2. Supply fan
opaque
surfaces
structure
transparent
fenestration
surfaces
Figure (I) Representation of Subsystem using control volume concept for
prototypical buildings
The thermal conditions and air humidity can be directly
controlled by the ventilation air flow.
A. Thermal Transmittance
The conditioned space temperature represents the principal
part of thermal building output.
To readily modeled the behavior of an overall HVAC
system, under thermal analysis, by applying conservation of
energy. Energy can enter and exit a subsystem control volume
by heat transfer. Energy also enters and exits with flowing
streams of matter which is dominant on HVAC process. The
rate of energy stored is defmed as:
dE . .
_
s
= L.i Ein - L.i Eout (1)
dt
where
dEs
is the rate of change in the total storage energy of the
dt
system and L.i Bin - L.i Bout is the rate of net energy transfer.
B. Moisture Transmittance
Moisture transfer processes are not only caused by internal
generation processes and air migration from outside but also by
the condition of the air being injected to the zone by an air
conditioning system. To follow amount variation of moisture in
air flow, must applied conservation of mass on subsystem
control volume. For a control volume concept with multi
dimensional flow at a multi-inlet and a multi-outlet system is.
(2)
The left side of this equation represents the time rate of
change of mass contained within the control volume, mi
denotes the mass flow rate at an inlet, and me is the mass flow
rate at an outlet.
C. Model Linearization
The building model can be represented by a large number
of nonlinear, partial differential equations. Most of the
equations are related to moisture flow and heat transfer
involving partial derivatives of time and space. Solution of a
set of these equations is very difficult and therefore some
simplifying assumptions have to be made. For purpose of
analysis, the AC system is divided into a number of sections,
and for each lumped parameter section, the humidity ration and
air temperature are assumed to vary only in the axial directions
and linearly with space. Linearising the partial differential
equations reduces these equations to ordinary linear differential
288
equations by applying small perturbation and difference
equation techniques.
III. MODEL DEVELOPMENT
The proposed model developed to determine the optimal
response for indoor temperature and humidity ratio. By taking
temperature and moisture transmission based on empirical
methods RLF. The main advantage of this hybrid model
approach is to get relationship between indoor and outdoor
variation data like temperature and humidity ratio. With the
RLF approach, the subsystem method treats outdoor air
temperature and humidity ratio as the independent variable in
the analysis.
The thermal mass of the building structure creates a load
leveling or flywheel effect on the instantaneous load. There are
two factors associated with the heat gain/losses to/from
building structure as a result of outdoor temperature and solar
radiation. These factors are related to opaque surface (walls,
ceilings, roofs and doors) and transparent fenestration surfaces
(windows, skylights and glazed doors).
A. Opaque Surfaces
The heat balances of Opaque surface as following the law
of conservation of energy can be written as:
(3)
where L.i Qin and L.i Qout= heat gain and loss through walls,
ceilings, and doors, (W) ,Mw1cpwl =heat capacitance of walls,
ceilings, and doors,( ]IK).
By applying RLF method on equation (3) to get transfer
function as follow:
TW1in(S) = [G1,1 G1,2 G1,3] [Tk:)]
Tr(s)
(4)
kl
G _ 1
G __k3_
where G1,1l =
('5s+1) , 1,12 - ('5S+1) ' 1,13 - ('5S+1)
MwlCPwl
LjAw.UjOFt
TS = kl = J
LA U OFt+LA h LjAw·UjOFt+LjAw·hi·] w] ] ] w] 'J
J J J
(function of thermal resistant and outside temperature),k2 =
LjAwjUjOFb+LjAwjUjOFrDR
-----"-------'----, (function of thermal resistant and
LjAwjUjOFt+LjAwjhij
solar radiation incident on the surfaces), (DC), k3 =
LjAw.hi·
____ J'----'J'----__ , (function of thermal resistant and
LjAwjUjOFt+LjAwjhij
convection heat transfer), Aw = net surface area, (m2), CF =
surface cooling factor,( W1m2), U = construction U-factor,
WICM2. K), M =cooling design temperature difference, (K)
OFt, OFb, OFr = opaque-surface cooling factors, DR = cooling
daily range, (K).
