Heat flow through concrete floor

16 October 2015
Finite Element Method – Flow Analysis
Prepared by
Maan Al-Haideri
Katharina Wagner
Amy Do

Table of content
1 Introduction 1
2 Theory 2
2.1 Strong and weak form of 2-dim transient heat flow 2
2.2 Spatial approximation of 2-dim transient heat flow: 2
2.3 Approximation in time 3
2.4 Convection 4
3 Method 5
3.1 Assumptions 5
3.2 Model 5
4 Results 6
4.1 Influence of time stepping parameters 6
4.2 Temperature distribution 8
4.3 Concrete surface 11
4.4 Supplied heat power 11
5 Discussion 12
6 References 13
7 Appendix 14
7.1 Code 14
7.2 compare time-stepping parameters 20

1
1 Introduction
To understand the heating mechanism and behaviour of a concrete floor slab with hollow cores,
a transient heat flow problem was modeled and solved. Hot air flows through every other
cylindrical cavity, and heats up the concrete surrounding it, and subsequently the air that is in
contact with the concrete. Figure 1, which is based on [1], illustrates the material and
dimensions of a cross section of the floor slab.
Figure 1: Geometry of the concrete floor problem, modified from [1]
Because the problem was symmetric, to simplify the solution, only one element, looking like
Figure 2, of the floor slab was considered.
Figure 2: Boundary conditions of the concrete floor problem, modified from [1]. Red shows the borders of the modelled system

2
A convection layer at the surface of the concrete was modelled using the boundary condition
qn = α(T-T∞), with α being the inverse of the internal flow resistance Rsi and T∞ being the
temperature far away from the concrete surface, i.e. room temperature. A finite element model
was set up and solved using MatLab with the geometry and boundary conditions illustrated
above. An explanation of the basic transient heat flow theory is provided to aid the
understanding of the solution.
2 Theory
2.1 Strong and weak form of 2-dim transient heat flow
This transient heat flow problem was solved using the finite element method. Similar to the one
dimensional transient heat flow equation, the strong form of the two dimensional transient heat
flow may be derived from inserting a heat capacity term and Fourier’s law to the heat-balance
equation.
The strong form of the two dimensional transient heat flow:
𝐝𝐢𝐯(𝑡ℎ 𝐃𝛁T) + 𝑡ℎ 𝐐 = 𝜌𝑡ℎ 𝐜
𝑑𝑇
𝑑𝑡
; 𝑖𝑛 𝑟𝑒𝑔𝑖𝑜𝑛 𝐴
𝑞 𝑛 = ℎ 𝑜𝑛 𝐿ℎ
𝑇 = 𝑔 𝑜𝑛 𝐿 𝑔
Where 𝑡ℎthe thickness, q is the flux, 𝜌 is the density of the material, c is the heat capacity, T is
the temperature, and Q is heat supply.
The weak form of the two dimensional transient heat flow may be derived from the strong form
by multiplying the balance equation with an arbitrary, time-independent weight function v(x,y)
and integrating over the region, and using the Green-Gauss theorem then using that q=-D𝛁𝑇.
This gives the weak form of two dimensional transient heat flow:
∫ 𝐴(𝛁𝑣)
𝑇
𝑡ℎ 𝑫𝛁T𝐝𝐀 + ∫ 𝐴 𝜌𝑡ℎ 𝐜
𝑑𝑇
𝑑𝑡
𝐝𝐀 = −∮ 𝐿ℎ
𝐯𝑡ℎℎ𝑑𝐿 − ∮ 𝐿 𝑔
𝑣𝑡ℎ 𝐪 𝐧 𝑑𝐿 + ∫ 𝐴 𝑣𝑡ℎ 𝐐𝐝𝐀
𝑇 = 𝑔 𝑜𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑆 𝑔
Since this equation contains time and also directional derivatives, the approximations for the
temperature must be made in both space and the time.
2.2 Spatial approximation of 2-dim transient heat flow:
The approximation for the temperature is separated in
𝑇(𝑥, 𝑦, 𝑡) = 𝐍(𝑥, 𝑦)𝐚(𝑡)
Where 𝐍(𝑥, 𝑦) describes the shape functions and 𝐚(𝑡) are the nodal temperatures.
Since the shape function is independent from the time and the temperature function is
independent of spatial coordinates, and the weak form of the heat flow equation includes a
derivative of time and another derivative of spatial coordinates it may be written
𝛁𝐓 = 𝛁𝐍𝐚 = 𝐁𝐚
𝑑𝑇
𝑑𝑡
= 𝐍
𝑑𝐚
𝑑𝑡
= 𝐍𝐚̇
A scalar weight function v is chosen using the Galerkin method
𝑣 = 𝐍𝐜 = 𝐜 𝐓
𝐍 𝐓
𝛁𝑣 = 𝐜 𝐓
𝐁 𝐓
where 𝐁 𝐓
= 𝛁𝐍 𝐓
When the approximation and the weight functions are inserted, and after some steps the
following equation is achieved:
Which also can be written as:

