Heat Conduction Formula

When two objects at different temperatures are brought into contact, heat flows from the object at a higher temperature to that at a lower temperature. The net flow is in the direction where the temperature is lower. Heat flow can occur in the following three possible ways: Conduction, Convection, and radiation.

Conduction

It is the transfer of heat from one part of a body to another, or from one body to another that is in physical contact with it, with no noticeable displacement of body particles. The flow of heat is constrained by conduction. Following are some examples of conduction

A solid-state Heat flows through a furnace’s brick wall, and metal conducts heat.
The metal wall of a heat exchanger tube and the sheet of a boiler
Temperature flows through the brick that is refractory of a furnace, the boiler’s metal sheet, and the steel wall surface of a heat exchanger tube.
It is thought to move by conduction if heat moves through human anatomy by the transference associated with the energy of specific atoms or particles without blending.

Fourier’s law is the physical governing law of thermal conduction. Fourier’s law states that the rate of temperature movement by conduction through a uniform and rial that is fixed directly proportional to the area of temperature transfer (the area normal towards the way of heat flow), the heat gradient in direction of temperature movement, and inversely proportional towards the amount of the path of heat flow. What the law states is relevant at any location and at any time.

The mathematical representation of Fourier’s law,

$\frac{dq}{dA}=-k\frac{dT}{dq}$

Where,

q = rate of heat flow in the way normal to the surface

A = area

x = distance measured normal to the surface

k = Coefficient of thermal conductivity of the material.

T = temperature

S.I unit of Heat Conduction is Watts per meter kelvin (W.m^-1K^-1)

Dimensional formula = M¹L¹T^-3Θ^-1

The general expressions of Fourier’s law for flow in all three directions in a material that is isotropic are given by,

$\frac{dq}{dA}=-k\frac{\partial T}{\partial q}+\frac{\partial T}{\partial y}+\frac{\partial T}{\partial y}=-k\bigtriangledown T$ ⇢ (1)

One Dimensional Steady State Heat Conduction

In the sense that the temperature does not change with time, steady-state heat conduction is a simpler example. T is a function of position inside the performing solid and is independent of time. The temperature gradient occurs only in one direction in one-dimensional steady-state heat transfer, making the movement unidirectional.

Plane Wall of Uniform Thickness

Consider a wall of A surface area of x thickness as shown. Let Q be the thermal transfer rate in X-direction and ‘k’ be the thermal conductivity of the material Form Fourier’s law of heat conduction equation.

Rate of heat transfer, $Q = -k.A.\frac{dt}{dq}$

Taking boundary conditions and integrating between them.

At x = 0, T = T₁

At x = x, T = T₂

$Q\int_{x}^{0}dx =-k.A\int_{T_1}^{T_2}dT$

Q.x = -k.A.(T₁ – T₂)

$Q=\frac{kA(T_1-T_2)}{x}$

q = $\frac{Q}{A}$

Plane Wall of Uniform Thickness

Thermal Conductivity In Different Shape And Sizes

Thermal conductivity in cylindrical coordinates. The Fourier’s law equation (Which is equation (1)) becomes,

$\frac{dq}{dx}= -k\left ( \frac{\partial T}{\partial r}+\frac{1}{r} \frac{\partial T}{\Theta }+\frac{\partial T}{\partial z}\right )=-k\bigtriangledown T$

Thermal conductivity in spherical coordinates. The Fourier’s law equation (Which is equation (1)) becomes,

$\frac{dq}{dA}=-k\left ( \frac{dq}{dA}+\frac{1}{r}\frac{\partial T}{\partial \Theta } +\frac{1}{rsin\Theta }\frac{\partial T}{\partial z}\right )=-k\bigtriangledown T$

Sample Problems

Problem 1: Determine the heat transfer rate per unit area via a copper dish 0.045 m thick whose one face is maintained at 370°C while the other face is at 40°C. The thermal conductivity of copper is 340 W/m°C.

Solution:

Given: Plate thickness, x = 0.045 m, Temperature 1: 340̇°C, Temperature 2: 40̇°C,

Thermal coefficient (k) = 370 W/m°C.

Conductivity transfer per unit area, q/A

$q=-kA\frac{(T_1-T_2)}{x}$

= $370\frac{(340-40)}{0.045}$

= 2466.667 × 10000 W/m²

= 2466.667 kW/m²

Problem 2: The plane slab of thickness δ = 60 cm is made up of material of thermal conductivity k = 16.5 W/m-deg. The side that is left of the slab absorbs an internet degree of radiant power through the radiant source, the price q = 540 watt/m². If the hand that is right of the slab is at a constant heat t₂ = 38°C, set up a manifestation for temp circulation inside the slab being a purpose of appropriate space coordinates. Therefore exercise the temp at the mid-plane for the slab together with optimum temp within the slab. It could be presumed that the temperature distribution is constant and there’s no heat generation.

