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Mathcad Enabled Content Copyright 2011 Knovel Corp.

Building Thermal AnalysisAndreas Athienitis 2011 Parametric Technology Corp.Chapter 1 Steady-state heat conduction

1.3 Walls with InternalHeat GenerationDisclaimer

While Knovel and PTC have made every effort to ensure that the calculations, engineering solutions,diagrams and other information (collectively Solution) presented in this Mathcad worksheet aresound from the engineering standpoint and accurately represent the content of the book on which theSolution is based, Knovel and PTC do not give any warranties or representations, express or implied,including with respect to fitness, intended purpose, use or merchantability and/or correctness oraccuracy of this Solution.

Array or ig in:

RIGIN 1

Heat may be generated within a wall. For example, electric radiant panels, which often form theinterior layer of a ceiling or a wall to provide radiant heat, contain electric resistance elements whichgenerate heat. This heat can be approximated as internal heat generation.

We often may assume one-dimensional heat conduction. If a steady-state analysis is performed thenthe relevant energy balance equation is

k d2T/dx2+Qg =0

where k is the thermal conductivity of the wall layer and Qg is the rate of internal heat generation.Consider, for example, a radiant panel of area 1 square meter and thickness 13 mm made of gypsumboard with electric resistance elements built into it with a total power output 250 W assumed uniformly

generated within the panel.

TRoom 68 F

h 0.6 attt2 F

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A 10 ft2 unit areaL 0.043 ft panel thickness

Volume A L

x , 0.003 ft 0.004 ft 0.043 ft distance from backside of panel

k 0.027 att

t F Qg 250 watt

Volume

Boundary conditions

1. Adiabatic at x =0:

d

dxT 0

2. Convective at x =L: kd

dxT h T TRoom

T x +TRoom

+

Qg L

hQg

L2 x 2

2 k

Examination of the graph on the right reveals that the maximum temperature is at the interfacebetween the insulation and the panel (x=0).

113.5

115.5

117.5

119.5

121.5

123.5

125.5

127.5

109.5

111.5

129.5

0.011 0.015 0.019 0.023 0.027 0.031 0.035 0.0390.003 0.007 0.043

T x F

x ftTherefore, maximum temperature is =T 0 ft 129.574 FMathcad Enabled ContentCopyright 2011 Knovel Corp.

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User Notices

Equations and numeric solutions presented in this Mathcad worksheet are applicable to thespecific example, boundary condition or case presented in the book. Although a reasonable effortwas made to generalize these equations, changing variables such as loads, geometries andspans, materials and other input parameters beyond the intended range may make some

equations no longer applicable. Modify the equations as appropriate if your parameters falloutside of the intended range.For this Mathcad worksheet, the global variable defining the beginning index identifier for vectorsand arrays, ORIGIN, is set as specified in the beginning of the worksheet, to either 1 or 0. IfORIGIN is set to 1 and you copy any of the formulae from this worksheet into your own, you needto ensure that your worksheet is using the same ORIGIN.Engineering and construction code values shown in US Customary units are converted fromoriginal values in Metric units. They are NOT obtained from US codes unless specified.

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