B. Transparent Fenestration Surfaces
Heat gain through a fenestration is consisting of two parts.
The first part is the simple heat transfer due to the difference
temperature of internal and external sides and the second part
is the heat transfer due to solar heat gains as shown in
equation (5).
3. (5)
Where CFren = UNFRC(M - O.46DR) + PXI x SHGCx
lAC x FFs, Qren = fenestration cooling load, (W), Aren =
fenestration area (including frame),( m2 ), CFren = surface
cooling factor, (W1m2), UNFRC = fenestration NFRC heating
V-factor, Wl(m2• K), NFRC = National Fenestration Rating
Council,M = cooling design temperature difference, (K), DR
= cooling daily range, (K), PXI = peak exterior irradiance,
including shading modifications, ( W1m2 ), SHGC =
fenestration rated or estimated NFRC solar heat gain
coefficient, lAC = interior shading attenuation coefficient, FFs
= fenestration solar load factor.
PXI is calculated as follows:
PXI = TXEt (unshaded fenestration) (6)
PXI = Tx[ Ed + (1 - Fshd)ED] (Shaded fenestration) (7)
where PXI = peak exterior irradiance, (W1m2), Ev Ed, ED =
peak total, diffuse, and direct irradiance, ( W1m2 ), Tx =
Transmission of exterior attachment (insect screen or shade
screen), Fshd = fraction of fenestration shaded by permanent
overhangs, fms, or environmental obstacles.
The fenestration inputs are outdoor temperature To(s) ,
indoor temperature Tr(s) and conditioned place locationfDR
, while output is inside glass temperature Tgin (5) as shown in
transfer function below.
(8)
where G =
Rgft
G =
1
1,14 (ftRg+1)(Tg S+1) ' 1,15 (ftRg+1)(Tg 5+1) ,
G -
-Rg _ Cag Rg
R _ 1
116 - ( )(
Tg - --
9 -, ftRg+l Tg 5+1) ftRg+l l.jAfenj hij
fDR = Lj Arenj UNFRCj x O.46DR , fl = Lj Arenj UNFRCj ,
(WIk).
C. Slab Floors
The slab floors of heat balances as following the law of
conservation of energy can be written as:
(9)
where Li Qin and Li Qout= heat gain and loss through slab
floor, (W) and MwlcpWI =heat capacitance of slab,( ]IK).
Wang [12] found that heat loss from an unheated concrete
slab floor is mostly through the perimeter rather than through
the floor and into the ground. Total heat loss is more nearly
proportional to the length of the perimeter than to the area of
the floor, and it can be estimated by the following equation for
both unheated and heated slab floors:
289
(10)
where QSlabout = heat loss through slab floors, (W), ft = heat
loss coefficient per meter of perimeter, W/(m'K), P =
perimeter or exposed edge of floor, (m), Tslabin = inside slab
floor temperature or indoor temperature, (DC ), To = outdoor
temperature, (DC).
Where ASHREA [10] calculated the input of cooling load
to slab floors as follows:
QSlabin = Aslab X Cfslab (11)
where Aslab = area of slab,( m2) Cfslab = slab cooling factor,
(W1m2).
The slab floors subsystem inputs are slab floors area
(Aslab) and outdoor temperature To, while output is inside slab
floors temperature Tslabin (5) as shown below.
Tslabin (s) = [G1,7 G1, 8] [A
¥:b] (12)
h G
(1.9-1.4hsrf) ftP Cslabw ere - G - T -1,17 - (Tslab5+1) , 1,18 - (Tslab5+1)' slab - ftP
hsrr = effective surface conductance.
D. Conditioned Space
The condition space is everything surrounded by walls,
windows, doors ceilings, roofs and slab floors that means
condition space include air space, furniture, occupant, lighting
and apparatus which emitting heating load as shown in figure
(2). By means of conditioned space control volume we analyze
temperature and humidity ratio effectiveness by applying
conservation of energy and mass by using RLF method. To
reduce the complexity of calculation, temperature and humidity
ratio will be separated to calculate the variation each of them.