3
𝐊𝐚 + 𝐂𝐚̇ = 𝐟 𝒃 + 𝐟𝒍
Where:
2.3 Approximation in time
The finite element equations have to be approximated in time as well. Different choices of the
time approximation may be made. Only a linear time approximation is considered. Since the
approximation in time is scalar, the temperature at the next time step 𝑎𝑖+1could be written
Where the average force f̅ is:
The weight parameter here is:
The point collocation method described in [2], is adopted to determine the weight parameter. A
Dirac’s function may be chosen for 𝜔 at different times in the time-step. Three common choices
were examined. 𝜏 = 0, 𝜏 =
∆𝑡
2
, 𝑎𝑛𝑑 𝜏 = ∆𝑡
Choosing 𝜔 = 𝛿(𝜏 − 0) gives:
, which is called forward Euler or explicit method. Here no inversion of K occurs.
Choosing 𝜔 = 𝛿(𝜏 − ∆𝑡) gives:
Being called the backward Euler or Implicit method. This requires an inversion of K.
Choosing 𝜔 = 𝛿(𝜏 −
∆𝑡
2
) gives:

4
, and this method is called Crank-Nicolson. Here an inversion of K is required.
Notable is that Θ is called α in MatLab and this nomenclature will be used throughout this
report.
2.4 Convection
Heat flow in a body can be written as
[
𝑞 𝑥
𝑞 𝑦
] = −𝑫 [
𝑑𝑇
𝑑𝑥
𝑑𝑇
𝑑𝑦
] − [
𝑘 𝑥 0
0 𝑘 𝑦
] [
𝑑𝑇
𝑑𝑥
𝑑𝑇
𝑑𝑦
]
, which leads to
𝑞 𝑥 = −𝑘 𝑥
𝑑𝑇
𝑑𝑥
𝑞 𝑦 = −𝑘 𝑦
𝑑𝑇
𝑑𝑦
( 1 )
Convection assumes that only flow normal to the boundaries occur. A typical model for
convection can be seen in Figure 3. The convection can be written as
𝑞 𝑛 = 𝛼(𝑇 − 𝑇∞)
Figure 3: modelling of convection [3]
If it is now assumed that the normal direction points in the y-direction, qx and qy become
𝑞 𝑥 = 0
𝑞 𝑦 = 𝛼(𝑇 − 𝑇∞) ( 2 )
The derivatives given in 𝑞 𝑦 = −𝑘 𝑦
𝑑𝑇
𝑑𝑦
( 1 ) can be changed to
𝑞 𝑥 = 0
𝑞 𝑦 = −𝑘 𝑦
𝑇∞−𝑇
𝑡
( 3 )
Comparing 𝑞 𝑦 = 𝛼(𝑇 − 𝑇∞) ( 2 ) with 𝑞 𝑦 = −𝑘 𝑦
𝑇∞−𝑇
𝑡
(
3 ) leads to
𝑘 𝑦
𝑡
= 𝛼
And therefore
𝐃 = [
0 0
0 𝛼𝑡
]

5
3 Method
The stated problem was solved using a finite element approach in MatLab. In MatLab,
CALFEM and a mesh module were added, which contain commands useful for the finite
element method.
3.1 Assumptions
To solve this problem a few assumptions were necessary. One assumption was that no
convection occurred in the cavities. For the heated hole this assumption was justified as long as
the heated air has a higher velocity. In the other cavities convection can be neglected since the
temperature gradients inside the holes were very small. Another assumption was that the
problem is symmetric, therefore only half a heated and unheated cavity were modelled. Since
these model borders are symmetry lines, no flow was passing these borders. The floor under
the concrete was also assumed to be impermeable, this holds true if the conductivity of the
concrete is by magnitudes bigger than the one from the other material. The last assumption was
that convection only occurred in a well-defined layer above the concrete and the heat only flows
in the normal direction. This method is commonly used to calculate convection, therefore it was
also used here.
3.2 Model
The first step of modelling was to define the geometry of the system. Since the geometry also
included circle segments, special attention had to be drawn to their modelling. Depending on
the program used the approach would be different. The inside of the heated cavity was not
modelled since the whole cavity had the same temperature for every time. Additionally, this
way it was easier to define the boundary conditions, since so the cavity temperature was just
set to be the boundary condition.
As soon as the geometry was defined, the number of elements on each segment had to be chosen
to define the mesh size for this problem. Since the problem was used for the temperature the
degrees of freedom per element were set as 1 and triangular elements were used to solve this
problem. After setting all the needed parameters the mesh was created and the nodal coordinates
were extracted.
In the next step the stiffness matrix and the force vector had to be defined. Therefore, the
thickness was chosen to be one meter to calculate the values per meter depth. The element
stiffness matrices and the element capacity matrices were separately calculated for every
material, namely concrete, air and the convection layer, and assembled in the according global
stiffness matrices. The force vector stayed empty since no flow was added inside the borders of
the system.
The boundary conditions were also defined. Along the border between the heated cavity and
the concrete the boundary condition was set to be 25°C and above the convection layer the
boundary condition was set to fit the room temperature of 22°C. The flow through the remaining
boundaries was 0 according to the assumptions.
Following, the initial values for the temperature had to be defined which were given to be 15°C
everywhere. The time stepping was defined and performed. Here several step-sizes as well as
weight parameter values were tested for their reliability. With the time-stepping it was possible
to obtain Tsnap, D and V. Tsnap stored the temperature at all nodes for certain time-steps, D
stored the temperature for all time-steps for certain nodes and V stored the temperature time
derivative for every time-step at certain nodes. Here all time-steps and nodes where selected so
Tsnap and D were the same.

