Your SlideShare is downloading. ×
  • Like
Building heat demand_and_moisture
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Building heat demand_and_moisture

  • 791 views
Published

By HERENA TORIO, an alumni from PPRE …

By HERENA TORIO, an alumni from PPRE
INTRODUCTION
– ENERGY SITUATION IN BUILDING SECTOR
• PHYSICAL PRINCIPLES
– HEAT TRANSFER
– MOISTURE TRANSFER
• ENERGY BALANCES
– STEADY STATE BEHAVIOR
– DYNAMIC BEHAVIOR - THERMAL INERTIA
• CALCULATION METHODS
– MONTHLY METHOD
– SIMPLIFIED METHOD

www.devi-renewable.com

Published in Technology
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
791
On SlideShare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
31
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. ESTIMATION OF HEATDEMAND IN BUILDINGS PPRE, SS 10/11 OLDENBURG HERENA TORIO
  • 2. CONTENTS • INTRODUCTION – ENERGY SITUATION IN BUILDING SECTOR • PHYSICAL PRINCIPLES – HEAT TRANSFER – MOISTURE TRANSFER • ENERGY BALANCES – STEADY STATE BEHAVIOR – DYNAMIC BEHAVIOR - THERMAL INERTIA • CALCULATION METHODS – MONTHLY METHOD – SIMPLIFIED METHOD05.05.2011 SS 10/11 2
  • 3. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION • ENERGY CONSUMPTION IN GERMANY Lighting, Domestic 5% Energy consumption by sectors (Germany) hot water demand, 13% Industry, 27% Space heating, 81% Households, Transport, 28% 45% Source: VDEW-Materialien: Endenergieverbrauch in Deutschland, 2002 05.05.2011 SS 10/11 3
  • 4. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION • FUELS USED IN GERMANY TO SUPPLY THE SPACE HEATING DEMAND Electricity 4% Others 8% Renewables are here! Distric heating 7% Natural gas 43% Carbon 2% Gasoil 36% Source: VDEW-Materialien: Endenergieverbrauch in Deutschland, 2002 05.05.2011 SS 10/11 4
  • 5. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION IN GERMANY EnEV - “Energie-einsparverordnung”: – Limits the maximal energy demand for buildings according to their constructive details – Establishes a calculation method for the energy demand of a building -> basis for comparison – Defines different building “categories” according to their energy consumption 05.05.2011 SS 10/11 5
  • 6. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION 05.05.2011 SS 10/11 6
  • 7. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION Heat demand: • Typical building: 80 – 300 kWh/m2a • “Low Energy house” : 40 – 79 kWh/m2a • “Three-liters house”: 16 – 39 kWh/m2a • “Passive house”: max. 15 kWh/m2a • “Zero-Energy house”: 0 kWh/m2a 05.05.2011 SS 10/11 7
  • 8. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY SITUATION RENEWABLE ENERGY HEAT STANDARD “EEWärmegesetz”: – Approved July08 -> Jan 09 – Application to new buildings • Biogas 30% • Solar: 15%, 0.03-0.04m2coll/m2living area • Others (biofuels, wood, geothermal or environmental heat) 50% 05.05.2011 SS 10/11 8
  • 9. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS GENERAL BALANCES ENERGY BALANCE IN A TYPICAL BUILDING 100% Transmission Biggest energy saving Internal gains potential!!! losses 80% Ventilation losses 60% 40% 20% Solar heat Heat supplied gains by heating 0% system Cold bridges % Ventilation losses % Transmission losses % 05.05.2011 SS 10/11 9
  • 10. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS GENERAL BALANCES • BUILDING ENVELOPE: Heat losses can amount up to 75% of total heat losses External walls 20% Roof 19% Moving to energy efficient buildings… Floor to crawl space 9% BASIC USED SOLUTION: Reduction of the major heat losses Windows 52% using better materials in Percentage of heat losses through different the building envelope constructive parts of the envelope 05.05.2011 SS 10/11 10
  • 11. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER • CONDUCTION Tin d layer [m2K/W] Rlayer = λlayerl Tout Rwall = ∑ Rlayer [m2K/W] λmaterial [W/mK] 1 1 dmaterial [m] U wall = = [W/m2K] dlayer Rwall ∑λ T layer QT , wall = U wall ⋅ Awall ⋅ (Tin − Tout ) 05.05.2011 SS 10/11 11
