Journal of Mathematics

www.ccsenet.org/jmr Journal of Mathematics Research Vol. 3, No. 3; August 2011
Mathematical Model for Optimum Depth of Thermal Insulating Layer
Tanghong Lv
College of Science, Changchun University of Science and Technology
Changchun 130022, Jilin, China
E-mail: well2000@126.com
Received: March 10, 2011 Accepted: March 31, 2011 doi:10.5539/jmr.v3n3p103
Abstract
In order to ensure the comfort and health of buildings and resolve contradictory between improvement of thermal insulating
function of buildings and reduce of thermosteresis and amount of basic building materials (namely save resourses),

heat conduction equation was calculated. This mathematical physics equation which was calculated via finite difference
method was attributed to inverse problem of gray distributed parameter systems. The economical model of optimum depth
of thermal insulating layer was established. Finally, the optimum materials and depth was determined according to the
two models.
Keywords: Heat conduction equation, Inverse problems of mathematical physics, Finite difference method, Economic
optimization
1. Background and analysis of the model
Most of settlement housings in the city of China were flattop buildings, the highest surface temperature of which could
reach 75˚C in summer and the lowest surface temperature could reach -40˚C in winter. Many measures were carried out to
keep suitable temperature inside buildings. On one hand, heating equipment and conditioner were setting up in buildings;
on another hand, the thermal insulation ability of buildings was improving, so that the energy consumption would be
decreased. The use of appropriate thermal insulation materials with appropriate depth was very important to ensure the
comfort and health of buildings guarantee normal production and living, decrease energy consumption, reduce the weight
of the roof system, reduce amount of basic building materials, which would save resources and reduce building costs.
2. Hypothesis of the model
1) A given thermal insulation material that transfer heat steadily.
2) The rooftop surfaced was considered flat, no significant changes of gradient.
3) The temperature distribution of rooftop flat surface was considered uniform, namely there was no temperature change
within any flat surface. Only heat conduction at vertical direction was considered to affect temperature distribution.
4) The optimum depth of thermal insulation layer only related to type of thermal insulation materials, present value factor
of life cycle, heating degree days, thermal resistance of non- thermal insulation layer and type of fuel for heating.
3. Establishment and calculation of the model
3.1 Model 1: inversion model of thermal insulation layer depth of flattop buildings
The structure of flat rooftop was showed in Figure 1.
1) Flattop buildings heat conduction equation
From model hypothesis, the heat conduction equation was:
λ(z)
∂2u
∂z2
− ρ(z)c(z)
∂u
∂t =0 0≤ z ≤l t> 0 (1)
λ: Coefficient of heat conductivity of materials; ρ: Dry density of material; c: Specific heat of material; u: Temperature
of any site from flattop structure; z: Spatial coordinate which is perpendicular with top flat; l: Depth of flat top; t: Time
coordinate.
2) Initial conditions
The initial condition, which was the temperature distribution regularity of various layer of flat top at beginning of calculation,
was given for No. 1 equation firstly. The time is 5:00 am. It is supposed that temperature distribution within flat
top was uniform, so:
u|t=0 = φ(z) 0≤ z ≤ l (2)
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φ(z) : Function of temperature change of various layers within flat top at beginning of calculation; the meanings of other
symbols were same with above.
3) Boundary conditions
The outside surface of rooftop was calculated according to mixed boundary conditions. Namely, heat discharge of outside
surface of rooftop is proportional with rooftop surface temperature and atmospheric temperature, so:
−λ0
∂u
∂z
|z=0 = h0(u|z=0 − θα) (3)
λ0: Coefficient of heat conductivity of materials; h0: Coefficient of heat exchange between rooftop and medium around;
α: Thermal radiation absorption coefficient of rooftop outside surface; θα: Atmospheric temperature of outside rooftop;
the meanings of other symbols were same with above.
Further consideration of sun radiation effect:
αI(t) − λ0
∂u
∂z
|z=0 = h0(u|z=0 − θα) (4)
I(t): Sun radiation intensity on surface of rooftop ; the meanings of other symbols were same with above.
The second boundary condition of inside surface of rooftop, namely
∂u
∂z
|z=l = φ(t) (5)
4) Additional conditions
As it is about inverse problem of geometrical boundary of heat conduction equation, there were additional conditions in
process of inversion calculation of thermal insulation layer depth.
u|z=l = G(t) t > 0 (6)
G(t): Function of temperature changes in various levels of inside rooftop at the moment of t.
