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A study on energy consumption, cooling load estimation and thermodynamic
analysis of solar assisted VARS of a building in a warm and humid climate
Arun Kumar Shukla1,Gopal Tripathi1, Debendra Chandra Baruah1
arunkumarshkl@gmail.com 1
gopaltripathi10@gmail.com 2
baruahd@tezu.ernet.in 3
Department of Energy, Tezpur University, Napaam, Sonitpur, Assam 784028, India
Abstract
Energy Performance Index (EPI) is a tool primarily used to assess energy performance of
building. According to the Indian Society of Heating, Refrigeration and Air Conditioning
Engineers (ISHRAE), EPI ranges between 200 to 400 kWh per square meter of built up area per
annum with major share attributed to heating, ventilating, and air conditioning (HVAC) for
residential, commercial and industrial applications (~40%). Therefore, HVAC remains the
primary focus area for energy savings. A building receives heat from several sources and there is
a wide variation of its amount depending upon the types of building, seasons and locations.
Optimal management of thermal load to provide a comfortable environment based on the
prevailing ambient conditions involves complex analysis. The present study considers a Seminar
room of an academic Department to analyse the thermal energy balance and finally estimating
the optimal plan of HVAC energy. Parameters, viz., orientation, location, size, capacity
(occupants) anddesign of the rooms are considered to estimate different components of thermal
load and hence the required cooling load using a standard procedure. About 15 ton of
refrigeration load for maintain a comfortable living environment during summer is expected to
be sufficient for the 110 occupants. Based on the activity and uses, sensible heat ratio of the
seminar room is estimated as 0.75 andheat gain from different direction isalso estimated.
Conservation of HVAC energy through change of orientation of room is also possible as
indicated by the study which is expected to be useful for planning of new building.
Keyword :Energy performance index, Heat gain, cooling load.
1. Introduction
The environmental problem is one of the most serious problem faces by humanity In present
days. Energy consumption by industries and buildings are also growing rapidly due to increase in
population and living standard of human being. Extensive uses of fossil fuel is also responsible
for global warming, which is increasing the earth temperature rapidly. About (~72%) of world
energy is consumed by infrastructure, industry, commercial buildings, residential houses, and
markets.[1] In a large building or complex, which is air‐conditioned, about (~60%) of the total
energy requirement in the building is allocated for the air‐conditioning plant installed to use the
cooling purpose.[2-7] so HVAC is an energy extensive process, There is lots of scope of energy
saving by exact prediction of the cooling and heating load, selection of proper sizing of the heat
ventilation air-conditioning (HVAC) system and optimal control of the HVAC systems are
important to minimize energy consumption. Cooling loads are affected by the external climates
such as outdoor temperature, solar radiation and humidity. [10] Because the energy consumption
in buildings depends on the climatic conditions and the performance of heating ventilating and
air conditioning (HVAC) systems changes with them as well, better design in building HVAC
applications that take account of the right climatic conditions will result in better comfort and
more energy efficient buildings. Calculation of thermal load of building is very essential to find
exact air-conditioning equipment and air handling unit, to achieve comfort operation and good
air distribution in the air-conditioned zone. The International Energy Agency has indicated that
energy efficiency is likely to supply more additional energy than oil through to 2030, supporting
energy-efficient innovations. The HVAC industry is also eyeing a huge opportunity through
energy efficiency for increased scale of business. HVAC systems, when run efficiently, can help
buildings achieve a better EPI of 100 to120 kWh/sq. m per year, which makes the buildings
energy efficient. Such concepts have brought a paradigm shift in HVAC technologies and trends,
[9,10] where majority of the changes are aimed at energy efficiency and higher energy
performance. HVAC systems are now designed to meet energy demand and also follow the
standards of Energy Conservation Building Codes (ECBC), American Society of Heating,
Refrigerating and Air Conditioning Engineers (ASHRAE) standards to achieve higher levels of
Green building Leadership in Energy and Environmental Design (LEED) rating.
Energy efficiency of the building are also improved by improving envelope design and
efficient mechanical equipment performance such as heating, cooling, domestic hot water etc.
The concept, which is mentioned for energy efficiency plays a large role in energy saving.
Buildings account for 35% of total final energy consumption in India today, and building energy
use is growing at 8% annually. Rawal et al.[8-10] In building, envelope and overall heat transfer
coefficient are the most important components with respect to total heat gain of the whole
building and the building envelope. In addition, orientation also plays a critical role on the
thermal comfort of occupants. Studies of thermal comfort show that the way indoor thermal
environment is evaluated and thus depends on the relationship between people, climate and
building; these can vary over time. Since larger proportion of energy are being consumed in
buildings, a way to alleviate the ever growing energy demand is to have more energy efficient
building designs and proper building energy conservation programmes. Kulkarni et
al[5].Optimized cooling load for a lecture theatre in a composite climate in India. The lecture
theatre had a dimension of 16m×8.4m×3.6m and was situated at Roorkee (28.58oN, 77.20oE) in
the northern region of India. The monthly, annual cooling load and cooling capacity of air
conditioning system was determined by a computer simulation program. They reported that the
use of false celling, ceramic tiles on roof and floor, electro chromic reflective colored, 13mm air
gap, clear glass gave the best possible retrofitting option. Suziyana et al[5]. Analyzed the heat
gain and calculated cooling load of a Computer Laboratory and Excellent Centre Rooms in the
Faculty of Mechanical Engineering, University Malaysia Pahang by using cooling load
calculation method and cooling load factor method based on 9 ASHRAE 1997 and then verified
by data provided by contractor of building. From this calculation, it was found that the highest
heat gain in the Computer Laboratory Room and in Excellent Centre Room is 20458.6 W and
33541.3 W respectively.[10-14]
Despite of enormous work carried out on building energy efficiency in the developed
world, a very few research work currently exist in the subject area in Tezpur Assam India
location. This study focused on the study of building energy consumption from doors, windows,
thermal mass, Infiltration losses and also load sharing by different direction of wall and part of
the building, which will be helpful for building energy saving and selection of proper air
conditioning system without compromising the thermal comfort of the occupants.
2. Estimation of Air-conditioning load
The building considered in this study is situated in Tezpur and located at Latitude:
26.6528° N Longitude: 92.7926° E in Sonitpur district of Assam, India at an elevation of about
48 meters above mean sea level. Tezpur is characterized with a hot and humid climate and
summer start from the month of April and end month of July. The average temperature during
this time of the year is usually 18oC to 36oC. But the highest temperature is recorded during the
month of May or early June, just before the start of the monsoons. Seminar hall present in the
Department of Energy, Tezpur University Assam, India, was selected as case study. The capacity
of the hall isare 110. The function mainly as students’ lecture halls. The maximum cooling loads,
sensible heat ratios and total dehumidify air quantities of the sample building (Tezpur seminar
hall) were calculated using energy balance method (sol-air temperature method).The method of
calculation were as per the standard methodology of American Society of Heating, Refrigeration
and Air-Conditioning Engineers (ASHRAE)) fundamental handbook guideline for cooling load
estimation
Cooling Load is function of temperature, relative humidity and inside occupants activity
so it varies with monthly and hourly basis. Climatic study (Solar Insolation, Ambient
temperature, Humidity,) of the given location has been done which is plotted in Fig 1. And Fig 2.
Cooling load calculation of the building mentioned above has been done for the month of June
which showed the maximum temperature rise., found to be sufficient for any time of the year.
2.2.Inside Design condition
The amount of cooling that has to be attainedto keep buildings comfortable in summer and
winter depends on the desired indoor conditions and the outdoor conditions on a given day.
These conditions are, respectively, called the “indoor design condition. For most of the comfort
systems, the recommended indoor temperature and relative humidity are as follows [1-2]
DBT (Dry bulb temperature) – 22.78 oC to 26.11 oC, and RH – 60% for summer
DBT(Dry bulb temperature) – 22.11 oC to 22.22 oC and RH – 20 to 30% for winter
The inside set temperature of seminar hall has been taken 23oC for cooling load calculation
2.3 Climate condition
1
19
37
55
73
91
109
127
145
163
181
199
217
235
253
271
289
307
325
343
361
0
500
1000
Iz
Days of a year
Solar Radiation (W/m2day)
Fig.1. Daily average solar radiation of year 2016
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
80
100 Humidity
Temperatu
re
Month
Fig 2.Monthly average Tezpur Temperature (oC) and Relative humidity(%) variation
2.2 Geometrical and physical properties of building
Different dimension of the building measured, Which is shown Fig3. and Fig4.
Information on Building wall glass and door material were calculated.
East (E) and South (S) walls are facing direct solar radiation, west (W) and north (N) do
not receive direct solar radiation.
Fig3. Top view of the seminar hall Fig 4 .Side view of seminar hall from North and south side
Wall material Concrete + brick Thermal conductivity of
brick (K)
0.8 W/m K
Thickness of doors 40 mm Thermal conductivity of 0.158 W/m K
Windo
ws
Doors
E
W
N
S
Ceilin
g
H
B
L
wood (K)
Doors material Wood Thermal conductivity of
glass (K)
0.78 W/m K
Convective heat transfer coefficient (h) 22 W/m2 K Absorptivity of wall(ε) 0.6
No of Windows 6 Transitivity of glass 0.83
No of doors 3 Thermal conductivity of
concrete
0.7 W/m K
Capacity of seminar hall 110 Length of seminar hall (L) 15.6 m
Area of seminar hall 120.12 m2Width of seminar hall (B) 7.7 m
Volume of seminar hall 408.4 m3Height of seminar hall (H) 3.4 m
Table1. Building dimensions and material properties
3. Estimation of heat gain
As we know the temperature variation throughout the year so cooling load is maximum in the
month of June. The total heat required to be removed from the space in order to bring it at the
desired temperature 23oC and relative humidity (60%) by the air conditioning equipment is
known as cooling load or conditioned load. This load consists of external and internal loads.
3.1 Radiation received by each wall and Sol-air temperature calculation
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
80
100 9AM
3PM
12A
M
Months
Elevation angle (α)
Fig 5Solar radiation on tilt surface [15]Fig 6. Elevation angle monthly variation
Elevation angle varies with month and hour so walls of the building which are taken for study
faces different radiation at different time of the day, which is measured at three time of a day
9AM, 12AM and 3PM, which is shown Fig 6. and corresponding solar radiation calculation has
been done for different directionEast, South and West /North.
α = Elevation angleβ (Tilt angle for wall) = 90oβ (Tilt angle of roof Fig4.) = 22o
Equation 1,2 and 3 written by using Fig 5. and Fig 6. [13-15]
Ihorizontal = Iincident
×sin α
(1)
Iaverage solar radiation on the wal l
= Iincident
α
(¿+β)
×sin ¿
(2)
Iaverage solar radiation on the wall
=
(α+β)
¿
horizontal ×sin ¿
I¿
¿
(3)
Sol-air temperature (Tsol-air) is a variable used to calculate cooling load of a building and
determine the total heat gain through exterior surfaces.
It combined the effect of ambient temperature and solar radiation.[2]
Tsol-air = Tamb +
ε × I average solar radiation on the wall
hair
(4)
Tsol-air is calculated by above equation which is function of ambient temperature, wall material,
solar radiation received by wall and convective heat transfer coefficient. Solar radiation received
by each wall calculation has been shown above and ambient temperature, wall material taken
from Fig.2.and Table 1.
Step wise load calculation is shown below.
