Calculate the heat flux (W/m^2) through a plane uniform wall whose thickness is much less than its width and height if the wall is made of moler bricks [lambda =0.11W/(m.^circ C)] wall thickness is delta =50mm. The temperatures on the wall surfaces are kept constant: t_(s1)=100^circ C and t_(s2)=90^circ C. 40 22 44 28

To calculate the heat flux through a plane uniform wall, we can use the formula:<br /><br />$q = \frac{k \cdot (T_{s1} - T_{s2})}{\delta}$<br /><br />where:<br />- $q$ is the heat flux (in $W/m^2$),<br />- $k$ is the thermal conductivity of the material (in $W/(m \cdot ^\circ C)$),<br />- $T_{s1}$ and $T_{s2}$ are the temperatures on the wall surfaces (in $^\circ C$),<br />- $\delta$ is the thickness of the wall (in $m$).<br /><br />Given:<br />- $k = 0.11 \, W/(m \cdot ^\circ C)$,<br />- $T_{s1} = 100^\circ C$,<br />- $T_{s2} = 90^\circ C$,<br />- $\delta = 50 \, mm = 0.05 \, m$.<br /><br />Substituting the given values into the formula:<br /><br />$q = \frac{0.11 \cdot (100 - 90)}{0.05} = \frac{0.11 \cdot 10}{0.05} = \frac{1.1}{0.05} = 22 \, W/m^2$<br /><br />Therefore, the heat flux through the plane uniform wall is $22 \, W/m^2$.