Zero Energy Home Case Study in Germany (1)

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Beuth University of Applied Sciences Berlin – MBA Renewables

RE and EE Systems and Concepts

Grid

‐

Connected Zero

‐

Energy Home

Group #05

Lecturer Dr.‐Ing. Volkhard Scholz

Group Members Guillaume Gerard Gatoil

Guillermo Alberto Chacon Giron

Wolf-Dietrich Kindt

Martin Kreuzer

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1. Introduction

The main objective of this assignment is to provide a solution with renewables sources

of energy for a chosen house in Germany in such a way that the balance of consumed

fossil energy and the generated renewable energy (REN) is zero over the year.

First, the chosen house will be introduced with its real energy consumption data. In

sequence, the environmental conditions will be analysed regarding the location of the

house.

Based on these information of energy consumption and climate data, three renewable

technologies from Table 1 will be chosen in order to achieve this assignment objective.

All calculations will be made to size the systems.

# Renewable Energy Technology

1 Photovoltaic (PV) Plant

2 Solar Water Heater (SWH)

3 Micro Wind Turbine

4 Micro Hydro Power Plant

5 Wood Boiler

6 Bio digester

Table 1: Renewable Energy Technologies considered for this assignment

2. Chosen house & Energy Consumption

The chosen house for this study was built in the 70’s and is situated in a little village in

Germany. The exact location is Lauterbacher Straße, 34516 Vöhl Obernburg, Hessen –

Germany and the geographical coordinates are 51°13'55.5"N 8°54'18.2"E. Two

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Figure 1: Google image from the chosen house

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5 The electricity and oil consumption is read out only once a year. The energy consumption

in 2013 for 2 residents is shown in the table 2.

Consumption 2013

Electric Energy Wel 3,000 kWh

Fuel Oil for central Heating (water and house)

3,000 ltr = 30,000 KWh (1KWh = 0.1L of petrol)

Table 2: Energy consumption for the chosen house in 2013 for 2 residents

Nowadays, the heating for the central heating and the warm water system is done by an

oil fired water boiler connected to a hot water storage tank.

The tank “Viessmann Vitocell γ00 B” is a twin coil cylinder, the upper one for a boiler and

the lower one for a solar water heating system (datasheet in annex). Currently only the

boiler is used.

3. Climate Data & Environmental Conditions

3.1. Solar Irradiation

The meteorological database used was obtained from Meteonorm 7.1. Using the city of

Vöhl, which is about 5 km from the chosen house and considering that the modules are

facing to the South West with an angle of 255° (with 0° being North), we obtain the solar

irradiation data as listed in Table 3.

Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Year

Specific Irradiation onto Horizontal Plane,

Irrhor [kWh/m²]

20.8 35.3 70.3 117 140 146 151 119 85.1 50.8 22.4 15.1 973

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6 3.2. Wind data

A wind station is located in Basdorf/Edersee, which is around 5 km from the house. For

this work we are assuming the same data for our chosen house. The wind data was

collected from the weather service of Windfinder and is shown in Figure 4 (access date:

31.11.2014). The data is accessible online at:

http://www.windfinder.com/windstatistics/basdorf_edersee.

Figure 3: Wind data of a wind station 5 km away from the house location

Average wind speed are null for 6 months of the year. For the other 06 months, wind

speed are between 2 to 3 m/s. Generally wind turbines start generating power from 3 to

4 m/s. Wind power is not appropriate for this site.

3.3. Hydrology data

A river is located around 350 m away from the house, as shown in Figure 5. No hydrology

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7 distribution system crossing over neighbours properties, wood and a road. Hydropower

plant is not appropriate for this site.

Figure 4: Distance from the house to the closest river and surroundings analysis

3.4. Temperature

The temperature data were taken from the Photovoltaic Geographical Information

System (PVGIS) of the European commission (available on the website

http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php#, access date: 31.11.2014) using the

coordinates of the house provided in section 2.

Table 4 shows the temperatures for the given location, where:

● TD: Average daytime temperature (°C)

● T24h: 24 hour average of temperature (°C)

Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Year TD 1.1 2.1 5.6 10.4 14.2 17.1 18.9 18.9 15.2 10.9 6 1.8 10.2 T24h 0.2 0.7 3.9 8.7 12.7 15.8 17.6 17.3 13.4 9.2 5 0.9 8.8

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4. Zero Energy Home

The objective of this assignment is to provide a solution with renewable sources of

energy for a chosen house in Germany in such a way that the balance of consumed

fossil energy and the generated renewable energy (REN) is zero over the year.