6
To obtain the temperatures at certain times, these had to be extracted from Tsnap. Also the
element flows and gradients were calculated for every examined time. Several plots showing
the temperature distribution in the whole model for a certain time were created with the before
extracted values. Also the temperatures along the concrete surface were extracted for certain
times to compare the temperature distribution for different times.
Since no heat was added inside the system 𝐊𝑎 + 𝐂𝑎̇ = 𝐟 𝐛 + 𝐟𝑙 = 𝐟 𝐛 + 𝟎 = 𝐫( 4 ) holds true
𝐊𝑎 + 𝐂𝑎̇ = 𝐟 𝐛 + 𝐟𝑙 = 𝐟 𝐛 + 𝟎 = 𝐫 ( 4 )
, where r describes the heat flow
And when this calculation is done for every time step:
𝐊 ∗ 𝑫 + 𝐂 ∗ 𝐕 = 𝐫 ( 5 )
Here r contains every heat flow for every element for every time step. To obtain the power
distribution into the room the flow out of the convection layer had to be sum up for every time
step and divided by the length and depth of the model to receive the power supply per square
meter.
4 Results
4.1 Influence of time stepping parameters
To decide the optimal parameters for the time-stepping operation, various combinations of dt
and α were tested. Figure 4-Figure 7 show two of these tested combinations with dt=3600s and
dt=360s and all possible values for α, so α=0, α=0.5 and α=1. Figure 4 shows clear differences
between all three curves, especially the one for α=0 clearly differs from the other two, since the
solution is oscillating. Also the temperatures in Figure 5 vary between the different α.
For dt=360s Figure 6 shows that the differences between α=0.5 and α=1 were very small and
only visible for the beginning of the heat flow. Here the curve for α=0 was not included, since
it oscillated even more and did therefore not give any useful information.
As Figure 5 and Figure 7 show, the simulation with α=0 gave temperatures outside the possible
range of 15°C-25°C, therefore these temperatures are not shown in the graph. Since it was not
able to get rid of the oscillation for α=0 for small time-steps as small as dt=3.6s, a dt was chosen
that gave similar results for α=0.5 and α=1. This time-step was selected to be dt=360s and α to
be α=1.

7
Figure 4: power supply at the concrete surface. The time-step was chosen to be dt=3600s. This figure depicts the different
outcomes for the power supply for varying α.
Figure 5: temperature distribution on the concrete surface for different times. The time-step was chosen to be dt=3600. The
dots represent the values for α=0, the circles represent α=0.5 and the x-marks represent α=1.

8
Figure 6: power supply at the concrete surface. The time-step was chosen to be dt=360. This figure depicts the different
outcomes for the power supply for varying α. The curve for α=0 was not plotted, since it oscillated even more than for Figure
4.
Figure 7: temperature distribution on the concrete surface for different times. The time-step was chosen to be dt=360. The
dots represent the values for α=0, the circles represent α=0.5 and the x-marks represent α=1.
4.2 Temperature distribution
In the beginning, heat flowed from the area of the heated cavity and from the room into the
hollow core and the concrete. Because of the curved geometry, different initial temperatures,
different materials and the behaviour of the convection layer, however, heat did not flow in one
single direction. Eventually, the variation between temperatures in the floor slab, excluding the
convection layer, should become smaller as temperature equilibrium was approached. The
result of the finite element analysis reflected these fundamental physical rules. To illustrate this
point, Figure 8 and Figure 9 of the temperature fluxes were produced. At 1 hour, heat flow was

