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ChE 516 – Numerical Methods in Chemical Engineering Fall 2011 Project

Subject: Numerical Solutions to Energy Balance in Trombe Walls

Submitted to Dr. Shuguang Deng

In partial fulfillment of course ChE 516

By XXXXXXXX

December 8, 2011

Table of Contents  Abstract

 Introduction

 Problem

 Assumptions

 Governing Equations

 Boundary and Initial Conditions

 Method of Solution

 MATLAB Program

 Graphs and Observations

 Conclusion

 References

Abstract Trombe walls are commonly used on south sides of homes to save solar energy

for transferring saved energy to the house. In this project a house in Las Cruces has

been considered whose south wall consists of a Trombe for heating the house using

solar energy. For calculating the temperature distribution along the thickness of the

wall, heat conduction in one dimension has been considered. Then, a numerical

technique for partial differential equations with finite difference scheme was

implemented. Finally, using MATLAB programming, the temperature distribution

along the thickness of the Trombe wall after 12, 24, 36, and 48 hours have been

calculated and the related plots are obtained using boundary and initial conditions.

In addition, the net amount of heat transferred to the house from the Trombe wall

during the first day and the second day have been determined.

Introduction

Since ancient times, people have used thick walls of adobe or stone to trap the

sun's heat during the day and release it slowly and evenly at night to heat their

buildings. Today's low-energy buildings often improve on this ancient technique by

incorporating a thermal storage and delivery system called a Trombe wall. Named

after French inventor Felix Trombe in the late 1950s, the Trombe wall continues to

serve as an effective feature of passive solar design. Trombe walls are dark painted

thick masonry walls which are commonly used on south sides of passive solar homes

to absorb solar energy, store it during the day, and release it to the house during the

night, as shown in Figure 1. The idea was proposed by E. L. Morse of Massachusetts

in 1881 and is named after Professor Felix Trombe of France, who used it extensively

in his designs in the 1970s.

Trombe walls have been integrated into the envelope of a recently completed

Visitor Center at Zion National Park and a site entrance building (SEB) at the

National Renewable Energy Laboratory’s (NREL’s) National Wind Technology

Center. The High Performance Building Initiative (HPBi) at NREL helped to design

these commercial buildings to minimize energy consumption, using Trombe walls as

an integral part of their design. A typical unvented Trombe wall consists of a 4- to

16-in (10- to 41-cm)-thick, south facing masonry wall with a dark, heat-absorbing

material on the exterior surface and faced with a single or double layer of glass. The

glass is placed from ¾ to 2 in. (2 to 5 cm) from the masonry wall to create a small

airspace. Heat from sunlight passing through the glass is absorbed by the dark

surface, stored in the wall, and conducted slowly inward through the masonry. High

transmission glass maximizes solar gains to the masonry wall. As an architectural

detail, patterned glass can limit the exterior visibility of the dark concrete wall

without sacrificing transmissivity.

Applying a selective surface to a Trombe wall improves its performance by

reducing the amount of infrared energy radiated back through the glass. The selective

surface consists of a sheet of metal foil glued to the outside surface of the wall. It

absorbs almost all the radiation in the visible portion of the solar spectrum and emits

very little in the infrared range. High absorbency turns the light into heat at the wall's

surface, and low emittance prevents the heat from radiating back towards the glass.

For an 8-in-thick (20-cm) Trombe wall, heat will take about 8 to 10 hours to reach

the interior of the building. This means that rooms receive slowly; even heating for

many hours after the sun sets, greatly reducing the need for conventional heating.

Rooms heated by a Trombe wall often feel more comfortable than those heated by

forced air because of the large warm surface providing radiant comfort. Architects

can use Trombe walls in conjunction with windows, eaves, and other building design

elements to balance solar heat delivery. Strategically placed windows allow the sun's

heat and light to enter a building during the day to help heat the building with direct

solar gains. At the same time, the Trombe wall absorbs and stores heat for evening

use.

