Fourier’s Law of Heat Conduction

Fourier's Law of Heat Conduction explains the heat transfer process through a material. It asserts that the heat transfer rate is proportional to the negative temperature gradient and the cross-sectional area through which heat is conducted. This law demonstrates that heat moves from regions of higher temperature to those of lower temperature. Understanding this principle is essential for analyzing and calculating heat conduction and is fundamental to thermodynamics and materials science.

1.0Conduction

• Conduction, or thermal Conduction, is a process of heat transfer in which heat is transferred from one particle to another without dislocating the particle from its equilibrium position.
• Heat flows from the hot end to the cold end. Particles of the medium oscillate but do not leave their place. The medium is necessary for Conduction, which is a process possible in all states of matter.
• It is a slow process. In solids, central conduction takes place.
• Transient State : In this state, the temperature of each part of the rod varies with time.
• Steady state : After a long time, when any part absorbs no heat, the temperature of every part is constant and decreases uniformly from the hotter end to the colder end. 

Note: In steady state, each point has a different temperature but remains constant.

2.0Statement of Fourier’s Law of Heat Conduction

Fourier's Law of Heat Conduction states that the heat transfer rate through an object is directly proportional to the negative temperature gradient and the cross-sectional Area through which heat flows and inversely proportional to the material's thickness.

3.0Mechanism of Thermal Conduction

• Solids conduct heat through the process of Conduction. When one end of a metal rod is heated, the molecules at the hot end vibrate with greater amplitude and thus possess higher average kinetic energy. These energized molecules collide with neighbouring molecules with lower kinetic energy, transferring some of their energy. As a result, the kinetic energy of the adjacent molecules increases. This energy transfer propagates from one layer to the next without the molecules moving from their average positions, effectively transferring heat to the colder end

4.0Derivation of  Fourier’s Law of Heat Conduction

In a given diagram, in steady state the heat flow through the rod is constant.The rate of heat flow \left(\frac{d Q}{d t}\right) depends on :-

1. Cross section Area (A) of rod
2. Rate of Change of temperature with distance \left(-\frac{d T}{d x}\right)= Temperature gradient

${\mathrm{i}}_{\mathrm{T}}=\frac{dQ}{dt}\propto \mathrm{A}$ ………..(1)

${\mathrm{i}}_{\mathrm{T}}=\frac{dQ}{dt}\propto \left(-\frac{dT}{dx}\right)$…….(2)

Combining equations (1) and (2) 

${\mathrm{i}}_{\mathrm{T}}=\frac{dQ}{dt}\propto A\left(-\frac{dT}{dx}\right)$

$\Rightarrow {\mathrm{i}}_{\mathrm{T}}=\frac{dQ}{dt}=-\mathrm{KA}\left(\frac{dT}{dx}\right)$

where, K= Coefficient of Thermal Conductivity

For object having uniform cross section and conductivity

Rate of heat flow ,

${\mathrm{i}}_{\mathrm{T}}=\frac{dQ}{dt}=\frac{KA}{L}\left(T_1-T_2\right)$

${\mathrm{i}}_{\mathrm{T}}=\frac{Q}{t}=\frac{KA}{L}\left(T_1-T_2\right)$

• K= Coefficient of Thermal Conductivity
• A=cross section area of rod
• L=Length of Rod
• $T_1-T_2$ = Difference in Temperature

5.0Thermal Conductivity and Thermal Resistance

Thermal Conductivity(K):

• Thermal conductivity measures a substance's ability to conduct heat. Materials with high thermal conductivity are effective at transferring heat, while those with low thermal conductivity are better at insulating and resisting heat flow.
• It depends on the nature of the material.
• Order of thermal conductivity (Ag>Cu>Au>Al).
• K for Ag maximum is (410 W/mK) and for Freon -12 minimum is (0.008 W/mK)
• For an ideal or perfect conductor of heat ,the value of K = ∞
• For an ideal or perfect bad conductor or insulator the value of K = 0
• Unit :SI $\to \frac{W}{mK}$ or J s^–1 m^–1 K^–1   ;  CGS ® Cal/cm – sec–°C                 
• Dimensions :[ M¹ L¹ T^–3 K^–1]

Thermal Resistance to Conduction:

• It is the obstruction offered to the flow of heat through the conductor when we compare the flow of heat to the flow of electricity.
• It is defined as the temperature difference ratio to the heat flow rate (rate of heat transfer).
• For a slab of cross-section A, Lateral thickness L and thermal conductivity K

${\mathrm{R}}_{\mathrm{T}}=\frac{T_1-T_2}{i_T}=\frac{T_1-T_2}{\frac{KA\left(T_1-T_2\right)}{L}}=\frac{L}{KA}$

6.0Electrical Analogy for Thermal Conduction

7.0Equivalent Thermal Conductivity Combinations

1. Series Combination

• In Series Combination

$R_{eq}=R_1+R_2$

$\Rightarrow \frac{\left(L_1+L_2\right)}{K_{eq}\cdot A}=\frac{L_1}{K_{1}A}+\frac{L_2}{K_{2}A}$

• Equivalent Thermal Conductivity is

$K_{eq}=\frac{L_1+L_2}{\frac{L_1+L_2}{K_1+K_2}}=\frac{\sum \left(L_i\right)}{\sum \frac{L_i}{K_i}}$

• If lengths of the rod are the same then $\left(L_1=L_2=L\right)$
• $K_{eq}=\frac{2K_1{K}_2}{K_1+K_2}$

2. Parallel Combination

• In Parallel Combination

$\Rightarrow \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{K_{eq}\cdot \left(A_1+A_2\right)}{L}=\frac{K_1{A}_1}{L}+\frac{K_2{A}_2}{L}$

Equivalent thermal conductivity is

$\Rightarrow K_{eq}=\frac{K_1{A}_1+K_2{A}_2}{A_1+A_2}=\frac{\sum \left(K_i{A}_i\right)}{\sum \left(A_i\right)}$

• If Area of both rods are the same then $\left(A_1=A_2=A\right)$

$K_{eq}=\frac{K_1+K_2}{2}$

8.0Temperature of Junction$(T_0)$

• To determine the temperature at the junction, we equate the heat flow rates through both sections.