This section will show the selected REN technologies and explain the reasons for the

choices. Calculations for each REN solution will also be presented in order to

demonstrate technical feasibility of a sustainable domestic energy supply with renewable

sources. Concrete costs for installation, operation and maintenance will not be a part of

this study.

4.1. Selected REN Technologies

Section 3 identifies wind and water as unfavourable options as energy sources, while

solar irradiation shows to be an available source of energy.

Therefore, the two first solutions proposed focus on technologies that uses the sun as

primary source, as follows:

1. Photovoltaic Plant (PV Plant) as an electrical energy source.

2. Solar Water Heater (SWH) as a heating source for domestic use.

Section 2 indicates a higher energy consumption with heating than electricity. The SWH

alone will cover hot water demand consumed by the residents but it is not enough to

generate energy for space heating. Therefore, a third solution is necessary to cover

complementary the demand for heating during cold seasons.

3. The chosen technology to supply space heating energy is the Wood Boiler.

The bio digester technology was declined due to insufficient organic waste availability to

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4.2. REN Technology 1 – Photovoltaic Plant

4.2.1. Position of the collectors and real solar irradiation

Due to the geographical location of the house on the northern hemisphere, the modules

should be oriented to face south. The optimal inclination is approximately the latitude of the house, which in our case is 51°. In details:

● solar angle ( ) at solar noon, when the irradiance of the sun is at its highest, are:

- β1st Jun: = 6β.β° (best inclination of the module is = 90° - 62.2° = 27.8°)

- 21st Dec: = 15.4° (best inclination of the module is = 90° - 15.4° = 74.6 degrees)

● mean solar angle during the year ( m) = 38.8°, then, the best inclination angle of

the module would be 90° - 38.8° = 51.2°

Figure 5: Best solar angles illustrations according to seasons for the area of Düsseldorf.1

For the chosen house, the modules will be installed on the roof, whose inclination is 30° (Table 5). Therefore, the modules will not be on the optimal position.

Optimal Real

Azimuth α [°] 0 75 Inclination [°] 51.2 30.0

Table 5:Optimal conditions for the PV system and the real conditions at the house

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10 For a module at an arbitrary tilt and orientation equation (1) 2_{describes how much}

irradiation the module receives:

Irrsol = Irrinc ·[cos( )· sin( ) · cos(α - ) + sin( ) ·cos( )] Equation (1)

where:

Irrsol: Irradiation at the surface of the module

Irrinc: Irradiation Incident to the module

: Solar Angle

: Inclination of the module (in our case, 30°) α: Azimuth of the module (in our case, 255°)

: Sun Azimuth = 180°

Knowing that:

Irrinc = Irrhor / sin( ) Equation (2)

where:

Irrhor: Irradiation on the horizontal plane (Table 3)

and replacing (2) in (1), we obtain:

Irrsol = Irrhor · [cos( ) · sin( ) · cos(α - ) + sin( ) · cos( )] /

sin( ) Equation (3)

Table 3, presented in section 3.1, gives the monthly irradiation on the horizontal plane

(Irrhor). Table 5 below shows the real solar irradiation on the surface of the module

considering the solar angle per month (Irrsol).

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year

Table 6: Real solar irradiation on the surface of the module considering the solar angle

per month Irrsol.

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11 With the modules installed with = γ0° of inclination and α = βηη° (SW) direction,

the real solar irradiation on the module surface is Irrsol = 970.44 [kWh/m²] per year.

4.2.2. Efficiency and PV Module Area

For our calculations, we will use the following data for the modules3_{and inverter:}

● Module efficiency ηmod = 21.15 %

● Area of one module unit = 1.63 m2

● Rated power = 345 Wp

● Inverter efficiency ηinv = 96 %

The PV plant must produce an electric energy of at least Etot = 3,000 [kWh] to supply the

current electricity demand of the house. Using the equation:

WPV, demand = Irrsol· Apv· mod· inv· (1 - Losses) Equation (5)

where

WPV, demand: Total energy = 3,000 kWh

Irrsol: Irradiation at the surface of the module = 970.44 kWh/m²

Apv: PV Module Area

mod: Module efficiency = 21.15 % inv: Inverter Efficiency = 96 %

Losses = installation losses assumed to be 10% = 0.10 This results in a required total module area of Apv = 16.91 m2.

