Presentation: Building Energy Efficiency Assessment

ENERGY EFFICIENCY ASSESMENT ON BUILDING BLOCK D (PART OF FACULTY OF TECHNOLOGY OF ARISTOTLE UNIVERSITY OF THESSALONIKI - GREECE)

GROUP 6

COMPLIES TO GREEK ENERGY EFFICIENCY STANDARIZATION “TOTEE-KENAK”

Makri Eleftheria Dourbois Grigoris Dimou Anastasios Tenekentzoglou Andreas Papadopoulos Athanasios

1.INTRODUCTION WHAT’S THE POINT OF THIS ASSESSMENT? A BETTER ENERGY MANAGEMENT PROFILE FOR BUILDING D

HOW DID WE DO THAT? Studied topographic site & architect building plans Inspected…inspected…and again inspected site & building Re-designed site & building to archicad to improve precision on calculations and make a better presentation ….and all time we studied greek national standards (TOTEE-KENAK) under the priceless guidance of google, michanicos.gr and of course our tutors(Gr. Papagiannis, Valia, Theofilos)

1.INTRODUCTION WHAT’S THE GREEK POINT OF VIEW OF E.E.A? WE CARE ABOUT THE BUILDING’S PROFILE, NOT THE USER’S PROFILES BUILDING PROFILES (=REFERENCE BUILDING) HAS MINIMUM REQUIREMENTS TO ALL ASPECTS AFFECTING BUILDING ENERGY EFFICIENCY WE STUDY ENERGY PERFORMANCE OF REAL BUILDING INPUT DATA OF REAL BUILDING ENERGY PERFORMANCE(E.P) ALONG WITH ITS PROFILE TO TEE-KENAK APPLICATION. TEE-KENAK COMPARES REAL BUILDING E.P. VS REFERENCE BUILDING E.P. TEE-KENAK OUTPUT IS THE E.E.A. REPORT. IF REAL E.P.<REF E.P. WE GOT ENERGY CLASS BELOW “B” (MINIMUM ALLOWED)

2.BUILDING DESCRIPTION SITE PLAN

SITE PLAN REFERENCE

Cyan: surrounding buildings Grey: Building D

2.BUILDING DESCRIPTION Ground level (Basement Floor Plan)

GROUND LEVEL REFERENCE

Zero level set to floor of Story 0 – Basement Floor is at -6.5m to zero lvl

2.BUILDING DESCRIPTION ď&#x192; Basement - Floor Plan

Basement - Floor Plan reference

2.BUILDING DESCRIPTION Story 0 - Floor Plan

Story 0 - Floor Plan reference

2.BUILDING DESCRIPTION Typical Story 1to9 - Floor Plan

Typical Story 1to9 - Floor Plan reference

2.BUILDING DESCRIPTION Typical Story 10 - Floor Plan

(based on measurements)

2.BUILDING DESCRIPTION Frontal View (SW elevation)

2.BUILDING DESCRIPTION ď&#x192; Perspective 3D View (SE & NE side)

(surrounds removed for better view point)

2.BUILDING DESCRIPTION ď&#x192; Perspective 3D View (NW & SW side)

(surrounds removed for better view point)

3.THERMAL ZONES & PROFILES BUILDING PROFILE: OFFICE TOTAL

HEATING

COOLING

A(m2) V(m3) A(m2) V(m3) A(m2) V(m3) 11925 46006 11640 44582 7842.2 28721

THERMAL ZONES

COOLING

A(m2)

V(m3)

A(m2)

V(m3)

9302

30061

7215

25252 A(m2) 285

ZONE 2

HEATING

UHA

ZONE 1

-ZONE 1(Z01): STORIES 0 TO 10, PROFILE: OFFICES -ZONE 2(Z02): BASEMENT FLOOR, PROFILE: OFFICES

V(m3) 1424

HEATING

A(m2) 2338

V(m3) 14521

COOLING

A(m2) 627

V(m3) 3469

UHA=UNCONDITIONED AREA (PART OF BASEMENT)