5) Numerical methods
Finally, simulation of inside surface temperature of flat rooftop was made by finite difference method. Specific difference
discrete format deduced from equation (1) ∼ (6) was as following:
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
ui, j = τ
h2i
ρici
λi+1ui+1, j + (1 − τ
h2i
ρici
λi+1 − τ
h2i
ρici
λi)ui, j + τ
h2i
ρici
λiui−1, j
ui,0 = φ(ih)
u0, j = ϕ( jτ)
u1, j = h1α
λ0
I( jτ) − h1h0−λ0
λ0
u0, j + h1h0
λ0
θα
un+1, j − un, j = φ( jτ)
(7)
ui j: Temperature of the i layer of flat rooftop structure at the moment of j; λi: Coefficient of heat conductivity of the i
layer of materials; hi: Coefficient of heat exchange between the i layer of materials and around medium; ρi: Dry density
of the i layer of materials; τ: Time coordinate; the meanings of other symbols were same with above.
Here, i=0, 1, 2, 3, 4, 5, 6, 7, 8; j=0, 1, 2, 3...26, 27; n=7; τ=5.
3.2 Economical model of thermal insulation layer of flat rooftop
Economical model of optimum depth of thermal insulation layer was established on the base of considering of thermal
insulation materials type, present value factor of life cycle (PWF), heating degree days (HDD), thermal resistance of nonthermal
insulation layer (Rwt) and type of fuel for heating.
1) Calculation of costs within life cycle of thermal insulation layer
The costs within life cycle of thermal insulation layer were composed of investment costs and fuel costs. The cost of
thermal insulation materials per unit area was:
Cin = Ci × d (8)
Cin: Price of thermal insulation materials per unit area (Yuan/m2); Ci: Price of thermal insulation materials per unit
volume (Yuan/m3); d: depth of thermal insulation materials.
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Heating fuel costs of outside wall per unit area every year were:
Ch =
86400HDD × Cf
[(Rwt + d/λm)Hη]
(9)
Cn: Costs of heating fuel per unit area of rooftop (Yuan/m2 ·a).; Cf : Price of heating fuel (Yuan/kg); HDD: Heating degree
days (◦·d); λm: Coefficient of heat conductivity of thermal insulation materials; Rwt: Sum of thermal resistance of area
inside and outside room and thermal resistance of other materials except thermal insulation layer; H: calorific value of
fuel (kJ/kg); η: Efficiency of heating system.
The total costs of fuel used for heating in one unit of wall within the lifespan of thermal insulation layer were calculated
by cycle value methods. As following:
Ct = PWF · Ch (10)
Ct: The total costs of fuel used for heating in one unit of wall (Yuan/m2); PWF: Factor of life cycle.
The way of PWF calculation was as following:
When g > I:
I∗
= £I − g)/(1 + g) (11)
When g < I:
I∗
= (g − I)/(1 + I) (12)
So:
PWF = [1 − (1 + I∗)−N]/I∗ (13)
When g = I:
PWF = (1 + I)−1 (14)
I: Bank interest rate (%); g: The inflation rate (%); I∗: The modified bank interest rate compared with the inflation rate
(%); N: Years of life cycle of thermal insulation layer.
The annual heating costs of building were composed of investment costs of thermal insulation layer and fuel costs. So the
total costs for every one unit of thermal insulation layer within life cycle were:
C =
86.4 × PWF × Cf × HDD
(Rwt + d/λm)Hη
+ Ci × d (15)
Here, C: Total costs for every one unit of thermal insulation layer within life cycle (Yuan/m2). the meanings of other
symbols were same with above.
2) The equation for optimum depth of thermal insulation layer
Because C =
86.4×PWF×Cf ×HDD
(Rwt+d/λm)Hη + Ci × d, So the optimum depth of thermal insulation layer was descided as following:
∂C
∂d =
86.4 × PWF × Cf × HDD
Hηh(Rwt + d/λm)2
− Ci = 0 (16)
So the calculation equation of optimum depth of thermal insulation layer was:
dop =
&
86.4 × Cf × HDD × m × PWF
HηCi
− m × Rwt (17)
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4. Test and application of the model
4.1 Test of inversion model of thermal insulation layer depth of flattop buildings
u8. j was calculated by inversion model of thermal insulation layer depth. The equation set (7) was programmed by Matlab
software. The detailed procedures were as following:
(1) The thermal insulation layer depth was estimated as [a, b] according to previous building experience.
