This document provides information to calculate the payback period of installing a small wind turbine. It details the costs of equipment, installation, annual maintenance, electricity production estimates, and revenue from electricity sold to the grid. The simple payback period is calculated to be 29.9 years, making the project unsustainable given the turbine lifespan is only 25 years. Additional information is provided on home electricity usage and average wind speeds at different heights.

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Energy Assignment 1
Q. Calculate the Payback Period of a small wind power installation
Feasibility Study
Beforehand
 Cost of site engineer €150 (may be refunded upon
purchase of turbine)
Cost Of Equipment  Turbotricity 2.5KW (includes 11.5m tower, grid-tie
inverter, controller and electronics) €13,500 inc. VAT
fully installed
 30m Armoured Cable (inc. in price)
 Includes a 5 year guarantee
 Transport Costs €?
 Hydraulic Ram for raising and lowering the mast €?
Cost Of
Installation
 Need 3 cubic metres of 40 Newton concrete for base
(1.2metres x 1.5metres) (Turbotricity.com) €400
 Hire digger for digging foundations €100
 Usually 10-20% of the turbine cost generally but here is
included in the price
 Grid-Tie Cost; €?
Total Installation
Cost
€14,150
Cost Of Annual
Maintenance
 As it is a hinged gin pole mast, maintenance costs will
be considerably less every year
 0.5-1% of original cost per year (according to Canada-
windpower.com). For this assignment I’ll take 0.75%
which is €106.13
Possible
Replacement Costs
 Inverter €1300
 Turbine may need to replace blade every 10 years but
it’s difficult to find prices for this (Turbotricity.com) €?
Average Wind
Speed at Location
(according to
SEAI)
(6m, 10m, 18m
figures came from
Sustainability.ie)
 6m Height; 4.8 m/s
 10m Height; 5.3 m/s
 Estimate (13m; 5.5 m/s)
 18m Height; 5.9 m/s
 50m Height; 7 m/s
 75m Height; 7.25-7.5 m/s
 100m Height; 7.75-8 m/s
Planning  As Tower is 11.5m high and rotor diameter is 3 meters
(radius is 1.5m), the turbine is exactly on the 13m
guideline needed for no planning permission
Annual Home
Electrical Use
 4,556KwH at 16c per kWh gives €728.96 a year of a bill
plus standing charge. Vat of 13.5% = €98.41. €90 of a
yearly standing charge. Total ESB = €917.37
Energy Cost
Inflation
 Theoretically 5% a year
Production by
Turbine
A. Capacity Factor; 25% (assumption)
B. Rated Power; 2.5 kW (assumption)

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C. Availability; 97% (assumption)
D. Number of hours in a year; 8,760
E. Estimated Annual Energy Production; (A x B x C x
D) = 5,948 kWh a year
Revenue F. Energy Used on Site (%); 40% (Assumption)
G. Your Electricity Price; 0.16c
H. Energy Price when sold;
Mean Wind Speed (metres/sec)
5.5 (25%
Capacity)
Annual Output (KwHr/Yr) 5,948
Units exported (60%) 3,568.8
Units exported at 19c 3,000 (€570)
Units exported at 9c 568.8 (€51.19)
Units used at home at 16c (40%) 2,379.2 (€380.67)
Value of units €1,001.86
NOTE; Still need to buy 2,176.8 kWh from ESB (4,556 is the
annual house requirement and only 2,379.2kW are available to
use). This would cost €348.29 annually. VAT at 13.5% =
€47.02. Standing charge at €28. Total needed = €423.16
I. Energy Value
J. Estimated Annual Revenue; 1,001.86 – 423.16=
€578.70
Costs K. Maintenance (0.5-1% of total cost); 0.75%
L. Annual Maintenance Costs; €106.13
M. Estimated Annual Income (J-L); €472.57
N. Estimated installed Cost; €14,150
Simple Payback O. (N/M) 29.9 years. It will take even longer than this
because my calculations do not factor in the fact that
after 5 years you no longer get 19c per unit for the
first 3,000 units produced, just 9c for every kW. This
will dramatically increase the payback period,
making the turbine unsustainable at this location as
it’s lifespan is 25 years
Energy Assignment 2
Annual kWhr from a solar system
Questions
1. Work out the annual kilowatt hours that a solar system of 2kWp will produce at that
location. Using the calculator looked at earlier:
http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php
2. Work out the area of panels that will be required comparing 2 different manufacturers
panels e.g. a Solarfun panel vs a Sunpower panel.
Answers
1. 2kWp at location in Gort = 1,690kW
2. Area of panels
a. Solarfun Panels; SF 160-24 Mono-crystalline 170W Panel

