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Theoretical considerations in the use of small passive-solar test-boxes to model the thermal performance of passively solar-heated building designs
So/at Energy. Vol. 22, PP. 343-350 © Pergamon Press Ltd., 1979. Printed in Great Britain
0038-092X/79/O401--0M3/$02.00/O
THEORETICAL CONSIDERATIONS IN THE USE OF SMALL PASSIVE-SOLAR TEST-BOXES TO MODEL THE THERMAL PERFORMANCE OF PASSIVELY SOLAR-HEATED BUILDING DESIGNS? D. P. GRIMMER Los Alamos ScientificLaboratory, Los Alamos, NM 87544, U.S.A. (Received 21 February 1978: revision accepted 3 November 1978)
Abstract--Theoretical considerations are presented for the thermal modellingof passively solar-heated building designs with passive-solartext boxes. Multi-roompassive solar buildings,passive solar buildingshaving realistically massive walls, thermocirculatingpassive-solar designs, air-infiltrationeffects, edge effect corrections, and microclimate shading effects are discussed. t. IN'mODUCrION Thermal modelling of passive-solar building designs with small test boxes could be a valuable tool for the architect, contractor and owner-builder[l]. The subject of dimensional analysis and scale factors in the modelling of physical phenomena, including heat transfer, is discussed extensively in the literature[2, 3]. Such analyses are usually done using dimensionless scalefactors while preserving geometric, kinematic, temporal and thermal similarity. The data gathered from the model must be converted to real-scale values by use of the appropriate scale-factor for these similar (non-distorted) models, The use of non-distorted models is not the subject of this paper, but rather the use of distorted models that present real-scale values for data collected. Specifically, the data gathered with the use of test-ceils consists essentially of ambient (outside) and interior (room) air temperatures as a function of time. It would be desirable for the non-technical experimenter to have a methodology of text-box construction that provides real temperatures and times of thermal waves moving through the building envelope. By using distorted models, the use of scale factors (such as Fourier and Blot numbers) can be avoided, In the following sections somewhat simplistic discussions will be presented on multi-room passive-solar buildings, passive solar buildings having realistically massive walls, thermocirculating passive-solar designs, air infiltration effects, edge-effect corrections and microclimate shading effects, 2. THERMAL MODELLING OF MULTI-ROOMPASSIVESOLAR BUILDINGSUSINGSMALLPA,~IVI~TEST BOXES Small passive test-boxes have been used to model the thermal performance of single-room passive solar building designs[l]. Test-box arrangements can also be constructed to thermally model multi-room passive-solar buildings. Since most real buildings are multi-room, the ability to model them with test-boxes would be quite useful. tWork performed under the auspices of the U.S. Department of Energy. SE VoL 22, No. ~----D
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In a typical multi-room passive-solar building design, some rooms will be solar heated (i.e. have south facing glazings) while others will not. A test-box thermal-model should have the same number of solar-heated rooms in approximately the same directional layout as the actual building design to get accurate modelling. As mentioned previously, the thermal model will not resemble a scaled down model of the actual building, except for the layout and direction of south facing windows, and the topological arrangement of the rooms. It is important to realize that volumetric considerations do not play an important part in thermal modelling: the volume (but not the wall area) of a room can essentially be ignored in heat-loss calculations. Stratification phenomena are related to room volume and can play a part in comfort zones in a room, but for an overall heat-loss calculation (using an average room temperature above ambient temperature) a test-box room should simulate an actual room well. This would be particularly true in cases where infiltration losses are low, and ceiling insulation is high. The procedure for thermal modelling a multi-room passive-solar design is simple. The total thermal conductance (the UA, in Btu/*F or W/m 2) is calculated for each room, with the conductance for exterior walls, ceiling and floor totalled separately from the conductance for walls, ceiling and floor common to other rooms. The thermal capacity is similarly totalled for each room. For those rooms containing solar-glazings, the room thermal conductance and room thermal capacitance are normalized to the area of the glazing, treating interior common walls, ceiling and floor separately as before. Thus, each solarheated room will have the thermal conductance and capacitance of the interior envelope normalized to that of the exterior envelope values (and of course the glazing area). Because the thermal conductance and capacitance of interior envelopes are common to adjoining rooms, the glazing areas of the solar-heated rooms will be in the same ratio for the test-box thermal model as for the real building. In summary then, for the test-box thermal-model, the ratios of the glazing areas for the solar-heated rooms will be the same as the ratios of the glazing areas for the real rooms. Also, the ratios of the thermal conductance and
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capacity of the exterior envelope to that of the interior envelope for each room will be the same for the test box rooms as for the building rooms. In the case of the test-box rooms, the interior and exterior envelope values will both of course be .normalized to the glazing areas of the solar heated rooms. As an example, let us consider the passive-solar building design shown in Fig. 1. Here we have a rather idealized case of essentially massless insulating walls, with the mass of the building contained in two storage walls, of thermal capacity C1 and C2, respectively. Specific thermal conductance UCI is that of a common (interior) wall and is much greater than the other (exterior) U values. The specific conductances U I1 and UC2 would be for storage walls and presumably would have a value comparable to UC1. Because of its mass, however, the thermal impedance of Wall No. 3 (with steady state conductance UC2) would be much greater than the impedance of the other common wall with steady-state specific conductance UCI. The U values for the side walls perpendicular to the plane of the figure are not shown, The UA's are summed for each room. Let (UA)~ be the sum for the ith room exclusive of the common walls, etc., for that room. So for Room No. ! we have that the total conductance is given by (UA), + UCI • W2, where W2 is the area of Wall No. 2. The heat capacity for Room No. 1 is given by CI. The conductance of Room No. 1 normalized to glazing area G I is given by [(UA), + UCI. W2]/GI and the normalized heat capacity CI/GI. Similarly, the normalized conductance of Room No. 2 is given by [ ( U A h + UCI • W2+ UC2. W3]/G2, where W3 is the area of Wall No. 3, and the normalized heat capacity by C2/G2. Finally, the total conductance of Room No. 3 is given by (UAh + UC2. W3 and the heat capacity by C2. If in the same test box rooms the simulated glazing areas GI' and G2' are kept in the same ratio as the real building's glazing areas (i.e. GI'/C2'= G IIG2 = C, where C is some constant), then the thermal conductance and capacitance for the common walls of the test-box rooms will be correctly matched: i.e. UC1 • W2/G1 = UCI'. W2'/GI and UCI • W2/G2 = UCI'. W2'/G2' only if GI'/G2'= GI/G2 (here primes denote test box values),
Double
The thermal model of a room without a solar glazing, such as Room No. 3 in Fig. 1, is quite simple, since volume does not play much of a role in thermal-loss calculations. Such a room can always be considered as an insulating wall for the solar heated room (here Room No. 2), and if the air temperature in Room No. 3 is not of interest, it could be replaced with a solid insulator whose overall series conductance would be given by 1 I U C 2 . A 3 + ~ + ~ where (UA),ir is the equivalent total conductance through the air. If we wish to measure the air temperature in Room No. 3 in this example, we must have at Wall No. 3 that UC2. W31G2 = UC2'. W3'/G2', or that (UC2'. WY)/(UC2. W3)= G2'/G2. That is, the total conductance through the common Wall No. 3 of the box model should be (G2'/G2)of the real total conductance through that wail. As a final note, remember that edge e~:ects must always be considered for small test-boxes. The log mean average of the thermal conductance through the inside and outside surface areas of a wall must be used to determine the correct total conductance for that wall: i.e. Ae~ = (A - A')/In (A/A')[4]. A detailed discussion of edge-effects is presented in Section 6. 3. "I'IU~MALTI[ST.BOXMODELLINGOF PASSIVESOLAR B~GS HAV~G ~ C ~ L V ~4LSSIV[WALLS The example discussed in Section 2 from Fig. I is essentially unrealistic in that (a) the insulating walls are assumed to have no thermal capacitance, and (b) the thermal capacitive reactances to transient temperature changes were not considered specifically for massive Walls Nos. 1 and 3. Instead of specific conductances U11 and UC2, we would have to consider overall thermal impedances involving thermal capacitive reactances. The situation of heat flow through a massive wall can be likened to current flowing through a transmission line. Lumped circuit elements can be used to describe the latter, and analogously the former[S]. To construct an accurate test-box thermal-model of a passive dwelling with massive walls not in direct sun, it
Double -~. Glazing Areu G2
.,L
J,
wo,,2 Wall I, Trombe Wnll
/f Wall 3,
/Woll 4 Direct Gain
Fig. 1. Schematic of multi-room passive-solar building design. The design has a solar-room with a Trombe-wall, a solar-room with direct gain on a storage wall, and a non-solar room. Symbols for thermal conductance and heat capacity are indicated.