4.BUILDING SHELL PROPERTIES 4.1.a. EXTERNAL WALLS BUILDING SHELL ANALYSIS RESULT TO 11 TYPES OF COMPLEX WALLS. MOST OF THEM HAVE UNSUFFICIENT THERMAL INSULATION PROPERTIES AS U-VALUE RANGES 0.544 TO 4W/Km2 WHEN MAXIMUM ALLOWED IS 0.45W/Km2

Z01 Z02

A(m2) Uav 2751.51 1.27 394 3.704

Umax 0.45 0.45

VERY HIGH U-VALUES!

4.1.b. UNDERGROUND WALLS Z02

A(m2) 634

U-value 4.762

Umax 0.8

OVER-EXCEEDING MAXIMUM ALLOWED!

4.2. EXTERNAL OPENINGS ZO1

A(m2) 1257.17

Uav 5.46

Gwav 0.53

INFLITRATION(m3/h)

7094.74

Umax=2.8W/Km2 Z02

A(m2) 188.47

Uav 6.18

Gwav 0.48

INFLITRATION(m3/h)

1297.17

OVER-EXCEEDING MAXIMUM ALLOWED U-VALUE! (DOORS INCLUDED AT CALCULATING AVERAGE VALUES)….we got nice G-values as reference is set to 0.55

(NOTE THAT MAXIMUM ALLOWED IS THE VALUE SET TO REFERENCE BUILDING)

4.3. SLABS ROOFS ZO2 ZO1

A(m2) U-value 1144.1 2.02 1260.7 0.448

Umax 0.4 0.4

FLOOR A(m2) U-value Umax ZO2 2339.1 3.39 0.75

OVER-EXCEEDING MAXIMUM ALLOWED!

4.4. BUILDING SHELL THERMAL INSULATION PROPERTIES (SUMMARY) WALLS OPENINGS ROOFS FLOOR SUM F/V

A 3146 1446 2405 2339 9335 0.20

Uav 1.57 5.55 1.20 3.39

Um

Ummax

2.55

1.05

FAILED!

4.3. SHELL COATING PROPERTIES

a : ABSORPTION COEFFICIENT TO RADIANT HEAT TRANSFER e : EMMITANCE COEFFICIENT TO RADIANT HEAT TRANSFER Z01: CONCRETE(NO PLASTER)  a=0.8, e=0.8 ROOFS DARK COLORED  a=0.9, e=0.8 (EXCEPT ROOF OF STORY 0  a=0.2, e=0.2 ) Z02: MEDIUM COLORED PLASTER WALLS  a=0.6, e=0.8 ROOFS LIGHT COLORED  a=0.65, e=0.8

4.4. SHADING FACTORS THERE ARE 3 MAIN TYPES OF SHADINGS WE STUDY 1)HORIZON SHADING(Fhor) 2)OVERHANG SHADING(Fov) 3)LATERAL SHADING(Ffin)

F_hwinter shading F_csummer shading

F=0 FULLY SHADED F=1NO SHADING

ANALYTIC CALULATIONS FOR OPENINGS AVERAGE FOR WALLS

TOTAL SHADING = Fhor*Fov*Ffin

4.4.1 HORIZON SHADING FACTORS Z01 -NE SIDE: LIGHT SHADING -SE SIDE:VERY LIGHT SHADING -NW SIDE:LIGHT SHADING -SW SIDE:VERY LIGHT SHADING

Z02 -NE SIDE: LIGHT SHADING -SE SIDE:LIGHT SHADING -NW SIDE:LIGHT SHADING -SW SIDE:LIGHT SHADING

LIGHT SHADING: Fhor_h=0.8 to 1 , Fhor_c=0.7 to 1 VERY LIGHT SHADING: most surface Fhor_h,c=1

