Question

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m${}^2$ and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 270${}^o$C and 0${}^o$C respectively.The thermal conductivity of glass is 0.8 and of air 0.08Wm${}^{-1}$K${}^{-1}$.

The rate of flow of heat through the window pane is nearly equal to:

• A. $1000J{s}^{-1}$
• B. $1224J{s}^{-1}$
• C. $3000J{s}^{-1}$
• D. $4000J{s}^{-1}$

Answer 1

The correct option is B $1224J{s}^{-1}$

Given: A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area $1m^2$ and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of ${270}^{o}C$ and $0^{o}C$ respectively.The thermal conductivity of glass is 0.8 and of air $0.08W{m}^{-1}{K}^{-1}$

To find the rate of flow of heat through the window pane.

Solution:

As per the given criteria,

Area of the each glass sheet, $A=1m^2$

Thermal conductivity of air, $k_a=0.08W{m}^{-1}{K}^{-1}$
Thermal conductivity of glass, $k_g=0.8W{m}^{-1}{K}^{-1}$

Temperature at room-glass interface, $T_1={270}^{\circ }C+273=543K$

Temperature at glass-outdoor interface, $T_4=0^{\circ }C+273=273K$

Thickness of the glass sheet, $d_g=0.01m$

Thickness of the air space, $d_a=0.05m$

By conduction, the equation of heat flow are:

$\frac{d{Q}_1}{dt}=\frac{k_{g}A(T_1-T_2)}{d_g}$

By substituting the values, we get

$\frac{d{Q}_1}{dt}=\frac{0.8\times 1\times (543-T_2)}{0.01}......\left(i\right)$

Similarly at second interface, we get

$\frac{d{Q}_2}{dt}=\frac{k_{a}A(T_2-T_3)}{d_a}$

By substituting the values, we get

$\frac{d{Q}_2}{dt}=\frac{0.08\times 1\times (T_2-T_3)}{0.05}......\left(ii\right)$

Similarly at third interface, we get

$\frac{d{Q}_3}{dt}=\frac{k_{g}A(T_3-T_4)}{d_g}$

By substituting the values, we get

$\frac{d{Q}_3}{dt}=\frac{0.8\times 1\times (T_3-273)}{0.01}......\left(iii\right)$

But in steady state,

$\frac{d{Q}_1}{dt}=\frac{d{Q}_2}{dt}=\frac{d{Q}_3}{dt}$

So equating eqn(i) and (ii), we get

$\frac{0.8\times 1\times (543-T_2)}{0.01}=\frac{0.08\times 1\times (T_2-T_3)}{0.05}⟹0.05(0.8\times (543-T_2\left)\right)=0.01(0.08\times (T_2-T_3\left)\right)⟹0.04(543-T_2)=0.0008(T_2-T_3)⟹(543-T_2)=\frac{0.0008}{0.04}(T_2-T_3)⟹543-T_2=0.02T_2-0.02T_3⟹1.02T_2-0.02T_3=\mathrm{543......}\left(iv\right)$

Now equating eqn(ii) and (iii), we get

$\frac{0.08\times 1\times (T_2-T_3)}{0.05}=\frac{0.8\times 1\times (T_3-273)}{0.01}⟹0.01(0.08\times (T_2-T_3\left)\right)=0.05(0.8\times (T_3-273)⟹0.0008(T_2-T_3)=0.04(T_3-273)⟹\frac{0.0008}{0.04}(T_2-T_3)=(T_3-273)⟹0.02T_2-0.02T_3=T_3-273⟹0.02T_2-1.02T_3=\mathrm{273......}(v)$

Multiply, eqn(iv) with $0.02$ and eqn(v) with $1.02$ and by subtracting them,we get

$0.0204T_2-0.0004T_3=10.86$

$\underset{–}{-(0.0204T_1-1.0404T_3=278.46)}⟹1.04T_3=-267.6⟹T_3=-257.3K$

Substituting the value of $T_3$ in eqn(v), we get

$0.02T_2-1.02(-257.3)=273⟹0.02T_2=273-262.446⟹T_2=527.7K$

So the rate of flow of the heat through the window pane is given by,

$\frac{d{Q}_1}{dt}=\frac{0.8\times 1\times (543-T_2)}{0.01}$

Substituting the value of $T_2$, we get

$\frac{d{Q}_1}{dt}=\frac{0.8\times 1\times (543-527.7)}{0.01}⟹\frac{d{Q}_1}{dt}=1224W$

is the required rate of flow of heat.