Small passive-splat test-boxes is best to consider the test-box as a model of a horizontal "slice" of finite thickness through the solar glazings of the real building. For example, consider that all walls of the building shown in Fig. l are made of masonry, and the floor and ceiling of insulated wood. Then the actual material used for the building walls should be used to construct the test box walls. The thermal conductance and capacitance of each wall is normalized to the glazing area and to each other. The proper normalization will determine the exact dimensions of the box, for some glazing area, Let Room No. 1 in Fig. l have all masonry walls (C=0.2Btu/Ib°F, or 825.7J/kg°C; U=O.SBtu/hrft°F, orO.87W/m°C;p=1351b/ft3or2189kg/m3) thatarelft (0.305 m) thick, and constructed in a rectangular shape 10ft high, by 15ft deep, by 20ft wide (facing south) (3.05mx4.58mx6.1m) outer dimensions, and 10ft high x 13 ft deep x 18 ft wide inner dimensions (3.05 m x 3.97mx 5.49m). Let the buildings ceiling and floor be made of insulated wood (U = 0.027 Btu/hr ft °F or 0.047 W/m °C) 7 in. thick (17.8 cm) and essentially massless. Then, the Trombe wall is l0 ftx 18 ft x 1 ft = 180 ft 3 (5.04m3), or 24,3001b (ll,178kg) of masonry, or 4860 Btu/°F (9.23 x l06 J/°C) of heat capacity; and l0 f t x 18ft= 180ft 2 (16.7 m 2) of collecting area, or 90Btu/hr°F (47.7 W/°C) of total conductance. Similarily, the side and back walls have a total of 440 f13 (12.3 m 3) or 59,400 lb (27,324 kg) of masonry or 11,880 Btu/OF (2.26 x l07 J/°C) of heat capacity; and 440 ft 2 (40.9 m 2) of area of 220Btu/hr°F (ll6.6W/°C) of total conductance. The ceiling and floor each has an area of 234ft 2 (21.76m 2) and a total conductance of 10.6 Btu/hr°F (5.62 W/°C). Note that edge effects have been neglected here for simplicity, Thus for the example room discussed here, normalized to glazing area (180 ft: or 16.7 m2), we have that: (1) for the Trombe Wall, C'=27.0Btu/°Fft~ 2 (5.5x 105J/°Cm~2)and U'=O.50Btu/°Fhrfq2(2.84W/°Cm,:): (2) for the side and back walls (total), C ' = 66.0 Btu/°Fftg 2 (1.35 x 106J/°C m~2) and U'= 1.22 Btu/°F hrftg 2 (6.92 W/°C m,2); (3) for the ceiling and floor (each), U' = 0.059 Btu/°F hr ft82 (0.33 W/°C mg2). If we now construct a passive test box using a glazing, say 2 ft x 2 ft = 4 ftg2 (0.37 m2), then we have in comparing the example room with the test-box that: (i) for the simulated Trombe wall, we use masonry 1 ft t h i c k x 2 f t x 2 f t (0.305mx0.610mx0.610m), of 4ft 3 (0.11 m 3) volume. This has the correct C' and U' values: (2) for the back wall we can use a masonry wall I ft thick x 2 ft x 2 ft (0.305 x 0.610 m x 0.610 m) (to make a rectangular box). This has 27.0 Btu/°Fft~: (5.5 x 105 J/°C m~:) and 0.50 Btu/°F hr ft 2 (2.84 W/°C m,2). The remainder of the C' and U' needed (39 Btu/°F ftg2 or 156 Btu/°F (2.96 x l03 J/°C) for the box, and 0.72 Btu/°F hr ft: or 2.88 Btu/°F hr (i.53 W/°C) for the box) is obtained by adjusting the length of the side walls to approx. 1.44 ft (0.44 m) neglecting edge effects. This will make each side wall have about 2.9ft 3 (0.08 m 3) volume, or about 78 Btu/°F (1.48 x l0 ~ J/°C) heat capacity: (3) for the ceiling and floor we now have from (2)
345
above an area of 2.88 ft 2 (0.27 m2). We need a U of 0.25 Btu/°F hr (0.13 W/°C) for the box ceiling and floor (each). If we are using the same insulation, with 0.27Btu/hr°F (0.14W/°C) then a thickness of 3.1in. (7.9 cm) will give the necessary conductance out the ceiling and floor. If the ceiling and floor were not considered massless it would have been more difficult to adjust thermal parameters to fit. Let us suppose that the floor was instead made of masonry 2 ft (0.61 m) thick; then the floor has 12636 Btu/°F (2.40 x 107 J/°C) of heat capacity and 59Btu/°Fhr (31.3W/°C) of thermal conductance. Hence for this floor, C'=70.2Btu/°Fftg, (1.43x 106H/°Cms 2) and U'=0.325Btu/°Fhrfts, (1.84W/ °Cruz2). It is obvious that a text box floor 2 f t x 1.44ftx2ft (0.61mx0.44mx0.61m) or 156Btu/°F (2.96 x 105 J/°C) heat capacity will not give the correct C' or U' with a 2 f t x 2 f t (0.61 rex0.61 m) glazing. How do we resolve this dilemma of normalizing floor and wall thermal values simultaneously? Apparently one can not use arbitrary dimensions for the glazing. The answer is suggested by the discussion in Section 2: use glazing dimensions in the same ratio as the real building glazing. That is, instead of 2 f t x 2 f t (0.61 mx0.61 m), make the glazing 1.491 ft high x 2.683 ft wide (0,45 m x 0.82 m) for the same glazing area (4 ft 2 or 0.37 m2). The Trombe and back wall will normalize correctly again, but the length of the side walls, floors and ceiling will be changed: they will now be (neglecting edge effects) 2.88ft2/l.491ft = 1.93ft (0.59m) long. Since the area of the side walls 2.88 ft 2 (0.27 m 2) is unchanged, the normalized thermal values for the side walls is correct. For the floor, the new area 1.93 ft x 2.68 ft = 5.2 ft ~ (0.48 m 2) will give the correct values for C' and U' for a floor 2 ft (0.61 m) thick: C = (280.8 Btu/°F)14 ftg2 and U' = (1.30 Btu/°F hr)14 ftg2. The necessary total conductance for the box ceiling is still 0.24Btu/°Fhr (0.132/°C) but a thickness of 7in. (17.8 cm) of insulation is now required. Note that this is the same thickness as the insulated ceiling in the real building. Note also that the approximate dimensions of the test box (2.68 ft wide, 1.49 ft high and 1.93 ft deep, or 0.82 m x 0.45 m x 0.59 m) are in the same ratio as the inner dimensions of the original building--18: 10: 13. Edge effects will change the box dimensions somewhat, such that the log mean average of the inside and outside box dimensions should be in the above ratio. The point of this whole discussion is that it is always possible to construct a test-box thermal-model of any real building with massive walls. It is always best from a design standpoint to scale the dimensions of the glazlag(s) in the same ratio as the building glazing dimensions. If the rest of the test box is constructed with the same materials used in the walls, floor and ceiling of the real building, the thicknesses of the walls, etc., should be approximately the same as the real building. Edge-effects will need to be considered as well. One can even model the thermal performance of multi-room passive-solar buildings with a realistically massive envelope: if both the ratio of the areas of the glazings for the solar-heated rooms and the ratio o[ the dimensions of these same glazings for the real building are used in the test-box,
346
D.P. GRIMMER
then the test-box can be an accurate thermal-model of the real building (see Section 2 for a discussion of thermal-modelling multi-room buildings).
4. THE SCALINCor rnV.aMOCmCOt.XT~Cp~.,~ivv.,soLw DESIGNSFOR TEST-BOXTHERMALMODELLING As mentioned previously, it is important to determine the scalability of the various thermal parameters involved in the test box thermal modelling of passive solar building designs, For example, in a thermocirculating passive-solar collector wall of height H and width W, the air-flow rate through vents at the top and bottom of the wall can be estimated by assuming the major flow resistance to be in the vents. The temperature distribution in the wall-glazing air space is assumed to be linear; also, the room temperature is assumed to be constant, so the inlet air space temperature is the room temperature. Then the volumetric air flow per unit glazing area (Ag) is given by ~'= Ca" A " ~ / ( g ' / 3 . I(Ts- T2)]" H)
where f' is volumetric air flow rate per unit area of glass; Ca the vent discharge coet~cient; A ' = A d A g = v e n t opening area per unit glazing area; H the wall height; g the gravitational acceleration; /'5 the arithmetic average of the inlet and outlet air space temperatures;/3 = I/T5 where T5 is in absolute temperature units; and /'2 the room temperature. All temperatures are in absolute units, In the case of a test-box thermal-model, the height H is scaled down considerably, so that it is necessary to increase the vent opening to maintain the volumetric air flow per unit glazing area. The factor A ' X / H must be held constant. Thus if h is the height of the test-box, then a" is the vent opening per unit glazing area for the test-box, so that a'~X/h = A'~/H, and Ca' a~ • X/(g./3 • [(TsT2)[. h) is the volumetric air flow per unit glazing area for the test box. The implicit assumption is that ternperatures /'2 and T5 are independent of scaling to first order, A fair question to ask is: What percentage error is involved in assuming that room temperature is constant so that the inlet air space temperature is the room temperature? Typical stratification temperature differences (AT) in a closed room are found to be on the order of 5°F. Lower A T's are found within a closed small test-box; however, the temperature gradient (AT/h) is about the same for a closed test-box as for a closed room. The principal effect of stratification will be to replace T2 in the expression for V with (T2-AT/2), so that the value of V when room air stratification is considered is given by f', where ~/( V, = Ca • A'~
[( g "~ •
~._T) I T5 -
T2 +
The ratio of I2 to I?~ becomes f'= ~/( I(Ts-Tz[ '~ ~'~ [(T5 T2 + A 1"/2)1/" -
) . H
.
For Ts.---l/2(Ti,j~,+ To~,,,~) equal to. say, 1/2 (522.5°R+610°R)=566.25°R (314.6°K), then for T_,= 522.50R (290.1°K) and A T = 5°R (2.8°K), f" x [ ( (566.25 - 525) '} 097 ~ , = ~\(566.25 - 525 + 2.5)/= " ' (Note here we have made the implicit assumption that Ti,~e,= - T2-ATI2). Hence, the error introduced by not considering stratification effects in the thermosiphoning vent modelling is about 3 per cent too low for test-rooms, and less than 3 per cent too low for test-boxes with their lower inherent stratification in a closed-wall configuration. In an open-wall, thermosiphoning mode, the value of AT will be dominated by the value of the outlet air space temperature mixing with the room temperature. Considering uncertainties in other assumptions (alluded to in Section 4), this error is not excessive. As an example, suppose we have a Trombe wall 10 ft high x 5 ft wide (3.05 m x 1.53 m) with vents at top and bottom that are each I ft z (0.093 m2) in area. Then A', = 1 ft~/50 ft2 = 0.02, and H = 10ft (3.05 m). So, A~,X/H = 0.063 ft tl~ (0.035 mr/2). If we now build a test box with h = 2 ft (0.61 m) and w = 1 ft (0.305 m), then ag = 2 ft: (0.19m2), a~=AX/(HIh)=O.045, and av=a~..a~= 0.089 ft ~ (8.3 x 10-3 m2). If the vents are each 8 in. across, then they are 1.6 in. high, top and bottom, to yield an area a~ = 0.089 ft 2 (8.3 x 10-3 m 2) for each vent. It should be noted that measurements of H (and h) for the thermocirculation air flow should be taken from the center of the vents. Also, bear in mind that making a space for a vent in the test-box Trombe wall will decrease the total heat capacity per unit glazing area, and adjustments should be made accordingly, either in glazing (e.g. by masking) or Trombe wall dimensions.
s. THE SCALINGOF AIR-INFILTRATIONEI~ECTSFOR TEST-SOXTHERMALMODELLING Heat losses due to outside air-infiltration can have a significant effect on the overall building thermal performance[6]. The usually accepted infiltration rate for fresh-air admittance is one interior-volume air-change per hr for residential buildings and higher air-changes per hr for commercial buildings or rooms with high occupancies (such as conference rooms). The heat loss due to infiltration can be regarded as a thermal load in addition to the load due to building skin conductance (the sum of the U x A values for the exterior areas of the building). This is the procedure used in computer simulations for developing techniques for estimating the performance of passive solar heating systems [7]. As discussed previously thermal loads can be normalized to the solar-collection glazing area, and this normalization can be done for infiltration loads also. The air-change rate will have to be varied in a test-box situation to obtain the correct normalized infiltration load. The use of a variable-speed exhaust-fan and adjustable dampers can obtain the appropriate normalized infiltration load value for a test-box model of a real building or room.
Small passive-solar test-boxes As indicated, the air infiltration rate is better expressed as a thermal-load rL--thermal power lost per indoorambient temperature difference (e.g. Btu/°F hr or W/°C)--rather than as a volumetric-air-change-per-unittime rv (e.g. ft3/hr or m3/s). The ratio of rL to r~ is equal to the heat capacity per unit volume Cv (e.g. Btu/°F ft 3 or J/°C. m3): rL = Cv. r~
347
ceiling heat loss is given by U ' c = 0.20 Btu/hr °F ft, 2 (or 1.13 W/°C ms2). The normalized floor heat loss is given by U,~= 0.05 Btu/hr °F fts 2 (or 0.269 W/°C ms2). The normalized air infiltration heat loss is given by U'~ = 0.180 Btu/hr °F ft~2 (or 1.02 W/°C m,2).
Here Cv = pc~ where p is the density (e.g. Ibs/ft3 or kg/m3), and c~ is the specific heat at constant volume (e.g. Btu/lb °F or J/kg °C). Hence rL r~
--~
pCv.
Expressed in terms of interior-volume air-changes per hr, the "infiltration rates" will in general be greater for a test-box than for the actual building being thermally modelled. As a simple example, consider a room with south-facing Trombe wall that is 10 ft long x 10 ft deep x 10ft high (3.05mx3.05mx3.05m) with R-20 (i.e. 0.05 Btu/hr °F ft 2, or 0.28 W/°C m 2, conductance C) walls, ceiling and floor. The heat loss through the insulated walls and ceiling (area Aw~) is given by
Let us now thermally-model the room with a scaleddown test-box. Let the test-box have a solar-collection glazing 2 ft x 2 ft (0.61 m x 0.61 m) or 4 fts 2 (0.37 ms2) glazing area as. The test-box wall and ceiling heat loss is then given by (Uwc)tes, box = U L~ • as = (0.20 Btu/hr °F fh 2) x 4 fts 2 = 0.80 Btu/hr °F(0.42 W/°C): Similarly, the test-box floor heat loss is given by (U:)'~'b°x=U'/'as=(O'O5Btu/hrOFfts2)x4fts2
= 0.20 Btu/hr °F (0.1 i W/°C): and the test-box air-infiltration by ( U~) ..... bo~ = U " as = (0.18 Btu/hr °F fq2). 4 fts 2
U ~ = C . Aw~
= (0.05 Btu/hr °F ft2)(3 x 10 ft x 10 ft + 10 ft. 10 ft)
= 0.72 Btu/hr °F (or 0.38 W/°C).
= 20 Btu/hr °F (or 10.6 W/*C).
The load U,,~ must be considered separately from U/ because the ambient air temperature is generally different from the ground temperature. Note that we have neglected edge-effects and are assuming perimeter ground insulation for simplicity.
The appropriate test-box dimensions to correctly scale the normalized thermal parameters are interior dimensions 2.0ft widex2.0ft deepx2.0ft high (0.61m× 0.61 m x 0.61 m). We will again neglect edge-effects for simplicity (See Section 6). The total area of the insulated walls and ceiling of the test-box is 16 ft 2 (1.49 m2), which with R-20 insulations yields the correct (Uw~),~,-box value. Similarly, the use of R-20 insulation in the 4 ft 2 (0.37 m2) floor yields the correct (U/) test-box value. However, since the interior-volume of the test-box is 8 ft 3 (or 0.22 m3), one interior-volume air-change per hr is
The interior-volume, V, of the room is 1000ft 3 (or 28.0 m 3) so that the total heat capacity of the room-air is
R, = 8 ft3/hr.
The heat loss through the floor is given by UI = (0.05 Btu/hr °F ft2) (100 ft 2) = 5 Btu/hr ° F (or 2.65 W/°C).
C ~ = C~. V = (0.018 Btu/°F ft3)(1000 ft 3) = 18.0 Btu/°F. If we allow an air infiltration rate of one interior-volume air-change per hr, then r~ = 1000ft3/hr and rL = 18.0 Btu/°F hr(or 9.54 W/°C). The value of RL is to be compared with Uw~ and U/. Let us define the air infiltration heat loss by U~ -= rL. If the solar-collection glazing is 10 ft x 10 ft (3.05 m × 3.05 m) or 100 fts2 (9.30 ms2) then the normalized wall and
This represents an infiltration heat-loss of Up = rL = (0.018 Btu/°F • ft ~) (8 ft3/hr) = 0.1~ Btu/hr° F (0.076 W/°C). This is a factor of 5 smaller than the value of (U~),~,,.~ = 0.72 Btu/hr °F (0.382 W/°C) desired. Hence for the test-box, 5 interior-volume air-changes per hr are required to correctly model the air-infiltration heat-loss of the example room. The scaling of infiltration losses is a problem in volumetric-flow losses vs essentially one-dimensional conduction losses, similar to the scaling of thermocirculation tsee Section 4).