9
greatest along the hot cavity, as heat was leaving this region rapidly. Some heat was also
flowing into the concrete from the room. Because concrete has a much higher thermal
conductivity than air, the heat flow through concrete was higher than that through the cold air
cavity. Much later in time, at 16 hours, flow had decreased across the floor, but there was still
some heat flow from the hot air towards the concrete surface and into the convection layer.
Figure 8: Heat flow through the model at 1 hour. The scaling factor used for the flow was 10-5

10
Figure 9: Heat flow through the model at 16 hours. The scaling factor used for the flow was 10-5
With this understanding of the heat flow, it was easier to interpret the colour plots of the
temperature distribution over time.
Figure 10 Colour plots of the temperature distributions over time

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In the earlier stage, there was a great difference between temperatures throughout the model,
corresponding to the larger heat fluxes. As time progressed, the temperature inside became
higher and more evenly distributed, corresponding to smaller heat fluxes, but the temperature
would never be the same everywhere throughout the model.
4.3 Concrete surface
The boundary of interest was the concrete surface, since this is where the floor comes in contact
with the air. The temperature was higher on the concrete directly above the heat source, so it
was warmer on the left of the model than on the right. Similar to the heat flow logic in 4.2, the
temperature variation between left and right decreased over time, but did never reach 0. As time
progressed, the temperature change with time also decreased, as can be seen in Figure 11 where
the temperatures for 8, 12 and 16 hours are very close together. The highest temperature reached
at 16 hours was 24.4ᵒC.
Figure 11: Temperature distribution along the top concrete surface over time
4.4 Supplied heat power
The supplied power was calculated according to 𝐊 ∗ 𝑫 + 𝐂 ∗ 𝐕 = 𝐫 ( 5 ) the change of power
supply per m2
floor over time can be seen in Figure 12.
The most power was supplied into the floor around 0.2h with a value around 45.5 𝑊
𝑚2⁄ and
the power remained positive until around 1.7 hours. Then the power became negative, which
meant a heat flow out of the floor, and plateaued to a steady value around -15.9 𝑊
𝑚2⁄ .

12
Figure 12: Total power supplied per square meter of floor area with time
5 Discussion
After the investigation of suitable stepping parameters, it was decided that a time step of 360s
with a weight parameter of 1 would give the most stable and reasonable results. Increasing the
time step gave less accurate results and the weighting parameter had a strong influence on the
outcome. Changing the weight parameter from 1 to 0 would have given unstable results
independent of the time-steps. With dt = 360 s, the values for α = 0.5 and α = 1 produces very
similar results, so one of those two was selected randomly to be combined with this time-step
size.
An even smaller time-step would only improve the accuracy slightly but at the cost of
processing power. So, as always for the FEM a compromise between accuracy and time had to
be agreed on. The investigation outlined in 4.1 is an adequate analysis to decide on dt and α,
providing a reasonable accuracy.
Considering the power supplied per square meter of the floor area as a function of time, it was
observed that power was positive until around 1.7 hours, then it became negative. Given the
way the model was defined, which included a layer of air convection, a positive power means
that heat is entering the model from surrounding areas. That means during the first
approximately 1.7 hours, since the air temperature (22°C) was higher than initial concrete
temperature (15°C), the air was transferring heat into the concrete floor. Then, once enough
heat was supplied to the floor slab (both from the room and from the hot air source), the concrete
started to become warmer than the air. This was when heat started leaving the model, supplying
heat to the air and therefore the power became negative. At approximately 11.5 hours, power
reached a steady value of -15.9 W/m2
. Also, comparing the figures of the flow for different
times (Figure 8 and Figure 9) it was seen that the flow inside the system decreased as it
approached steady-state conditions.

13
6 References
[1] E. Serrano, „Hand-in assignment 2b - Transient Heat Flow,“ Lund, 2015.
[2] K. Persson, „Transient Flow,“ Lund, 2015.
[3] K. Persson, „FE formulation-two-and three-dimensional heat flow,“ Lund, 2015.
[4] „Wikipedia,“ 7 October 2015. [Online]. Available:
https://en.wikipedia.org/wiki/Properties_of_concrete. [Zugriff am 15 October 2015].
[5] „The Engineering ToolBox,“ [Online]. Available:
http://www.engineeringtoolbox.com/concrete-properties-d_1223.html. [Zugriff am 15
December 2015].
http://www.engineeringtoolbox.com/air-properties-d_156.html. [Zugriff am 15 October
2015].
[7] „Wikipedia,“ 7 October 2015. [Online]. Available:
https://en.wikipedia.org/wiki/Thermal_transmittance . [Zugriff am 15 October 2015].
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html . [Zugriff am 15
December 2015].
[9] N. P. H. Ottosen, Introduction to the Finite Element Method, Prentice Hall, 1992.