3.2. Sensible heat load
Heat gain through wall, window, roof and door due to temperature difference between
inside (Ti = 23oC) and outside (Tsol-air) is known as sensible heat load.[4-8]
Q= U× A × (Tsol-air– 23 ) (5)
U=
1
1
h0
+∑X
K+1
hi
X= Thickness of wall and plaster
U value calculated by using above formula which is function of material and dimension of the
given component of building, taken from Table 1.
3.2. Internal heat generation
Heat generated by occupants, lighting and other electrical devices are known as internal heat
generation.For electronic equipment W is total wattage [3,4]
Q= W× Fu× Fsa × CLF(6)
Fu = 1, Use factorFsa = special allowance factor, taken as 1.25
For Internal heat generation by occupant depends on the type of activity carried out by them
For a person seated at rest would dissipate = 105 W , Which is supported by Table2. Given by
ASHRAE as standard for different activity
Table 2. Heat gain from occupants at conditioned space [2-4]
Degree of activity Typical
application
Total heat (watt) Sensible
heat
(W)
Latent
heat
(W)
Adult
Male
Adjusted
Seated at Theater Theater, matinee 115 95 65 30
Seated, very light
work
Offices, hotels,
apartment
130 115 70 45
Moderately active
office work
Offices, hotels,
apartment
140 115 75 55
Standing, light
work:walking
Department store:
retail store
160 130 75 55
Walking, Standing Drug store; retail
store
160 145 75 55
Sedentary work Restaurants 145 160 80 80
Light bench work Factory 235 220 80 140
Moderate dancing Dance hall 265 250 90 160
3.3 Infiltration load
Outside air infiltrate through cracks of window, door and also air infiltrate due to opening and
closing of door so conditioning of this air some load is required, which is known as infiltration
load.Infiltration load calculation was done by empirical procedure recommended by carrier
handbook. Handbook has given extensive set of tables for evaluation of infiltration in m3/h per
meter crack length through windows, doors and at various velocity or pressure drops. The data is
presented in table,[4]
Infiltration rate function of wind velocity and crack length so for this study wind velocity was
assumed as 25 Km/h.Crack length is one dimensional parameter measurement for window and
door is shown bellow.
Fig.7. Crack length calculation of door and windows
H
W
Single door
crack = 2W + 2H
Double door
crack length = 4W
+ 3H
Infiltration load due to door opening calculated based on the assumption is given bellow,
Maximum traffic rate condition was considered as 70 person/ hour going out and in for
subsequent lecture.So number of door opening per hour (Traffic rate) would be 140
For calculation of latent heat load required, This is due to fresh air the requirement for each
individual. The minimum value from the table in handbook [2-4] is 0.33 cmm(m3/min) per
person for given inside condition.
Load estimation for the given infiltration rate is done by formula given bellow
Sensible heat gain due to infiltration (OASH) =20.4× Infiltration rate (m3/min)×(ho- hi) (7)
Latent heat gain due to infiltration (OALH) = 50000×Infiltration rate(m3/min) × (wo - wi)
(8)Where,h = Enthalpy and w = Specific humidity which is calculated by standard psychometric
chart SHR (Sensible heat ratio) =
Sensible heat gain due
Sensible heat gain due
Latent heat gaindue
OASH (¿infiltration)
OASH (¿infiltration)+OALH (¿infiltration)
(9)
3.4. Load sharing by different wall and part of the building
Heat gain from the different direction of wall has been calculated which will give energy share
by each wall individually of total energy required. Energy sharing by thermal mass, internal heat
generation and infiltration are also calculated .