O . P . GRIMMER
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6. EDGE-EFFECTCORRECTIONSFOR THERMALMODELLING wrrn SMALLPASS~F~SOLARTF~r Boxrs Edge-effect corrections result from the inability to accurately scale down thermal conductances through edges or corners where walls, ceilings or floor interface, It can be shown that the log mean average of the wall areas must be used to determine the correct total thermal conductance for that wall[4]. That is, the effective area used in the conductance calculations is given by:
the wall is adobe and about I ft (0.305 m) thick, then Uwa, ~ 0.5 Btu/°F ft 2 (1.02 x 10" JI°C m2) is the conductance through that wall. If urethane foam with k = 0.14 Btu. in./°Fhr ft 2 (0.021 W/m °C) is used for the cornet block, then a I ft (0.305 m) thick foam block has Utoam---0.012 Btu/°F hr ft: (0.068 W/°C m2) conductance, which is negligible compared to Uwa. + Ufoam-~ Uwa,. The amount of highly-insulating foam can be adjusted at the corner to meet this condition as precisely as desired. Probably 5 per cent precision is acceptable: or
Ae~ = (A2- A J/In (A2/AI) Ufo,m ~<0.05 Uw~,. where A2 and A~ are the outside and inside wall areas, respectively. This is used to calculate the effective U • A values for the design, and also the total conductance normalized to solar glazing area. To properly scale down the normalized thermal conductance and simultaneously the normalized heat capacity, it is necessary to preserve the ratio of the outside and inside wall areas (AdAJ since
A~
\A~-1
7. THE USE OF THERMAL-MODEL TEST-BOXES TO MEASURE MICROCLIMATE SHADING EFFECTS ON THE DAILY
COLLEL"nONOV SOLARENERGY Test-cell thermal-models of a proposed passive-solar design can be built on-site to investigate various microclimate effects, such as a south-facing slope. It is iraportant to determine whether the microclimate effect to be studied can be scaled down to the physical size of a small test-box. The effect of shading (e.g. by trees) on the performance of a passive design may not be easily scaled down from a full-size passive collector to a small test-ceil, depending on the size and placement of the shading object relative to the test-box. Consider the schematic passive-solar collector shown in Fig. 3. The collector has width W and height h to intercept the solar radiation (assume due south orienration). Directly in front of the collector a distance L is a shading object of width D and height greater than h. For a low altitude angle winter sun, let us ignore variations in solar altitude for simplicity, and let the solar flux on a vertical south-facing passive collector vary as Q cos ~b, where Q = f(t) is the direct normal solar radiation as a function of time, and ~b varies as 0.262 rad per hr (15°lhr): if t = time in hr, ¢i = 0.262t where t = 0 at solar noon. Thus, the total energy intercepted by the W x h collector is given approximately by
In(A2/AI)
and this correction ratio would be used in the normalized total conductance calculations. However, the thickness of the test-box wall is kept identical to that of the building wall, and hence the test-box is not geometrically similar to the building. The wall thickness must be kept identical to insure the same temporal behavior of the heat wave. It is not possible to preserve the (A2/A1) ratio and have solid, closed corners in the small-test boxes (see Appendix A). Figure 2 illustrates how the corners of a test-box must appear to have (AdAm) in the same ratio as the real wall and maintain the reall wall thickness. This will allow the heat capacities to normalize properly, and will give the correct normalized total thermal conductance through the open corners is resolved by filling them with a block of low-conductance foam-board as suggested by the dashed lines in Fig. 2(b). The foam insulation in the corners will act as a small parallel conductance with the rest of the solid wall and can be treated as negligible, depending on the U values of the foam insulation and the wall material. As an example, if
Q' = 2
ft-t
h • W. Q . cos (0.262t) dt J,-,,,-~ (W/2L)IO.262
f #'tan-I(g/12L)lO'262 + 2 ~j,.o h • [ W - D . sec (0.262t)] • Q . cos (0.262t) • dt. A2
=
A'~
=
Fig. 2. (a) A real wall with closed, solid corners; (b) the equivalent wall, with open corners and (A2/Aj)= (A~/AI). The wall thickness is the same for the real and equivalent (tesbbox) walls.
Small passive-solar test-boxes The first term in the above expression gives the unshaded integrated flux, while the second term gives the integrated flux for when the collector is being shaded by the object. The length of the day is given by 2to. Letting
the passive solar building. Then Q~ = 1653.5 -
= 0.262t, then gives
Q,
If we normalize this to the area of collector, h × W we get 2
(TDD~°~_l ( W ~
~. \w/\/~"
Note that this formula is approximate in that it ignores the case of when the shadow is only partially falling on the collector area. Let us now put some reasonable numbers for size of collector, and size and distance of shading object. Let the shading object be a tree whose branches are 15 ft (4.6 m) in diameter, L = 30 ft (9.15 m) away from the passive collector (e.g, a large evergreen). Let the shadow now fall upon a real, passive solar building of W = 50 ft (15.25 m) width. Also, let Q = 250 Btu/ft 2 hr (787.5 W/m e) be a constant for a 2to= 8hr winter day. Then the integrated and normalized (to glazing area) daily solar energy on that passive collector is given approximately by Q,,
~
•
• (250). tan -t
= 1653.5- 476.7 = 1176.8 Btu (1.24 x 106 J).
2hWQ t 2hDqtan-' =~ sin (0.262 o ) - 0."0~(~L)"
,, 2Q Q = ~. sin (0.262to)-
349
2(250) sin (0.262x4)- 2 15 (250)tan-' ( ~ ) = 0.262 " 0.262 40 = 1653.5-420.8 = 1232.7 Btu (1.30 × 106J).
Now consider a test-box of W = 3 ft (0.92 m) in place of Sun's Rays
Thus, the use of a passive test-box to predict total daily solar energy collected (normalized to area of glazing) for a passive solar building design is in error by only about - 5 per cent in this particular example. The results of this somewhat crude calculation tell us what may be intuitively obvious: that a small test-box can be used to give an accurate prediction of the effects of certain kinds of shading on the daily solar energy intercepted by a passive solar collector. The actual shading phenomenon will be quite dilferent for the building than for the test-box, however. The building will intercept a shadow small relative to its dimension W for a large part of the day, whereas the test cell will have a period of total shadow for a small part of the day (the thermal capacitance of the test cell will smooth out this perturbation a bit). The total daily energy intercepted will be very nearly the same for the passive solar building and the test box. A work of caution. Every building/site has to be analyzed individually for shading/scaling problems. In particular, one does have to be careful about making sure the test box would intercept approximately the same amount of normalized energy as a real building. For example, ff a shadow from an object does not fall on the full height of the building, it would nevertheless completely shadow a test-box placed on the ground for some part of the day. In this situation a test-box would intercept considerably less normalized energy than the equivalent building/collector.
\
D
/
/
l/ s°
of Shodow Cut- Off
Angle
,/,"
J
~
.TAN.I(wI2L)
w
Passive
Solar
Areo of Shadow
• O Sec (ib • D/Cos
Collector
Fig. 3. Schematic of a vertical passive solar collectorof width W facing due south, with a shading object of diameter D located a distance L from the middle of the collector face. The collector is of arbitrary size in the calculations, and can either be a real passive-solar building or a passive solar thermal-model test-box. Area of shadow on collector face, and angle of shadow cut-off are indicated.
350
D . P . GRIMMER 8. CONCLUSIONS
Theoretical considerations have been presented for the test-box thermal modelling of passive-solar building designs. Rather simplistic arguments have been used, and experimental confirmations are necessary. However, the mathematical arguments are straightforward and indicate reasonable quantitative agreement should be expected,
Acknowledgements--The author would like to thank Drs. C.A. Bankston and J. D. Balcomh for helpful discussions on similitude and heat flow.
I.
2. 3. 4. 5.
6. 7.
REFERENCES D. P. Grimmer, J. D. Balcomb and R. D. McFarland, The use of small passive-solar test-boxes to model the thermal performance of passively solar-heated building designs. LASL Report LAUR-77-1323, presented at the 1977 Annual Meeting Am. Section ISES, Orlando, Florida (6-10 June, 1977). R. C. Pankhurst, Dimensional Analysis and Scale Factors. Butler and Tanner, London (1964). G. Murphy, Similitude in Engineering. Ronald Press Co., New York (1950). W. H. McAdams, Heat Transmission. McGraw-Hill, New York(1954). J. D. Balcomb, J. C. Hedstrom and R. D. McFarland, Simulation analysis of passive solar heated buildings--preliminary results. Solar Energy 19, 277 (1977). B. Anderson and M. Riordan, The Solar Home Book. Cheshire Books, Harrisville, New Hampshire (1976). J. D. Balcomb and R. D. McFarland, A simple technique of estimating the performance of passive solar heating systems. LASL Report LA-UR-78-1571, presented at the 1978 U.S. ISES meeting, August 1978, Denver, Colorado (1978).
are given by
k (A2-AI) 1 • ln(A21Ai) Ag'
u' and
c , = ~ (AI+Az)¢ 2 AR For a small test box wall with parameters analogous to the real wall given by A~, A~, A~, r', but with the same values of k, p, c. we can write that
k A2-AI k A~-AI ~ " Ag In (A2IA z) = r' A'~ In (A~IA ~) and
pc (At + A2)¢ pc (Ai + A~)~-' 2
As
-- 2
As
(2)
in comparing the real and test box walls. Rewriting these equations we have that
~' A~-A't In(A21AI).A_t -~= ln(A~lAi) A2- A~ A'~
(3)
and
~" = ( A, + A2~ ~ ¢ \AI + A~/ A s"
(4)
Then for (3) and (4) to be simultaneously true, we must have that A_~2 A~- A[
A ~ In(A~/A[) APPENDIX A The use of eHective wall areas in the calculation and scale-down of normalized thermal conductances and heat capacities for test boxes
(1)
In (A21AI)
Ai + A2 A2-A~ -A[+A~"
So, then
A~2- AI 2 A22- A~2 A 's2. In (A ~IA 'I) • As2. In (A 21A j)
It is known that the effective wall area for heat flow is given by the log mean of the outside and inside wall areas:
(A~IA's)2- (AI/A~)2 "(A21As)2-(AI/Ar) 2. In (A~IAI) In (A21A I)
A2- AI Ac~ In (A21A~) where A2 and A, are the outside and inside wall areas, respectively. For a given real wall, the total thermal conductance (u') and heat capacity (c') are given by Ut=
k A,.-AI ~"In (A2/AI)
where k is the thermal conductivity and r the wall thickness: and
Identifying term by term, the above is true if A~ = A2. Ai = Ai. A~ = AI A~ As ' A[ As ' AI A,' or A__.~= A_~. Ai = A~. A_~= A_~
A2
A~" AI
A~ = A~ = A_~
A2 c'=
~-(A, + Az)¢
where p is the density and c the specific heat. Then the total conductance and heat capacity, normalized to glazing area AR,
A s' A2 A (
Thus, AI
As
and hence for a given A'rlAs chosen: it is not possible to have A21A~ = A~/AI unless the massive wall does not extend all the way to the corned': the corners must be left open and filled with a low conductance, low thermal capacity material.
