q=−k⋅A⋅ΔTLq equals the fraction with numerator negative k center dot cap A center dot cap delta cap T and denominator cap L end-fraction = Heat transfer rate (Watts, W) = Thermal conductivity ( = Heat transfer surface area ( m2m squared ΔTcap delta cap T = Temperature difference ( = Thickness of the material (m) Conduction Example Problem A concrete wall ( ) is 0.2 meters thick. The inside surface temperature is 22∘C22 raised to the composed with power C and the outside surface temperature is 2∘C2 raised to the composed with power C . The total wall area is . Calculate the rate of heat loss through the wall. Step-by-Step Solution: Identify the variables: Apply Fourier's Law:
Try modifying the numbers: add a contact resistance, change the emissivity, or switch to a different fluid. That’s where the real learning happens.
[ L_c = \fracD6 = \frac0.026 = 0.00333 , \textm ] [ Bi = \frach L_ck = \frac20 \cdot 0.00333401 \approx 1.66 \times 10^-4 \ll 0.1 ] Valid. heat transfer example problems
[ R_total = 0.03183 + 0.00193 + 0.2653 = 0.2991 , \textm·K/W ]
First, compute the thermal resistances per unit area: [ R_A = \frac0.21.2 = 0.1667 , \textm²·K/W ] [ R_B = \frac0.10.15 = 0.6667 , \textm²·K/W ] [ R_total = 0.1667 + 0.6667 = 0.8334 , \textm²·K/W ] Calculate the rate of heat loss through the wall
q=1.3⋅15⋅200.2q equals the fraction with numerator 1.3 center dot 15 center dot 20 and denominator 0.2 end-fraction
[ \fracQL = \frac200 - 250.2991 = \frac1750.2991 \approx 585 , \textW/m ] [ L_c = \fracD6 = \frac0
For steady-state 1D conduction without heat generation: