Transcript Calculation of Energy Performance of Buildings

Calculation of Energy Performance of
Buildings - Lithuanian Case
Dr. Jurate Karbauskaite,
dr. Edmundas Monstvilas,
prof. V.Stankevicius,
Institute of Architecture and Construction,
Kaunas Technology University, Lithuania
Building Energy Efficiency in the Baltics
(BENEFIT-2006), Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
At the point of view of long-term energy policy, the buildings
ought to conform to the minimal requirements of energy
performance in them in respect to the local climate.
The factors which serve to the growth of energy efficiency,
must be taken into advantage
The biggest part in
environment pollution
is formed by heating
of buildings
27
41
18
5
(IEA data)
Heating
Ventilation
4
Light
5
Cooling
Equipment
BENEFIT-2006, Riga, 25 October, 2006
Computers
Energy consumption for heating, kWh/m2 per year
Calculation of Energy Performance of
Buildings - Lithuanian Case
450
420
400
Small houses
Big houses
350
300
250
250
220 220
200
180
170
150 150
150
120
100
70
130
70
50
0
tll 1992
West Europe
Skandinavia
Economical
house
STR
2.05.01:1999
RSN 143:91
Comparison of energy consumption in Lithuania building stock
with the European trends
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
35,0
30,0
Amount of buildings, %
25,0
20,0
15,0
10,0
5,0
0,0
0 - 5
5 -1 0
10 - 15
15 - 20
20 - 25
25 - 30
30 - 35
35 - 40
40 - 45
45 - 50
50 - 55
55 - 60
Bins, kW h/m 2 per m onth
Energy consumption for heating during the recent 1997-2001 has been decreased by 15 20 %. In big apartment buildings mean energy consumption value can be assumed as 145 240 kWh/m2 at 3790 degree days at indoor air temperature of 18 °C.
Energy consumption in a big part, about 30 % of them, is significantly lower, mainly due to
lowered indoor temperature, as well as other 30 % are consuming more than mean value. .
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
The National Building Code STR 2.01.09.2005 (BUILDING
TECHNICAL REGULATION) BUILDING ENERGY
PERFORMANCE CERTIFICATION states, that building
energy efficiency is assessed only by calculation of the
building energy consumption according to the method
presented in the obligatory Annex.
Character of the tenants behavior and
state of the building envelope are not taken into
consideration in recent normative document
Development in this direction shall be provided in the future stage
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
The energy performance is evaluated by classification indicator value,
which shall be determinated for a considered building due to the
total building normative QN.sum, reference QR.sum, and calculated Qsumvalues
of building energy consumption for 1 m2 of building heated area.
classification indicator value is expressed by equation:
Qsum
1
QN .sum.
if
if
Qsum
1
QR.sum.
in the other cases
Qsum
C
QN .sum.
Qsum
C  1
QR.sum.
C  1
Qsum  QN .sum.
QR.sum.  QN .sum.
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
The performance class, depending on the value of the
classification indicator C :
- Class A, if C  0,5;
- Class B, if 0,5  C  1;
- Class C, if 1  C  1,5;
- Class D, if 1,5  C  2;
- Class E, if 2  C  2,5;
- Class F, if 2,5  C  3;
- Class G, if C  3.
THE BUILDINGS COULD BE
ATTRIBUTED TO ONE OF THE
7CLASSES:
A.B, C, D, E, F, G.
CLASS A IS AT THE TOP AND SUCH A
BUILDING IS VERY ENERGY
EFFICIENT WITH LOW ENERGY
CONSUMPTION IN IT.
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Heating season values
Transmission heat losses
Heat losses due to door
opening
Ventilation heat losses
Heat losses due over-infiltration
of external air through windows
-
Solar heat gains
Internal heat gains
Domestic hot water supply
(annual value)
Electricity (annual value)
Heat losses through the external
walls;
Heat losses through roof;
Heat losses through building ceilings,
which are in contact with external air;
Heat losses through building ceilings
over unheated basement and crawls;
Heat losses through building
elements on ground;
Heat losses through the windows;
Heat losses through external doors,
excluding the heat losses due the
door opening;
Heat losses through thermal bridges
in building;
Heat losses due to opening of
external door
Building energy
performance
components
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Roofs
Walls
Doors
Ventilation
40,0
Building heat losses, %
35,0
Ceilings
Windows
Thermal bridges
30,0
25,0
20,0
15,0
10,0
5,0
0,0
till 95
95
96
97
98
99
2000
2001
2002
2003
2004
2005
2006
Lay-out of basic heat losses in the residential buildings according to the construction year
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation of the energy used for the ventilation of the building
Ventilation systems could be attributed to one of the three types:
-natural ventilation;
-mechanical ventilation without heat recovery;
-mechanical ventilation with heat recovery.