0038-092X/79/O401--0M3/$02.00/O
THEORETICAL CONSIDERATIONS IN THE USE OF SMALL PASSIVE-SOLAR TEST-BOXES TO MODEL THE THERMAL PERFORMANCE OF PASSIVELY SOLAR-HEATED BUILDING DESIGNS? D. P. GRIMMER Los Alamos ScientificLaboratory, Los Alamos, NM 87544, U.S.A. (Received 21 February 1978: revision accepted 3 November 1978)
Abstract--Theoretical considerations are presented for the thermal modellingof passively solar-heated building designs with passive-solartext boxes. Multi-roompassive solar buildings,passive solar buildingshaving realistically massive walls, thermocirculatingpassive-solar designs, air-infiltrationeffects, edge effect corrections, and microclimate shading effects are discussed. t. IN'mODUCrION Thermal modelling of passive-solar building designs with small test boxes could be a valuable tool for the architect, contractor and owner-builder[l]. The subject of dimensional analysis and scale factors in the modelling of physical phenomena, including heat transfer, is discussed extensively in the literature[2, 3]. Such analyses are usually done using dimensionless scalefactors while preserving geometric, kinematic, temporal and thermal similarity. The data gathered from the model must be converted to real-scale values by use of the appropriate scale-factor for these similar (non-distorted) models, The use of non-distorted models is not the subject of this paper, but rather the use of distorted models that present real-scale values for data collected. Specifically, the data gathered with the use of test-ceils consists essentially of ambient (outside) and interior (room) air temperatures as a function of time. It would be desirable for the non-technical experimenter to have a methodology of text-box construction that provides real temperatures and times of thermal waves moving through the building envelope. By using distorted models, the use of scale factors (such as Fourier and Blot numbers) can be avoided, In the following sections somewhat simplistic discussions will be presented on multi-room passive-solar buildings, passive solar buildings having realistically massive walls, thermocirculating passive-solar designs, air infiltration effects, edge-effect corrections and microclimate shading effects, 2. THERMAL MODELLING OF MULTI-ROOMPASSIVESOLAR BUILDINGSUSINGSMALLPA,~IVI~TEST BOXES Small passive test-boxes have been used to model the thermal performance of single-room passive solar building designs[l]. Test-box arrangements can also be constructed to thermally model multi-room passive-solar buildings. Since most real buildings are multi-room, the ability to model them with test-boxes would be quite useful. tWork performed under the auspices of the U.S. Department of Energy. SE VoL 22, No. ~----D
343
In a typical multi-room passive-solar building design, some rooms will be solar heated (i.e. have south facing glazings) while others will not. A test-box thermal-model should have the same number of solar-heated rooms in approximately the same directional layout as the actual building design to get accurate modelling. As mentioned previously, the thermal model will not resemble a scaled down model of the actual building, except for the layout and direction of south facing windows, and the topological arrangement of the rooms. It is important to realize that volumetric considerations do not play an important part in thermal modelling: the volume (but not the wall area) of a room can essentially be ignored in heat-loss calculations. Stratification phenomena are related to room volume and can play a part in comfort zones in a room, but for an overall heat-loss calculation (using an average room temperature above ambient temperature) a test-box room should simulate an actual room well. This would be particularly true in cases where infiltration losses are low, and ceiling insulation is high. The procedure for thermal modelling a multi-room passive-solar design is simple. The total thermal conductance (the UA, in Btu/*F or W/m 2) is calculated for each room, with the conductance for exterior walls, ceiling and floor totalled separately from the conductance for walls, ceiling and floor common to other rooms. The thermal capacity is similarly totalled for each room. For those rooms containing solar-glazings, the room thermal conductance and room thermal capacitance are normalized to the area of the glazing, treating interior common walls, ceiling and floor separately as before. Thus, each solarheated room will have the thermal conductance and capacitance of the interior envelope normalized to that of the exterior envelope values (and of course the glazing area). Because the thermal conductance and capacitance of interior envelopes are common to adjoining rooms, the glazing areas of the solar-heated rooms will be in the same ratio for the test-box thermal model as for the real building. In summary then, for the test-box thermal-model, the ratios of the glazing areas for the solar-heated rooms will be the same as the ratios of the glazing areas for the real rooms. Also, the ratios of the thermal conductance and
344
D, P. GRIMMER
capacity of the exterior envelope to that of the interior envelope for each room will be the same for the test box rooms as for the building rooms. In the case of the test-box rooms, the interior and exterior envelope values will both of course be .normalized to the glazing areas of the solar heated rooms. As an example, let us consider the passive-solar building design shown in Fig. 1. Here we have a rather idealized case of essentially massless insulating walls, with the mass of the building contained in two storage walls, of thermal capacity C1 and C2, respectively. Specific thermal conductance UCI is that of a common (interior) wall and is much greater than the other (exterior) U values. The specific conductances U I1 and UC2 would be for storage walls and presumably would have a value comparable to UC1. Because of its mass, however, the thermal impedance of Wall No. 3 (with steady state conductance UC2) would be much greater than the impedance of the other common wall with steady-state specific conductance UCI. The U values for the side walls perpendicular to the plane of the figure are not shown, The UA's are summed for each room. Let (UA)~ be the sum for the ith room exclusive of the common walls, etc., for that room. So for Room No. ! we have that the total conductance is given by (UA), + UCI • W2, where W2 is the area of Wall No. 2. The heat capacity for Room No. 1 is given by CI. The conductance of Room No. 1 normalized to glazing area G I is given by [(UA), + UCI. W2]/GI and the normalized heat capacity CI/GI. Similarly, the normalized conductance of Room No. 2 is given by [ ( U A h + UCI • W2+ UC2. W3]/G2, where W3 is the area of Wall No. 3, and the normalized heat capacity by C2/G2. Finally, the total conductance of Room No. 3 is given by (UAh + UC2. W3 and the heat capacity by C2. If in the same test box rooms the simulated glazing areas GI' and G2' are kept in the same ratio as the real building's glazing areas (i.e. GI'/C2'= G IIG2 = C, where C is some constant), then the thermal conductance and capacitance for the common walls of the test-box rooms will be correctly matched: i.e. UC1 • W2/G1 = UCI'. W2'/GI and UCI • W2/G2 = UCI'. W2'/G2' only if GI'/G2'= GI/G2 (here primes denote test box values),
Double
The thermal model of a room without a solar glazing, such as Room No. 3 in Fig. 1, is quite simple, since volume does not play much of a role in thermal-loss calculations. Such a room can always be considered as an insulating wall for the solar heated room (here Room No. 2), and if the air temperature in Room No. 3 is not of interest, it could be replaced with a solid insulator whose overall series conductance would be given by 1 I U C 2 . A 3 + ~ + ~ where (UA),ir is the equivalent total conductance through the air. If we wish to measure the air temperature in Room No. 3 in this example, we must have at Wall No. 3 that UC2. W31G2 = UC2'. W3'/G2', or that (UC2'. WY)/(UC2. W3)= G2'/G2. That is, the total conductance through the common Wall No. 3 of the box model should be (G2'/G2)of the real total conductance through that wail. As a final note, remember that edge e~:ects must always be considered for small test-boxes. The log mean average of the thermal conductance through the inside and outside surface areas of a wall must be used to determine the correct total conductance for that wall: i.e. Ae~ = (A - A')/In (A/A')[4]. A detailed discussion of edge-effects is presented in Section 6. 3. "I'IU~MALTI[ST.BOXMODELLINGOF PASSIVESOLAR B~GS HAV~G ~ C ~ L V ~4LSSIV[WALLS The example discussed in Section 2 from Fig. I is essentially unrealistic in that (a) the insulating walls are assumed to have no thermal capacitance, and (b) the thermal capacitive reactances to transient temperature changes were not considered specifically for massive Walls Nos. 1 and 3. Instead of specific conductances U11 and UC2, we would have to consider overall thermal impedances involving thermal capacitive reactances. The situation of heat flow through a massive wall can be likened to current flowing through a transmission line. Lumped circuit elements can be used to describe the latter, and analogously the former[S]. To construct an accurate test-box thermal-model of a passive dwelling with massive walls not in direct sun, it
Double -~. Glazing Areu G2
.,L
J,
wo,,2 Wall I, Trombe Wnll
/f Wall 3,
/Woll 4 Direct Gain
Fig. 1. Schematic of multi-room passive-solar building design. The design has a solar-room with a Trombe-wall, a solar-room with direct gain on a storage wall, and a non-solar room. Symbols for thermal conductance and heat capacity are indicated.