As one module has an area of 1.63 m2_{, a total of 11 modules is required to cover the}

yearly electricity consumption of the house4_.

The resulting PV system will have a total area of 17.93 m2_{with a module}

efficiency of mod = 21.15 %.

3_{See Annex β “PV technological selection” & Annex γ “SunPower X-Series β1% commercial datasheet”.}

4_{“ee A e PV ields si ulatio fro soft are PV“ol for alidatio of al ulated data a d e a ple}

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Figure 6: 3D simulation of the PV modules on the house with real slope and azimuth orientation. Source: PVSol

4.2.3. Performance ratio

To calculate the performance ratio we will use the following equation:

PR = (Yf / Yr) _{Equation (6)}

where:

Yf: Energy delivered to the grid per year and KWp (specific Annual Yield) Yr: Theoretically available energy

Considering that the PV system will have 11 modules and each module produces 345

Wp, the PV plant produces a total nominal power of 3.795 KWp.

The energy delivered is calculated by equation (5):

WPV = 970.44 · 17.93 · 0.2115 · 0.96 · 0.90 = 3,179.60 kWh/y

The energy delivered to the grid per year and kWp (Yf) is then equal to:

Yf = 3,179.6 / 3.795 = 837.84 [h/year]

The theoretically available energy (Yr) is the total in-plane irradiance in kWh/m2_{per year}

divided by the reference irradiance G, which is 1 [kW/m2_].

Yr = 970.44/1= 970.44 [h/year]

Performance ratio is then calculated by equation (6):

PR = Yf / Yr = 837.84 / 970.44 = 0.86

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13 4.2.4. Power Generation and solar fraction

The total power generation WPV, as calculated in section 4.2.3, is 3,179.60 kWh per year.

The annual solar fraction fPV is calculated dividing this amount by the total demand of our

project, which is 3,000 kWh:

fPV = WPV / WPV, demand = 3,179.6 / 3,000 = 1.0599

fPV = 105.99%

Table 65_{shows the monthly electrical power generated by the PV system}_W

PV and per 1

m² module area WPV,A, as well as the monthly solar fraction fPV. The monthly demand of

electricity WPV, demand results from average proportions of consumption through the days

of the year in German households and is taken from the simulation software package

PVSol.

Jan 24.39 79.90 4.46 305.96 -226.05 26.12%

Feb 38.21 125.20 6.98 260.05 -134.85 48.15%

Mar 72.06 236.11 13.17 270.16 -34.06 87.39%

Apr 115.01 376.82 21.02 241.31 135.51 156.15%

May 133.64 437.85 24.42 230.69 207.16 189.80%

Jun 136.43 447.01 24.93 213.78 233.23 209.10%

Jul 144.88 474.70 26.47 200.40 274.30 236.87%

Aug 117.43 384.75 21.46 217.84 166.91 176.62%

Sept 87.32 286.08 15.95 222.07 64.02 128.83%

Oct 54.94 180.00 10.04 256.09 -76.09 70.29%

Nov 26.24 85.96 4.79 274.77 -188.81 31.28%

Dec 19.90 65.21 3.64 306.88 -241.66 21.25%

Year 970.44 3,179.60 177.33 3,000.00 179.61 105.99%

Table 7: Power Generation and solar fraction for the PV system

5_{“ee A e PV s ste desig spreadsheet a d origi al E el file for o er ie of the o plete}

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14 The proposed PV system is supplying the annual electricity demand of the house

and generating a surplus of 179.61 kWh to the grid, with an annual power

generation of WPV = 3,179.60 kWh and WPV, A = 177.33 kWh per m².

4.3. REN Technology 2 – Solar Water Heater

4.3.1. Heat demands for warm water

Solar Water Heater (SWH) is the second REN technology chosen to supply the energy

demand and the first one for heating purposes. The warm water consumption assumed

is 40 litres at 60°C per person a day. Considering 2 residents in the house and an

average of 30.5 days per month, the monthly average warm water consumption is: 40 x

2 x 30.5 = 2440 l/month or 2.44 m3_/month.

The daily energy demand is calculated with the formula:

Equation (7)

c: Specific Heat Capacity of Water = 1,16Wh/kg∙K

ΔT: Temperature Difference Between Hot and Cold Water = 60°C - TCold Water

The temperatures of the cold water are the average 24h temperatures (T24h) listed in

Table 4. The resulting monthly energy demand for warm water QH2O is listed in Table 8.