J) Thermal Transmission: Sensible heat gain can be
evaluated by applying thermal balance equation on
conditioned space to get components thermal load. The most
critical components affecting the conditioning space are:-
(1) through opaque surfaces (walls, roofs, ceilings, and doors),
(2) through transparent fenestration surfaces (windows,
skylights, and glazed doors), (3) because of occupants,
lighting, and appliance, (4) caused by infiltration, (5) caused
by ventilation, (6) through slab floors and (7) caused by
furnishing and air conditioning space capacitance.
2) Moisture Transmission:
The rate of moisture change in conditioned space is the
result of three predominant moisture sources: outdoor air
(infiltration and ventilation), occupants, and miscellaneous
sources, such as cooking, laundry, and bathing. We Applied
conservation of mass on the components of conditioning space
to get general formula as following.
rate of moisture change
= rate of moisture transfer
+ rate of moisture generation
4. d moisture
=
dt
Li input moisture rate -
Le output moisture rate +
Lgen. moisture generation rate
���3mv.n.'�..TO.l . � '::' ::rt �
w � .
o Inside heat gain
(l3)
Figure (2) Illustrate heat and humidity flow in/out of conditioned space
A complete description of the plant behavior for the two
main output components is given by combining space model
equations with building model equations. The whole
subsystems model equation of conditioned space is presented
in Eqn. (14) which shown at the bottom of the page.
where G -
k
wl
G -
1
G _ kSlb
1,9 - f2('6S+1)
, 1,10 - f2Rg(�6S+1)' 1,11 - f2(�6S+1)
G - To G -
1
G - 0 G - 01,12 - f2(�6S+1)' 1,13 - h(T6S+1)' 1,14 - , 1,15 - ,
G2,9 = 0 , G2,10 =
0 , G2,11 = 0 , G2,12 = 0 , G2,13 = 0
G =_
1
_ G =
1
k ="· A h.2,14 (TrS+1)' 2,15 hfgmexh(TrS+1) wi L..J Wj Ij ,
kSlb =
LJ· Aslb·hi· , f3 =
Cs X AL X IDF + mvenCPa , (Wjk),) )
(function of the mass flow rate of ventilation supply air) Cs =
air sensible heat factor, W/(L's'K), AL = building effective
leakage area, cm2
, IDF = infiltration driving force,Lj(s, cm2
)
1
12 =
Lj Awjhij +
R
g + Lj Aslbjhij + Cs X AL X IDF +
mvenCPa Wjk , T6 =
�:f, (sec.) , Caf = heat capacitance of
indoor air and furniture, mven= mass flow rate of ventilation
supply air, (kg/s) rllinf= infiltration air mass flow rate, (kgjs),
mexh =
mven + min!, f4 =
ffen+ 136 + 2.2Acf + 22Noc ,W,
ffen= direct radiation, ,( W), Wo= humidity ratio of outdoor,
(Kgw/Kgda), Qig,l = latent cooling load from internal gains,
(W).
To integrate of building structure (opaque surfaces,
transparent fenestration surfaces and slab floor) and
G1,10(S)
G2,1O(S)
G1,11(S)
G2,11(S)
G1,12(S)
G2,12(s)
G1,13(S)
G2,13(S)
290
conditioned space into integration part is represented by figure
(3). From figure (3) the input variables are (1) k2 =
perturbations due to thermal resistance and solar radiation
incident of building envelope, (2) To (s) = perturbations in
outside temperature, eC), (3) IDR= location factor, (4) ASlb =
slab floors area, (m2), (5)wo(s)= perturbations in outside air
humidity ratio, (6) Tr(s)= Indoor temperature, (DC), (7) Qi9,l =
perturbations of internal latent heat gain, (w), (8) f3= function
of the mass flow rate of ventilation supply air, (W/K) and (9) 14
= perturbations of internal sensible heat gain due to occupants.
Output variables are (1) Tr(s) =
Room temperature or
conditioned space temperature and (2) wr(s) =
Room
humidity ratio or conditioned space humidity ratio.