  • 12. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER TRANSMISSION LOSSES: Conduction + convection 1 1 U wall = = [W/m2K] Rwall 1 d layer 1 +∑ + hi λlayer he Superficial heat transmission coefficient: [0 - 100 W/m2K] T • Floor to unheated basement • Roof under winter conditions! • Roof in summer conditions 05.05.2011 SS 10/11 12
  • 13. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER THERMAL BRIDGES • DEFINITION: Places on the envelope where, during the heating period, higher heat flows and lower inner surface temperatures occur. • CAUSES: Material caused thermal bridge Geometric thermal bridge Source: Maas 05.05.2011 SS 10/11 13
  • 14. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER THERMAL BRIDGES • CHARACTERISATION: Ψ = Coefficient of losses through thermal bridge, [W/mK] f = (superficial) Temperature factor , [-] Θsi= surface temperature inside wall Source: Maas Θe = exterior temperature Θi = indoor temperature f=0 -> exterior temperature f=1 -> indoor air temperature 05.05.2011 SS 10/11 14
  • 15. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER THERMAL BRIDGES Source: Maas 05.05.2011 SS 10/11 15
  • 16. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER THERMAL BRIDGES Source: Maas 05.05.2011 SS 10/11 16
  • 17. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER VENTILATION (CONVECTION) LOSSES 05.05.2011 SS 10/11 17
  • 18. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER VENTILATION LOSSES • Definition: Energy losses due to the exchange of an air flow between the building and the surroundings • Characterization: measured in h-1 = represents the portion of the total (heated) building volume exchanged in one hour • Causes: – Air leakages in the building envelope: constructive problem / solution – Health and Safety reasons: necessary to allow pollutants leave the living space According to building typology (residential, office buildings, hospitals…) minimum air exchange rates have to be assured 05.05.2011 SS 10/11 18
  • 19. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER VENTILATION LOSSES Air exchange Tight envelope (n50<3h-1) Untight envelope (n50> 5h-1) Regulable Ventilation units Window open up without cross ventilation Window open up with cross ventilation Window open without cross ventilation Window open with cross ventilation Source: Recknagel 05.05.2011 SS 10/11 19
  • 20. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS HEAT TRANSFER VENTILATION LOSSES TYPICAL VALUES for a non efficient old building: 1,5 – 2 h-1 or even higher (through air leakages in envelope) According to EnEV (Energieeinsparverordnung) in Germany: Non efficient Efficient (proven tight) Efficient building building building without with mechanical mechanical ventilation ventilation system system Values allowed Air leakages: 0,7 Air leakages: 0,6 Mech.vent.: 0,4 in EnEV, h-1 Air leakages: 0,2 05.05.2011 SS 10/11 20
  • 21. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER 1 - 2 liters/day person • Transport mechanisms: 2 people house: – DIFFUSION ca. 2liters/day – CONVECTION person (ventilation) – (SORPTION) 4 people house: ca. 4liters/day person Source: Maas 05.05.2011 SS 10/11 21
  • 22. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS Maximal water content in air MOISTURE TRANSFER 10°C 9.4g 7.9g Air temperatureSource: Maas 05.05.2011 SS 10/11 22
  • 23. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODSCARRIER (MOLIERE) DIAGRAM 05.05.2011 SS 10/11 23
  • 24. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODSCARRIER (MOLIERE) DIAGRAM 1.- 100%RH, 20°C,14.5g/kg 2.- 100%RH, 10°C,7.5g/kg 3.- 70%RH, 20°C, 7g/kg 4.- 85%RH,17°C, 7g/kg Air density ≈ 1.2kg/m3 Dew point temperature 05.05.2011 SS 10/11 24
  • 25. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER Rel. humidity Air temperatureDew point temperature Dew point temperature Rel. humidity Source: Maas Air temperature 05.05.2011 SS 10/11 25
  • 26. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER SUPERFICIAL TEMPERATURE THERMAL BRIDGES Outdoor air temp. -15°C Surface temperatures Max. relative humidity Indoor air temp. 20°C 70%RH External wall - corner 05.05.2011 SS 10/11 26
