Temperature gradient

What is temperature gradient in heat transfer?

For the given direction, the temperature gradient is the rate of change of temperature with respect to the displacement.

It means that the temperature gradient is the ratio of the temperature difference between two points to the distance between these two points.

The term temperature gradient gives the direction as well as the rate of temperature change in a particular direction.

The term temperature gradient is denoted by the symbol of ∇T.


Temperature gradient example:

What is temperature gradient in heat transfer

For the pipe shown in the figure the formula for temperature gradient can be written as,

`\nabla T=\frac{dT}{dx}=\frac{t_{b}-t_{a}}{d}`


Unit of temperature gradient:

1] SI unit:

In the SI system the unit of temperature is Kelvin and the unit of distance is m.

`\therefore \nabla T=\frac{dT}{dx}=\frac{K}{m}`

∴ SI unit of the temperature gradient is K/m

2] FPS unit:

In the FPS system unit of temperature is °F and the unit of distance is ft.

`\therefore \nabla T=\frac{dT}{dx}=\frac{ °F }{ft}`

∴ FPS unit of the temperature gradient is °F/ft.


What is the relationship between thermal conductivity and temperature gradient?

By the Fourier law of heat conduction,

`\therefore q=-K(\frac{dT}{dx})`

Hence for attaining a constant amount of heat flux,

1) For higher thermal conductivity, the temperature gradient is smaller.
2) For lower thermal conductivity the temperature gradient is higher.


Numerical on temperature gradient:

1) For the spoon dipped in hot oil the temperature at one end is 573 K and at another end is 325 K if the length of the spoon is 0.3 then find the temperature gradient present in a spoon.

Given: T1= 573 K
T2= 325 K
x= 0.3 m

By using the formula of temperature gradient,

∇T = dT/dx = (T2 – T1)/x

∇T = (325 – 573)/0.3

∇T = – 826 K/m


2) For the brick wall with heat flux of 200 w/m2, the thermal conductivity is 0.9 w/mK. find the temperature gradient in brick wall.

Given:
q= 200 w/m2
K= 0.9 w/mK

The temperature gradient is given by,

q = K.(dT/dx)

200 = 0.9 × (dT/dx)

(dT/dx) = 222.22 K/m

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