# Conductive Heat Transfer

## Heat transfer takes place as conduction if there is a temperature gradient in a solid or fluid

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Conduction will take place if there exist a temperature gradient in a solid (or stationary fluid) medium.

Energy is transferred from more energetic to less energetic molecules when neighboring molecules collide. Conductive heat flow occur in direction of the decreasing temperature since higher temperature are associated with higher molecular energy.

**Fourier's Law** express conductive heat transfer as

q = k A dT / s (1)

where

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

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

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

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

s = material thickness (m, ft)

### Example - Heat Transfer by Conduction

A plane wall constructed of solid iron with thermal conductivity *70 W/m ^{o}C, *thickness

*50 mm*and with surface area

*1 m by 1 m,*temperature

*150*on one side and

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

^{o}CConductive heat transfer can be calculated as:

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

= 98,000 (W)

= 98 (kW)

### Heat Transfer through Plane Walls In Series

Heat conducted through several walls in good 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 a

*1.2 cm*thick stainless steel inner layer covered with a

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

*800 K*and the outside surface temperature of the asbestos is

*350 K*. The thermal conductivity for stainless steel is

*19 W/m.K*and for asbestos board

*0.7 W/m.K*.

The conductive heat transport through the wall can be calculated as

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

*) + ((0.05 m)*

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

^{2})*))*

* = 6245 (W)*

### Thermal Conductivity and Common 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)

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