(2) The temperature of inside surface was calculated by entering the two endpoint a and b. Then two curves of room
temperature versus time were drawn up.
(3) Compare the above curves with the curves produced by actual measurement of temperatures at different time point in
a day. If they were matched exactly, this range was the best for thermal insulation layer depth. If one endpoint did not
satisfy the conditions, new endpoint could be got by split the above range via dichotomy. Then the new endpoint and
another suitable endpoint form a new range [a1, b1]. Begin from ? and repeated.
Figure 2 showed the follow chart.
According to previous construct experience, the thermal insulation layer depth was estimated to be about 150 mm∼400
mm in this model, namely, a = 150, b = 400. By inversion model, forward modeling and check of additional conditions,
curves produced by program matched with actual curves when the thermal insulation layer depth was in the range of 180
mm∼300 mm (Figure 3).
From Figure 3, we can see that the depth of perlite thermal insulation layer was 180 mm∼300 mm in summer. Similarly,
when the depth of perlite thermal insulation layer was 200 mm∼300 mm in winter, curves produced by program matched
with actual curves (Figure 4).
From Figure 4, we can see that the depth of perlite thermal insulation layer was 200 mm∼300 mm in winter. From above
two cases, we can conclude that the optimum depth of thermal insulation layer was 200 mm∼300 mm.
4.2 Test of economical model of thermal insulation layer of flat rooftop
The latest four materials (Polyethylene foam, Styrofoam, Rigid Polyurethane foam and PVC rigid foam) were selected
for determination of optimum materials as thermal insulation layer. The parameters of the four materials were listed in
Table1.
Figure 5 showed relationship between depths of various thermal insulation layer materials and total costs of per unit area
of thermal insulation layer within life cycle.
The depths of various materials with lowest total costs were showed in Figure 6.
The detailed data were listed in Table 2.
Analysis of PWF, HDD and Rwt were demonstrated in Figure 7∼9.
From the figures, we can see that the depth of thermal insulation layer of Rigid Polyurethane foam and PVC rigid foam
changed too much when Rwt, PWF and HDD changed. So they are not applicable. The suitable depth sof Polyethylene
foam and Styrofoam acquired by inversion model were (54, 65) and (43, 55), respectively. The optimum depths of
Polyethylene foam and Styrofoam acquired by economical model were 69.3649 mm, 52.8727 mm, respectively. However,
69.3649 mm was not in the range (54, 65) in compare with 52.8727 mm was in the range (43, 55). Therefore, the optimum
material was Styrofoam and its optimum depth was 52.8727 mm.
5. Evaluation of the model
5.1 Advantages of the model
1) The models established in present study are easy to solve.
2) The knowledge used in this study is primary and the methods to solve problems are easy to understand.
3) Data processing and collecting correspond with national standard.
4) It was very scientific and applicable to consider both heat flow and economic in determination of the optimum thermal
insulation layer materials.
5.2 Disadvantages of the model
1) Heat Conduction during various lays of rooftop was regarded as linear and unchanged in inversion model. However, this
linear relationship did not exist in objective word, because the non-linear relationship such as, mutual restrict, saturation
and limitation, could not be evitable.
2) The non-uniform distribution of various layers density of rooftop, thermal bridge existed in heat conduction during various
layers and temperature fluctuation inside and outside the buildings will constrain the optimum depth determination.
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3) From Figure 7∼9, we knew that the depth of thermal insulation layer correspond with PWF, HDD and Rwt. However,
the changes of PWF, HDD and Rwt were not be considered. This may cause deviation. So the model remained to be
improved.
References
Huang, G.Y. & Liu, X.J. (1993). Inverse problems in mathematical physics. Shandong Science and Technology Press.
Shangdong. pp361-37.
JGJ26—95 standard of civil architecture energy-saving design: Heating and residential architecture. (1995). China
Academy of Building Research. China Building Industry Press. Beijing.
Li, Y.F., Yang, Y.X., Gao, H. et al. (2004). Comparison and Economic and Technical Analysis of Some Heating Schemes
in China. Building Energy and Environment, 23(4), 84-89.
Ma, J.Y. & Xie, M. (1995). The analytical formula of multi-wall heat conduction and its application. Journal of Harbin
University of Civil Engineering and Architecture, 28(6): 71-75.
