Answer:
The heat flow from the composite wall is 1283.263 watts.
Explanation:
The conductive heat flow through a material, measured in watts, is represented by the following expression:
[tex]\dot Q = \frac{\Delta T}{R_{T}}[/tex]
Where:
[tex]R_{T}[/tex] - Equivalent thermal resistance, measured in Celsius degrees per watt.
[tex]\Delta T[/tex] - Temperature gradient, measured in Celsius degress.
First, the equivalent thermal resistance needs to be determined after considering the characteristics described below:
1) B and C are configurated in parallel and in series with A and D. (Section II)
2) A and D are configurated in series. (Sections I and III)
Section II
[tex]\frac{1}{R_{II}} = \frac{1}{R_{B}} + \frac{1}{R_{C}}[/tex]
[tex]\frac{1}{R_{II}} = \frac{R_{B}+R_{C}}{R_{B}\cdot R_{C}}[/tex]
[tex]R_{II} = \frac{R_{B}\cdot R_{C}}{R_{B}+R_{C}}[/tex]
Section I
[tex]R_{I} = R_{A}[/tex]
Section III
[tex]R_{III} = R_{D}[/tex]
The equivalent thermal resistance is:
[tex]R_{T} = R_{I} + R_{II}+R_{III}[/tex]
The thermal of each component is modelled by this:
[tex]R = \frac{L}{k\cdot A}[/tex]
Where:
[tex]L[/tex] - Thickness of the brick, measured in meters.
[tex]A[/tex] - Cross-section area, measured in square meters.
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius degree.
If [tex]L_{A} = 0.03\,m[/tex], [tex]L_{B} = 0.08\,m[/tex], [tex]L_{C} = 0.08\,m[/tex], [tex]L_{D} = 0.05\,m[/tex], [tex]A_{A} = 0.01\,m^{2}[/tex], [tex]A_{B} = 3\times 10^{-3}\,m^{2}[/tex], [tex]A_{C} = 7\times 10^{-3}\,m^{2}[/tex], [tex]A_{D} = 0.01\,m^{2}[/tex], [tex]k_{A} = 150\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{B} = 25\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{C} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex] and [tex]k_{D} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex], then:
[tex]R_{A} = \frac{0.03\,m}{\left(150\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{A} = \frac{1}{50}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{B} = \frac{0.08\,m}{\left(25\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (3\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{B} = \frac{16}{15}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{C} = \frac{0.08\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (7\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{C} = \frac{4}{21}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{D} = \frac{0.05\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{D} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{I} = \frac{1}{50} \,\frac{^{\circ}C}{W}[/tex]
[tex]R_{III} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{II} = \frac{\left(\frac{16}{15}\,\frac{^{\circ}C}{W} \right)\cdot \left(\frac{4}{21}\,\frac{^{\circ}C}{W}\right)}{\frac{16}{15}\,\frac{^{\circ}C}{W} + \frac{4}{21}\,\frac{^{\circ}C}{W}}[/tex]
[tex]R_{II} = \frac{16}{99}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{1}{50}\,\frac{^{\circ}C}{W} + \frac{16}{99}\,\frac{^{\circ}C}{W} + \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex]
Now, if [tex]\Delta T = 400\,^{\circ}C - 60\,^{\circ}C = 340\,^{\circ}C[/tex] and [tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex], the heat flow is:
[tex]\dot Q = \frac{340\,^{\circ}C}{\frac{2623}{9900}\,\frac{^{\circ}C}{W} }[/tex]
[tex]\dot Q = 1283.263\,W[/tex]
The heat flow from the composite wall is 1283.263 watts.
