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

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

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

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

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: , if if in the other cases 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 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

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

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

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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

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

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

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

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

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

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

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

Calculation of Energy Performance of Buildings - Lithuanian Case Energy consumption lay-out for 5-story apartment building 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