Properly sized roof overhangs shade the Trombe wall during the summer when

the sun is high in the sky. Shading the Trombe wall can prevent the wall from getting

hot during the time of the year when the heat is not needed. These Trombe wall

design concepts were applied to the low-energy design of the Visitor Center at Zion

National Park in Utah and to NREL’s Wind Site SEB in Colorado. Figure 2 shows

the Trombe wall locations in the NREL SEB (a), and the Zion Visitor Center (b).

a) NREL SEB b) Zion Visitor Center

Figure 2. Trombe wall design, (Torcellini and Pless, 2004).

Trombe Wall Energy Performance

The energy performance of the Zion Visitor Center was monitored and analyzed

over a two-year period. The analysis consisted of measured electrical end uses,

Trombe wall temperature profiles, and thermographic pictures to determine the

performance of this Trombe wall (Torcellini, 2004). Similar measurements were

taken at the SEB over a one-year period.

Figure 3 shows the thermal distribution of the Zion Trombe wall at 8:30 p.m. on

December 16, 2000. The interior surface temperature is generally homogeneous,

ranging from 90-96ºF (32-36ºC). The wall temperature typically peaks between 8-9

p.m. The reduced wall temperature at the far right section of Trombe wall is due to

shading. The building shades a portion of the Trombe wall in the afternoon, resulting

in reduced interior temperatures (Torcellini and Pless, 2004).

Figure 3. Infrared pictures of a) Zion Trombe wall December 16, 8:30 p.m. and b)

NREL SEB Trombe Wall January 21, 8:00 p.m. (Torcellini and Pless, 2004).

Problem

In this project the problem is defined as follows:

A house in Las Cruces, New Mexico, whose south wall consists of a 1-ft-thick

Trombe wall whose thermal conductivity is K = 0.40 Btu/h ft. °F and whose thermal

diffusivity is α= 4.78 * 10^6 (ft2/s) is considered. The variation of the ambient

temperature, T_out and the solar heat flux q′′ (Btu/h ft2) incident on a south-facing

vertical surface throughout the day for a typical day in January is given in Table 1 in

3-h intervals. The Trombe wall has single glazing with an absorptivity-transmissivity

product of k= 0.77 (that is, 77 percent of the solar energy incident is absorbed by the

exposed surface of the Trombe wall), and the average combined heat transfer

coefficient for heat loss from the Trombe wall to the ambient is determined to be

h_out = 0.7 Btu/hft2 °F. The interior of the house is maintained at Tin = 70°F at all

times, and the heat transfer coefficient at the interior surface of the Trombe wall is

h_in = 1.8 Btu/h ft2 °F. Also, the vents on the Trombe wall are kept closed, and thus

the only heat transfer between the air in the house and the Trombe wall is through

the interior surface of the wall. Assuming the temperature of the Trombe wall to vary

linearly between 70°F at the interior surface and 30°F at the exterior surface at 7 AM

and using the explicit finite difference method with a uniform nodal spacing of dx =

0.2 ft., determine and plot the temperature distribution along the thickness of the

Trombe wall after 12, 24, 36, and 48 h. Also, determine the net amount of heat

transferred to the house from the Trombe wall during the first day and the second

day. Assume the wall is 10 ft. high and 25 ft. long.

Table 1. The hourly variation of monthly average ambient temperature and solar heat flux

incident on a vertical surface for January in Las Cruces, New Mexico

Assumptions

1- The heat transfer is considered as one dimensional in wall thickness direction.

2- The temperature in the house is constant at 70°F.

3- Based on finite difference scheme, the temperature in every 25 cm interval along

the wall is considered same for that interval.