IV. ApPLICATION TO RESIDENTIAL BUILDING
The model transfer function for residential building is
shown in figure (3). In order to be able to implement model
transfer function, all the parameters of Eqns. (4), (8), (12) and
(14) are determined. So that we have to describe the building
structure that we are applied on this modeling approach. The
geometry of the building figure (4) is identical to one in
ASHRAE [10] used to investigate the parameters of the
developed model. The building construction characteristics are
documented in Table (1).
V. SIMULATION RESULT AND DISCUSSION
The building house which we used as a test house to verify
the model, it is typical one-story house has a simple structure.
,
----- ...
Figure (3) Subsystem model transfer function relations
G1,14(S) G1,lS(S)]
G2,14(S) G2,lS(S)
TWl. (s)m
Tgin (s)
TSlbin(s)
13
14
WoeS)
Qi9,l
(14)
5. The general hose characteristics are overall area is 248.6
m2 while overall area less garage area is 195.3 m2, the gross
windows and wall exposed area are 126.2 m2 while the net
wall exterior area is 108.5 m2, the overall house volume less
garage volume is 468.7 m3 and construction characteristics
are documented in table (I). From the house characteristics we
deduce that it has heavy thermal and moisture masses and this
indicate to existing thermal and moisture lag which is depicted
in figure (5) where figure (5) illustrate the house inside
temperature and humidity responding due to outdoor
temperature and humidity random variation.
A. Model validation
To validate the derived models, a comparison test was
carried out by the indoor model conditions and other different
calculation method. The building properties and weather data
obtained for the Kuala Lumpur city have been used to
calculate indoor condition. By means of natural ventilation
applied on building model, where outside climate condition
was only affected on the indoor condition. The behaving of
indoor temperature and humidity ratio were obtained as shown
in figures (6) and (7). The aim of simulation modeling is to
represent as closely as possible the underlying physical laws
and other principles of cooling load calculation by using
software. According to house characteristics the indoor
temperature and humidity ratio are calculated every one hour
for 24 hours by using software (simulation in the building
design professions) [13]. The data result of comparison
between building simulation output and calculation result by
software program, is shows partial agreement as shown in
figure (6) and (7). Because we concern with model behavior
we did two types of test, one nightly and the other daytime.
Both of tests are applied natural ventilation mode, the outside
house air supply due ventilation and infiltration are 41 Lis and
17 Lis respectively [II].
I' --'*1'"--
7 3-----=1
-"""'N
Figure (4) The geometry of the building has been chosen to get model
Sample No.
Figure (5) Indoor temperature and humid response to random outdoor
variation
291
36
35 -cucklor'lmperlltU'l
32
0 Indoor calculated tllmp8flll�.
-"
t,.k 30
� ;:
� 27
� 26
� 25 .
E 24 .
{! 23
22
21
20
"
"0 -
Tlml.(hour)
Figure (6) Indoor temperature variation due to outdoor temperature variation
TABLE 1. MATERlAL PROPERTIES OF MODEL BUILDING CONSTRUCTIONS
Component
Roof/ceiling
Exterior
walls
Doors
Floor
Windows
Construction
Description
Flat wood frame ceiling (insulated
with R-5.3 fiberglass) beneath
vented attic with medium asphalt
shingle roof.
Wood frame, exterior wood
sheathing, interior gypsum board,
R-2.3 fiberglass insulation.
Wood, solid core.
Slab on grade with heavy carpet
over rubber pad; R-0.9 edge
insulation to I m below grade
Clear double-pane glass in wood
frames. Half fixed, half operable
with insect screens (except living
room picture window, which is
fixed). 0.6 m eave overhang on
east and west with eave edge at
same height as top of glazing for
all windows. Allow for typical
interior shading, half closed.