Solution:

The vitality absorbed from the radiant source equals the rate at which it is carried out through the slab under stipulations of steady-state with no heat generation.

Heat flux, $q=k\frac{(T_1-T_2)}{\delta }$

540=16.5\frac{t1-38}{0.6} $540=16.5\frac{t1-38}{0.6}$

Therefore temperature at the left side of slab,

$t_1= 38+\tfrac{540\times0.6}{16.5}$

= 57.63°C

This also represents the utmost temperature inside the slab. From the expression for steady state temperature distribution.

$t=t_1+\frac{t_1-t_2}{\delta}a$

At mid plane a= 30 cm

$57.63+\tfrac{(38-57.63)}{0.6}0.3$

t = 47.815°C

Problem 3: A plane wall has a thickness of 15 cm and a surface area of 4.5 m². The wall has a thermal conductivity of 9.5 W/mK. The temperature of the wall’s inner and exterior surfaces is kept at 125°C and 35°C, respectively. Determine,

The rate of heat flow across the wall.
Temperature gradient in the direction of heat movement and
Surface temperatures at 5 cm and 10 cm from the inner surface

The plane slab

Solution:

Given: wall thickness, x = 15 cm = 0.15 m; Area (A) = 4.5 m; Thermal conductivity (k) = 9.5 W/mK;

Temperature 1 (T₁) =125 Temperature (T₂) =35

1. The rate of heat flow across the wall.

$Q=\frac{kA(T_1-T_2)}{x}$

= $\frac{9.5\times 4.5(125-35)}{0.15}$

Q = 25650 W

2. Temperature gradient in the direction of heat movement, dT/dx

Fouriers law heat conduction,

$Q=-kA\frac{dT}{dx}$

= $\frac{dT}{dx}=\frac{Q}{-kA}$

= $\frac{25650}{(9.5\times 4.5)}$

= -600 °C/m

3. Surface temperatures at 5 cm and 10 cm from the inner surface.

Let T₁ and T₂ be the temperatures at corresponding surfaces at distances of x₁and x₂, respectively. Under steady-state conditions, the heat transfer rate Q is constant throughout, implying that,

$Q=\frac{kA(T_1-T_{x_1})}{x_1}$

$25650=\frac{9.5\times 4.5(125-Tx1)}{0.05}$

T_x1 = 125°C

$Q=\frac{kA(T_1-T_{x_2})}{x_2}$

$25650=\frac{9.5\times 4.5(125-Tx2)}{0.1}$

T_x2 = 90°C

Problem 4: The quantity $\frac{dt}{Q}$ for thermal conduction through a body that is spherical is given by?

Solution:

We get this equation by integrating the heat equation,

$\frac{dq}{dx}=-k\frac{dT}{dx}$

(Q) = -k A dt/dx from limits r₁ to r₂ and T₁ to T₂.

Thermal conductivity in spherical coordinates. The Fourier’s law equation (Which is equation 1) becomes,

$\frac{dq}{dA}=-k\left ( \frac{dq}{dA}+\frac{1}{r}\frac{\partial T}{\partial \Theta } +\frac{1}{rsin\Theta }\frac{\partial T}{\partial z}\right )=-k\bigtriangledown T$

Problem 5: The temperature of the 1.6 m² hot plate is kept at 280°C. The plate is blown over by 15°C air. Calculate the rate of convective heat transmission if the convective heat transfer coefficient is 18W/m²K.

Solution:

Plate temperature, T_w= 280°C; A = 1.6 m;

Film temperature (air), T = 15°C; h = 18W/m²K.

Rate of convective heat transfer,

Q = h-A (T_w– T)

=18 × 1.6 (280 – 15)

Q = 7632 W

Problem 6: A 36-centimeter-thick oven wall has a thermal conductivity of 0.6 W/mK. The temperature within the oven is kept at 700°C, while the temperature outside the oven is kept at 180°C. The oven’s wall has a total surface area of 2 m². Calculate the thermal resistance, heat flux, and heat flow rate.

Solution:

Thickness, x = 36 cm = 0.36 m;

Thermal conductive, k = 0.6 W/mK;

T₁ = 700°C; T₂ = 180°C; A= 2 m².

Thermal resistance, R;

$R =\frac{x}{kA}$

$=\frac{0.36}{0.6\times 2}$

= 0.3 K/W

Heat flow rate, Q;

$Q=\frac{(T_1-T_2)}{\frac{x}{kA}}$

$=\frac{T_1-T_2}{R}$

$=\frac{700-180}{0.3}$

Q= 1733.33 W

Heat flux, q;

$q=\frac{Q}{A}$

$q=\frac{1733.33}{2}$

q = 866.665 W/m²

Reffered: https://www.geeksforgeeks.org

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