REFERENCE BUILDING Fhor= same as calculated *FOR ANALYRIC VALUES CHECK CALCULATION SHEETS

4.4.2 OVERHANG SHADING FACTORS Z01: SW,NE CURTAIN WALL OPENINGS Fov_c=0.6 to 0.7, Fov_h=0.76

Z02 - NW SIDE: Fov_c=0.9, Fov_h=0.9

REFERENCE BUILDING OPENINGS: F_c= 0.7 TO 0.75, F_h= same as calculated WALLS: Fc,h=0.9

4.4.3 LATERAL SHADING FACTORS Z01: SMALL AREA(story0,1) NW SIDE(right) has lateral shading - Ffin_c=0.66, Fov_h=0.81 SW SIDE CURTAIN WALLS OPENINGS(left) Ffin_c=0.99, Fov_h=0.97

NE SIDE CURTAIN WALLS OPENINGS(left) Ffin_c=0.96, Fov_h=0.95

5.1 HEATING SYSTEM

Description: The heating system consists of two boilers/burners of maximal thermal power 2000kW each one. Every office is heated through radiators and in some cases, with local air-cooling heat pumps (air conditioners). Air conditioners are also used for cooling of offices.

First thermal zone The average power in which each boiler operates is 600kW .That gives us a sum of 1200kW. The real power of the boiler that corresponds to the first thermal zone is the percentage of the total volume that the zone represents ,it is 68.84% of the building.

So the real power is 1200*0.6884=826.08kW. For the control of oversizing (it is needed for determining the factor ng1) we apply the formula 4.1 of Τ.Ο.Τ.Ε.Ε. 20701-1: Where: A=4410.07 m2.

Um=0.95 W/m2*K

ΔΤ = 23°C

So Second thermal zone The second thermal zone is consisted of the basement. The basement is the 31.15% of the total volume of the building so the real power of the boiler in this zone is 1200*0.3115=373.8kW. For the control of oversizing (it is needed for determining the factor ng1) we apply the formula 4.1 of Τ.Ο.Τ.Ε.Ε. 20701-1: So Now A=569.15 m2.

For the factors ngm,ng1 and ng2 stand the same for both zones.

So ngm=0.88 (from gas analysis) ng1=0.75 (T.O.T.E.E. 20701-1 Table 4.3 ) ng2=1 (T.O.T.E.E. 20701-1 Table 4.4) Thus, the overall efficiency of the heating system is: nge=ngm*ng1*ng2= 0.88*0.75*1=0.66(66%)

The final real thermal power of the boiler that goes to the distribution network is: Zone 1

Zone 2

Because of the boiler the two parameters efficiency and COP will be: nge=0.66 and COP=1

Local air-cooling units (Heat Pump) We have heating: First zone •In all offices (in 9 floors and ground floor) •In the computer lab and •In the conference room.

Second zone In all offices (18 offices)

We have an air conditioner in each office (23 offices/per floor, 21 offices/ground floor) size of 9000 BTU/hr and 12000BTU/hr (we take them by half). There are three air conditioners of 12000BTU/hr in the conference room and also Two of the same power in the computer lab. We have an air conditioner in each office(18 offices/basement) size of 9000 BTU/hr and 12000BTU/hr (we take them by half).

Conversion of BTU/hr in kW:

The degree performance (COP) for the calculations of the energy efficiency of the under study building is taken COP=2.2(10 y.o. systems).

So the installed air-cooling power is: First zone

Second zone

Distribution network The power of the distribution network is 826.08*0.66=545.213kW.(first zone)

The power of the distribution network is 373. 8*0.66=246.708kW.(second zone) From the table 4.11 of T.O.T.E.E. _fix1_feb11i the percentage of heat loss of the distribution system is 7%(first zone) and 10.5%(second zone).

Heating efficiency is:

1-0.07=0.93(Zone 1) 1-0.105=0.895(Zone 2)

The air conditioners are not included in the network because their production is local.