The calculated efficiency factor of the mechanical ventilation
system with heat recovery r
Description of the heat recovery system
Mechanical ventilation without heat recovery
The efficiency factor of mechanical ventilation
with heat recovery is unknown
r
0
0,5
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation of the heat losses because of extra normative
infiltration of external air through windows and external doors
Normative heat losses QN.inf. (kWh/(m2per year)) during the heating season
because of extra normative infiltration of external air through windows and
external doors, is set, that there should be no higher infiltration of external air than
it’s needed for the ventilation of the building, so QN.inf.=0.
Reference QR.inf. (kWh/(m2per year)) heat losses during the heating season because of
extra normative infiltration of external air through windows and external doors shall be
calculated with respect to new and old window and door amount according to the equations
in the Annex 2 of the Regulation.
Calculated Qinf (kWh/(m2per year)) heat losses during the heating season because of extra
normative infiltration of external air through windows and external doors shall be
calculated with respect to new and old window and door amount according to the equations
in the Annex 2 of the Regulation.
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation of the heat losses through the external door due to
doors opening
The heat losses Qd1 (kWh/(m2 per year)) during the heating season through the external
door due to door opening shall be calculated according to the equation:
Qd1.  2,15 
1
 k d1  k d 2  (1  0,2  h)
Ao
where: Ao – the area for one occupant (m2). Selected from table 2.4;
kd1 – the correction coefficient, evaluating the frequency of opening of external doors of different types of the
buildings, selected from table 2.11 of the Regulation.
kd2 – the correction coefficient, evaluating the type of external doors. The value of the coefficient shall be selected
from table 2.12 of the Regulation with respect to the type of the external doors: according to the door, that is most
frequently used or according to the door, that generalized corresponds to all types of external doors of the building;
h – the height of the building (m). That is the distance from the ground level to the highest point of the heated
room, located at the upper part of the building.
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Type of building
Correction coefficient for the doors kd1
Correction coefficient for the doors kd2
Description of the type of external
doors
One door without tambour
kd2
1
Two doors without tambour between
1,1
Two doors with tambour between
0,6
Three doors with tambour between
0,4
Swinging-doors
0,75
Doors with air curtain
0,1
One automatic door without tambour
0,9
One automatic doors with tambour
0,5
kd1
One and two storey residential buildings
7
Apartment buildings
7
Administrative
10
Educational
5
Health care
14
Restaurants
14
Trade
2
Sport, except swimming pools
9
Swimming pools
7
Cultural
3
Garage, manufacturing and industry
5
Storage
5
Hotels
7
Service
18
Transport
50
Recreation
10
Special
5
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation of energy use for domestic hot water
Annual energy use for hot water Qh.w. (kWh/(m2per year)):
Qh.w.
 h.w.

 h.w.
The efficiency factor of the hot water supply system h.w.
where:
ψh.w. – the annual energy demand for hot water
per unit of the building area (kWh/(m2per
year)). Selected from table 2.4;
h.w. – the efficiency factor of the hot water
supply system. Selected from table 2.4.
Description of the hot water preparation and
regulation system
h.w.
Central heat substation
0,6
Building heat substation + manual temperature control
0,8
Building heat substation + automatic temperature control
0,95
Building boiler-house + manual temperature control
0,7
Building boiler-house + automatic temperature control
0,8
Gas heater in the apartment
0,75
Electric heater in the apartment
0,3
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation of sum energy consumption of the building
Normative sum QN.sum. (kWh/(m2per year)), reference sum QR.sum. (kWh/(m2
per; year)) and calculated sum
Qsum (kWh/(m2per year)) :
( )
QN .sum. 
(QN .env.  QN .vent.  Qd1  Qe  Qi )
 QE  Qh.w.
QR.sum. 
(QR.env.  QR.vent.  Qd1  QR. inf  Qe  Qi )
 QE  Qh.w.
Qsum 
 N .h.s.
 R.h.s.
(Qenv  Qvent  Qd1  Qinf  Qe  Qi )
 QE  Qh.w.
 h.s.
h.s.  1 2
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Casec
The calculated efficiency factor of the heat source 2
Description of the heat source
District heating, manual temperature control
District heating, automatic temperature control
2
0,9
Calculated efficiency factor of the temperature
control devices of the heating system 1
1
Gas boiler, manual temperature control
0,8
Gas boiler, automatic temperature control
0,94
Description of the regulation devices
1
No temperature control devices in the building
heating system
0,88
Temperature control in the premises of the
building, but only thermostatic valves on the
heating devices or only internal or external
thermostats are installed.