  • 27. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER MOLD GROWTH 05.05.2011 SS 10/11 27
  • 28. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER MOLD GROWTH HUMIDITY TEMPERATURE Probability of growth Probability of growth Relative humidity, % Surf. temperature, °C Source: Maas 05.05.2011 SS 10/11 28
  • 29. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER WATER CONDENSATION Rel. Humidity MOLD GROWTH Source: Maas 05.05.2011 SS 10/11 29
  • 30. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER EXAMPLE Mold growth is more restrictive condition Rel. Humidity Source: Maas 05.05.2011 SS 10/11 30
  • 31. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER DIFFUSION Source: Maas Description Unit Description Unit Temperature Partial vapor pressure Heat transm. Coeff. Material transm. Coeff. Heat conductivity Vapor diffusivity Thermal resistance Resistance to vapor diffusion Heat flow Vapor diffussion flow 05.05.2011 SS 10/11 31
  • 32. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER DIFFUSION Air Bitumen Metal Insulation d air Concrete Z air δ air δ air μ= = = Z material d material δ material [-] δmaterial dmaterial=dair Source: Maas 05.05.2011 SS 10/11 32
  • 33. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER g DIFFUSION - EXAMPLE Material μ Concrete 70-150 1086 Insulation g = 0.421 g/m2h Kork 5-10 PU foams 30-100 Alu-foil Tight 472 (100000000) 281 Wood 40 (50/400) Source: Maas [m h Pa / kg] 05.05.2011 SS 10/11 33
  • 34. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER CONVECTION - EXAMPLE Source: Maas 05.05.2011 SS 10/11 34
  • 35. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER CONVECTION - EXAMPLE Vh,buil=50m3 ; n=0.8 h-1 Ps = 1170 Pa Vvent=40m3/h (=Vh,buil*n) R = 462 J/kgK Ti=20°C, RH=50% Ps = 139 Pa Te=-10°C, RH=80% and -10°C and 1.15 263.15 1.15 304.3 Source: Maas 05.05.2011 SS 10/11 35
  • 36. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER COMPARISON CONVECTION - DIFFUSION g Aint,walls=22.5 m2 g = 0.421 g/m2h n = 0.8 h-1 Outside: 80% RH, -10°C 304.3 g/h 9.47 g/h Inside: 50% RH, 20°C Source: Maas 05.05.2011 SS 10/11 36
  • 37. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MOISTURE TRANSFER CONVECTION: Air exchange Required air exchange Rel. humidity Humidity production Source: Maas 05.05.2011 SS 10/11 37
  • 38. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY BALANCES STEADY STATE Transmission Internal gains losses Ventilation QT losses Qv Solar heat In order to keep gains Heat supplied by heating the room temperature system at a constant acceptable value Energy Supplied = Heat Losses - Energy Gains “Active gains” “Passive gains” 05.05.2011 SS 10/11 38
  • 39. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS THERMAL LOSSES TRANSMISSION LOSSES env,i = walls, floor, roof, windows QT ,env = ΣU env,i ⋅ Aenv,i ⋅ (Tin − Tout ) [W] (separately for each of them) – INCLUDING THERMAL BRIDGES H T ,building = ΣU env ,i ⋅ Aenv ,i + ΔU tb ⋅ Aenvelope [W/K] – TOTAL TRANSMISSION LOSSES QT ,buil = (ΣU env,i ⋅ Aenv,i + ΔU tb ⋅ Aenv ) ⋅ (Tin − Tout ) [W] 05.05.2011 SS 10/11 39
  • 40. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS THERMAL LOSSES VENTILATION LOSSES n = air exchange rate [h-1] H V = Vh,buil ⋅ 0.34 ⋅ n [W/K] HEATED Heat volume of the capacity of air [Wh/m3K] building [m3] QV = H V ⋅ (Tin − Tout ) According to the German regulation EnEV, can be simplified: Vh ,buil = Vbrutto ⋅ 0.76 05.05.2011 SS 10/11 40
  • 41. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS THERMAL LOSSES TOTAL LOSSES (TRANSMISSION+VENTILATION) – Transmission Losses QT ,buil = (ΣU env,i ⋅ Aenv,i + ΔU tb ⋅ Aenv ) ⋅ (Tin − Tout ) =HT ⋅ (Tin − Tout ) [W] – Ventilation Losses QV = H V ⋅ (Tin − Tout ) [W] – Total Losses Qlosses = ( H T + H V ) ⋅ (Tin − Tout ) [W] 05.05.2011 SS 10/11 41
  • 42. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS THERMAL LOSSES + GAINS WINDOWS – Upane = 3 – 0.6 [W/m2K] -> great influence on heat demand – SHGC, g = 0.5 – 0.8 [-] -> great influence on cooling demand – ε = 0.84 – εlow = 0.2 !!! 05.05.2011 SS 10/11 42