The highlighted cells show the months with an average daily temperature above 10°C.

Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Year

T24h [°C] 0.2 0.7 3.9 8.7 12.7 15.8 17.6 17.3 13.4 9.2 5 0.9 8.8 QH2O [KWh] 170 168 159 146 134 126 121 121 132 144 156 168 1745

Table 8: Monthly energy demand for warm water.

4.3.2. Position of the collectors and real solar irradiation

The collector will be installed on the same roof as the PV modules. Thus, the best angle

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15 for the PV plant in section 4.2.1. We can consider the optimal inclination of the solar

collectors ( ) of 51.β° and anazimuth angle (α) of 0° (True South). But the real conditions

of the house roof, where the collectors will be installed, are different, as shown in table

8 below.

Optimal Real

Azimuth (α°) 0 75

Inclination (tilt °) 51 30

Table 9: Optimal conditions for the SWH system and the real conditions for the house

The solar irradiation on the solar collectors Irrsol is the same as for the PV modules,

calculated using Equation 3. Resulting values are listed in Table 10.

Irrsol

[kWh/m²] 24,39 38,21 72,06 115,01 133,64 136,43 144,88 117,43 87,32 54,94 26,24 19,90 970,44

Table 10: Solar irradiation on the solar collectors Irrsol, calculated by using Equation 3.

4.3.3. Collector area

In winter time, the SWH energy generation is low, due to poor solar irradiation. For this

reason the SWH system will be installed in parallel with a second heating technology

(section 4.4). The second system will be used when average daily temperature is below

10°C. Therefore, the SWH will be calculated for the summer months (May to September)

when the average daily temperature is above 10°C.

The SWH system will be sized with the worst case scenario in order to assure enough

hot water. The highest heat demand for the SWH system is in September with 132 kWh

(Table 8).

The required collector area for warm water supply ASWH in the summer month will be

calculate using the following equation:

Equation (8)

Using:

Average monthly consumption = monthly energy demand for September from Table

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16 Solar fraction for the summer month = 120 % (considering 20% of losses in energy storage)

Average monthly solar irradiation = monthly real solar irradiation Irrsol of September

from Table 10 = 87.32 KWh/m2_{per month}

System Efficiency = 50% (assumption)

This results in a required collector area of ASWH = (132 x 120%) / (87.32 x 50%) = 3.62 m2.

According to the products available on the distribution market (module dimensions of 1m

x 2m = 2m2_{/unit), two modules will be required.}

The chosen solar collector will have an area of ASWH = 4.0 m2.

4.3.4. Heat yields and solar fraction

As the system efficiency is 50%, the monthly heat yields per m2_(Q

SWH,A) is always half

of the monthly real solar irradiation Irrsol of Table 10. The monthly heat yield of the

installed SWH system of the house (QSWH) is then given by multiplying the heat yield per

m2_{by the collector area, which is 4.0 m}2_{. The monthly solar fraction (f}

H2O,SWH) is obtained

Table 11: Resulting heat yields QSWH and solar fraction fSWH of the SWH system.

6_{“ee A e “WH & Wood Boiler h rid s ste desig spreadsheet a d origi al E el file for a al sis}

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4.4. REN Technology 3 – Wood Boiler

The wood boiler will directly replace the current oil boiler. The system is connected to

the same water storage tank7_{as the solar water heating systems. Since Space heating}

and domestic hot water are interconnected in the same storage of heat energy, systems

can complement one another for space and water heating functions.

4.4.1. Heat demands for space The wood boiler will cover:

- Space heating (Qroom) during the month when it is necessary

o Heating is necessary in months with an average temperature of <10°C,

as in the assumptions for the SWH. Hence heating is necessary in Oct,

Nov, Dec, Jan, Feb, Mar and Apr.

o Monthly Qroom are calculated according to the formula with the

assumption that the house is well insulated:

Qroom = 1.2 kWh * (15°C – Tair)/°C per m². Equation 9

- Warm water supply (QH2O), complementally to the SWH when needed

o In the month of April the SWH will provide extra heat for the space

heating.

o During colder season the wood boiler will support the SWH because sun

irradiation is poor at those times.

