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

`\dot Q=350.438\ W`

Step-by-step explanation:

Given:

the thermal resistance in the form of

`R_1=(x_1)/(k_1) =0.095\ m^2.^(\circ)C.W^(-1)`

`R_2=(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.(dT)/(x)`

`\dot Q={dT}/ {(x)/(k.A) }`

in series the resistances get add up

`\dot Q=dT/ ((x_1)/(k_1.A)+(x_2)/(k_2.A) )`

`\dot Q=(24-10)/ ((0.095 )/(20)+ (0.704 )/(20) )`

`\dot Q=350.438\ W`