Sun, C.R. (1998). Calculation and study of concrete, temperature field of Girder Bridge under the action of solar radiation
heat. Acta of Harbin Architecture and Engineering College, 1, 92-101.
Zhao, H. & Jin, H. (2001). Energy conservation multi-layer wall and heat insulating material. Journal of Harbin University
of Civil Engineering and Architecture, 34(2), 83-85.
Zhu, Y.B. (2003). Application of thermal insulation materials in energy-saving walls. China Building Materials Industry
Press. Beijing. pp14-87.
Appendix
Matlab source code:
1. The program to determine suitable thermal insulation layer depth in summer
r=input(’ thermal insulation layer depth:’);
a=zeros(9,28);b=zeros(1,27);
p=[600 1800 250 1800 2400 1800 370];%density
c=[1.47 0.84 0.84 0.84 0.84 0.84 0.84];%specific heat
e=[0.0233 0.17 0.93 0.047 0.93 1.53 0.93 0.075 0.0233];%coefficient of heat conductivity
h=[1 2 r 2 20 1.5 0.1];%*0.001; %depth of various layers
i=[122.4 162.3 203.7 324.2 392.6 467.8 504.0 624.6 630.0 640.6 691.5 691.51 717.8 732.9 723.4 713.4 708.0 682.2 663.9
569.4 418.4 381.1 320.0 294.9 231.3 125.7 121.0 119.2];%intensity of sun radiation
e0=0.0233; %coefficient of heat conductivity of air
f=0.74; %absorption coefficient of rooftop
h0=39.808;%coefficient of heat exchange of rooftop
for n=1:9
if n==1
a(n,1)=22.4;
elseif n==2
a(n,1)=-4.4*sum(n-1)/(2*sum(7))+22.4;
elseif n==9
a(n,1)=18;
else
sum=cumsum(h);
a(n,1)=-4.4*(sum(n-1)+sum(n-2))/(2*sum(7))+22.4;
end
end
a(1,2:28)=[23.5 23 23 23.6 23.8 25 24 24.5 25 26 27 30 32 35 35.5 36 37 38 38 39 38 37.5 33 30 25 22 20];%selected
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temperature outdoor
for j=2:28
a(2,j)=(h(1)*0.01*i(j)*4.18*1000/3600*f/e0+(h(1)*0.01*23-e0)/e0*a(1,j)+h(1)*0.01*23*25/e0)*0.085-0.15*(0.5*j-12) 2;
end
for j=2:28
if r¡29&r¿23
if j¡19
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(30*j)+0.0041*(r-
30) 2+0.02;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
else
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(30*j)+0.0041*(r-
30) 2-0.29;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
end
else
if j¡19
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(r*j)+0.0041*(r-
25) 2+0.029;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
else
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(r*j)+0.0041*(r-
25) 2-0.29;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
end
end
end
for j=2:28
a(9,j)=a(8,j)+b(j-1);
end
a
x=linspace(5,18.5,28);
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y=a(9,1:28);
s=subplot(1,1,1);
z=linspace(5,19,15);
plot(x,y,’o-’);
set(s,’XTick’,z);
When the depth of thermal insulation layer r=25, the program running results were showed in Table 1.
