1. Calculation of Energy Performance of Buildings - Lithuanian Case. Dr
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

2. 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
(IEA data)
BENEFIT-2006, Riga, 25 October, 2006

3. Calculation of Energy Performance of Buildings - Lithuanian Case
Comparison of energy consumption in Lithuania building stock with the European trends
BENEFIT-2006, Riga, 25 October, 2006

4. Calculation of Energy Performance of Buildings - Lithuanian Case
Energy consumption for heating during the recent has been decreased by %. In big apartment buildings mean energy consumption value can be assumed as 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

5. Calculation of Energy Performance of Buildings - Lithuanian Case
The National Building Code STR (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

6. 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: , if if
in the other cases
BENEFIT-2006, Riga, 25 October, 2006

7. 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

8. Calculation of Energy Performance of Buildings - Lithuanian Case
Heating season values
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
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)
Building energy
performance components
Electricity (annual value)
BENEFIT-2006, Riga, 25 October, 2006

9. Calculation of Energy Performance of Buildings - Lithuanian Case
Lay-out of basic heat losses in the residential buildings according to the construction year
BENEFIT-2006, Riga, 25 October, 2006

10. Description of the heat recovery system
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 r
Mechanical ventilation without heat recovery
The efficiency factor of mechanical ventilation with heat recovery is unknown
0,5
BENEFIT-2006, Riga, 25 October, 2006

11. 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

12. 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year)) during the heating season through the external door due to door opening shall be calculated according to the equation:
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:
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

13. Calculation of Energy Performance of Buildings - Lithuanian Case
Type of building
kd1
One and two storey residential buildings 7
Apartment buildings
Administrative 10
Educational 5
Health care 14
Restaurants
Trade 2
Sport, except swimming pools 9
Swimming pools
Cultural 3
Garage, manufacturing and industry
Storage
Hotels
Service 18
Transport 50
Recreation
Special
Correction coefficient for the doors kd1
Correction coefficient for the doors kd2
Description of the type of external doors
kd2
One door without tambour 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
BENEFIT-2006, Riga, 25 October, 2006

14. Description of the hot water preparation and regulation system
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)):
The efficiency factor of the hot water supply system h.w.
Description of the hot water preparation and regulation system
hh.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
Gas heater in the apartment
0,75
Electric heater in the apartment
0,3
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.
BENEFIT-2006, Riga, 25 October, 2006

15. 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)) :
; ( )
BENEFIT-2006, Riga, 25 October, 2006

16. Calculation of Energy Performance of Buildings - Lithuanian Casec
The calculated efficiency factor of the heat source 2
Calculated efficiency factor of the temperature control devices of the heating system 1
Description of the heat source 2
District heating, manual temperature control
0,9
District heating, automatic temperature control 1
Gas boiler, manual temperature control
0,8
Gas boiler, automatic temperature control
0,94
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
Heating by electricity, automatic temperature control
Heat pump
1,1
Stoves
0,5
Fireplaces
0,4
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
BENEFIT-2006, Riga, 25 October, 2006

17. 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
 Rh.s. 1
One and two storey residential buildings
0,7
0,83 2
Apartment buildings
0,90 3
Administrative 4
Educational 5
Health care
0,84 6
Restaurants
0,86 7
Trade 8
Sport, except swimming pools
0,81 9
Swimming pools 10
Cultural
0,82
No. 
Type of building [3.2] 
 N.h.s.
 Rh.s. 11
Garage, manufacturing and industry
0,7
0,83 12
Storage
0,84 13
Hotels
0,9 14
Service
0,88 15
Transport 16
Recreation
0,82 17
Special
0,85
BENEFIT-2006, Riga, 25 October, 2006

18. 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

19. Description of energy consumption type
Calculation of Energy Performance of Buildings - Lithuanian Case
Description of energy consumption type
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
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.
BENEFIT-2006, Riga, 25 October, 2006

20. Calculation of Energy Performance of Buildings - Lithuanian Case
Energy consumption lay-out for 5-story apartment building
BENEFIT-2006, Riga, 25 October, 2006

21. 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