Good
A. Natural ventilation of day
Factors
U = 0.031 18 (W/
(m2.K))
aroo!=0.85
U=51 W/(m2.K))
U=2.3W/(m2.K)
Rcvr = 0.21 Cm2• K)/
W)
Fp=85W/Cm2.K)
Fixed: U 2.84
W/Cm2.K); SHGC =
0.67
Operable: U = 2.87
WI(m2K); SHGC
0.57;
Tx=0.64
lACe!=0.6
Aul= 1.4 cm21m2
Daytime natural ventilation test was applied in Kuala
Lumpur city were last year maximum mean temperatures was
33 'c, where the mean humidity ratio for the same maximum
temperatures was 0.0209 Kgw/Kg. (Kg water vapor/Kg dry
air). According to house characteristics the indoor temperature
and humidity ratio are calculated by using software simulation
in the building design professions as 20 'c and 0.00805
Kgw/Kg. respectively [13]. The indoor temperature affected by
many types of effectiveness some of them doing to rise
temperature like indoor outdoor difference temperature,
ventilation and filtration supply airflow rate, windows and
wall exposed area and incident solar radiation intensity while
the other effectiveness are doing to suppress rIsmg
temperature like opaque envelop thermal capacitance, weight
of air and furniture inside house and slab floors area. While
humidity ratio is a function of ventilation, filtration, outside
humidity ratio and internal latent load gains as proved in eqn.
(14). Figure (8) demonstrate high rising gradient at initial state
then reduced due to alteration of effectiveness with time. The
simulation process perceIvmg that the output indoor
temperature was greatly affected at input outside temperature
while the other inputs were barely affected.
B. Natural ventilation of night
In this test we figured out the minimum mean temperatures
of Kuala Lumpur city for last year was 18 (C). Whereas the
minimum mean humidity ratio for the same minimum
temperatures was 0.0085 Kgw/Kg.. According to house
characteristics the indoor temperature and humidity ratio are
6. Figure (7) Indoor humidity ratio variation due to outdoor humidity variation
Murll wntllU on response of d.y
r---.-------.------,--�,__-_.__'-___,:-____.-____,--_.__-_,'O'022
�;
E
=:===r::�:�b���+��.O.0206
'""t..
; ::::�:��.................,.. .........-1'0.0164.1
.................,...............-I'O .Ol li lC
O.0I36�
_-=_:::, �c;:_::-;""=....,;:;;.,c-:••:;;:.""O.D122 i
····· lndoorI'MrMy rltlo 0.0108 E
............ .................:..
...............:..
..............)................. ....
.............:................. ............
-
oudoort..,.,IIIIUfI 0.0094
.E
.!--+---+---+-----;e--±--+--+.-��--�1---�·��·�2 :8·008
Figure (8) Indoor temperature and humidity ratio response to natural of day
calculated by using software simulation in the building design
professions as 27 (C) and 0.018 Kgw/Kg. respectively
[13].We plugged in the input data on model to get indoor
temperature and humidity ratio as shown in figure (9).
Although there was no incident solar radiation subjected to
model, declining temperature was very sluggish. That's
because of indoor outdoor temperature difference and thermal
storage for opaque envelope, furniture, internal walls and slab
floors all these factors influence a model's behave which
illustrated in figure (9).
C. psychrometric process line analyses
The psychrometric chart is widely used to illustrate and
analyze the change in properties and the thermal
characteristics of the air-conditioning process and cycles [14].
Figure (10) represent process line of natural ventilation of a
night and a day. From figure (10) we observe that both of the
two processes are intervening change in temperatures and
humidity ratio. But both of the processes are equal of airflow
rate induced by ventilation and filtration also the processes
time duration and temperatures difference are almost equal.
Hence the processes heating load gains at daytime grater than
cooling load at nighttime. This is a rational illustration
because existing of the solar radiation at daytime.
;;:_ 27�. . . .. , . .. . ......
' 25.5 1"···········,·················
� 24 ."',,:...••,.................
::: 22.Sf-·······-s.,J-:::····..····
� " f- ··········""'c
=
t 19.5 f-.... ..........;.........;:,-
� "1---+---
IS,
- CIUdo«I�l.h..
· · · ·r· · ·· i===�O.0116 �
-----Indoor�fIIIio O.Ol64 t
................+ .............0.0152.1
.................j.............. ;. ·············0.014 :.::.
.............O.0128ii
a.o11s i
-+---+-----;----+---+0.0104 §
O.0092
J:.