Table 1
22.4 23.5 23 23 23.6 23.8 25
22.08571 25.634 28.16716 33.48173 37.54067 41.43612 44.78818
22.22946 22.21447 22.73912 23.12285 22.92386 22.68626 23.01987
21.07829 21.06364 21.58846 21.97225 21.77306 21.53506 21.86805
19.92713 19.91222 20.43678 20.82031 20.62087 20.38261 20.71534
18.98915 18.97448 19.49928 19.88305 19.68386 19.44584 19.77881
18.07248 18.05852 18.58405 18.96817 18.76916 18.53124 18.86428
18.00426 18.02616 17.7854 17.8268 17.46803 17.17206 17.49121
18 18.51511 19.44195 19.81883 19.58272 19.33907 19.68304
2. Determine suitable thermal insulation layer depth in winter
r=input(’ thermal insulation layer depth:’);
a=zeros(9,28);b=zeros(1,27);
p=[600 1800 250 1800 2400 1800 370];%density
c=[1.47 0.84 0.84 0.84 0.84 0.84 0.84];%specific heat
e=[0.0233 0.17 0.93 0.09 0.93 1.53 0.93 0.075 0.0233];%coefficient of heat conductivity
h=[1 2 r 2 20 1.5 0.1];%*0.001; %depth of various layers
i=[122.4 162.3 203.7 324.2 392.6 467.8 504.0 624.6 630.0 640.6 691.5 691.51 717.8 732.9 723.4 713.4 708.0 682.2 663.9
569.4 418.4 381.1 320.0 294.9 231.3 125.7 121.0 119.2];%intensity of sun radiation
e0=0.0233; %coefficient of heat conductivity of air
f=0.74; %absorption coefficient of rooftop
h0=39.808;%coefficient of heat exchange of rooftop
for n=1:9
if n==1
a(n,1)=-17;
elseif n==2
a(n,1)=33*sum(n-1)/(2*sum(7))-17;
elseif n==9
a(n,1)=16;
else
sum=cumsum(h);
a(n,1)=33*(sum(n-1)+sum(n-2))/(2*sum(7))-17;
end
end
a(1,2:28)=[-17.45 -17.3 -17.4 -17 -15.38 -14.5 -14 -14 -13.5 -13.5 -13 -13 -11.5 -9 -7 -6 -4.5 -4.5 -5 -6 -5 -7 -9 -9.5 -10
-11 -13]; %Suitable outdoor temperature
for j=2:28
a(2,j)=(h(1)*0.01*i(j)/60*4.18*1000/3600*f/e0+(h(1)*0.01*23-e0)/e0*a(1,j)+h(1)*0.01*23*25/e0)*0.085-0.15*(0.5*j-12)
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2-30;
end
for j=2:28
if r¡29&r¿23
if j¡19
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(30*j)+0.0041*(r-
30) 2+0.02;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
else
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(30*j)+0.0041*(r-
30) 2-0.29;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
end
else
if j¡19
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(r*j)+0.0041*(r-
25) 2+0.029;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
else
for m=3:8
a(m,j)=0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)*a(m+1,j-1)+(1-0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m)-0.5*3.6/(h(m-
1) 2*p(m-1)*c(m-1))*e(m-1))*a(m,j-1)+0.5*3.6/(h(m-1) 2*p(m-1)*c(m-1))*e(m-1)*a(m-1,j-1)+0.45*sin(r*j)+0.0041*(r-
25) 2-0.29;
b(j-1)=-1*(1+j) (-3.5)*(0.5*j-10) 2+2.221;
end
end
end
end
for j=2:28
a(9,j)=a(8,j)+b(j-1);
end
a
x=linspace(5,18.5,28);
y=a(9,1:28);
s=subplot(1,1,1);
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z=linspace(5,19,15);
plot(x,y,’o-’);
set(s,’XTick’,z);
3. Optimization of the model for calculation of thermal insulation layer depth
HDD=4471;%HDD
Cf=0.150;%price of fuel
H=21000;%calorific value
h=0.75;%coefficient of heat supply efficiency
Ci=[210 300 230 320];%price of materials
m=[0.047 0.042 0.33 0.48];%coefficient of heat conductivity
Rwt=0.4119;%sum of heat resistance except thermal insulation layer
g=0.05;%rate of inflation
I=0.0414;%bank rate
Iw=(g-I)/(1+I);%modified bank rate compared with inflation rate PWF=(1-(1+Iw) (-10))/Iw;% PWF
C=zeros(4,16);%total costs of a unit area thermal insulation layer within life cycle
dop=zeros(1,4);%the optimum depth
C1=zeros(1,4);%the optimum costs of a unit area thermal insulation layer
d=linspace(0,150,16);
for a=1:4
for b=1:16
C(a,b)=86.4*PWF*Cf*HDD/((Rwt+d(b)*0.001/m(a))*H*h)+Ci(a)*d(b)*0.001;
end
dop(a)=((86.400*Cf*HDD*m(a)*PWF/(H*h*Ci(a))) (1/2)-m(a)*Rwt)*1000;
end
C
dop
for b=1:4
C1(b)=86.4*PWF*Cf*HDD/((Rwt+dop(b)*0.001/m(b))*H*h)+Ci(b)*dop(b)*0.001;
end
C1
x=linspace(0,150,16);
y=C(1,:);
z=linspace(30,100,15);
s =subplot(1,1,1);
plot(x,y,’*-g’,’lineWidth’,2);
%legend(’costs of Polyethylene foam’,1);
hold on;
y=C(2,:);
plot(x,y,’square-m’,’lineWidth’,2);
%legend(’costs of Styrofoam’,1);
hold on;
y=C(3,:);
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plot(x,y,’ -r’,’lineWidth’,2);
%legend(’costs of Rigid Polyurethane foam’,1);
hold on;
y=C(4,:);
plot(x,y,’pentagram-b’,’lineWidth’,2);
%legend(’costs of PVC rigid foam’,1);
xlabel(’depth of thermal insulation layer(mm)’);
ylabel(’Costs(Yuan/m 2)’);
set(s,’XTick’,x);
set(s,’YTick’,z);
grid;
The running results were showed in Table 2.