14
7 Appendix
7.1 Code
clear all;
close all;
%hole diameter
d = 125;
%hole radius
ra = d/2;
%distance between hole centers
b = 190;
%plate heigth
H = 200;
%define coordinates for points on the circle
x1 = ra*cos(67.5*pi/180);
x2 = ra*cos(45*pi/180);
x3 = ra*cos(22.5*pi/180);
x4 = b - ra*cos(22.5*pi/180);
x5 = b - ra*cos(45*pi/180);
x6 = b - ra*cos(67.5*pi/180);
y1 = H/2 -ra*sin(67.5*pi/180);
y2 = H/2 - ra*sin(45*pi/180);
y3 = H/2-ra*sin(22.5*pi/180);
y4 = H/2 + ra*sin(22.5*pi/180);
y5 = H/2 + ra*sin(45*pi/180);
y6 = H/2 + ra*sin(67.5*pi/180);
L=10;
%coordinates of nodes multiply by 1/1000 because the
values are in mm but
%m is needed
Vertices=1/1000*[
0 0;%1
x2 0;
x5 0;
b 0;
b 37.5;%5
b 70;
b H/2;
b 130;
b 162.5;
b H;%10
x5 H;
x2 H;
0 H;
0 162.5;
0 200+L;%15
0 37.5;
x2 y2;
x5 y2;
ra H/2;
b-ra H/2;%20
x2 y5;
x5 y5;
160 130;
160 H/2;
160 70;%25
x1 y1;
x3 y3;
x3 y4;
x1 y6;
x6 y6;%30
x4 y4;
x4 y3;
x6 y1;
x2 200+L;
x5 200+L;%35
190 200+L];
%define the line segments from vertex numbers
Segments=[
1 2; %1
2 3;
3 4;
4 5;
5 6;%5
6 7;
7 8;

15
8 9;
9 10;
10 11;%10
11 12;
12 13
13 14;
12 34;
15 34;%15
16 1;
2 17;
3 18;
17 18;
19 20;%20
21 22;
21 12;
22 11;
14 21;
21 19;%25
19 17;
17 16;
18 25;
20 24;
18 5; %30
20 18;
20 22;
22 9;
22 23;
23 8; %35
24 7;
25 6;
24 25;
24 23;
13 15;%40
34 35;
35 36;
11 35;
10 36];
%define the surfaces from the segment numbers
Surfaces=[
1 17 27 16; %1
2 18 19 17;
3 4 30 18;
11 43 41 14;
19 31 20 26; %5
28 38 29 31;
30 5 37 28;
37 6 36 38;
12 14 15 40;
20 32 21 25; %10
29 39 34 32;
36 7 35 39;
24 22 12 13;
21 23 11 22;
33 9 10 23; %15
35 8 33 34;
10 44 42 43];
%define the number of elements on each segment
Seed=ones(1, 44);
Seed(1, 2)=3; %the middle ones are made smaller
Seed(1, 19)=3;
Seed(1, 20)=3;
Seed(1, 21)=3;
Seed(1, 11)=3;
Seed(1, 41)=3;
Seed=3*Seed; %scaling of the mesh
Seed(1, 40)=1; %convection layer only one mesh broad
Seed(1, 14)=1;
Seed(1, 43)=1;
Seed(1, 44)=1;
%define the curved lines
iso8=zeros(1, 44);
iso8(27)=26; iso8(26)=27; iso8(25)=28; iso8(24)=29;
iso8(30)=33; iso8(31) = 32; iso8(32)=31; iso8(33)=30;
% COMBINE SEED AND CURVED LINE DEFINITION

16
Segp(1:2:88)=Seed;
Segp(2:2:88)=iso8;
nen=3;
dofsPerNode=1;
mp=[dofsPerNode, nen];
%number the segments
figure(1)
geomdraw2(Vertices, Segments, Surfaces, Segp,mp);
%generate the element mesh
[Coord Edof Dof meshdb]=strMeshgen(Vertices, Segments,
Surfaces, Segp, mp);
%generate the element coordinates
[Ex, Ey]=coordxtr(Edof,Coord,Dof,nen);
%Draw the element mesh
figure(2)
eldraw2(Ex, Ey, [1 4 0]);
%Create element stiffness, assemble to global
stiffness, create bc's and
%solve the equation system
nel=size(Edof,1);
ndof=length(Dof);
K=sparse(ndof,ndof);
C=sparse(ndof,ndof);
f=sparse(ndof,1);
%% Create element stiffness and assemble for concrete
surfaces
for j=[1 2 3 5 10 13 14 15]
t=1; %thickness
rho=2400; % [4]
c=750; % [5]
ep=[t,rho,c];
k=1.7; % [6]
Dc=k*[1 0; 0 1];
elems=extrSurf(j,meshdb); %extract elements on
surface j
for l=elems' %Loop over elements in elems
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Dc);
K=sparse_assem(Edof(l,:),K,Ke);
C=sparse_assem(Edof(l,:),C,Ce);
end
end
%Create element stiffness and assemble for air in hole
%values chosen for 20C
%http://www.engineeringtoolbox.com/air-properties-
d_156.html
for j=[6 7 8 11 12 16]
t=1;
rho=1.205; % [6]
c=1005; % [6]
ep=[t,rho,c];
k=0.0257; % [6]
Da=k*[1 0; 0 1];
surface j
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Da);
end
end
%Convection layer
for j=[4 9 17]
t=1;
rho=1.205; % [6]
c=1005; % [6]
Rsi = 0.13; % [7]
k=1/Rsi;
ep=[t,rho,c];
Dr=k*[0 0; 0 1*L/1000];

