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

The heat flow from the composite wall is 1283.263 watts.

Step-by-step explanation:

The conductive heat flow through a material, measured in watts, is represented by the following expression:

`\dot Q = (\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

`(1)/(R_(II)) = (1)/(R_(B)) + (1)/(R_(C))`

`(1)/(R_(II)) = (R_(B)+R_(C))/(R_(B)\cdot R_(C))`

`R_(II) = (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 = (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* 10^(-3)\,m^(2)`,
`A_(C) = 7* 10^(-3)\,m^(2)`,
`A_(D) = 0.01\,m^(2)`,
`k_(A) = 150\,(W)/(m\cdot ^(\circ)C)`,
`k_(B) = 25\,(W)/(m\cdot ^(\circ)C)`,
`k_(C) = 60\,(W)/(m\cdot ^(\circ)C)` and
`k_(D) = 60\,(W)/(m\cdot ^(\circ)C)`, then:

`R_(A) = (0.03\,m)/(\left(150\,(W)/(m\cdot ^(\circ)C) \right)\cdot (0.01\,m^(2)))`

`R_(A) = (1)/(50)\,(^(\circ)C)/(W)`

`R_(B) = (0.08\,m)/(\left(25\,(W)/(m\cdot ^(\circ)C) \right)\cdot (3* 10^(-3)\,m^(2)))`

`R_(B) = (16)/(15)\,(^(\circ)C)/(W)`

`R_(C) = (0.08\,m)/(\left(60\,(W)/(m\cdot ^(\circ)C) \right)\cdot (7* 10^(-3)\,m^(2)))`

`R_(C) = (4)/(21)\,(^(\circ)C)/(W)`

`R_(D) = (0.05\,m)/(\left(60\,(W)/(m\cdot ^(\circ)C) \right)\cdot (0.01\,m^(2)))`

`R_(D) = (1)/(12)\,(^(\circ)C)/(W)`

`R_(I) = (1)/(50) \,(^(\circ)C)/(W)`

`R_(III) = (1)/(12)\,(^(\circ)C)/(W)`

`R_(II) = (\left((16)/(15)\,(^(\circ)C)/(W) \right)\cdot \left((4)/(21)\,(^(\circ)C)/(W)\right))/((16)/(15)\,(^(\circ)C)/(W) + (4)/(21)\,(^(\circ)C)/(W))`

`R_(II) = (16)/(99)\,(^(\circ)C)/(W)`

`R_(T) = (1)/(50)\,(^(\circ)C)/(W) + (16)/(99)\,(^(\circ)C)/(W) + (1)/(12)\,(^(\circ)C)/(W)`

`R_(T) = (2623)/(9900)\,(^(\circ)C)/(W)`

Now, if
`\Delta T = 400\,^(\circ)C - 60\,^(\circ)C = 340\,^(\circ)C` and
`R_(T) = (2623)/(9900)\,(^(\circ)C)/(W)`, the heat flow is:

`\dot Q = (340\,^(\circ)C)/((2623)/(9900)\,(^(\circ)C)/(W) )`

`\dot Q = 1283.263\,W`

The heat flow from the composite wall is 1283.263 watts.