ptc_ce_bta_1.1_us_mp

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PTC_CE_BTA_1.1_us_mp.mcdx

Mathcad Enabled Content Copyright 2011 Knovel Corp.

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

1.1 Heat Conduction in Multilayered Walls

Disclaimer

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

The following example demonstrates calculation of the thermal resistance and temperature distributionwithin a wall assuming one-dimensional steady-state heat transfer. Note that in some cases differentparts of the wall may have different layers, such as wood studs providing structural support. Todetermine a correct wall R-value in such cases, we need to calculate the correct value through eachheat flow path and determine the overall R-value based on the relative area of each path.

Basic Equations

Heat flow q through a wall layer of thickness L, surface area A and thermal conductivity k is given by

q T hot T cold

Rwith R

L

k A

Heat flow q by convection or radiation described in approximate terms by a heat transfer coefficient h(convective, radiative, or combined):

q T hot T cold R

with R 1h A

Thermal resistance Ra through a path a is given by the sum of the resistances of the interior film,each wall layer and the exterior film. If a different path has thermal resistance Rb then the total wallresistance based on parallel heat flow paths is given by

R A

+ Aa Ra

Ab Rb

where A + Aa Ab

Mathcad Enabled ContentCopyright 2011 Knovel Corp.

All rights reserved.

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In the example below, consider a wall with three layers:

1. Gypsum board2. Insulation3. Brick

Input parameters:

hi 0.465 att

t 2 Finterior heat transfer(film) coefficient

ho 1.032 att

t 2 Fexterior heat transfer coefficient

A 10 ft 2 heat transfer area (surface)

L1

0.043 ft Li =thickness of layer i

k1

0.027 attt F

ki =thermal conductivity of layer i

L2

0.164 ft

k2

0.004 attt F

L3

0.328 ft

k3

0.254 attt F



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N 3 N =number of layers (idenotes layer)

i 1 N

RI 1hi A RI =interior film resistance

RO 1ho A

RO =exterior film resistance

Ri

L

i

ki

ARi =resistance of wall layer i

Rtot ++ RI i

Ri RO

= Rtot 4.7 F

wattRtot =total resistance of wall

Calculation of heat flow Q from inside (temperature TI) to outside (temperature TO):

TO 4 F TI 68 F

m 2 N Q TI TO

Rtot

Twi TI Q RI Twi =wall room side surfacetemperature

T 1

Twi Q R1

T m

T m 1

Q Rm

=Q 15.318 watt =Twi 64.706 F

=i

1

23

=T i

62.266

0.5382.516

F



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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 and

spans, materials and other input parameters beyond the intended range may make someequations 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. If ORIGIN 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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