4- The only heat transfer between the air in the house and the Trombe wall is through

the interior surface of the wall

Governing Equations

The governing equations describing the heat transfer, unsteady, conduction in one

dimension is as follows:

Using energy balance for each of the internal intervals in Figure 4.:

Figure 4. The intervals for writing energy balance

Input-output+generation-consumption=accumulation (1)

𝐾𝐾𝐾𝐾 𝑇𝑇𝑚𝑚+1 𝑝𝑝 −𝑇𝑇𝑚𝑚

𝑝𝑝

∆𝑥𝑥 + 𝐾𝐾𝐾𝐾

𝑇𝑇𝑚𝑚−1 𝑝𝑝 −𝑇𝑇𝑚𝑚

𝑝𝑝

∆𝑥𝑥 = 𝜌𝜌𝐾𝐾∆𝑥𝑥𝑥𝑥 𝑇𝑇𝑚𝑚

𝑝𝑝+1−𝑇𝑇𝑚𝑚 𝑝𝑝

∆𝑡𝑡 (2)

Fo = α∆t/∆x^2 (3)

So by simplifying the equation (2) and using equation (3) it can be written for nods,

2, 3 4 and 5 as interior nods:

Tm p+1 = Fo�Tm−1

p + Tm+1 p � + (1 − 2Fo)Tm

p (4)

Energy balance for node 6 as the external node:

kq′′A + hA(Tin − T6 p + KA

T5 p−T6

∆x = ρA∆x/2c

T6 p+1−T6

∆t (5)

∆x

• • • • • 1

3 4 5

•

Tin

qsola’’

Bi = h∆x K

(6)

Therefore using equation (6), equation (5) can be written in its final form for external

nod, 6:

T6 p+1 = 2Fo �Biout × Tout + T5

p + kq ′′

K ∆x� + (1 − 2Fo − 2BioutFo)T6

p (7)

And using energy balance around nod 1as the internal node equation (8) can derived

as follows:

AK � T2 p−T1

∆x � − hA�T1

p − Tin� = ρA ∆x 2

c T1 p+1−T1

∆t (8)

Then the equation (8) can be rearranged in the form of equation (9):

T1 p+1 = (1 − 2BiinFo − 2Fo)T1

p + 2Fo(T2 p + BiinTin) (9)

Boundary and Initial Condition

During writing equation related to each nod, all boundary conditions are

implemented. Table 1 shows the needed values for solar energy and out temperature

during the day. Also, the inner temperature of home is at 70oF whole the day.

Initial Condition:

At t=7 O’clock as an initial point the node temperatures are as follows:

T1=70

T2=60

T3=50

T2=40

T1=30

As the equation shows the explicit finite difference method is used for finding the

temperature distribution along Trombe wall. So, a MATLAB program is prepared

for solving the mentioned equations.

MATLAB CODE: clear all; clc % One Dimensional Unsteady Heat Transfer Simulation Using Explicit Finite Difference Method % Matlab using to solve numerically a heat transfer % Leila Karimi, New Mexico State University, November 2011. % L -m Length of 1D space % n -"" Number of nodes % del_x -m Spacing of nodes % t0 -Seconds Starting time % tmax -Seconds Ending time % k -W/(m K) Thermal conductivity - conduction ciefficient % alpha -m^2/s Diffusivity % h -W/m^2 K Convection coefficient % del_t -hour Change in time each iteration % T1 -hour Initial time %h_out average combined heat transfer coefficient for heat loss from %the Trombe wall to the ambient is determined to be h_out=0.7 Btu/(h*ft^2F)

%h_in heat transfer coefficient at the interior surface of the %Trombe wall is h_in=1.8 Btu/(h*ft^2) %%%% Trombe Wall Properties %%%%%%%%%%%%% L= input(' Please input the thickness of the wall, (L=1) = '); del_x= input(' Please input the uniform nodal spacing, (del_x=0.2) = '); n=L/del_x+1; %n=6; T1= input(' Please input the internal surface of the wall, (T1(f)=70) = '); %T1=70 T(n)= input(' Please input the external surface of the wall, (T(n)(f)=30) = '); %T(n)=30; %del_x=0.2; k= input(' Please input conductivity coefficient,(K=0.4),= '); %k=0.4; t0= input(' Please input the minimum time (h), for this problem t0(h),(t0=7) = '); %t0=7; tmax = input(' Please input the maximum time (h), for two days=t0+48, (55) = '); %tmax=55; alpha_1 = input('Please input the thermal diffusivity, a(ft^2/s), (?= 4.78 x 10-6 ft2/s) = '); alpha=alpha_1*3600; %L=xmax-xmin; %del_x= L/(n-1); A=250; %%%%%%%% del_t should be less than 0.5*del_x^2/alpha %%%%%%%% del_t = 0.2*del_x^2/alpha ;