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2,000÷17 = 11.8 (need 12 panels)
Dimensions= 1.580 x .808metres = 1.276 = 1.28m2 for 1 panel
1.28 x 12 = 15.31m2 needed to produce 2kW
b. SunPower Panels; 315 Solar Panel 315W
2,000÷315 = 6.35 (need 7 panels)
Dimensions = 1.05 x 1.56 = 1.638m2 for 1 panel
1.638 x 7 = 11.47m2 needed to produce 2kW
Energy Assignment 3
Take a lookat yourown homeand describethe energy featuresof the house
 Insulation
7 years ago each of the bedrooms were dry lined, insulated and DPC was used also.
Last year the house was re-roofed and the attic insulated properly with Kingspan
300mm thick. This had a great effect on keeping the house sealed and got rid of any
drafts coming down from the attic. We also have a lagging jacket on the cylinder.
 Quality of windows
About 7 years ago we replaced our old wooden sash windows with PVC double-
glazing. The new insulation in the windows immediately made a big difference. The
seals in the windows were replaced last year again making a big difference.
 Heating System
Our house is heated by an oil range, which is about 15 years old, so probably isn’t that
efficient. It heats the house and the water. To the best of my knowledge there isn’t any
modern thermostat on it.
 Heating controls
We have valves on each radiator in the house.
 Air tightness
As the house is nearly 100 years old there isn’t much air tightness. On windy and cold
days you can feel a draft going through the house.
 Lighting
We have fluorescent lighting in our kitchen and energy efficient light bulbs in the rest
of the house.
 Energy rating of your appliances
Most of our appliances such as washing machines etc are B and A rated machines
Overall I think our house wouldn’t be far off a high C rating in the BER. It is an old house,
but as well insulated as possible without sacrificing the already small rooms. On average we
use 4,000kWh a year which doesn’t seem to be too bad, and the oil tank is filled twice a year
at most. We also use turf for about a month a Christmas for an open fire in the house. Overall
it doesn’t seem to use too much energy. I think it is relatively energy efficient.
Describe what elements you would consider upgrading in order to make the house more
energy efficient (taking cost into account).
1. I would install thermostat controls and timers on the heaters throughout the house.
These are cheap and would have a short payback period.
2. In 3 bedrooms the radiator is mounted below a window, so I would install a projecting
window-board or shelf above the radiator which will direct warm air into the room,
reducing heat loss through the window. This would pay for itself in a few short
months.

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3. If the windows were to be replaced in the near future, I would install double glazing
with Argon fill and low-emissivity glass, or triple-glazed windows.
4. I would replace the hot water cylinder, with a cylinder with factory applied insulation.
Such insulation is more effective at retaining heat than a lagging jacket, is less easily
damaged and cannot be pulled out of place. The cylinder at home is 15 years old and
has a lagging jacket but would not be as efficient as the factory applied insulation.
5. A reflective foil, backed by insulation if space permits could be fixed behind radiators
mounted on external walls.
6. A modern wood burning stove can achieve efficiencies of up to 80% compared to
only 20-30% for a traditional open fire, so this could be a possible replacement to the
open fire. Costs start from about €5,000
7. I would fit draught excluders around windows and doors, where no draught protection
is in place. This are cheap and would have a short payback.
8. I would install a Back Boiler in the fire. By trapping more of the fire’s heat energy
and using it to provide domestic hot water and space heating, a high output back
boiler increases the efficiency to approximately 40–50%.
Energy Assignment 4
1. What advice would you give a friend on building a low energy home, taking into
account cost, payback and building regulations.
If money is not a major concern this is what I would advise;
a. Insulation. Put in more than the building regulations require, and as much as
you can afford. This includes the insulation value of your windows & doors
too. 200-300 mm of insulation would be ideal; however after this the law of
diminishing return makes it unnecessary, and air tightness would be better
improved than insulation.
i. Insulating a 50 sq.m. attic space of a typical house costs around €400
and would pay for itself in about 3 years. Cavity wall insulation cost
depends on a number of factors, including the width of the cavity, but
is in the region of €5 -7 per square metre. For a typical house, this
gives a total cost of about €550–700. The payback period will be about
4-7 years. Internal insulation costs more to install than cavity
insulation. It is most cost-effective to install internal insulation when
the inside of the house is being refurbished.
b. Control of the heat emitters and Domestic Hot Water. That means zoning,
and clever and more expensive controls to make sure that heating is on only
when and where it's needed.
i. Heat controls are relatively cheap and will pay for themselves in a few
short years. Typically homeowners can reduce their energy usage by
up to 20% by installing easy to use heating controls in their home and
using these controls in an efficient manner.
ii. Irish Building regulations specifically call for controls on space
heating and water heating systems with control systems.
c. Heat source. Energy efficient heating systems are a priority. I would
recommend installing solar panels and combine this with a wood pellet boiler.
i. The cost of installing solar thermal panels ranges from €5-6,000 for a
4-bed house. Payback could be 5-10 years. Payback will depend on
how efficiently the system is being used, how much it costs and how
well it was fitted.