4. Results and discussion
The details of calculation that has been described in previous section discusses a method
how to determine heat gain and infiltration losses in particular room. Solar radiation received by
each wall, Sol-air temperature and Sol-air temperature difference has been shown in Fig.8,
Fig.9 and Fig.10 respectively , which is dynamic parameter changes throughout the year so heat
gain in a building also changes but after observing the trend it was found that sol-air temperature
difference maximum for month of June, Heat gain that has been calculated for month of June
2016 from the different sources thermal mass, windows , doors and roof and infiltration which
is shown in the Table3.,Table4.and Table5 given bellow.Heat gain due to internal sources are
fixed. That is due to constant heat produce by appliances such as computer and projector, and
also constant heat that has been produce by lighting and people as well. Since the heat gain is
dependent on external heat gain because of internal components of heat gain are constant. There
is not much work that can be done in order to decrease the heat gain in order to controlpower
consumption of air conditioning system. Heat load share from different direction and part has
been shown in Fig.11. and Fig.12, which will be helpful for building energy auditing. A
suggestion can be addressed here that possible to be implemented here which is reduce overall
heat transfer coefficient of the wall, Which is done by putting proper insulation on the outer wall
faces direct solar radiation. Share of heat gain due to infiltration is 28% , which is shown in
Fig.12.Infiltration load could be reduced by putting proper ceiling material and controlling the
traffic rate.
01234567891011
0
2
4
6
8
10
12
North/west
east
south
ceiling
Month
solar radiation recieved by wall (w/m2 )
Fig 8.Solar radiation received by each wall
0 2 4 6 8 10 12
0
2
4
6
8
10
12 north/
west
east
wall
south
wall
Months
Tsol-air oC
Fig 9.Sol-air temperature monthly variation
0 2 4 6 8 10 12
0
5
10
15
20
25
30 east wall
south wall
north/west
wall
ceiling
Months
∆Tsol (oC)
Fig 3. Monthly variation in Temperature difference for each wall and ceiling
Table3. Sensible heat gain through wall Window, Glass and Roof
Component Area
(m2)
Temperature
Different (oC)
Factor
U(W/m2 K)
Total Load
(Watt)
Door (N) 3 12 2.43 87
Glass (E) 12 13 6 948
Glass (S) 5 13 6 421
Door (W) 6 11 2.43 160
Wall (N) 23.38 20.12 3.27 1538
Wall (E) 41 25.21 3.27 3380
Wall (S) 21.18 27 3.27 1870
Wall (W) 50.24 20.12 3.27 3305
Roof 120 18.3 0.48 1055
Table 5.Infiltration load calculation from different component
Component Crack length
(m)
Infiltration
rate(m3/min)
Infiltration load (Watt)
Sensible heat
gain
Latent heat
gain
Doors 10.5 1.8 1254 540
Windows 21.1 3.8 2647 1140
Door opening (Traffic rate) -14.7 10240 4410
Latent heat load -23.1 16493 6930
Finally total sensible heat load , total latent heat load , grand total heat load and SHR (sensible heat
ratio) were found out 39143.4 W 13020 W, 52163 W (14.9 TR).
Table4. Internal heat generation
Component No of
component
Heat
generation
per (W)
Total heat
Generation
(W)
People 110 80 8800
CFL 44 20 880
FAN 15 25 375
Equip.Load –W 00 100 100
33.81%
26.41%
12.27%
19.66%
7.85%
Total load = 52163 Watt
EAST
WEST
NORTH
SOUTH
ROOF
Fig.11 . Heat gain from different direction of the room
19.47%
27.55%
28.65%
24.33%
Total Load = 52163 Watt
INTER HEAT
GENERATION
INFILTRATION LOAD
ROOM LATENT HEAT
LOAD
OTHERLOAD(WALL+RO
OF+WINDOW)
Fig.12 . Heat gain from different part of the building
5.Conclusion
This paper analyzed different type heat gain loss of given building situated in warm
humid climate and cooling load estimation of a building in Department of Energy at the Tezpur
University India. The maximum cooling loads, sensible heat ratios and total dehumidify air
quantities of the sample building (Tezpur seminar hall) calculated using energy balance method
(sol-air temperature method). The maximum cooling load details for seminar hall at summer
(Month of June) has been calculated. Heat gain from different direction and sources are also
shown which will be expected to be helpful for reduction in heat gain.
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