# Heat Transfer by Conduction through a Plane Wall

on . Posted in Thermodynamics

## Heat Transfer by Conduction through a Plane Wall Formula

$$\large{ Q_c = \frac {- k \; A_c \; \left( T_2 \;-\; T_1 \right) } { l } }$$
Symbol English Metric
$$\large{ Q_c }$$ = heat transfer by conduction $$\large{\frac{Btu}{hr}}$$ $$\large{W}$$
$$\large{ A_c }$$ = area cross-section $$\large{in^2}$$ $$\large{mm^2}$$
$$\large{ T_1 }$$ = temperature of one surface of the wall $$\large{F}$$ $$\large{K}$$
$$\large{ T_2 }$$ = temperature of the other surface of the wall $$\large{F}$$ $$\large{K}$$
$$\large{ k }$$ = thermal conductivity $$\large{\frac{Btu-ft}{hr-ft^2-F}}$$ $$\large{\frac{W}{m-K}}$$
$$\large{ l }$$ = thickness of material $$\large{in}$$ $$\large{mm}$$

## Heat Transfer by Conduction through a Cylindrical Wall formula

$$\large{ Q_c = \frac { 2 \; \pi \; k \; l \; \left( T_1 \;-\; T_2 \right) } { ln \; \left( \frac {r_2 }{ r_1 } \right) } }$$
Symbol English Metric
$$\large{ Q_c }$$ = heat transfer by conduction $$\large{\frac{Btu}{hr}}$$ $$\large{W}$$
$$\large{ A_c }$$ = area cross-section $$\large{in^2}$$ $$\large{mm^2}$$
$$\large{ \pi }$$ = Pi $$\large{3.141 592 653 ...}$$
$$\large{ r_1 }$$ = radius of one surface of the wall $$\large{in}$$ $$\large{mm}$$
$$\large{ r_2 }$$ = radius of the other surface of the wall $$\large{in}$$ $$\large{mm}$$
$$\large{ T_1 }$$ = temperature of one surface of the wall $$\large{F}$$ $$\large{K}$$
$$\large{ T_2 }$$ = temperature of the other surface of the wall $$\large{F}$$ $$\large{K}$$
$$\large{ k }$$ = thermal conductivity $$\large{\frac{Btu-ft}{hr-ft^2-F}}$$ $$\large{\frac{W}{m-K}}$$
$$\large{ l }$$ = thickness of material $$\large{in}$$ $$\large{mm}$$

Tags: Heat Transfer Heat