Figure (9) Indoor temperature and humidity ratio response to natural of night
£�IIO.�__
--.>---
"
T�ur··"Cl1W
Figure (10) cooling and heating load process for night and day natural
ventilation
292
II. CONCLUSION
From analyzing internal house condition (dry bulb
temperature and relative humidity) for two processes line as
shown in figure (10). Obviously both processes line are
passing through comfort zone in the psychrometric chart [15].
So we infer there is interval period of time cannot use air
conditioning, and this period depends on type of process. For
natural ventilation of day based on process line in figure lOwe
can figure out the time from figure 8. Therefore we need air
conditioning after 1.5 hour when outside temperature gets
maximum value. While natural ventilation of night depending
on figure 9 and figure lOwe cannot use air-conditioning after
2 hours when outside temperature gets minimum value.
Where mechanical intervention is necessary to control
indoor air quality, by which can greatly extend time period
cannot use air-conditioning or reduce time period of using air
conditioning by controlling on variable-supply airflow rate for
natural ventilation by using variable speed fan.
REFERENCES
[I) D.G. Stephenson and G.P. Mitalas (1967), 'Cooling Load Calculations
by Thermal Response Factor Method', ASHRAE Trans. 73 (I) 508-515.
[2) T. Kusuda (1976), NBSLD: The Computer Program for Heating and
Cooling Loads in Buildings, NBS Building Science Series No. 69,
National Bureau of Standards, Washington USA.
[3) J.A. Clarke (1977), Environmental Systems Performance, PhD Thesis,
University of Strathclyde, Glasgow UK.
[4) 1. ed. Lebrun (1982), Proc. Int. Con! on System Simulation in Buildings,
Commission of the European Communities, Lie'ge, Belgium.
[5) S.O. Jensen, Ed. (1993), Validation of Building Energy Simulation
Programs, Part I and II, Research Report PASSYS Subgroup Model
Validation and Development, CEC, Brussels, EUR 15115 EN.
[6) R. ludkoff and 1. Neymark (1995), International Energy Agency
Building Energy Simulation Test (BESTEST) and Diagnostic Method,
lEA Energy Conservation in Buildings and Community Systems
Programme Annex 21 Subtask C and lEA Solar Heating and Cooling
Programme Task 12 Subtask B.
[7) ASHRAE (1998), Standard Method of Test for the Evaluation of
Building Energy Analysis Computer Programs: ASHRAE Standard
140P, Working Draft 98/2, American Society of Heating, Refrigerating,
and Air-Conditioning Engineers, Atlanta USA.
[8) K.J. Lomas, H. Eppel, C. Martin, and D. Bloomfield (1994), Empirical
Validation of Thermal Building Simulation Programs Using Test Room
Data, Volume 1: Final Report, lEA Energy Conservation in Buildings
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and Cooling Programme Task 12.
[9) C.S. Barnaby, J.D. Spitler, and D. Xiao., 2004, 'Updating the
ASHRAE/ACCA residential heating and cooling load calculation
procedures and data' (RP-1199). ASHRAE Research Project.
[10) C.S. Barnaby, J.D. Spitler, and D. Xiao. 2005, 'The residential heat
balance method for heating and cooling load calculations (RP-1199).
ASHRAE Transactions 111(1):308-319.
[I I) ASHRAE, 2009, 'residential cooling and heating load calculations'
handbook-fundamentals, chp. 17, American Society of Heating,
Refrigerating, and Air-Conditioning Engineers.
[12) F.S. Wang 1979. "Mathematical modeling and computer simulation of
insulation systems in below grade applications". ASHRAEIDOE
Conference on Thermal Performance of the Exterior Envelopes of
Buildings, Orlando, FL
[13) Cooling load calculation software available from URL:
[14)
[15)
httpllwww.carmelsoft.com. Accessed March 24, 2010.
T. Grondzik Walter, 2008 'Air Conditioning System Design Manual'
second edition, Ashrae Special Publications NE, Atlanta
ASHRAE. 2004. ANS/IASHRAE Standard 55-2004, "Thermal
Environmental Conditions for Human Occupancy". Atlanta: Inc.