eea presentation

49
ENERGY EFFICIENCY ASSESMENT ON BUILDING BLOCK D (PART OF FACULTY OF TECHNOLOGY OF ARISTOTLE UNIVERSITY OF THESSALONIKI - GREECE) GROUP 6 Makri Eleftheria Dourbois Grigoris Dimou Anastasios Tenekentzoglou Andreas Papadopoulos Athanasios COMPLIES TO GREEK ENERGY EFFICIENCY STANDARIZATION “TOTEE-KENAK”

Upload: andreas-teneketzoglou

Post on 07-Mar-2016

227 views

Category:

Documents


1 download

DESCRIPTION

EEA presentation

TRANSCRIPT

Page 1: EEA presentation

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

GROUP 6Makri EleftheriaDourbois GrigorisDimou AnastasiosTenekentzoglou AndreasPapadopoulos Athanasios

COMPLIES TO GREEK ENERGY EFFICIENCY STANDARIZATION “TOTEE-KENAK”

Page 2: EEA presentation

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 plansInspected…inspected…and again inspected site & buildingRe-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

Page 3: EEA presentation

WHAT’S THE GREEK POINT OF VIEW OF E.E.A?

WE CARE ABOUT THE BUILDING’S PROFILE, NOT THE USER’S PROFILESBUILDING PROFILES (=REFERENCE BUILDING) HAS MINIMUM REQUIREMENTS TO

ALL ASPECTS AFFECTING BUILDING ENERGY EFFICIENCYWE STUDY ENERGY PERFORMANCE OF REAL BUILDINGINPUT 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)

1.INTRODUCTION

Page 4: EEA presentation

2.BUILDING DESCRIPTIONSITE PLAN

SITE PLAN REFERENCE

Cyan: surrounding buildingsGrey: Building D

Page 5: EEA presentation

2.BUILDING DESCRIPTIONGround level (Basement Floor Plan)

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

GROUND LEVEL REFERENCE

Page 6: EEA presentation

2.BUILDING DESCRIPTIONBasement - Floor Plan

Basement - Floor Plan reference

Page 7: EEA presentation

2.BUILDING DESCRIPTIONStory 0 - Floor Plan

Story 0 - Floor Plan reference

Page 8: EEA presentation

2.BUILDING DESCRIPTIONTypical Story 1to9 - Floor Plan

Typical Story 1to9 - Floor Plan reference

Page 9: EEA presentation

2.BUILDING DESCRIPTIONTypical Story 10 - Floor Plan

(based on measurements)

Page 10: EEA presentation

2.BUILDING DESCRIPTIONFrontal View (SW elevation)

Page 11: EEA presentation

2.BUILDING DESCRIPTIONPerspective 3D View (SE & NE side)

(surrounds removed for better view point)

Page 12: EEA presentation

2.BUILDING DESCRIPTIONPerspective 3D View (NW & SW side)

(surrounds removed for better view point)

Page 13: EEA presentation

3.THERMAL ZONES & PROFILES

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

BUILDING PROFILE: OFFICE

THERMAL ZONES

TOTAL HEATING COOLING

A(m2) V(m3) A(m2) V(m3) A(m2) V(m3)

11925 46006 11640 44582 7842.2 28721

ZON

E 1 HEATING COOLING

A(m2) V(m3) A(m2) V(m3)

9302 30061 7215 25252

ZON

E 2 HEATING COOLING

A(m2) V(m3) A(m2) V(m3)

2338 14521 627 3469

UH

A A(m2) V(m3)

285 1424

UHA=UNCONDITIONED AREA (PART OF BASEMENT)

Page 14: EEA presentation

4.BUILDING SHELL PROPERTIES

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

4.1.a. EXTERNAL WALLS

VERY HIGH U-VALUES!A(m2) Uav Umax

Z01 2751.51 1.27 0.45Z02 394 3.704 0.45

4.1.b. UNDERGROUND WALLS

A(m2) U-value UmaxZ02 634 4.762 0.8 OVER-EXCEEDING MAXIMUM ALLOWED!

Page 15: EEA presentation

4.2. EXTERNAL OPENINGS

ZO1A(m2) Uav Gwav INFLITRATION(m3/h)

1257.17 5.46 0.53 7094.74

Z02A(m2) Uav Gwav INFLITRATION(m3/h)

188.47 6.18 0.48 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

Umax=2.8W/Km2

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

Page 16: EEA presentation

4.3. SLABS

OVER-EXCEEDING MAXIMUM ALLOWED!

ROOFS A(m2) U-value UmaxZO2 1144.1 2.02 0.4ZO1 1260.7 0.448 0.4

FLOOR A(m2) U-value UmaxZO2 2339.1 3.39 0.75

A Uav Um Ummax

WALLS 3146 1.57

2.55 1.05OPENINGS 1446 5.55ROOFS 2405 1.20FLOOR 2339 3.39SUM 9335F/V 0.20

4.4. BUILDING SHELL THERMAL INSULATION PROPERTIES (SUMMARY)

FAILED!