Heating terminal units First zone: Radiators and Splits Second zone: Radiators,Heaters and Splits •Radiators

According to Τ.Ο.Τ.Ε.Ε. 20701-1 the procedure for determining the efficiency of every terminal unit is described below.

According to the standard ΕΛΟΤ ΕΝ 15316.2.1:2008 the efficiency (nem,t) of the terminal units (heat emission) of the heating system, is estimated using the following formula:

The factors frad ,fim ,fhydr , get values according to tables from paragraph 4.4.2.

Based on the table 4.12 of T.O.T.E.E. 20701-1 we take the emission efficiency ηem of terminal units equal to 0.89.

So for the radiators: nem ,t1

nem f rad * f im * f hydr

0.89 0.92 1*0.97 *1

Splits The T.O.T.E.E. 20701-1 refers in page 104 that for local heat pumps emission efficiency of the indoor units in the calculations shall be equal to 0.93 (nem). For the same frad,fim,fhydr as the radiatorsâ&#x20AC;&#x2122;: nem,t 2

nem f rad * fim * f hydr

0.93 0.96 1*0.97 *1

Heaters These are the other terminal units of the boiler. According to the previous nem f rad * fim * f hydr

nem ,t1,2

0.89 0.917 1*0.97 *1

Where frad is 1 because they are not radiation units and fim is one because these terminal units are controlled by switches.

For the final calculation of the efficiency So Firt zone

nem,t

1* nem,t1 0* nem,t 2

0.92

Second zone

nem,t

1*(

nem,t1 nem,t1,2 2

) 0* nem,t 2

0.918

Auxiliary Units According to the help of the program TEE KENAK the filling of the form will be the following: Type: The type of ancillary units is determined.

Number: The number of each type is imported.

Power (kW): The rated power of each secondary unit type is imported. If there are none auxiliary units in the heating system then their power is zero (0).

First zone Our heating system has as auxiliary units three boiler circulator pumps. The power of each one is 470 W, so there is a total of 1.41kW,and for the zone is 1.41*0.6884=0.97kW. Second zone Our heating system has as auxiliary units three boiler circulator pumps. The power of each one is 470 W, so there is a total of 1.41kW,and for the zone is 1.41*0.3115=0.439kW

5.2 COOLING SYSTEM First zone •In all offices (in all floors and ground floor) •In the computer lab and •In the conference room.

Second zone In all offices (18offices)

We have an air conditioner in each office (23 offices/per floor, 21 offices/ground floor) size of 9000 BTU/hr and 12000BTU/hr (we take them by half). There are three air conditioners of 12000BTU/hr in the conference room and also Two of the same power in the computer lab. We have an air conditioner in each office(18 offices/basemant) size of 9000 BTU/hr and 12000BTU/hr (we take them by half).

So the installed power is: First zone

Second zone

Air-Cooling heat pump In both zones,for the two parameters, efficiency and EER, because of the heat pump, paramaters will be efficiency=1 and EER=2(10 y.o. systems).

Distribution network There is none in both zones, as the cooling systems are local.

Cooling terminal units The analysis is the same for both zones.

The efficiency (nem,t) of cooling terminal units is given from the following formula:

The factors fim ,fhydr, get values according to tables from paragraph 4.4.3. Ρem is the emission efficiency From the table 4.14 of T.O.T.E.E. 20701-1, the emission efficiency Ρem of terminal units is 0.93.

So

Auxiliary systems There are none.

6.IMPROVING PROJECTS 1. External Wall Insulation (EWI)

The current condition of the building as captured with a thermal imaging camera…

Aspect of North- East facade

Aspect of South – West facade

6.IMPROVING PROJECTS An EWI…

• Minimizes the heat transfer.

• Ensures coverage of thermal bridges especially in concrete slab, beams and columns.

• Protects the exterior wall surfaces from contractions and expansions.