Small passive-splat test-boxes is best to consider the test-box as a model of a horizontal "slice" of finite thickness through the solar glazings of the real building. For example, consider that all walls of the building shown in Fig. l are made of masonry, and the floor and ceiling of insulated wood. Then the actual material used for the building walls should be used to construct the test box walls. The thermal conductance and capacitance of each wall is normalized to the glazing area and to each other. The proper normalization will determine the exact dimensions of the box, for some glazing area, Let Room No. 1 in Fig. l have all masonry walls (C=0.2Btu/Ib°F, or 825.7J/kg°C; U=O.SBtu/hrft°F, orO.87W/m°C;p=1351b/ft3or2189kg/m3) thatarelft (0.305 m) thick, and constructed in a rectangular shape 10ft high, by 15ft deep, by 20ft wide (facing south) (3.05mx4.58mx6.1m) outer dimensions, and 10ft high x 13 ft deep x 18 ft wide inner dimensions (3.05 m x 3.97mx 5.49m). Let the buildings ceiling and floor be made of insulated wood (U = 0.027 Btu/hr ft °F or 0.047 W/m °C) 7 in. thick (17.8 cm) and essentially massless. Then, the Trombe wall is l0 ftx 18 ft x 1 ft = 180 ft 3 (5.04m3), or 24,3001b (ll,178kg) of masonry, or 4860 Btu/°F (9.23 x l06 J/°C) of heat capacity; and l0 f t x 18ft= 180ft 2 (16.7 m 2) of collecting area, or 90Btu/hr°F (47.7 W/°C) of total conductance. Similarily, the side and back walls have a total of 440 f13 (12.3 m 3) or 59,400 lb (27,324 kg) of masonry or 11,880 Btu/OF (2.26 x l07 J/°C) of heat capacity; and 440 ft 2 (40.9 m 2) of area of 220Btu/hr°F (ll6.6W/°C) of total conductance. The ceiling and floor each has an area of 234ft 2 (21.76m 2) and a total conductance of 10.6 Btu/hr°F (5.62 W/°C). Note that edge effects have been neglected here for simplicity, Thus for the example room discussed here, normalized to glazing area (180 ft: or 16.7 m2), we have that: (1) for the Trombe Wall, C'=27.0Btu/°Fft~ 2 (5.5x 105J/°Cm~2)and U'=O.50Btu/°Fhrfq2(2.84W/°Cm,:): (2) for the side and back walls (total), C ' = 66.0 Btu/°Fftg 2 (1.35 x 106J/°C m~2) and U'= 1.22 Btu/°F hrftg 2 (6.92 W/°C m,2); (3) for the ceiling and floor (each), U' = 0.059 Btu/°F hr ft82 (0.33 W/°C mg2). If we now construct a passive test box using a glazing, say 2 ft x 2 ft = 4 ftg2 (0.37 m2), then we have in comparing the example room with the test-box that: (i) for the simulated Trombe wall, we use masonry 1 ft t h i c k x 2 f t x 2 f t (0.305mx0.610mx0.610m), of 4ft 3 (0.11 m 3) volume. This has the correct C' and U' values: (2) for the back wall we can use a masonry wall I ft thick x 2 ft x 2 ft (0.305 x 0.610 m x 0.610 m) (to make a rectangular box). This has 27.0 Btu/°Fft~: (5.5 x 105 J/°C m~:) and 0.50 Btu/°F hr ft 2 (2.84 W/°C m,2). The remainder of the C' and U' needed (39 Btu/°F ftg2 or 156 Btu/°F (2.96 x l03 J/°C) for the box, and 0.72 Btu/°F hr ft: or 2.88 Btu/°F hr (i.53 W/°C) for the box) is obtained by adjusting the length of the side walls to approx. 1.44 ft (0.44 m) neglecting edge effects. This will make each side wall have about 2.9ft 3 (0.08 m 3) volume, or about 78 Btu/°F (1.48 x l0 ~ J/°C) heat capacity: (3) for the ceiling and floor we now have from (2)
345
above an area of 2.88 ft 2 (0.27 m2). We need a U of 0.25 Btu/°F hr (0.13 W/°C) for the box ceiling and floor (each). If we are using the same insulation, with 0.27Btu/hr°F (0.14W/°C) then a thickness of 3.1in. (7.9 cm) will give the necessary conductance out the ceiling and floor. If the ceiling and floor were not considered massless it would have been more difficult to adjust thermal parameters to fit. Let us suppose that the floor was instead made of masonry 2 ft (0.61 m) thick; then the floor has 12636 Btu/°F (2.40 x 107 J/°C) of heat capacity and 59Btu/°Fhr (31.3W/°C) of thermal conductance. Hence for this floor, C'=70.2Btu/°Fftg, (1.43x 106H/°Cms 2) and U'=0.325Btu/°Fhrfts, (1.84W/ °Cruz2). It is obvious that a text box floor 2 f t x 1.44ftx2ft (0.61mx0.44mx0.61m) or 156Btu/°F (2.96 x 105 J/°C) heat capacity will not give the correct C' or U' with a 2 f t x 2 f t (0.61 rex0.61 m) glazing. How do we resolve this dilemma of normalizing floor and wall thermal values simultaneously? Apparently one can not use arbitrary dimensions for the glazing. The answer is suggested by the discussion in Section 2: use glazing dimensions in the same ratio as the real building glazing. That is, instead of 2 f t x 2 f t (0.61 mx0.61 m), make the glazing 1.491 ft high x 2.683 ft wide (0,45 m x 0.82 m) for the same glazing area (4 ft 2 or 0.37 m2). The Trombe and back wall will normalize correctly again, but the length of the side walls, floors and ceiling will be changed: they will now be (neglecting edge effects) 2.88ft2/l.491ft = 1.93ft (0.59m) long. Since the area of the side walls 2.88 ft 2 (0.27 m 2) is unchanged, the normalized thermal values for the side walls is correct. For the floor, the new area 1.93 ft x 2.68 ft = 5.2 ft ~ (0.48 m 2) will give the correct values for C' and U' for a floor 2 ft (0.61 m) thick: C = (280.8 Btu/°F)14 ftg2 and U' = (1.30 Btu/°F hr)14 ftg2. The necessary total conductance for the box ceiling is still 0.24Btu/°Fhr (0.132/°C) but a thickness of 7in. (17.8 cm) of insulation is now required. Note that this is the same thickness as the insulated ceiling in the real building. Note also that the approximate dimensions of the test box (2.68 ft wide, 1.49 ft high and 1.93 ft deep, or 0.82 m x 0.45 m x 0.59 m) are in the same ratio as the inner dimensions of the original building--18: 10: 13. Edge effects will change the box dimensions somewhat, such that the log mean average of the inside and outside box dimensions should be in the above ratio. The point of this whole discussion is that it is always possible to construct a test-box thermal-model of any real building with massive walls. It is always best from a design standpoint to scale the dimensions of the glazlag(s) in the same ratio as the building glazing dimensions. If the rest of the test box is constructed with the same materials used in the walls, floor and ceiling of the real building, the thicknesses of the walls, etc., should be approximately the same as the real building. Edge-effects will need to be considered as well. One can even model the thermal performance of multi-room passive-solar buildings with a realistically massive envelope: if both the ratio of the areas of the glazings for the solar-heated rooms and the ratio o[ the dimensions of these same glazings for the real building are used in the test-box,
346
D.P. GRIMMER
then the test-box can be an accurate thermal-model of the real building (see Section 2 for a discussion of thermal-modelling multi-room buildings).
4. THE SCALINCor rnV.aMOCmCOt.XT~Cp~.,~ivv.,soLw DESIGNSFOR TEST-BOXTHERMALMODELLING As mentioned previously, it is important to determine the scalability of the various thermal parameters involved in the test box thermal modelling of passive solar building designs, For example, in a thermocirculating passive-solar collector wall of height H and width W, the air-flow rate through vents at the top and bottom of the wall can be estimated by assuming the major flow resistance to be in the vents. The temperature distribution in the wall-glazing air space is assumed to be linear; also, the room temperature is assumed to be constant, so the inlet air space temperature is the room temperature. Then the volumetric air flow per unit glazing area (Ag) is given by ~'= Ca" A " ~ / ( g ' / 3 . I(Ts- T2)]" H)
where f' is volumetric air flow rate per unit area of glass; Ca the vent discharge coet~cient; A ' = A d A g = v e n t opening area per unit glazing area; H the wall height; g the gravitational acceleration; /'5 the arithmetic average of the inlet and outlet air space temperatures;/3 = I/T5 where T5 is in absolute temperature units; and /'2 the room temperature. All temperatures are in absolute units, In the case of a test-box thermal-model, the height H is scaled down considerably, so that it is necessary to increase the vent opening to maintain the volumetric air flow per unit glazing area. The factor A ' X / H must be held constant. Thus if h is the height of the test-box, then a" is the vent opening per unit glazing area for the test-box, so that a'~X/h = A'~/H, and Ca' a~ • X/(g./3 • [(TsT2)[. h) is the volumetric air flow per unit glazing area for the test box. The implicit assumption is that ternperatures /'2 and T5 are independent of scaling to first order, A fair question to ask is: What percentage error is involved in assuming that room temperature is constant so that the inlet air space temperature is the room temperature? Typical stratification temperature differences (AT) in a closed room are found to be on the order of 5°F. Lower A T's are found within a closed small test-box; however, the temperature gradient (AT/h) is about the same for a closed test-box as for a closed room. The principal effect of stratification will be to replace T2 in the expression for V with (T2-AT/2), so that the value of V when room air stratification is considered is given by f', where ~/( V, = Ca • A'~
[( g "~ •
~._T) I T5 -
T2 +
The ratio of I2 to I?~ becomes f'= ~/( I(Ts-Tz[ '~ ~'~ [(T5 T2 + A 1"/2)1/" -
) . H
.