Months Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year

Table 12: Monthly balance heat demands of wood boiler Qwood for water QH2O,WB and space Qroom.

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18 4.4.2. Wood Specifications

The wood boiler will be fired by wood pellets according to the standard EN 14961-2.

According to this standard, water content w in wood pellets has to be less than 10% with

a pellet bulk density of ≥ 600kg/m³. The lower energy content LHWFMhas to be ≥ 16.5

MJ/kg (for the A1 quality defined by EN 14961-2). For the calculation we assume LWHFM

= 16,5 MJ/kg (4.44kWh/kg).

4.4.3. Consumption of wood and storage

We consider the efficiency of the wood boiler at 80% to convert wood energy from

pellets to heat energy in the water storage tank. The wood consumption per month is

calculated from the heat demand for the wood boiler according to the formula:

Qwood = (Qroom + QH2O,WB) Equation 10

The resulting amount of wood pellets mFM is calculated using the energy content of

LWHFM = 4.44 kWh per kg:

mFM = Qwood /( • LWHFM) Equation 11

We can calculate storage volume Vstore accordingly:

Vstore = mFM / pellets density Equation 12

Resulting data are listed in Table 13.

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19 the winter months (QH2O, WB), when the SWH system is not sufficient, and cover to 100%

the space heating demands (Qroom). Therefore, the annual heat fraction for space heating

supplied by the wood boiler results to froom,wood = 100%. The heat fractions of the wood

boiler for water heating fH2O, WB are listed in Table 148. Furthermore, the total energy

needed for heating (Qtotal = QH2O, total + Qroom), the total energy supplied by the wood boiler

Qwood and the total heating fraction of the wood boiler ftotal wood are listed.

In addition, the total heat fraction of the SWH system is given in the table, taking the

fraction of the SWH for the total heat energy demand into account.

Month QH2O,tot

Table 14: Listed are the total heat demand Qtotal, the heat demand covered by wood boiler Qwood and the heat fractions of the wood boiler fwood.

The annual heat fraction of the wood boiler for warm water supply results in an

average of fH2O,WB = 23 % and the annual heat fraction for the wood boiler for the

total heat demand results to an average of ftotal wood = 94%.

8_{“ee A e “WH & Wood Boiler h rid s ste desig spreadsheet a d origi al E el file for a al se}

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5. Discussion

5.1. Summary of technical choices

Renewable energy technologies were investigated (Table 1), in order to identify three

convincing technologies for the renovation of a chosen 40 years old house. The house

was built in the 70’s and depends until now exclusively on fossil energy sources and

electricity from the national grid system. The addressed task was to find a site specific

concept for providing all consumed energy by renewable energy sources.

Considering the location of the analysed house, a wind turbine and hydro power plant

are not applicable as renewable energy sources. The average wind speed is not

sufficient to provide energy from wind turbines. A hydropower plant cannot be installed

as a river is not accessible from the premises of the house. Concerning a bio-digester,

not enough organic waste is available on the premises.

Despite low average sun radiation and the inadequate orientation of the roof, the energy

demand can partly be meet by a photovoltaic plant and a solar water heater system. The

roof space available allows for the installation of the required 17.93 m2_photovoltaic

modules, producing 3180 kWh year which exceeds slightly the annual electricity usage

of 3000 kWh.

Solar water heater collectors can be installed with an area of 4 m² to provide 1941 kWh

per year for warm water heating. The energy conversion from solar energy to heat energy

works well during the summer months (April – September) at the house location. In order to provide enough energy for warm water and space heat during the winter

months (October – March) a wood boiler needs to be installed. The wood boiler works in perfect symbiosis with the solar water heater system.

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21 5.2.1. Pros

Electricity and Petrol prices are increasing consequently and the world realises that

energy becomes valuable. The homeowner will see his electricity bill balance very small

(connection fees only) and wood pellets more affordable than petrol.

The house increases its degree of energy self-sufficiency and limits impact on the

environment.

5.2.2. Cons

The usage of wood is very convenient as it is easily storable. However burning wood,

even in a high efficient manner, is still emitting CO2. The supply of raw material is also

controversial: cutting trees and using fields that could serve food supply. Some

ecologists do not consider wood a sustainable source of energy.

The photovoltaic plant is providing energy generated during the day to the grid and

consumes energy from the grid during the night. Therefore, the PV plant does not

constantly self-consume electricity and the homeowners depend on the grid.