17
surface j
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Dr);
end
end
%Determine degrees of freedom along segments for
boundary conditions
bc1=extrSeg([24 25 26 27]',Segments,
Segp,Vertices,Dof,Coord,1); %warm air
bc2=extrSeg([15 41 42]',meshdb,1); %room temperature
bc3=extrSeg([12 11 10]',meshdb,1); %concrete surface
%Define boundary conditions
bc=[bc1, ones(length(bc1),1)*25; bc2,
ones(length(bc2),1)*22];
%define initial temperature at time t=0
a0=ones(ndof,1)*15;
%%
n=10; %time-steps per hour
%define and perform time stepping for 36 hours (given
in seconds)
%times at which the results is extracted
ti=[0:3600/n:3600*36];
%define input for the time-stepping command
ip=[3600/n 3600*36 1 [length(ti) ndof ti Dof']];
%Perform time-stepping
[Tsnap,D,V]=step1(K,C,a0,ip,f,bc);
%Extract element temperatures after 1 hour
Ed1=extract(Edof,Tsnap(:,n));
%Calculate element heat fluxes after 1 hour
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end
end
elemens=extrSurf(j,meshdb);
for i=elemens'
[Es1(i,:),Et1(i,:)]=flw2ts(Ex(i,:),Ey(i,:),Dm,Ed1(i,:))
;
end
end
%Plot flux vectors after 1 hour
figure(3)
eldraw2(Ex,Ey, [1 3 0]);
elflux2(Ex, Ey, Es1, [1, 1], 10e-5);
xlabel('length in m')
ylabel('height in m')
%Extract element temperatures after 2 hours
Ed2=extract(Edof,Tsnap(:,2*n));
%Calculate element heat fluxes after 2 hours
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end

18
end
for i=elemens'
;
end
end
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end
end
for i=elemens'
;
end
end
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end
end
for i=elemens'
;
end
end
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end
end
for i=elemens'
[Es12(i,:),Et12(i,:)]=flw2ts(Ex(i,:),Ey(i,:),Dm,Ed12(i,
:));
end
end

19
for j=1:17
if(j==1)|
(j==2)|(j==3)|(j==5)|(j==10)|(j==13)|(j==14)|(j==15)
Dm=Dc;
else
if (j==9)|(j==4)|(j==17)
Dm=Dr;
else
Dm=Da;
end
end
for i=elemens'
[Es16(i,:),Et16(i,:)]=flw2ts(Ex(i,:),Ey(i,:),Dm,Ed16(i,
:));
end
end
%Plot flux vectorsafter 16 hours
figure(4)
eldraw2(Ex,Ey, [1 3 0]);
elflux2(Ex, Ey, Es16, [1, 1], 10e-5);
ylabel('heigth in m')
%% Colour plots of the temperature distribution
figure(11)
subplot(2,2,1)
fill(Ex',Ey',Ed1')
caxis([15 25])
axis('equal')
title('t=1 hour')
subplot(2,2,2)
fill(Ex',Ey',Ed2')
caxis([15 25])
axis('equal')
title('t=2 hours')
subplot(2,2,3)
fill(Ex',Ey',Ed8')
caxis([15 25])
axis('equal')
title('t=8 hours')
subplot(2,2,4)
fill(Ex',Ey',Ed16')
caxis([15 25])
axis('equal')
title('t=16 hours')
colorbar
colormap('jet')
%%Extract element temperatures along bc3 after 1, 4, 8
and 16 hours
figure (12)
X1=Coord(bc3,1);
T1 = Tsnap(bc3, n);
T2 = Tsnap(bc3, 2*n);
T12= Tsnap(bc3, 12*n);
T16 = Tsnap(bc3, 16*n);
hold on
plot(X1, T1, '.r');
plot(X1, T2, '.g');
plot(X1, T4, '.b');
plot(X1, T8, '.m');
plot(X1, T12, '.c');
plot (X1, T16, '.k');