For Ts.---l/2(Ti,j~,+ To~,,,~) equal to. say, 1/2 (522.5°R+610°R)=566.25°R (314.6°K), then for T_,= 522.50R (290.1°K) and A T = 5°R (2.8°K), f" x [ ( (566.25 - 525) '} 097 ~ , = ~\(566.25 - 525 + 2.5)/= " ' (Note here we have made the implicit assumption that Ti,~e,= - T2-ATI2). Hence, the error introduced by not considering stratification effects in the thermosiphoning vent modelling is about 3 per cent too low for test-rooms, and less than 3 per cent too low for test-boxes with their lower inherent stratification in a closed-wall configuration. In an open-wall, thermosiphoning mode, the value of AT will be dominated by the value of the outlet air space temperature mixing with the room temperature. Considering uncertainties in other assumptions (alluded to in Section 4), this error is not excessive. As an example, suppose we have a Trombe wall 10 ft high x 5 ft wide (3.05 m x 1.53 m) with vents at top and bottom that are each I ft z (0.093 m2) in area. Then A', = 1 ft~/50 ft2 = 0.02, and H = 10ft (3.05 m). So, A~,X/H = 0.063 ft tl~ (0.035 mr/2). If we now build a test box with h = 2 ft (0.61 m) and w = 1 ft (0.305 m), then ag = 2 ft: (0.19m2), a~=AX/(HIh)=O.045, and av=a~..a~= 0.089 ft ~ (8.3 x 10-3 m2). If the vents are each 8 in. across, then they are 1.6 in. high, top and bottom, to yield an area a~ = 0.089 ft 2 (8.3 x 10-3 m 2) for each vent. It should be noted that measurements of H (and h) for the thermocirculation air flow should be taken from the center of the vents. Also, bear in mind that making a space for a vent in the test-box Trombe wall will decrease the total heat capacity per unit glazing area, and adjustments should be made accordingly, either in glazing (e.g. by masking) or Trombe wall dimensions.
s. THE SCALINGOF AIR-INFILTRATIONEI~ECTSFOR TEST-SOXTHERMALMODELLING Heat losses due to outside air-infiltration can have a significant effect on the overall building thermal performance[6]. The usually accepted infiltration rate for fresh-air admittance is one interior-volume air-change per hr for residential buildings and higher air-changes per hr for commercial buildings or rooms with high occupancies (such as conference rooms). The heat loss due to infiltration can be regarded as a thermal load in addition to the load due to building skin conductance (the sum of the U x A values for the exterior areas of the building). This is the procedure used in computer simulations for developing techniques for estimating the performance of passive solar heating systems [7]. As discussed previously thermal loads can be normalized to the solar-collection glazing area, and this normalization can be done for infiltration loads also. The air-change rate will have to be varied in a test-box situation to obtain the correct normalized infiltration load. The use of a variable-speed exhaust-fan and adjustable dampers can obtain the appropriate normalized infiltration load value for a test-box model of a real building or room.
Small passive-solar test-boxes As indicated, the air infiltration rate is better expressed as a thermal-load rL--thermal power lost per indoorambient temperature difference (e.g. Btu/°F hr or W/°C)--rather than as a volumetric-air-change-per-unittime rv (e.g. ft3/hr or m3/s). The ratio of rL to r~ is equal to the heat capacity per unit volume Cv (e.g. Btu/°F ft 3 or J/°C. m3): rL = Cv. r~
347
ceiling heat loss is given by U ' c = 0.20 Btu/hr °F ft, 2 (or 1.13 W/°C ms2). The normalized floor heat loss is given by U,~= 0.05 Btu/hr °F fts 2 (or 0.269 W/°C ms2). The normalized air infiltration heat loss is given by U'~ = 0.180 Btu/hr °F ft~2 (or 1.02 W/°C m,2).
Here Cv = pc~ where p is the density (e.g. Ibs/ft3 or kg/m3), and c~ is the specific heat at constant volume (e.g. Btu/lb °F or J/kg °C). Hence rL r~
--~
pCv.
Expressed in terms of interior-volume air-changes per hr, the "infiltration rates" will in general be greater for a test-box than for the actual building being thermally modelled. As a simple example, consider a room with south-facing Trombe wall that is 10 ft long x 10 ft deep x 10ft high (3.05mx3.05mx3.05m) with R-20 (i.e. 0.05 Btu/hr °F ft 2, or 0.28 W/°C m 2, conductance C) walls, ceiling and floor. The heat loss through the insulated walls and ceiling (area Aw~) is given by
Let us now thermally-model the room with a scaleddown test-box. Let the test-box have a solar-collection glazing 2 ft x 2 ft (0.61 m x 0.61 m) or 4 fts 2 (0.37 ms2) glazing area as. The test-box wall and ceiling heat loss is then given by (Uwc)tes, box = U L~ • as = (0.20 Btu/hr °F fh 2) x 4 fts 2 = 0.80 Btu/hr °F(0.42 W/°C): Similarly, the test-box floor heat loss is given by (U:)'~'b°x=U'/'as=(O'O5Btu/hrOFfts2)x4fts2
= 0.20 Btu/hr °F (0.1 i W/°C): and the test-box air-infiltration by ( U~) ..... bo~ = U " as = (0.18 Btu/hr °F fq2). 4 fts 2
U ~ = C . Aw~
= (0.05 Btu/hr °F ft2)(3 x 10 ft x 10 ft + 10 ft. 10 ft)
= 0.72 Btu/hr °F (or 0.38 W/°C).
= 20 Btu/hr °F (or 10.6 W/*C).
The load U,,~ must be considered separately from U/ because the ambient air temperature is generally different from the ground temperature. Note that we have neglected edge-effects and are assuming perimeter ground insulation for simplicity.
The appropriate test-box dimensions to correctly scale the normalized thermal parameters are interior dimensions 2.0ft widex2.0ft deepx2.0ft high (0.61m× 0.61 m x 0.61 m). We will again neglect edge-effects for simplicity (See Section 6). The total area of the insulated walls and ceiling of the test-box is 16 ft 2 (1.49 m2), which with R-20 insulations yields the correct (Uw~),~,-box value. Similarly, the use of R-20 insulation in the 4 ft 2 (0.37 m2) floor yields the correct (U/) test-box value. However, since the interior-volume of the test-box is 8 ft 3 (or 0.22 m3), one interior-volume air-change per hr is
The interior-volume, V, of the room is 1000ft 3 (or 28.0 m 3) so that the total heat capacity of the room-air is
R, = 8 ft3/hr.
The heat loss through the floor is given by UI = (0.05 Btu/hr °F ft2) (100 ft 2) = 5 Btu/hr ° F (or 2.65 W/°C).
C ~ = C~. V = (0.018 Btu/°F ft3)(1000 ft 3) = 18.0 Btu/°F. If we allow an air infiltration rate of one interior-volume air-change per hr, then r~ = 1000ft3/hr and rL = 18.0 Btu/°F hr(or 9.54 W/°C). The value of RL is to be compared with Uw~ and U/. Let us define the air infiltration heat loss by U~ -= rL. If the solar-collection glazing is 10 ft x 10 ft (3.05 m × 3.05 m) or 100 fts2 (9.30 ms2) then the normalized wall and
This represents an infiltration heat-loss of Up = rL = (0.018 Btu/°F • ft ~) (8 ft3/hr) = 0.1~ Btu/hr° F (0.076 W/°C). This is a factor of 5 smaller than the value of (U~),~,,.~ = 0.72 Btu/hr °F (0.382 W/°C) desired. Hence for the test-box, 5 interior-volume air-changes per hr are required to correctly model the air-infiltration heat-loss of the example room. The scaling of infiltration losses is a problem in volumetric-flow losses vs essentially one-dimensional conduction losses, similar to the scaling of thermocirculation tsee Section 4).