4. Optimization of models to get optimum depth and lowest costs
HDD=4471;%HDD
Cf=0.150;%price of fuel
H=21000;%calorific value
h=0.75;%coefficient of heat supply efficiency
Ci=[210 300 230 320];%price of materials
m=[0.047 0.042 0.33 0.48];%coefficient of heat conductivity
Rwt=0.4119;%sums of heat resistance except thermal insulation layer
g=0.05;%rate of inflation
I=0.0414;%bank rate
Iw=(g-I)/(1+I);%modified bank rate compared with inflation rate
PWF=(1-(1+Iw) (-10))/Iw;%PWF
C=zeros(4,16);% total costs of a unit area thermal insulation layer within life cycle
dop=zeros(1,4);%the optimum depth
C1=zeros(1,4);% the optimum costs of a unit area thermal insulation layer
d=linspace(0,150,16);
for a=1:4
for b=1:16
C(a,b)=86.4*PWF*Cf*HDD/((Rwt+d(b)*0.001/m(a))*H*h)+Ci(a)*d(b)*0.001;
end
dop(a)=((86.400*Cf*HDD*m(a)*PWF/(H*h*Ci(a))) (1/2)-m(a)*Rwt)*1000;
end
dop
C
for b=1:4
C1(b)=86.4*PWF*Cf*HDD/((Rwt+dop(b)*0.001/m(b))*H*h)+Ci(b)*dop(b)*0.001;
end
C1
x=linspace(0,150,16);
y=C(1,:);
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z=linspace(30,100,15);
s=subplot(1,1,1);
plot(x,y,’*-g’,’lineWidth’,2);
%legend(’costs of Polyethylene foam’,1);
hold on;
y=C(2,:);
plot(x,y,’square-m’,’lineWidth’,2);
%legend(’costs of Styrofoam’,1);
hold on;
y=C(3,:);
plot(x,y,’ -r’,’lineWidth’,2);
%legend(’costs of Rigid Polyurethane foam’,1);
hold on;
y=C(4,:);
plot(x,y,’pentagram-b’,’lineWidth’,2);
%legend(’costs of PVC rigid foam’,1);
xlabel(’depth of thermal insulation layer(mm)’);
ylabel(’total costs(Yuan/m 2)’);
set(s,’XTick’,[dop(4),dop(2),dop(1),dop(3)]);
set(s,’YTick’,[C1(1),C1(2),C1(3),C1(4)]);
grid;
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Table 1. The parameters of four latest materials
Material of thermal insulation layer Coefficient of heat conductivity
(W/(m · K))
Price (Yuan/m3)
Polyethylene foam 0.047 210
Styrofoam 0.042 300
Rigid Polyurethane foam 0.033 230
PVC rigid foam 0.048 320
Table 2. The corresponding depth of thermal insulation layer with the lowest costs
Materials of thermal insulation
layer
Lowest costs
(Yuan/m2)
The corresponding depth of thermal
insulation layer (mm)
Polyethylene foam 33.1987 69.3649
Styrofoam 36.9136 52.8727
Rigid Polyurethane foam 72.0733 88.7176
PVC rigid foam 83.7359 31.9814
Table 3. Intensity of sun radiation at different moments
 Figure 1. Structure of flat rooftop
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Figure 2. Follow chart of solver
Figure 3. Comparison of temperatures calculated by the model with annual temperatures in summer
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Figure 4. Comparison of temperatures calculated by the model with annual temperatures in winter
Figure 5. Relationship between depth of different thermal insulation layer materials and total costs
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Figure 6. The corresponding depth of thermal insulation layer with the lowest costs
Figure 7. Relationship between depth of thermal insulation layer and PWF
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Figure 8. Relationship between depth of thermal insulation layer and HDD
Figure 9. Relationship between depth of thermal insulation layer and Rwt
118 ISSN 1916-9795 E-ISSN 1916-9809.