20
legend ('1 hour', '2 hours','4 hours', '8 hours','12
hours', '16 hours');
ylabel('temperature in °C')
hold off
%power as a function of time: r = Ka + Ca
r=K*D+C*V;
%divide by square meter of floor
r2=r/(1*b/1000);
figure(13)
tt = linspace(1/n,36+1/n,length(ti)); %right scale
plot(tt,sum(r2(bc2,:)))
xlabel('time in h')
ylabel('power supply in W/m^2')
7.2 compare time-stepping parameters
clear all;
close all;
%hole diameter
d = 125;
%hole radius
ra = d/2;
%distance between hole centers
b = 190;
%plate heigth
H = 200;
%define coordinates for points on the circle
x1 = ra*cos(67.5*pi/180);
x2 = ra*cos(45*pi/180);
x3 = ra*cos(22.5*pi/180);
x4 = b - ra*cos(22.5*pi/180);
x5 = b - ra*cos(45*pi/180);
x6 = b - ra*cos(67.5*pi/180);
y1 = H/2 -ra*sin(67.5*pi/180);
y2 = H/2 - ra*sin(45*pi/180);
y3 = H/2-ra*sin(22.5*pi/180);
y4 = H/2 + ra*sin(22.5*pi/180);
y5 = H/2 + ra*sin(45*pi/180);
y6 = H/2 + ra*sin(67.5*pi/180);
L=10;
%coordinates of nodes multiply by 1/1000 because the
values are in mm but
%m is needed
Vertices=1/1000*[
0 0;%1
x2 0;
x5 0;
b 0;
b 37.5;%5
b 70;
b H/2;
b 130;
b 162.5;
b H;%10
x5 H;
x2 H;
0 H;
0 162.5;
0 200+L;%15
0 37.5;
x2 y2;
x5 y2;
ra H/2;
b-ra H/2;%20
x2 y5;
x5 y5;
160 130;
160 H/2;
160 70;%25
x1 y1;
x3 y3;
x3 y4;
x1 y6;
x6 y6;%30
x4 y4;
x4 y3;
x6 y1;

21
x2 200+L;
x5 200+L;%35
190 200+L];
%define the line segments from vertex numbers
Segments=[
1 2; %1
2 3;
3 4;
4 5;
5 6;%5
6 7;
7 8;
8 9;
9 10;
10 11;%10
11 12;
12 13
13 14;
12 34;
15 34;%15
16 1;
2 17;
3 18;
17 18;
19 20;%20
21 22;
21 12;
22 11;
14 21;
21 19;%25
19 17;
17 16;
18 25;
20 24;
18 5; %30
20 18;
20 22;
22 9;
22 23;
23 8; %35
24 7;
25 6;
24 25;
24 23;
13 15;%40
34 35;
35 36;
11 35;
10 36];
%define the surfaces from the segment numbers
Surfaces=[
1 17 27 16; %1
2 18 19 17;
3 4 30 18;
11 43 41 14;
19 31 20 26; %5
28 38 29 31;
30 5 37 28;
37 6 36 38;
12 14 15 40;
20 32 21 25; %10
29 39 34 32;
36 7 35 39;
24 22 12 13;
21 23 11 22;
33 9 10 23; %15
35 8 33 34;
10 44 42 43];
%define the number of elements on each segment
Seed=ones(1, 44);
Seed(1, 2)=3; %the middle ones are made smaller
Seed(1, 19)=3;
Seed(1, 20)=3;
Seed(1, 21)=3;
Seed(1, 11)=3;

22
Seed(1, 41)=3;
Seed=3*Seed; %scaling of the mesh
Seed(1, 40)=1; %convection layer only one mesh broad
Seed(1, 14)=1;
Seed(1, 43)=1;
Seed(1, 44)=1;
%define the curved lines
iso8=zeros(1, 44);
iso8(27)=26; iso8(26)=27; iso8(25)=28; iso8(24)=29;
iso8(30)=33; iso8(31) = 32; iso8(32)=31; iso8(33)=30;
% COMBINE SEED AND CURVED LINE DEFINITION
Segp(1:2:88)=Seed;
Segp(2:2:88)=iso8;
nen=3;
dofsPerNode=1;
mp=[dofsPerNode, nen];
%number the segments
figure(1)
geomdraw2(Vertices, Segments, Surfaces, Segp,mp);
%generate the element mesh
[Coord Edof Dof meshdb]=strMeshgen(Vertices, Segments,
Surfaces, Segp, mp);
%generate the element coordinates
[Ex, Ey]=coordxtr(Edof,Coord,Dof,nen);
%Draw the element mesh
figure(2)
eldraw2(Ex, Ey, [1 4 0]);
%Create element stiffness, assemble to global
stiffness, create bc's and
%solve the equation system
nel=size(Edof,1);
ndof=length(Dof);
K=sparse(ndof,ndof);
C=sparse(ndof,ndof);
f=sparse(ndof,1);
%Create element stiffness and assemble for concrete
surfaces
%literature in the file
for j=[1 2 3 5 10 13 14 15]
t=1; %thickness
rho=2400; % [4]
c=750; % [5]
ep=[t,rho,c];
k=1.7; % [8]/ value: concrete, stone
Dc=k*[1 0; 0 1];
surface j
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Dc);
end
end
%Create element stiffness and assemble for air in hole
%values chosen for 20C
%http://www.engineeringtoolbox.com/air-properties-
d_156.html
for j=[6 7 8 11 12 16]
t=1;
rho=1.205; % [6]
c=1005; % [6]
ep=[t,rho,c];
k=0.0257; % [6]
Da=k*[1 0; 0 1];
surface j
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Da);

