# Conductive Heat Transfer

## Heat transfer takes place as conduction in a soilid if there is a temperature gradient

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Conduction as heat transfer takes place if there is a temperature gradient in a solid or stationary fluid medium.

With conduction energy transfers from more energetic to less energetic molecules when neighboring molecules collide. Heat flows in direction of decreasing temperatures since higher temperatures are associated with higher molecular energy.

Conductive heat transfer can be expressed with "**Fourier's Law**":

q = k A dT / s (1)

where

q = heat transfer (W, J/s, Btu/hr)

A = heat transfer area (m^{2}, ft^{2})

k = thermal conductivity of material (W/m K or W/m^{o}C, Btu/(hr^{o}F ft^{2}/ft))

dT = temperature gradient - difference - in the material (K or^{o}C,^{o}F)

s = material thickness (m, ft)

#### Example - Heat Transfer by Conduction

A plane wall is constructed of solid iron with thermal conductivity *70 W/m ^{o}C. *Thickness of the wall is

*50 mm*and surface length and width is

*1 m by 1 m.*The temperature is

*150*on one side of the surface and

^{ o}C*80*on the other.

^{o}CThe conductive heat transfer through the wall can be calculated

q= (70 W/m^{o}C) (1 m) (1 m) ((150^{o}C) - (80^{o}C)) / (0.05 m)

= 98000 (W)

= 98 (kW)

### Heat Transfer through Plane Walls or Layers in Series

Heat conducted through several walls or layers in thermal contact can be expressed as

*q = (T*_{1}* - T*_{n}*) / ((s*_{1 }*/ k*_{1 }*A) + (s*_{2 }*/ k*_{2 }*A) + ... + (s*_{n }*/ k*_{n }*A)) (2)*

*where *

*T*_{1}* = temperature inside surface (K or *^{o}*C, *^{o}*F)*

*T*_{n}* = temperature outside surface (K or *^{o}*C, *^{o}*F)*

#### Example - Heat Transfer through a Furnace Wall

A furnace wall of *1 m ^{2}* consist of

*1.2 cm*thick stainless steel inner layer covered with

*5 cm*outside insulation layer of insulation board. The inside surface temperature of the steel is

*800 K*and the outside surface temperature of the insulation board is

*350 K*. The thermal conductivity of the stainless steel is

*19 W/(m K)*and the thermal conductivity of the insulation board is

*0.7 W/(m K)*.

The conductive heat transport through the layered wall can be calculated as

*q = ((800 K)** - (350 K)**) / (((0.012 m)** / (19 W/(m K)) (1 m ^{2})*

*) + ((0.05 m)*

*/ (0.7 W/(m K)) (1 m*

^{2})*))*

* = 6245 (W)*

### Thermal Conductivity Units

*Btu/(h ft*^{2}^{o}F/ft)*Btu/(h ft*^{2}^{o}F/in)*Btu/(s ft*^{2}^{o}F/ft)*MW/(m*^{2}K/m)*kW/(m*^{2}K/m)*W/(m*^{2}K/m)*W/(m*^{2}K/cm)*W/(cm*^{2}^{o}C/cm)*W/(in*^{2}^{o}F/in)*kJ/(h m*^{2}K/m)*J/(s m*^{2}^{o}C/m)*kcal/(h m*^{2}^{o}C/m)*cal/(s cm*^{2}^{o}C/cm)

*1 W/(m K) = 1 W/(m*^{o}C) = 0.85984 kcal/(h m^{o}C) = 0.5779 Btu/(ft h^{o}F) = 0.048 Btu/(in h^{o}F)

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