0,93
Temperature control devices in all the
premises of building. Thermostatic valves on
the heating devices and internal or external
thermostats are installed.
0,98
Temperature control is settled in part of the
building.
0,90
Gas radiant heating device
Liquid fuel boiler, manual temperature control
0,75
Liquid fuel boiler, automatic temperature control
0,87
Solid fuel boiler, manual temperature control
0,7
Solid fuel boiler, automatic temperature control
0,85
Heating by electricity, manual temperature control
0,9
Heating by electricity, automatic temperature control
1
Heat pump
1,1
Stoves
0,5
Fireplaces
0,4
BENEFIT-2006, Riga, 25 October, 2006
1
Calculation of Energy Performance of
Buildings - Lithuanian Casec
Normative N.h.s. and reference  Rh.s. heating system efficiency factors, different
types of buildings
No.
Type of building [3.2]
N.h.s
1
One and two storey residential
buildings
0,7
2
Apartment buildings
0,7
3
Administrative
4
 Rh.s.
No.
0,83
Type of building [3.2]
 N.h.s.
 Rh.s.
11
Garage, manufacturing and
industry
0,7
0,90
12
Storage
0,7
0,84
0,7
0,83
13
Hotels
0,7
0,9
Educational
0,7
0,83
14
Service
0,7
0,88
5
Health care
0,7
0,84
15
Transport
0,7
0,84
6
Restaurants
0,7
0,86
16
Recreation
0,7
0,82
7
Trade
0,7
0,86
17
Special
0,7
0,85
8
Sport, except swimming pools
0,7
0,81
9
Swimming pools
0,7
0,83
10
Cultural
0,7
0,82
BENEFIT-2006, Riga, 25 October, 2006
0,83
Calculation of Energy Performance of
Buildings - Lithuanian Case
Translation of the indices
•Building address
Building destination
Heated area of building
Building energy performance
classification indicator
Calculated summed energy
consumption related to 1 m2 heated
area of the building (building part)
Main heat supply source
Certificate issue date
Building (building part) certificate
expiry date
Name and surname of expert
 Expert certificate registration
number
Expert signature
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Description of energy consumption type
Calculation example:
Apartment building, 3
staircases, 5 stories
Input data:
-Heated area 2700 m2;
-Window area – 15 % from
building heated area;
-Ventilation natural;
-Domestic hot water supply
with manual control;
-Temperature control in
building heating system is
absent;
-Energy source - district
heating, manual temperature
control.
kWh/(m2.per
year)
Transmission heat losses through the external wall
63,67
Transmission heat losses through the roof
21,77
Transmission heat losses through the ceilings over unheated basements
and crawls
9,14
Transmission heat losses through the windows
36,28
Transmission heat losses through the doors, except opening losses
0,59
Transmission heat losses through the thermal bridges
13,94
Transmission heat losses due to external door opening
0,90
Energy losses doe to ventilation
24,04
Heat losses due to over-infiltration of external air
29,19
Solar heat gains
19,31
Internal heat gains
14,12
Electricity consumption in building
21,00
Energy consumption for domestic hot water supply
25,00
Building sum energy consumption without heating system efficiency
assessment
212,09
Building sum energy consumption with of heating system efficiency
assessment
255,71
BENEFIT-2006, Riga, 25 October, 2006
2
Domestic
hot water
Sum
without
Sum with
heating
Electricity
Air overinfiltration
Solar heat
gains
Internal
heat gains
Ventilation
Thermal
bridges
Door
opening
Doors
150
Windows
200
Ceilings in
contact
Ceilings
over crawls
Roof
Walls
Energy consumption, kWh/(m per year)
Calculation of Energy Performance of
Buildings - Lithuanian Case
Energy consumption lay-out for 5-story apartment building
300
250
Normative
Reference
Calculated
100
50
0
BENEFIT-2006, Riga, 25 October, 2006
Calculation of Energy Performance of
Buildings - Lithuanian Case
Calculation program window with the part of input data and calculation results.
Building data:
Destination – residential
Heated area- 175,22 m2
Height – 4,2 m
Walls, windows, doors by
facades:
External door description
DHW supply system- gas heater
Energy consumption – 213,76
Classification indicator value –
0,80
Energy performance class - B
BENEFIT-2006, Riga, 25 October, 2006