23
end
end
%Convection layer
for j=[4 9 17]
t=1;
rho=1.205; % [6]
c=1005; % [6]
Rsi = 0.13; % [7] /values for inside surface
k=1/Rsi;
ep=[t,rho,c];
Dr=k*[0 0; 0 1*L/1000];
surface j
[Ke, Ce]=flw2tt(Ex(l,:),Ey(l,:),ep,Dr);
end
end
%Determine degrees of freedom along segments for
boundary conditions
bc1=extrSeg([24 25 26 27]',Segments,
Segp,Vertices,Dof,Coord,1); %warm air
bc2=extrSeg([15 41 42]',meshdb,1); %room temperature
bc3=extrSeg([12 11 10]',meshdb,1); %concrete surface
%Define boundary conditions
bc=[bc1, ones(length(bc1),1)*25; bc2,
ones(length(bc2),1)*22];
%define initial temperature at time t=0
a0=ones(ndof,1)*15;
%define and perform time stepping for 36 hours (given
in seconds)
%times at which the results is extracted
n=1; %time-steps per hour n=1 and n=10
ti=[0:3600/n:3600*36];
%% alpha=0
ip0=[3600/n 3600*36 0 [length(ti) ndof ti Dof']];
[Tsnap0,D0,V0]=step1(K,C,a0,ip0,f,bc);
%Extract element temperatures along bc3 after 1, 4, 8
and 16 hours
T10 = Tsnap0(bc3, n);
T20 = Tsnap0(bc3, 2*n);
T40 = Tsnap0(bc3, 4*n);
T80 = Tsnap0(bc3, 8*n);
T120= Tsnap0(bc3, 12*n);
T160 = Tsnap0(bc3, 16*n);
r0=K*D0+C*V0;
r20=r0/(1*b/1000);
%% alpha=0.5
ip5=[3600/n 3600*36 0.5 [length(ti) ndof ti Dof']];
and 16 hours
T25 = Tsnap5(bc3, 2*n);
T45 = Tsnap5(bc3, 4*n);
T85 = Tsnap5(bc3, 8*n);
T125= Tsnap5(bc3, 12*n);
T165 = Tsnap5(bc3, 16*n);

24
r5=K*D5+C*V5;
r25=r5/(1*b/1000);
%% alpha=1
ip1=[3600/n 3600*36 1 [length(ti) ndof ti Dof']];
and 16 hours
T21 = Tsnap1(bc3, 2*n);
T41 = Tsnap1(bc3, 4*n);
T81 = Tsnap1(bc3, 8*n);
T121= Tsnap1(bc3, 12*n);
T161 = Tsnap1(bc3, 16*n);
r1=K*D1+C*V1;
r21=r1/(1*b/1000);
%%
figure(6)
tt = linspace(1/n,36+1/n,length(ti)); %right scale
hold on
plot(tt,sum(r20(bc2,:)), 'r');
plot(tt,sum(r25(bc2,:)), 'b');
plot(tt,sum(r21(bc2,:)), 'g');
legend('alpha=0','alpha=0.5','alpha=1');
axis([0,37,-100,100]);
xlabel('time in h');
ylabel('power supply in W/m^2');
hold off
figure(7)
hold on
X1=Coord(bc3,1);
plot(X1, T11, 'xr');
plot(X1, T21, 'xg');
plot(X1, T41, 'xb');
plot(X1, T81, 'xm');
plot(X1, T121, 'xc');
plot (X1, T161, 'xk');
plot(X1, T15, 'or');
plot(X1, T25, 'og');
plot(X1, T45, 'ob');
plot(X1, T85, 'om');
plot(X1, T125, 'oc');
plot (X1, T165, 'ok');
plot(X1, T10, '.r');
plot(X1, T20, '.g');
plot(X1, T40, '.b');
plot(X1, T80, '.m');
plot(X1, T120, '.c');
plot (X1, T160, '.k');
legend ('1 hour', '2 hours','4 hours', '8 hours','12
hours', '16 hours');
hold off
xlabel('length in m');
ylabel('temperature in °C');
axis([0,0.2,15,25]);

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