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ii. It is also a requirement in Building Regulations Part L to have a
renewable energy system
2. List 4 types of heating controls and how they affect efficiency;
a. Thermostatic Radiator Valves (TRVs) are cheap and provide a very simple
way of controlling heat in every room of the house. They are a totally
automatic process and don’t require any electrical power. These are
mechanical devices installed at the water inlet to the radiator. The valve has an
adjustable thermostat. As the room temperature increases and approaches the
temperature set at the thermostat the valve shuts off the water flow into the
radiator which prevents the room getting any warmer. When the room
temperature falls the valve allows water to flow and the room heats up again.
b. Thermostats measure the heat in a room or on a pipe and normally provides
an electrical signal to a controller to do something such as shutting off a valve.
The thermostat, controlling the heat output from the boiler to the house, should
be installed in an area that is indicative of the general heat conditions of the
house.
c. Motorised valves. Motorised valves allow for pipes to be shut off or opened.
For example a heating system normally sends heat through a coil in the hot
water cylinder. However if the water in the cylinder is only required at
55degrees then there’s no need heating it to 70degrees. A thermostat and
motorised valve on the pipework can automatically stop the flow through the
cylinder when it reaches 55 degrees.
d. Zone controller. Nowadays different areas of a house or building are
normally on different zones. For radiators this might be three zones (upstairs,
downstairs, water) or for an underfloor system it might be 6 or more zones. A
zone controller allows each zone to be individually controlled to switch on or
off. For example the heating in the upstairs of a house may only be required to
be turned on at about 10pm whereas the downstairs heating may be on from
6pm.
3. Explain what a u-value is and list the u-value requirements for each building element
in current Irish building regulations.
U-value is the measure of the rate at which heat is lost. The lower the U-value the
better it is for your home comfort.
a. The current building regulations part L gives maximum u-values for each
element of a building.
b. For walls the u-value in the building regulations are 0.27
c. The u-value to external doors and windows and roof lights is 2.20
d. The u-value of 0.16 for a pitched roof (insulation horizontal at ceiling level)
can be achieved by many different combinations of materials. Flat roofs have
a u-value of 0.22.
e. Ground floors and other exposed floors should have a u-value of 0.25
4. Explain why the BER system was introduced. Do you think BERs will have an impact
on the energy efficiency of buildings? Explain your answer.
The BER was introduced to promote energy efficiency in buildings as a result of the
Kyoto Protocol. The reasons for wanting to promote energy efficiency were as
follows;

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a. The CO2 emissions related to buildings is a significant proportion of all the
CO2 emissions in the country. Houses alone account for 25% of all energy
related emissions
b. The cost of building new power stations is expensive, more energy efficient
buildings means not having to build new stations
c. Most fuel for heating and electricity is imported. Making buildings more
energy efficient reduces the amount of money leaving the country
A BER is an indication of the energy performance of a home, with ratings operating
on a scale of A1, A2, A3, B1, B2 etc down to G. A rating of A1 represents the best
performance.
I think that the BER should focus people’s minds on the amount of energy that
a house uses. It is a clear indication on how energy efficient a house is, and a rating
will undoubtedly have an influence when buying or renting a house, as it is an
indicator of how much it will cost to heat the house in question based on its rating.
This is one reason why I believe that BERs will have an impact on the energy
efficiency of buildings.