T_out=30; T_in=70; h_in= input(' Please input the heat transfer coefficient of internal fluid, h_in (1.8) = '); %h_in=1.8; h_out=input(' Please input the heat transfer coefficient of external fluid, h_out (0.7) = '); %h_out=0.7; Fo=alpha*del_t/(del_x^2); B_in=h_in*del_x/k; B_out=h_out*del_x/k; ii=1 t_total=t0:del_t:tmax; T_out=zeros(ii,6) q=zeros(ii,6) for ii=1:length(t_total) t = t_total(ii); if t0<t & t<(t0+3) T_out(ii,6)=33 ; q(ii,6)=0.77*114; else if (t0+3)<=t & t<(t0+6) T_out(ii,6)=43 ; q(ii,6) =0.77*242; elseif (t0+6)<=t & t<(t0+9) T_out(ii,6)=45 ; q(ii,6) =0.77*178; elseif (t0+9)<=t & t<(t0+12) T_out(ii,6)=37 ; q(ii,6) =0; elseif (t0+12)<=t & t<(t0+15) T_out(ii,6)=32 ; q(ii,6) =0; elseif (t0+15)<=t & t<(t0+18) T_out(ii,6)=27 ; q(ii,6) =0; elseif (t0+18)<=t & t<(t0+21) T_out(ii,6)=26 ; q(ii,6) =0; elseif (t0+21)<=t & t<(t0+24) T_out(ii,6)=25 ; q(ii,6) =0; elseif (t0+24)<=t & t<(t0+27) T_out(ii,6)=33 ; q(ii,6)=0.77*114; else if (t0+27)<=t & t<(t0+30) T_out(ii,6)=43 ; q(ii,6) =0.77*242; elseif (t0+30)<=t & t<(t0+33)

T_out(ii,6)=45 ; q(ii,6) =0.77*178; elseif (t0+33)<=t & t<(t0+36) T_out(ii,6)=37 ; q(ii,6) =0; elseif (t0+36)<=t & t<(t0+39) T_out(ii,6)=32 ; q(ii,6) =0; elseif (t0+39)<=t & t<(t0+42) T_out(ii,6)=27 ; q(ii,6) =0; elseif (t0+42)<=t & t<(t0+45) T_out(ii,6)=26 ; q(ii,6) =0; elseif (t0+45)<=t & t<(t0+48) T_out(ii,6)=25 ; q(ii,6) =0; end end end end T_out; q; for ii=0 for i=1:n T(1,i)=T1-((T1-T(n))/(n-1))*(i-1); end end T t=t_total; for ii=1:length(t) for i = 1:n if i == 1 T(ii+1,i) = (1-2*B_in*Fo- 2*Fo)*T(ii,1)+2*Fo*(T(ii,i+1)+B_in*T_in); elseif i == n

T(ii+1,i) = 2*Fo*(B_out*T_out(ii,6)+T(ii,n- 1)+q(ii,6)*del_x/k)+(1-2*Fo-2*B_out*Fo)*T(ii,i); else T(ii+1,i) = Fo*(T(ii,i-1)+T(ii,i+1))+(1- 2*Fo)*T(ii,i); end end end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Calculating the node temperatures after 12, 24, 36 and 48 hours T for i=1:6 Temp12(1,i)=T(length(t)/4,i); Temp24(1,i)=T(length(t)/2,i); Temp36(1,i)=T(78,i); Temp48(1,i)=T(length(t),i); end Temp12 Temp24 Temp36 Temp48 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Plotting the node temperatures after 12, 24, 36 and 48 hours Temp=[Temp12;Temp24;Temp36;Temp48];

