Question

Find the heat flow from the composite wall as shown in figure. Assume one dimensional flow KA=150 W/m°C , KB=25 W/m°C, KC=60 W/m°C , KD=60 W/m°C

Answer 1

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:

$\dot Q = \frac{\Delta T}{R_{T}}$

Where:

$R_{T}$ - Equivalent thermal resistance, measured in Celsius degrees per watt.

$\Delta T$ - 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

$\frac{1}{R_{II}} = \frac{1}{R_{B}} + \frac{1}{R_{C}}$

$\frac{1}{R_{II}} = \frac{R_{B}+R_{C}}{R_{B}\cdot R_{C}}$

$R_{II} = \frac{R_{B}\cdot R_{C}}{R_{B}+R_{C}}$

Section I

$R_{I} = R_{A}$

Section III

$R_{III} = R_{D}$

The equivalent thermal resistance is:

$R_{T} = R_{I} + R_{II}+R_{III}$

The thermal of each component is modelled by this:

$R = \frac{L}{k\cdot A}$

Where:

$L$ - Thickness of the brick, measured in meters.

$A$ - Cross-section area, measured in square meters.

$k$ - Thermal conductivity, measured in watts per meter-Celsius degree.

If $L_{A} = 0.03\,m$, $L_{B} = 0.08\,m$, $L_{C} = 0.08\,m$, $L_{D} = 0.05\,m$, $A_{A} = 0.01\,m^{2}$, $A_{B} = 3\times 10^{-3}\,m^{2}$, $A_{C} = 7\times 10^{-3}\,m^{2}$, $A_{D} = 0.01\,m^{2}$, $k_{A} = 150\,\frac{W}{m\cdot ^{\circ}C}$, $k_{B} = 25\,\frac{W}{m\cdot ^{\circ}C}$, $k_{C} = 60\,\frac{W}{m\cdot ^{\circ}C}$ and $k_{D} = 60\,\frac{W}{m\cdot ^{\circ}C}$, then:

$R_{A} = \frac{0.03\,m}{\left(150\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}$

$R_{A} = \frac{1}{50}\,\frac{^{\circ}C}{W}$

$R_{B} = \frac{0.08\,m}{\left(25\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (3\times 10^{-3}\,m^{2})}$

$R_{B} = \frac{16}{15}\,\frac{^{\circ}C}{W}$

$R_{C} = \frac{0.08\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (7\times 10^{-3}\,m^{2})}$

$R_{C} = \frac{4}{21}\,\frac{^{\circ}C}{W}$

$R_{D} = \frac{0.05\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}$

$R_{D} = \frac{1}{12}\,\frac{^{\circ}C}{W}$

$R_{I} = \frac{1}{50} \,\frac{^{\circ}C}{W}$

$R_{III} = \frac{1}{12}\,\frac{^{\circ}C}{W}$

$R_{II} = \frac{\left(\frac{16}{15}\,\frac{^{\circ}C}{W} \right)\cdot \left(\frac{4}{21}\,\frac{^{\circ}C}{W}\right)}{\frac{16}{15}\,\frac{^{\circ}C}{W} + \frac{4}{21}\,\frac{^{\circ}C}{W}}$

$R_{II} = \frac{16}{99}\,\frac{^{\circ}C}{W}$

$R_{T} = \frac{1}{50}\,\frac{^{\circ}C}{W} + \frac{16}{99}\,\frac{^{\circ}C}{W} + \frac{1}{12}\,\frac{^{\circ}C}{W}$

$R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}$

Now, if $\Delta T = 400\,^{\circ}C - 60\,^{\circ}C = 340\,^{\circ}C$ and $R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}$, the heat flow is:

$\dot Q = \frac{340\,^{\circ}C}{\frac{2623}{9900}\,\frac{^{\circ}C}{W} }$

$\dot Q = 1283.263\,W$

The heat flow from the composite wall is 1283.263 watts.