$\Rightarrow {\left(\frac{dQ}{dt}\right)}_1={\left(\frac{dQ}{dt}\right)}_2$

$\Rightarrow \frac{K_{1}A}{L_1}\left(T_1-T_0\right)=\frac{K_{2}A}{L_2}\left(T_0-T_2\right)$

$\Rightarrow T_0=\frac{K_1{L}_2{T}_1+K_2{L}_1{T}_2}{K_1{L}_2+K_2{L}_1}$

9.0Growth of Ice on Lake

In winter atmospheric temperature falls below 0°C and water in the lake starts freezing. Let at time ‘t’ thickness of ice on the surface of the lake = x and air temperature = –T° C.

• The temperature of the water in contact with the underside of the ice is 0°C.
• Let area of the Lake = A
• Heat escaping through the ice over a time interval  dt is $\mathrm{dQ}=\mathrm{KA}\frac{[0-(-T)}{x}dt$
• Due to escape of this heat increasing extra thickness of ice =dx
• Mass of this extra thickness of ice is $\mathrm{m}=\rho \mathrm{V}=\rho A\cdot dx$

$\mathrm{dQ}=\mathrm{mL}=(\rho A\cdot dx)|\Rightarrow dt=\frac{\rho L}{KT}xdx$

So, time taken by ice to grow a thickness x is $\mathrm{t}=\frac{\rho L}{KT}{\int }_0^{x}xdx=\frac{1}{2}\frac{\rho L}{KT}{x}^2$

So, time taken by ice to grow from thickness x_1 to thickness x_2 is

$t=t_2-t_1=\frac{1}{2}\frac{\rho L}{KT}\left(x_2^2-x_1^2\right)$

$t\propto \left(x_2^2-x_1^2\right)$

• Time taken to double and triple the thickness ratio 

t₁ : t₂ : t₃ :: 1² : 2² : 3²

t₁ : t₂ : t₃ :: 1 : 4 : 9

•  Ratio of time taken to form thickness

$(0\to x):(x\to 2x):(2x\to 3x)$

$\Delta t_1:\Delta t_2:\Delta t_3::\left(x^2-0^2\right):\left(2x^2-x^2\right):\left(3x^2-2x^2\right)$

$\Delta t_1:\Delta t_2:\Delta t_3::1:3:5$

10.0Sample Questions on Fourier’s Law of Heat Conduction

Q-1.Given two metallic rods made of the same material, with diameters in the ratio 4:1 and lengths in the ratio 1:2, if the temperature difference across each rod is the same, what will be the ratio of the rates of heat flow through the two rods?

Solution:

$\mathrm{H}=\frac{dQ}{dt}=\frac{KA(\Delta T)}{L}$

$\mathrm{H}\propto \frac{A}{l}\propto \frac{r^2}{l}$

$\frac{H_1}{H_2}={\left(\frac{r_1}{r_2}\right)}^2\times \left(\frac{l_2}{l_1}\right)={\left(\frac{4}{1}\right)}^2\left(\frac{2}{1}\right)=\frac{16\times 2}{1}=\frac{32}{1}$

Q-2.Given a rod with a length of 0.5 meters and a temperature gradient of 80°C per meter, if the temperature at the hotter end of the rod is 30°C, what is the temperature at the cooler end?

Solution:

$\frac{\Delta T}{\Delta x}=80$

$\frac{30-T}{0.5}=80$

T=-10 ⁰C

Q-3. Find the Equivalent Conductivity (K_eq)

Solution:

$K_{eq}=\frac{L_{eq}}{\frac{L_1}{K_1}+\frac{L_2}{K_2}+\frac{L_3}{K_3}}==\frac{8L}{\frac{L}{K}+\frac{4L}{2K}+\frac{3L}{K}}=\frac{16K}{2+4+6}=\frac{4}{3}K$ 

Q-4.Hot end of a brass rod 2m long and having 1 cm radius is maintained at 250°C. When a steady state is reached, the rate of heat flow across any cross–section is 0.5 cal s^–1. What is the temperature of the other end K = 0.26 cal s^–1 cm^–1 °C^–1.

Solution:             

$\frac{Q}{t}$=0.5 cal s^-1_,r=1 cm

Area (A) =$\pi r^2=3.142\times 1{\mathrm{cm}}^2=3.142{\mathrm{cm}}^2$

L= 2 m =200 cm ,T₁=250⁰C, T₂=?

$\frac{Q}{t}=\frac{KA}{L}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)$

$\left(T_1-T_2\right)=\frac{Q}{t}\times \frac{L}{KA}=\frac{0.5\times 200}{0.26\times 3.142}=122.4^0$

$T_2=250-122.4=127.6^{\circ }\mathrm{C}$

Q-5 Find the temperature of the junction for the following Figure.

Solution:                  

$i_1=i_2$

${\left(\frac{Q}{t}\right)}_1={\left(\frac{Q}{t}\right)}_2$

$(\therefore \frac{Q}{t}=\frac{KA}{L}\left({\mathrm{T}}_1-{\mathrm{T}}_2\right)$

${\left(\frac{KA\Delta T}{L}\right)}_1={\left(\frac{KA\Delta T}{L}\right)}_2$

$\frac{KA(100-T)}{L}=\frac{3KA(T-20)}{3L}$

100-T=T-20

2T=120

T=60⁰C