Question

Bricks and insulation are used to construct the walls of a house. The
insulation has R1-value of 0.095 m2.cº/W. The bricks have an R2-value of
0.704 m².C°/W, see Fig. 12.38. In the steady-state condition, the temperature
inside the house is Th = 24 °C and the outside temperature is Tc=10°С. Find
the rate of heat loss through such a wall, if its area is 20 m².​

Answer 1

Answer:

$\dot Q=350.438\ W$

Explanation:

Given:

the thermal resistance in the form of

$R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}$

$R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}$

where:

$x_1\ \&\ x_2$ are the thickness of the respective bricks

$k_1\ \&\ k_2$ are the respective coefficient of conductivity

temperature inside the house, $T_h=24\ ^{\circ}C$

temperature outside the house, $\ T_c=10^{\circ}C$

area of the wall, $A=20\ m^2$

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

Using Fourier's law:

$\dot Q=k.A.\frac{dT}{x}$

$\dot Q={dT}\div {\frac{x}{k.A} }$

in series the resistances get add up

$\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )$

$\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )$

$\dot Q=350.438\ W$