  • 43. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS ENERGY - DYNAMIC ENERGY BALANCE - DYNAMIC BEHAVIOR Transmission Internal gains losses Ventilation QT losses Qv Solar heat gains Heat supplied by heating system Energy Supplied = Heat Losses - Energy Gains +- Energy Stored 05.05.2011 SS 10/11 43
  • 44. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS 5000 4000 Specific heat capacity ENERGY - DYNAMIC THERMAL MASSc [J/kgK] 3000 2000 1000 0 Aluminium Foam glass Glass Brick Wood Sand Concrete Water insulation Mineral 3000 density 2500 2000 rho [kg/m3] Csto = ci ⋅ ρ i ⋅ Ai ⋅ d i 1500 1000 500 0 Aluminium Glass Brick Sand Concrete Foam glass Wood Water insulation Mineral Source: Wikipedia 05.05.2011 SS 10/11 44
  • 45. lambda [W/mK] 0 1 2 3 4 5 Wood Glass05.05.2011 INTRODUCTION Source: Wikipedia Mineral insulation Foam glass THERMAL MASS Sand Brick PHYS. PPLES. 237 Aluminium c [J/kgK] Concrete 0 1000 2000 3000 4000 5000 rho [kg/m3] Water 0 500 1000 1500 2000 2500 3000 WoodSS 10/11 Wood Glass Glass Mineral insulation Mineral insulation Foam glass ENERGY BALANCESFoam glass Sand Sand Brick Brick Aluminium Aluminium Concrete Concrete45 THERMAL MASS Water Water CALC. METHODS
  • 46. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS Specific heat capacity ENERGY - DYNAMIC THERMAL MASS Concrete Insulation Temperature Temperature Thickness 05.05.2011 SS 10/11 46 Source: Maas
  • 47. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS Specific heat capacity ENERGY - DYNAMIC U-Value Mass THERMAL MASS [W/m2K] [kg/m2] 6cm Outdoor Temperature40cm Energy flow43.5cm Solar radiation 26cm Time of day Outdoor Temperature radiation Solar 05.05.2011 SS 10/11 47 Source: Maas
  • 48. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS TOTAL LOSSES (TRANSMISSION+VENTILATION) – Transmission Losses QT ,buil = (ΣU env,i ⋅ Aenv,i + ΔU tb ⋅ Aenv ) ⋅ (Tin − Tout ) =H T ⋅ (Tin − Tout ) [W] – Ventilation Losses QV = H V ⋅ (Tin − Tout ) [W] For which time-step Depends on the data we have for the do we apply this equation? outdoor – Total Losses temperature… Qlosses = ( H T + H V ) ⋅ (Tin − Tout ) [W] Tin is the indoor desired temperature: regarded as a CONSTANT value, typically set between 19 and 21°C for the heating period. 05.05.2011 SS 10/11 48
  • 49. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD TOTAL LOSSES Q losses = Σ ( H T + H V ) ⋅ (Tin − Tout ) ⋅ t M ⋅ 24 [Wh/a] months • Tout represents MONTHLY mean values • tM represents the number of days of the month considered 05.05.2011 SS 10/11 49
  • 50. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD SOLAR HEAT GAINS QSolar , windows = Σ Awindows ⋅ g i ⋅ FF ⋅ Fs ⋅ Gwindow [Wh/a] months • gi represents the energy transmissivity of the window glass; typically is around 0.6 • FF represents the % of glass against frame in the window area; typically is around 0.7 • Fs represents the % of shadowing over the glass • Gwindow represents the incident solar radiation onto the window, in Wh/m2 05.05.2011 SS 10/11 50
  • 51. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS THERMAL LOSSES + GAINS WINDOWS, 52% of total losses: Does not require 1. Avoid heat losses -> Better insulation materials: much more planning - Uw= 3 - 0.6 W/m2K Three pane window Single or two effort. filled with Ar/Kr pane window Typ. In efficient houses 2. Increase solar heat gains -> Orientation - Highest solar irradiation on the south façade, Requires integral planning of high potential for solar heat gains -> maximize glazed surface facing south the building integrated into - North façade receives very few solar its environment for solar gains and irradiation, low potential high heat losses through windows -> minimize Typ. Approach passive houses glazed surfaces Yearly variation of solar path in the sky 05.05.2011 SS 10/11 51