i=[1:1:6]; plot(i,Temp) xlabel('node number ') ylabel('Temperature (.F)') title('Temperature changes for nodes after 12, 24, 36 and 48 hours') legend('Temp12','Temp24','Temp36','Temp48',1) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% % Calculating the net heat transfer to the home %heat1 is for the first and heat2 is for the second day heat1=zeros(1,length(t/2)); M=zeros(1,length(t/2)); for ii=1:length(t)/2 heat1(1,ii)=ii; M(1,ii)=h_in*A*(T(ii,1)-T_in); end heat1=trapz(heat1,M) heat2=0; for ii=length(t)/2:length(t) heat2(1,ii)=ii; M(1,ii)=h_in*A*(T(ii,1)-T_in); end heat2=trapz(heat2,M) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 % Plotting the temperatures distribution in the form of surface for whole % the two days continously along the wall figure(1)

hold on figure(2) colormap hot plot(T) surf(T); figure (gcf) xlabel('i (Node Number (Position through Wall)') ylabel('ii (Time)') zlabel('T (.F)') title('Temperature profile changes within 48 hr') shading interp colorbar

Graphs and Observations

Input and Results:

Please input the thickness of the wall, (L=1) = 1

Please input the uniform nodal spacing, (del_x=0.2) = 0.2

Please input the internal surface of the wall, (T1(f)=70) = 70

Please input the external surface of the wall, (T(n)(f)=30) = 30

Please input conductivity coefficient,(K=0.4),= 0.4

Please input the minimum time (h), for this problem t0(h),(t0=7) = 7

Please input the maximum time (h), for two days=t0+48, (55) = 55

Please input the thermal diffusivity, α(ft^2/s), (α= 4.78 x 10-6 ft2/s) = 4.78e-6

