Building Energy Efficiency in the Baltics ( BENEFIT-2006 ) , Riga, 25 October, 2006

1. Calculation of Energy Performance of Buildings - Lithuanian CaseDr. 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 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

5. 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

6. Calculation of Energy Performance of Buildings - Lithuanian Case The energy performance is evaluated by classification indicator value, which shall bedeterminated for a considered building due to the total building normative QN.sum, reference QR.sum and calculated Qsumvalues of building energyconsumption for 1 m2 of buildingheated area. classification indicator valueis 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. 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 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 lossesQN.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 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 Correction coefficient for the doors kd1 Correction coefficient for the doors kd2 BENEFIT-2006, Riga, 25 October, 2006

14. 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. 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 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 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. Calculation of Energy Performance of Buildings - Lithuanian Case • 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