  • 52. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD INTERNAL HEAT GAINS • Internal heat gains depend on the use pattern of the building: office, hospital, residential… • For residential buildings: constant hourly value of 5 W/m2, per m2 useful area in the building Qint_ gains = Σ 5 ⋅ AN ⋅ 24 ⋅ tM [Wh/a] months 05.05.2011 SS 10/11 52
  • 53. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD ENERGY DEMAND – Simplification: Qh = Qlosses − QSolar , windows − Qint_ gains [Wh/a] – Actually, not all energy gains can be “used”: Qh = Qlosses − η (QSolar , windows + Qint_ gains ) [Wh/a] • η depends on the heat storage capacity of the building structure and its materials, which is a function of ρ [kg/m3], c [Wh/kgK], d [m], A [m2] of the material: Csto = ci ⋅ ρ i ⋅ Ai ⋅ d i [W/K] 05.05.2011 SS 10/11 53
  • 54. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD ENERGY DEMAND – Types of building constructions according to its heat capacity • LIGHT – Csto/A < 50 Wh/m2K • HEAVY – Csto/A > 130 Wh/m2K Qh = Qlosses − η (QSolar , windows + Qint_ gains ) [Wh/a] – η = 0.9 for light buildings [-] – η = 0.95 for heavy buildings [-] 05.05.2011 SS 10/11 54
  • 55. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS MONTHLY METHOD ENERGY DEMAND – BUILDING WITH ZONES AT DIFFERENT TEMPERATURES: • German Norm: gives correction factors, Fx, that have to be applied to obtain the HT corrected of the building Building part Fx [-] Outside wall, window, roof, floor 1 H T = ΣU wall ⋅ Awall + ΔU tb ⋅ Aenvelope Walls and roofs to unheated rooms 0.5 - Floor to ground 0.6 - Walls and floor to unheated crawl space H T = ΣU wall ⋅ Awall ⋅ Fx + ΔU tb ⋅ Aenvelope 05.05.2011 SS 10/11 55
  • 56. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS DEGREE-DAYS METHOD TOTAL HEATING DEMAND DEGREE-DAYS z [Kd/a] Gt 20 /15 = ∑ (Tin − Tout ) 1 [°Cd/a] • Tout represents mean DAILY values • Sets up a “heating limit” (15°C), above which no space heating is required. For this conditions (Tin-Tout)=0 • Below the “heating limit”, (Tin-Tout) is calculated and added up to give a value of the “degrees-day” 05.05.2011 SS 10/11 56
  • 57. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS DEGREE-DAYS METHOD TOTAL HEATING DEMAND [Wh/a] Qlosses = Σ ( H T + H V ) ⋅ Gt Q losses = Σ ( H T + H V ) ⋅ (Tin − Tout ) ⋅ t M months days [Wh/a] QSolar , windows = orientation Awindows ⋅ g i ⋅ FF ⋅ Fs ⋅ Gwindow Σ QSolar , windows = Σ Awindows ⋅ g i ⋅ FF ⋅ Fs ⋅ Gwindow months [kWh/a] Qint_ gains = 22 ⋅ AN Qint_ gains = Σ 5 ⋅ AN ⋅ 24 ⋅ tM months [Wh/a] Qlosses = Qh − η (Qsolar − Qint ernal ) Qh = Qlosses − η (Qsolar , windows − Qint ernal ) 05.05.2011 SS 10/11 57
  • 58. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS DYNAMIC TOOLS – The equations for steady state conditions are not valid here!!!-> Energy stored in the building structure plays a role – FREEware available (Hourly simulations):DOE2, eQUEST, ePLUS (http://www.doe2.com/ ) Much more accurate results × Require the description of the HVAC system as INPUT × Time demanding to learn how to work with them: weather data for Stüdl Hütte, etc… may not be in database 05.05.2011 SS 10/11 58
  • 59. INTRODUCTION PHYS. PPLES. ENERGY BALANCES CALC. METHODS STATIONARY METHODS • STATIC (simplified) METHODS & SOFTWARE: – Based on the steady-state simple equation -> quite simple calculations Depends only on (rough) CLIMATIC data and the BUILDING ENVELOPE -> Does not require the description of the HVAC system as INPUT × Much more rough results – Examples: “DEGREE-DAY Method” and Monthly simplified method in EnEV http://www.uni- kassel.de/fb6/bpy/de/index.html 05.05.2011 SS 10/11 59
  • 60. THANK YOU FOR YOUR THANKS FOR YOUR ATTENTION!!! ATTENTION!!!!!!05.05.2011 SS 10/11 60
  • 61. EXAMPLE 3.0m• AN = 147m2 ; Vbrutto = 580 m3 3.0m 10m• Awalls = 209.34 m2; Afloor = 88.2 m2; Aroof = 88.2 m2 7.35m• Awindows: S 15 m2; E/W 10m2; N 5.5 m2• Uwalls = 0.45 W/m2K (walls); • G19/10 = 2750 °Cd/a (Hamburg) Ufloor-roof = 0.3 W/m2K (floor and roof); Orientation Solar radiation Uwindows = 1.4 W/m 2K (windows) [j] [kWh/m²] Nord 136• Utb = 0.1 W/m 2K Süd 349• n =0.6 h-1 Ost 220 West 220• Windows: Ff= 0.7; Fs=0.9;g=0.58;• Heavy building05.05.2011 SS 10/11 61 T 19°C