Please input the heat transfer coefficient of internal fluid, h_in (1.8) = 1.8

Please input the heat transfer coefficient of external fluid, h_out (0.7) = 0.7

T =

70 62 54 46 38 30

T =

70.0000 62.0000 54.0000 46.0000 38.0000 30.0000

66.8000 62.0000 54.0000 46.0000 38.0000 29.0000

66.0320 61.3600 54.0000 46.0000 37.8000 50.7160

65.5917 60.8224 53.8720 45.9600 42.0232 60.6254

65.2710 60.3862 53.6797 46.7550 46.5310 66.8729

65.0195 60.0218 53.6361 48.0952 50.6442 71.5500

64.8134 59.7442 53.8050 49.7131 54.3155 75.3467

64.6529 59.5702 54.1745 51.4520 57.6013 78.5617

64.5448 59.5076 54.7091 53.2264 60.5635 102.4669

64.4938 59.5553 55.3723 54.9903 67.4768 114.6482

64.5006 59.7064 56.1325 57.5640 74.4138 123.0169

64.5627 59.9505 57.1336 60.6477 80.7644 129.6413

64.6752 60.3096 58.3998 63.9682 86.5164 135.2287

64.8459 60.8007 59.8954 67.3642 91.7493 140.0998

65.0833 61.4287 61.5702 70.7474 96.5423 134.8576

65.3915 62.1879 63.3774 74.0710 99.0464 134.3634

65.7691 63.0665 65.2782 76.9273 101.1147 135.1377

66.2112 64.0494 67.1657 79.4350 103.0819 136.3213

66.7104 65.1050 68.9963 81.7105 105.0004 137.6525

67.2525 66.2043 70.7609 83.8256 106.8728 139.0323

67.8223 67.3253 72.4625 85.8221 108.6953 140.4160

68.4075 68.4521 74.1070 87.7248 110.4648 113.2495

68.9987 69.5742 75.6996 89.5493 106.4737 101.4607

69.5893 70.6842 77.2444 90.1642 102.0862 94.4414

70.1751 71.7773 78.5163 89.9647 98.1729 89.4575

70.7529 72.8046 79.4582 89.3166 94.7882 85.5996

71.3026 73.7250 80.0992 88.4393 91.8561 82.4711

71.8026 74.5153 80.4924 87.4546 89.2958 79.1592

72.2388 75.1682 80.6894 86.4304 86.9002 76.6115

72.6046 75.6866 80.7334 85.3762 84.7485 74.4814

72.8997 76.0795 80.6526 84.3221 82.8206 72.6408

73.1277 76.3582 80.4719 83.2879 81.0849 71.0230

73.2939 76.5348 80.2123 82.2841 79.5131 69.5846

73.4045 76.6221 79.8912 81.3155 78.0816 68.2942

73.4659 76.6324 79.5222 80.3839 76.7709 66.4280

73.4848 76.5771 79.1166 79.4890 75.4249 65.0452

73.4672 76.4665 78.6832 78.6017 74.1618 63.8708

73.4187 76.3100 78.2235 77.7300 72.9916 62.8253

73.3445 76.1144 77.7421 76.8810 71.9060 61.8762

73.2485 75.8860 77.2444 76.0582 70.8950 61.0055

73.1340 75.6302 76.7355 75.2628 69.9498 60.0605

73.0042 75.3520 76.2199 74.4947 69.0345 59.2478

72.8618 75.0560 75.7013 73.7477 68.1692 58.5078

72.7092 74.7462 75.1815 73.0227 67.3526 57.8213

72.5487 74.4259 74.6627 72.3205 66.5804 57.1788

72.3820 74.0978 74.1469 71.6409 65.8481 56.5744

72.2108 73.7645 73.6359 70.9835 65.1519 56.0035

72.0364 73.4280 73.1311 70.3477 64.4886 55.3224

71.8599 73.0903 72.6338 69.7325 63.8271 54.7437

71.6825 72.7529 72.1449 69.1317 63.1915 54.2130

71.5050 72.4172 71.6639 68.5463 62.5839 53.7146

71.3281 72.0841 71.1910 67.9773 62.0025 53.2422

71.1524 71.7543 70.7269 67.4251 61.4454 52.7924

70.9783 71.4284 70.2720 66.8895 60.9108 71.0387

70.8062 71.1071 69.8268 66.3703 64.1321 79.2181

70.6363 70.7909 69.3916 66.6139 67.5969 84.2692

70.4691 70.4801 69.1159 67.3661 70.7348 87.9786

70.3046 70.2050 69.0388 68.3898 73.5098 90.9401

70.1551 69.9917 69.1422 69.5436 75.9718 93.4123

70.0339 69.8545 69.3924 70.7490 78.1743 95.5344

69.9499 69.7980 69.7561 71.9627 80.1612 118.5035

69.9072 69.8200 70.2058 73.1611 86.1900 129.8641

69.9057 69.9146 70.7197 75.1758 92.3190 137.5015

69.9432 70.0738 71.4499 77.7132 97.9269 143.4663

70.0159 70.3229 72.4274 80.5033 102.9920 148.4533

70.1330 70.6824 73.6217 83.3859 107.5865 152.7733

70.3049 71.1604 74.9867 86.2732 111.7838 147.0223

70.5373 71.7545 76.4787 89.1180 113.7294 146.0558

70.8308 72.4559 78.0617 91.5124 115.2724 146.3894

71.1818 73.2521 79.6307 93.5743 116.7438 147.1601

71.5844 74.1137 81.1437 95.4194 118.1931 148.1031

72.0258 75.0139 82.5928 97.1190 119.6204 149.1167

72.4917 75.9320 83.9823 98.7141 121.0194 121.6218

72.9708 76.8540 85.3186 100.2288 116.6788 109.5338

73.4546 77.7703 86.6077 100.5367 111.9598 102.2371