Page 17: EEA presentation

4.3. SHELL COATING PROPERTIES

Z02: MEDIUM COLORED PLASTER WALLS a=0.6, e=0.8 ROOFS LIGHT COLORED a=0.65, e=0.8

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 )

a : ABSORPTION COEFFICIENT TO RADIANT HEAT TRANSFER

e : EMMITANCE COEFFICIENT TO RADIANT HEAT TRANSFER

Page 18: EEA presentation

4.4. SHADING FACTORS

THERE ARE 3 MAIN TYPES OF SHADINGS WE STUDY1)HORIZON SHADING(Fhor)2)OVERHANG SHADING(Fov)3)LATERAL SHADING(Ffin)

TOTAL SHADING = Fhor*Fov*Ffin

ANALYTIC CALULATIONS FOR OPENINGSAVERAGE FOR WALLS

F=0 FULLY SHADEDF=1NO SHADING

F_hwinter shadingF_csummer shading

Page 19: EEA presentation

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

*FOR ANALYRIC VALUES CHECK CALCULATION SHEETS

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 1VERY LIGHT SHADING: most surface Fhor_h,c=1

REFERENCE BUILDING Fhor= same as calculated

Page 20: EEA presentation

4.4.2 OVERHANG SHADING FACTORS

Z01: SW,NE CURTAIN WALL OPENINGS Fov_c=0.6 to 0.7, Fov_h=0.76

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

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

Page 21: EEA presentation

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

Page 22: EEA presentation

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

5.1 HEATING SYSTEM

Page 23: EEA presentation

First thermal zoneThe 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 zoneThe 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.

Page 24: EEA presentation

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 andCOP=1

Page 25: EEA presentation

Local air-cooling units (Heat Pump)

We have heating:First zone Second zone•In all offices (in 9 floors and ground floor) In all offices (18 offices)•In the computer lab and •In the conference room.

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 alsoTwo 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:

Page 26: EEA presentation

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

Page 27: EEA presentation

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.

Page 28: EEA presentation

Heating terminal units

First zone: Radiators and SplitsSecond 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:

, 1

0.890.92

* * 1*0.97*1

emem t

rad im hydr

nn

f f f

Page 29: EEA presentation

SplitsThe 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 radiators’:

Heaters

These are the other terminal units of the boiler. According to the previous

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 efficiencySo Firt zone

Second zone

, 2

0.930.96

* * 1*0.97*1

emem t

rad im hydr

nn

f f f

, , 1 , 21* 0* 0.92em t em t em tn n n

, 1,2

0.890.917

* * 1*0.97*1

emem t

rad im hydr

nn

f f f

, 1 , 1,2

, , 21*( ) 0* 0.9182

em t em t

em t em t

n nn n

Page 30: EEA presentation

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 zoneOur 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 zoneOur 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

Page 31: EEA presentation

First zone Second zone•In all offices (in all floors and ground floor) In all offices (18offices)•In the computer lab and •In the conference room.

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 alsoTwo 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

5.2 COOLING SYSTEM

Page 32: EEA presentation

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 networkThere 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.

ηem is the emission efficiencyFrom the table 4.14 of T.O.T.E.E. 20701-1, the emission efficiency ηem of terminal units is 0.93.So

Auxiliary systemsThere are none.

Page 33: EEA presentation

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

Page 34: EEA presentation

6.IMPROVING PROJECTS

• 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.

An EWI…

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.

Page 35: EEA presentation

6.IMPROVING PROJECTSWhat 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 €

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

Energy needs reduction

Page 36: EEA presentation

6.IMPROVING PROJECTS

2. PV panels

Installation on the building’s roof

Total roof area: 800 m2

Total covered area: 313.36 m2 of monocrystalline PVs

Page 37: EEA presentation

6.IMPROVING PROJECTS

Estimated cost: 4000 € / kW

Installed power: 39.84 kW

Total estimated cost: 159,360 €

What will it cost?

Existing Building Building with PVs on Roof

Renewable Sources of Energy (kWh / m2 )

0 23.3

Energy gain

Page 38: EEA presentation

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

Page 39: EEA presentation

6.IMPROVING PROJECTS

Unfortunately Colt Ind. didn’t reply to our inquiries..

What will it cost?

Existing Building Building with PVs on Roof

Renewable Sources of Energy (kWh / m2 )

0 23.3

Heating Needs (kWh / m2 ) 63.9 70.1

Cooling Needs (kWh / m2 ) 48.2 43.9

Energy gain

Page 40: EEA presentation

6.IMPROVING PROJECTS4.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 .

Existing Building Building with management lighting system

Lighting needs(kWh / m2 ) 68.6 54.9

Page 41: EEA presentation

6.IMPROVING PROJECTS5. 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

Page 42: EEA presentation

6.IMPROVING PROJECTS6. Results

Combining all the aboveinterventions, the classificationof the under study buildingimproves from D to B, whichwas our goal.

Page 43: EEA presentation

6.IMPROVING PROJECTS

7. Green WallThe idea: growing plants vertically on the building’s wallsHow: - 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.

Page 44: EEA presentation

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

AlthoughA green wall would highlight the building as a landmark of the city!

Page 45: EEA presentation

7.PROSPECTS ON THIS PROJECT

E.E.A BASED ON AMERICAN STANDARDS

ECODESINER ANALYSIS ALMOST READY...we just input calculated values!

Page 46: EEA presentation

ECOTECT ANALYSIS WE JUST SIMPLIFY ARCHICAD MODEL

FOR BETTER PROCESS TIME(guides has been found)

Page 47: EEA presentation

MORE GREEN DESIGN IDEAS

Page 48: EEA presentation

Youtube screen recordings

Page 49: EEA presentation

Muchas Gracias!

Thank you for your attention!