O . P . GRIMMER
348
6. EDGE-EFFECTCORRECTIONSFOR THERMALMODELLING wrrn SMALLPASS~F~SOLARTF~r Boxrs Edge-effect corrections result from the inability to accurately scale down thermal conductances through edges or corners where walls, ceilings or floor interface, It can be shown that the log mean average of the wall areas must be used to determine the correct total thermal conductance for that wall[4]. That is, the effective area used in the conductance calculations is given by:
the wall is adobe and about I ft (0.305 m) thick, then Uwa, ~ 0.5 Btu/°F ft 2 (1.02 x 10" JI°C m2) is the conductance through that wall. If urethane foam with k = 0.14 Btu. in./°Fhr ft 2 (0.021 W/m °C) is used for the cornet block, then a I ft (0.305 m) thick foam block has Utoam---0.012 Btu/°F hr ft: (0.068 W/°C m2) conductance, which is negligible compared to Uwa. + Ufoam-~ Uwa,. The amount of highly-insulating foam can be adjusted at the corner to meet this condition as precisely as desired. Probably 5 per cent precision is acceptable: or
Ae~ = (A2- A J/In (A2/AI) Ufo,m ~<0.05 Uw~,. where A2 and A~ are the outside and inside wall areas, respectively. This is used to calculate the effective U • A values for the design, and also the total conductance normalized to solar glazing area. To properly scale down the normalized thermal conductance and simultaneously the normalized heat capacity, it is necessary to preserve the ratio of the outside and inside wall areas (AdAJ since
A~
\A~-1
7. THE USE OF THERMAL-MODEL TEST-BOXES TO MEASURE MICROCLIMATE SHADING EFFECTS ON THE DAILY
COLLEL"nONOV SOLARENERGY Test-cell thermal-models of a proposed passive-solar design can be built on-site to investigate various microclimate effects, such as a south-facing slope. It is iraportant to determine whether the microclimate effect to be studied can be scaled down to the physical size of a small test-box. The effect of shading (e.g. by trees) on the performance of a passive design may not be easily scaled down from a full-size passive collector to a small test-ceil, depending on the size and placement of the shading object relative to the test-box. Consider the schematic passive-solar collector shown in Fig. 3. The collector has width W and height h to intercept the solar radiation (assume due south orienration). Directly in front of the collector a distance L is a shading object of width D and height greater than h. For a low altitude angle winter sun, let us ignore variations in solar altitude for simplicity, and let the solar flux on a vertical south-facing passive collector vary as Q cos ~b, where Q = f(t) is the direct normal solar radiation as a function of time, and ~b varies as 0.262 rad per hr (15°lhr): if t = time in hr, ¢i = 0.262t where t = 0 at solar noon. Thus, the total energy intercepted by the W x h collector is given approximately by
In(A2/AI)
and this correction ratio would be used in the normalized total conductance calculations. However, the thickness of the test-box wall is kept identical to that of the building wall, and hence the test-box is not geometrically similar to the building. The wall thickness must be kept identical to insure the same temporal behavior of the heat wave. It is not possible to preserve the (A2/A1) ratio and have solid, closed corners in the small-test boxes (see Appendix A). Figure 2 illustrates how the corners of a test-box must appear to have (AdAm) in the same ratio as the real wall and maintain the reall wall thickness. This will allow the heat capacities to normalize properly, and will give the correct normalized total thermal conductance through the open corners is resolved by filling them with a block of low-conductance foam-board as suggested by the dashed lines in Fig. 2(b). The foam insulation in the corners will act as a small parallel conductance with the rest of the solid wall and can be treated as negligible, depending on the U values of the foam insulation and the wall material. As an example, if
Q' = 2
ft-t
h • W. Q . cos (0.262t) dt J,-,,,-~ (W/2L)IO.262
f #'tan-I(g/12L)lO'262 + 2 ~j,.o h • [ W - D . sec (0.262t)] • Q . cos (0.262t) • dt. A2
=
A'~
=
Fig. 2. (a) A real wall with closed, solid corners; (b) the equivalent wall, with open corners and (A2/Aj)= (A~/AI). The wall thickness is the same for the real and equivalent (tesbbox) walls.
Small passive-solar test-boxes The first term in the above expression gives the unshaded integrated flux, while the second term gives the integrated flux for when the collector is being shaded by the object. The length of the day is given by 2to. Letting
the passive solar building. Then Q~ = 1653.5 -
= 0.262t, then gives
Q,
If we normalize this to the area of collector, h × W we get 2
(TDD~°~_l ( W ~
~. \w/\/~"
Note that this formula is approximate in that it ignores the case of when the shadow is only partially falling on the collector area. Let us now put some reasonable numbers for size of collector, and size and distance of shading object. Let the shading object be a tree whose branches are 15 ft (4.6 m) in diameter, L = 30 ft (9.15 m) away from the passive collector (e.g, a large evergreen). Let the shadow now fall upon a real, passive solar building of W = 50 ft (15.25 m) width. Also, let Q = 250 Btu/ft 2 hr (787.5 W/m e) be a constant for a 2to= 8hr winter day. Then the integrated and normalized (to glazing area) daily solar energy on that passive collector is given approximately by Q,,
~
•
• (250). tan -t
= 1653.5- 476.7 = 1176.8 Btu (1.24 x 106 J).
2hWQ t 2hDqtan-' =~ sin (0.262 o ) - 0."0~(~L)"
,, 2Q Q = ~. sin (0.262to)-
349
2(250) sin (0.262x4)- 2 15 (250)tan-' ( ~ ) = 0.262 " 0.262 40 = 1653.5-420.8 = 1232.7 Btu (1.30 × 106J).
Now consider a test-box of W = 3 ft (0.92 m) in place of Sun's Rays
Thus, the use of a passive test-box to predict total daily solar energy collected (normalized to area of glazing) for a passive solar building design is in error by only about - 5 per cent in this particular example. The results of this somewhat crude calculation tell us what may be intuitively obvious: that a small test-box can be used to give an accurate prediction of the effects of certain kinds of shading on the daily solar energy intercepted by a passive solar collector. The actual shading phenomenon will be quite dilferent for the building than for the test-box, however. The building will intercept a shadow small relative to its dimension W for a large part of the day, whereas the test cell will have a period of total shadow for a small part of the day (the thermal capacitance of the test cell will smooth out this perturbation a bit). The total daily energy intercepted will be very nearly the same for the passive solar building and the test box. A work of caution. Every building/site has to be analyzed individually for shading/scaling problems. In particular, one does have to be careful about making sure the test box would intercept approximately the same amount of normalized energy as a real building. For example, ff a shadow from an object does not fall on the full height of the building, it would nevertheless completely shadow a test-box placed on the ground for some part of the day. In this situation a test-box would intercept considerably less normalized energy than the equivalent building/collector.
\
D
/
/
l/ s°
of Shodow Cut- Off
Angle
,/,"
J
~
.TAN.I(wI2L)
w
Passive
Solar
Areo of Shadow
• O Sec (ib • D/Cos
Collector
Fig. 3. Schematic of a vertical passive solar collectorof width W facing due south, with a shading object of diameter D located a distance L from the middle of the collector face. The collector is of arbitrary size in the calculations, and can either be a real passive-solar building or a passive solar thermal-model test-box. Area of shadow on collector face, and angle of shadow cut-off are indicated.
350
D . P . GRIMMER 8. CONCLUSIONS
Theoretical considerations have been presented for the test-box thermal modelling of passive-solar building designs. Rather simplistic arguments have been used, and experimental confirmations are necessary. However, the mathematical arguments are straightforward and indicate reasonable quantitative agreement should be expected,
Acknowledgements--The author would like to thank Drs. C.A. Bankston and J. D. Balcomh for helpful discussions on similitude and heat flow.
I.
2. 3. 4. 5.
6. 7.
REFERENCES D. P. Grimmer, J. D. Balcomb and R. D. McFarland, The use of small passive-solar test-boxes to model the thermal performance of passively solar-heated building designs. LASL Report LAUR-77-1323, presented at the 1977 Annual Meeting Am. Section ISES, Orlando, Florida (6-10 June, 1977). R. C. Pankhurst, Dimensional Analysis and Scale Factors. Butler and Tanner, London (1964). G. Murphy, Similitude in Engineering. Ronald Press Co., New York (1950). W. H. McAdams, Heat Transmission. McGraw-Hill, New York(1954). J. D. Balcomb, J. C. Hedstrom and R. D. McFarland, Simulation analysis of passive solar heated buildings--preliminary results. Solar Energy 19, 277 (1977). B. Anderson and M. Riordan, The Solar Home Book. Cheshire Books, Harrisville, New Hampshire (1976). J. D. Balcomb and R. D. McFarland, A simple technique of estimating the performance of passive solar heating systems. LASL Report LA-UR-78-1571, presented at the 1978 U.S. ISES meeting, August 1978, Denver, Colorado (1978).
are given by
k (A2-AI) 1 • ln(A21Ai) Ag'
u' and
c , = ~ (AI+Az)¢ 2 AR For a small test box wall with parameters analogous to the real wall given by A~, A~, A~, r', but with the same values of k, p, c. we can write that
k A2-AI k A~-AI ~ " Ag In (A2IA z) = r' A'~ In (A~IA ~) and
pc (At + A2)¢ pc (Ai + A~)~-' 2
As
-- 2
As
(2)
in comparing the real and test box walls. Rewriting these equations we have that
~' A~-A't In(A21AI).A_t -~= ln(A~lAi) A2- A~ A'~
(3)
and
~" = ( A, + A2~ ~ ¢ \AI + A~/ A s"
(4)
Then for (3) and (4) to be simultaneously true, we must have that A_~2 A~- A[
A ~ In(A~/A[) APPENDIX A The use of eHective wall areas in the calculation and scale-down of normalized thermal conductances and heat capacities for test boxes
(1)
In (A21AI)
Ai + A2 A2-A~ -A[+A~"
So, then
A~2- AI 2 A22- A~2 A 's2. In (A ~IA 'I) • As2. In (A 21A j)
It is known that the effective wall area for heat flow is given by the log mean of the outside and inside wall areas:
(A~IA's)2- (AI/A~)2 "(A21As)2-(AI/Ar) 2. In (A~IAI) In (A21A I)
A2- AI Ac~ In (A21A~) where A2 and A, are the outside and inside wall areas, respectively. For a given real wall, the total thermal conductance (u') and heat capacity (c') are given by Ut=
k A,.-AI ~"In (A2/AI)
where k is the thermal conductivity and r the wall thickness: and
Identifying term by term, the above is true if A~ = A2. Ai = Ai. A~ = AI A~ As ' A[ As ' AI A,' or A__.~= A_~. Ai = A~. A_~= A_~
A2
A~" AI
A~ = A~ = A_~
A2 c'=
~-(A, + Az)¢
where p is the density and c the specific heat. Then the total conductance and heat capacity, normalized to glazing area AR,
A s' A2 A (
Thus, AI
As
and hence for a given A'rlAs chosen: it is not possible to have A21A~ = A~/AI unless the massive wall does not extend all the way to the corned': the corners must be left open and filled with a low conductance, low thermal capacity material.