Using only solar power would require storing the electricity generated during the day in

some type of battery bank in order to become autonomous. However the cost of such

energy back-up system is doubling the investment from the original solar plant.

5.3. Conclusion & Recommendations

With the described renewable energy technologies photovoltaic, solar water heater and

wood boiler, the energy demand for the specified client can be met completely. In a next

step the installation cost for the suggested systems should be calculated, in order to

provide a conclusive picture of the profitability and feasibility of the proposed systems.

In addition, the reduction of energy can be achieved by reducing the energy

consumption. It is recommended to consider investing in low consumption items first, as

LED lights or insulation of the walls. Ultimately the REN systems can be designed smaller

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6. List of symbols / acronyms

α Azimuth

Inclination

mod Module efficiency

Apv Photovoltaic module area

ASWH Solar water heater area

fPV Solar fraction

fSWH Heat fraction produced of solar water heater

fwood Heat fraction produced of wood boiler

Irrinc Irradiation Incident to the module

Irrhor Irradiation on the horizontal plane

Irrsol Irradiation at the surface of the module

mFM Total mass of wood pellets

Qroom Heat demand for space heating

QH2O Total heat demand for warm water

QH2O,WB

Heat demand for warm water, covered by wood boiler

QSWH Heat produced by solar water heater

QSWH,A Heat produced by solar water heater per m²

Qtotal Total Heat demand

Qwood Total heat produced by wood boiler

REN renewable energy PV Photovoltaic PR Performance ratio SWH Solar water heater

TD Average daytime temperature

T24h 24 hour average of temperature

Vstore Wood pellet storage volume

WPV, demand

Electricity demand

WPV Electricity generated by Photovoltaics

WPV,A Electricity generated by m²

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7. List of figures

Figure 1: Google image from the chosen house

Figure 2: Front view from the chosen house

Figure 3: Wind data of a wind station 5 km away from the house location

Figure 4: Distance from the house to the closest river and surroundings analysis

Figure 5: Best solar angles illustrations according to seasons for the area of Dusseldorf.

Figure 6: 3D simulation of the PV modules on the house with real slope and azimuth

orientation. Source: PVSol

8. List of tables

Table 1: Renewable Energy Technologies considered for this assignment

Table 2: Energy consumption for the chosen house in 2013 for 2 residents

Table 3: Solar Irradiation Irrhor data of the city of Vöhl used for the calculations for the

chosen house. Source: Meteonorm 7.1

Table 4: Temperatures for the house location

Table 5:Optimal conditions for the PV system and the real conditions at the house

Table 6: Real solar irradiation on the surface of the module considering the solar angle

per month Irrsol.

Table 7: Power Generation and solar fraction for the PV system

Table 8: Monthly energy demand for warm water.

Table 9: Optimal conditions for the SWH system and the real conditions for the house

Table 10: Solar irradiation on the solar collectors Irrsol, calculated by using Equation 3.

Table 11: Resulting heat yields QSWH and solar fraction fSWH of the SWH system.

Table 12: Monthly balance heat demands of wood boiler Qwood for water QH2O,WB and

space Qroom.

Table 13: Wood pellet consumption

Table 14: Listed are the total heat demand Qtotal, the heat demand covered by wood

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9. Annexes

9.1. Annex #1: PV System Design Spreadsheet

* Original version file attached “ZeroEnergyHome

-Group05_CalculationTables_Rev0β.0.xls”, Sheet “PV System”.

9.2. Annex #2: PV Technology Selection

Different PV technologies exists. The most common PV modules are made from crystalline solar cells. They are efficient under direct sunlight but their performance dramatically decreases when thick clouds are passing by. In the present location, irradiation under 1000 KWh/m2/year is sign of recurrent overcast weather.

CdTe or commonly called thin-film modules keeps sun efficient energy conversion into diffuse exposure. That is a good option but they take more space for the same nominal power installed. We must consider the roof space limitation and physical installation. Roof already integrate skylights where it will be difficult to work around. We must also keep some space for the SWH system.

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26 9.4. Annex #4: PV Energy Yield Simulation from software PVSol

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27 9.5. Annex #5: Storage Tank Viessmann Vitocell 300B

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28 9.6. Annex #6: SWH & Wood boiler Hybrid System Design Spreadsheet

* Original version file attached “ZeroEnergyHome-Group05_CalculationTables_Rev0β.0.xls”,