• Does not reduce usable living space.

What we should do?  Installation of an external thermal insulation system on the North-East and South-West facades.  Thermal Insulating Material: Expanded polystyrene (EPS) boards of 4 cm width.

6.IMPROVING PROJECTS What will it cost?

Estimated cost: 35 € / m2 Surface to be covered: North – East facade: 1110 m2 South – West facade: 1053 m2

Total estimated cost: 75,700 €

Energy needs reduction Existing Building

Building with EWI

Energy spend on heating (kWh / m2 )

63.9

51.6

Energy spend on cooling (kWh / m2 )

48.2

45.7

6.IMPROVING PROJECTS 2. PV panels

Installation on the buildingâ&#x20AC;&#x2122;s roof Total roof area: 800 m2 Total covered area: 313.36 m2 of monocrystalline PVs

6.IMPROVING PROJECTS What will it cost?

Estimated cost: 4000 € / kW Installed power: 39.84 kW Total estimated cost: 159,360 €

Energy gain Existing Building Renewable Sources of Energy (kWh / m2 )

0

Building with PVs on Roof 23.3

6.IMPROVING PROJECTS 3. Shading system - Shadovoltaics What’s a Shadovoltaic?

• a patent of Colt Industries • combines Louvers and PV panels • Fixed Louvers at 20 degrees.

It offers: • Shading reduces solar heat gain • PV panels contribute to the improvement of the building’s efficiency.

But..

• During Winter months  increase of heating needs

6.IMPROVING PROJECTS What will it cost?

Unfortunately Colt Ind. didnâ&#x20AC;&#x2122;t reply to our inquiries..

Energy gain Existing Building

Renewable Sources of Energy (kWh / m2 )

Building with PVs on Roof

0

23.3

Heating Needs (kWh / m2 )

63.9

70.1

Cooling Needs (kWh / m2 )

48.2

43.9

6.IMPROVING PROJECTS 4.Lighting system management Our suggestion…

The installation of a semi-automatic lighting control system using motion sensors.

• manual switch on • automatic switch off

What will it cost?

It’s impossible to define the exact cost of a such system.

But… It’s a relatively cheap intervention .

Lighting needs(kWh / m2 )

Existing Building

Building with management lighting system

68.6

54.9

6.IMPROVING PROJECTS 5. Other Improvements In addition we suggest: The installation of a second row of windows to the openings of the North-East facade with single glass windows .

What will it cost?

Cheaper than replacing the old ones with new windows Existing Building

2nd row of windows

Heating needs (kWh / m2 )

63.9

59.6

Cooling needs (kWh / m2 )

48.2

48.2

6.IMPROVING PROJECTS 6. Results Combining all the above interventions, the classification of the under study building improves from D to B, which was our goal.

6.IMPROVING PROJECTS 7. Green Wall The idea: growing plants vertically on the building’s walls How: - on vertical soil pallets - hydroponically on felt “pockets” Why: • reduction of heat island effect by lowering the temperature of the building’s microclimate • the green layer acts as extra insulation • reduction of the building’s cooling loads by shading it • plants trap carbon dioxide and produce oxygen • aesthetic factor • decreases the building’s temperature in the summer up to 7 degrees.

6.IMPROVING PROJECTS But..

• The green wall‘s components aren’t produced widely on an industrial scale yet. expensive!

• There is no method for calculating the exact impact on the building’s energy efficiency Although A green wall would highlight the building as a landmark of the city!

7.PROSPECTS ON THIS PROJECT E.E.A BASED ON AMERICAN STANDARDS ECODESINER ANALYSIS  ALMOST READY ...we just input calculated values!

ECOTECT ANALYSIS WE JUST SIMPLIFY ARCHICAD MODEL FOR BETTER PROCESS TIME (guides has been found)

MORE GREEN DESIGN IDEAS

Youtube screen recordings

Thank you for your attention!

Muchas Gracias!