundings. ( ${K_{cu}} = 401\,W/m - K,{L_f} = 3.35 \times {10^5}\,J/kg$ ).
Solution: Thermal resistance of rod $$\begin{equation} \begin{aligned} R = \frac{l}{{KA}} = \frac{{1.0}}{{\left( {401} \right)\left( {4 \times {{10}^{ - 4}}} \right)}} = 6.23\,K/W \\ Heat\;current = \frac{Temperature\ difference}{Thermal \ resistance} \\ = \frac{{\left( {100 - 0} \right)}}{{6.23}} = 16\,Watt \\\end{aligned} \end{equation} $$Heat transferred in 1 hour, $$\begin{equation} \begin{aligned} Q = it \\ = \left( {16} \right)\left( {3600} \right) = 57600\,J \\\end{aligned} \end{equation} $$Now, let $m$ be the mass of ice melted in 1 hour, then$$\begin{equation} \begin{aligned} m = \frac{Q}{L} \\ = \frac{{57600}}{{3.35 \times {{10}^5}}} = 0.172\,kg \\\end{aligned} \end{equation} $$
Question 3. A copper rod $2\ m$ long has a circular cross section of radius $1\ cm$. One end is kept at ${100^o}C$ and the other end at ${0^o}C$, and the surface is insulated such that no heat flows out through the surface. Find
(a) the thermal resistance of the bar
(b) thermal current
(c) temperature gradient
(d) temperature 25$cm$ from the hot end
( ${K_{copper}} = 401\,W/m - K$ )
Solution: (a) Thermal resistance $$\begin{equation} \begin{aligned} R = \frac{l}{{KA}} = \frac{l}{{K\left( {\pi {r^2}} \right)}} \\ R = \frac{{\left( 2 \right)}}{{\left( {401} \right)\left( \pi \right){{\left( {{{10}^{ - 2}}} \right)}^2}}} = 15.9\,K/W \\\end{aligned} \end{equation} $$
(b) Thermal current$$\begin{equation} \begin{aligned} i = \frac{{\Delta T}}{R} = \frac{{100}}{{15.9}} \\ = 6.3\,Watts \\\end{aligned} \end{equation} $$
(c) Temperature gradient$$ = \frac{{0 - 100}}{2} = - 50K/m = - {50^o}C/m$$
(d) Let $T$ be the temperature at 25$cm$ from the hot end,$$\begin{equation} \begin{aligned} \left( {T - 100} \right) = \left( {temperature\,gradient} \right) \times \left( {distance} \right) \\ T - 100 = \left( { - 50} \right)\left( {0.25} \right) \\ T = {87.5^O}C \\\end{aligned} \end{equation} $$