73.9372 78.6746 87.6260 100.0355 107.7306 96.9930

74.4148 79.5174 88.3177 99.0927 104.0441 92.8890

74.8665 80.2570 88.7126 97.9279 100.8228 89.5266

75.2707 80.8700 88.8646 96.6639 97.9846 86.6913

75.6130 81.3491 88.8255 95.3681 95.4618 83.5519

75.8867 81.6971 88.6387 94.0783 93.0611 81.0986

76.0917 81.9234 88.3383 92.7870 90.8720 79.0098

76.2314 82.0400 87.9451 91.5143 88.8826 77.1733

76.3115 82.0593 87.4779 90.2741 87.0670 75.5327

76.3385 81.9935 86.9534 89.0734 85.4016 74.0519

76.3186 81.8545 86.3854 87.9151 83.8660 72.0045

76.2583 81.6535 85.7852 86.7993 82.3035 70.4485

76.1634 81.4008 85.1617 85.6973 80.8317 69.1077

76.0395 81.1055 84.5166 84.6171 79.4600 67.9022

75.8917 80.7745 83.8545 83.5656 78.1799 66.7990

75.7238 80.4139 83.1807 82.5462 76.9808 65.7795

75.5393 80.0293 82.5005 81.5600 75.8536 64.8309

75.3411 79.6255 81.8181 80.6068 74.7904 63.8037

75.1321 79.2072 81.1373 79.6858 73.7563 62.9058

74.9146 78.7782 80.4610 78.7902 72.7721 62.0792

74.6908 78.3420 79.7903 77.9208 71.8372 61.3053

74.4626 77.9014 79.1267 77.0779 70.9475 60.5753

74.2316 77.4587 78.4719 76.2616 70.0992 59.8836

73.9991 77.0159 77.8272 75.4712 69.2885 59.0861

73.7661 76.5748 77.1937 74.7059 68.4846 58.3950

73.5338 76.1369 76.5724 73.9592 67.7109 57.7556

73.3029 75.7034 75.9626 73.2322 66.9695 57.1519

73.0740 75.2751 75.3647 72.5257 66.2585 56.5777

72.8478 74.8528 74.7790 71.8401 65.5758 56.0291

72.6246 74.4370 74.2060 71.1750 64.9193 55.5037

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Calculating the node temperatures after 12, 24, 36 and

48 hours

Temp12 =

70.7529 72.8046 79.4582 89.3166 94.7882 85.5996

Temp24 =

71.3281 72.0841 71.1910 67.9773 62.0025 53.2422

Temp36 =

74.8665 80.2570 88.7126 97.9279 100.8228 89.5266

Temp48 =

72.8478 74.8528 74.7790 71.8401 65.5758 56.0291

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Calculating the net heat transfer to the home

%heat1 is for the first and heat2 is for the second day

heat1 =

-2.6487e+004

heat2 =

1.0749e+005

Figure 5. Node temperatures after 12, 24, 36 and 48 hours

Figure 6. Temperatures distribution along the wall during two days

Also, the results were exported to Excel and the temperature distribution along the

wall was plotted in Excel as follows:

Figure 7. Node temperatures after 12, 24, 36 and 48 hours using Excel

Conclusion

The results show that Trombe wall can be used for house heating efficiently. At

the beginning the house may lose some heat due to its lower temperature, but after

a time period the wall starts to work as a heating source and the all absorbed solar

energy during the day releases. According to observations during the first 12 hour

period the wall has absorbed lots of solar energy and its internal parts show energy

storage because of their high temperatures. In addition, the wall will release its

saved energy during the night; so, after 24 hours its temperature decreases. This

100

120

1 2 3 4 5 6 7

After 12 hours-7 PM-First day

After 24 hours-7 AM-Second day

After 36 hours-7 PM-Second day

After 48 hours-7 AM-Third day

operation repeats during the next days. Also, the calculated transferred net heat for

first and second day show that it takes some time the energy to be saved in Trombe

wall and 264870000 Btu heat goes out from the house through the Trombe wall

during 12 hours after beginning; while after 48 hours the 1074900000 BTU is

transferred to the house via Trombe wall.

References

P. Torcellini and S. Pless, “Trombe Walls in Low-Energy Buildings: Practical

Experiences” , 2004 , NREL/CP-550-36277, World Renewable Energy Congress

VIII and Expo Denver, Colorado, August 29–September 3, 2004.

Table of Contents

� Abstract

� Introduction

� Problem

� Assumptions

� Governing Equations

� Boundary and Initial Conditions

� Method of Solution

� MATLAB Program

� Graphs and Observations

� Conclusion

� References