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What is Energy?
Session 1
1
© FEC Services Ltd 2009
Energy management training 2009
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A word about multiplication factors
• Kilo (k) = 1,000
• Mega (M) = 1,000,000
• Giga (G) = 1,000,000,000
e.g. 1kW = 1,000W
1kWh = 1,000Wh … etc
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© FEC Services Ltd 2009
Energy management training 2009
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Converting J to kWh
1 kilowatt-hour (kWh)
=
Energy consumed by
1 kW in 1 hour
1,000 J/s
x
3,600 s
= 3,600,000J
or 3.6 MJ
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Energy management training 2009
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Common units for energy buying
• kWh is the main unit for energy utilities
• But we can convert
– Litres of oil
– Tonnes of coal
– Therms of gas
– ………………..to kWh using standard tables
5
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Energy management training 2009
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B y W e ig h t
S o lid fu e ls
C oke
C o a l ( w e ig h te d a v e r a g e )
L iq u id fu e ls
C r u d e O il ( w e ig h te d a v e r a g e )
H e a v y F u e l o il
M e d iu m fu e l o il
G a s /d ie s e l o il
K e ro s e n e
P e tr o l
LPG
E th a n e
k W h / to n n e
8 ,2 7 8
7 ,4 4 4
k W h / to n n e
1 2 ,6 9 4
1 1 ,9 9 9
1 2 ,2 5 2
1 2 ,6 6 6
1 2 ,8 3 3
1 3 ,0 8 3
1 3 ,7 2 2
1 4 ,0 8 3
G a s e o u s f u e ls
S o lid R e n e w a b le s
1 0 .7
1 1 .8
1 1 .3
1 0 .8
1 0 .3
9 .6
7 .4
5 .2
1 1 .0
1 0 .7
1 0 .7
k W h / to n n e
8 ,8 8 8
5 ,1 9 4
4 ,4 4 4
4 ,1 6 6
3 ,8 8 9
3 ,3 0 5
2 ,7 7 8
2 ,6 3 9
2 ,4 4 4
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k W h / litr e
kW h / m 3
N a tu r a l g a s
L a n d fill g a s
Sewage gas
T yre s
R e fu s e d e r iv e d w a s te
G e n e r a l in d u s tr ia l w a s te
S tr a w
H o s p ita l w a s te
In d u s tr ia l w o o d ( p a lle ts e .t.c )
D o m e s tic w o o d ( lo g s )
M u n ic ip a l s o ild w a s te
P o u ltr y litte r
B y V o lu m e
Energy management training 2009
C a u t io n :
D epends on
c a lo r if ic v a lu e
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Conversions can vary
• The quality of fuel (calorific value) changes
slightly depending on the ‘batch’
• Gas calorific value is published for each Transco
region each month and appears on the bill
• Gas calorific value expressed in MJ/m3
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o
or r th e
th
r
Th n
am
N
es
or
th
W
es
t
S
co
tl a
S
ou nd
th
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as
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ou
th
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o u e rn
th
W
W
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t M es
t
id
la
nd
W
al
s
es
N
W
or
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th
es
S
ou
th
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or
th
an
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l
te
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tM
as
E
as
E
Ea
ds
4 0 .6
4 0 .4
4 0 .2
4 0 .0
3 9 .8
3 9 .6
3 9 .4
3 9 .2
3 9 .0
3 8 .8
3 8 .6
n
Calorific value
Variation in calorific value by region
R e g io n
Up to 3.2% difference
8
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Energy management training 2009
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Measuring gas
• Gas meters don’t read kWh
– They measure volume
– m3, ft3 or 100ft3 (HCF)
• Need to correct for
– Temperature changes
– Pressure changes
– ‘Correction factor’
• Calorific value then applied
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Energy management training 2009
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Large gas meters
• Have ‘correctors’ next to the ‘meter’
• These adjust the volume for temperature and pressure
10
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Your gas bill…..
Shows consumption (or corrected consumption)
in ft3 or m3
– May be a multiplier (e.g. x10, x100)
– Applies a standard correction factor
• if you have an uncorrected meter
– Uses a ‘calorific value’ to work out kWh
11
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Gas meter reading
Gas corrector reading
100 ft3 to m3 (x 2.83)
100 ft3 to m3 (x 2.83)
Multiplier (x10, x100)
Multiplier (x10, x100)
Correction factor
(for temp/ pressure - from bill)
Calorific value (MJ/m3)
div by 3.6 = kWh
Calorific value (MJ/m3)
div by 3.6 = kWh
12
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Example gas bill – what can you tell me?
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Energy management training 2009
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Example – calculate kWh from ft3
• Meter readings show consumption of 12,565 ft3
• How many kWh is this?
–
–
–
–
Conversion of 100 ft3 to m3 = 2.83
Temperature and pressure conversion factor = 1.12
Calorific value = 39.5 MJ/m3
1 kWh = 3.6 MJ
14
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Answer
• Convert to HCF
– 12,565 ft3 / 100 = 125.65 HCF
• Convert to m3
– 125.65 HCF x 2.83 = 355.89 m3
• Correction for temp and press
– 355.89 m3 x 1.12 = 398.26 ‘standard’ m3
• Use calorific value
– 398.26 x 39.5 = 15,731.28 MJ
• Convert to kWh
– 15,731.28 MJ / 3.6 = 4,370 kWh
15
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Energy management training 2009
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About different oils
Buying different oils
you must consider
• Price
• Energy value
• Suitability
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Energy management training 2009
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Comparing the cost of different fuels
Use conversion factors to calculate which fuel is cheapest
Units of
purchase
Electricity
kWh
Cost per
unit
supply
Cost per
kWh
1
8p
8p (day)
4p (night)
Diesel / fuel oil
Litre
35p
LPG
Litre
26p
Therms
75p
Mains gas
kWh per
unit of
purchase
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4p
10.8
3.2p
7.4
3.5p
29.31
2.7p
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Application efficiency
You must also consider how efficiently the fuel is
used to meet your needs
100%
55%?
Boiler loss
25% ?
Application
efficiency loss
10%?
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Primary
pipe
Loss 10%?
Distribution
system
Loss 10%?
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So which fuel is the cheapest?
I want to heat water - Should I use oil @ 35.6p / litre or
electricity @ 4p / kWh?
• Electricity
– 100% efficiency for water heating
– 4 / 1 = 4p / kWh supplied
• Oil
– 35.6p / litre / 10.8 = 3.2p / kWh
– 55% efficiency for water heating
– 3.2 / 0.55 = 5.8p / kWh supplied
In this case electricity is cheaper
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Energy management training 2009
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The efficiency rule
Real price
=
Purchase price
Efficiency of use (as a decimal)
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Run a generator or buy electricity?
Diesel @ 36p / litre or electricity @ 8p / kWh?
• Generator efficiency = 30%
• Cost for generated electricity
– Convert litres to kWh = 36p / litre / 10.8 = 3.33p / kWh
– Apply efficiency factor = 3.33p / 0.3 = 11.1p / kWh
• Excludes capital & maintenance costs
• Don’t generate – buy!
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Remember
When choosing fuels ……………..
………………it is not just the cost of the fuel
It is the cost of getting the result you want
– a tank of hot water?
– a greenhouse temperature of 20oC?
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Energy management training 2009
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Calculating savings
• We buy energy
– kWh or units = kW x hours
• Energy = power x time
– A 10 kW motor running for 5 hours = 50 kWh
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Energy management training 2009
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Calculating savings
Saving = [current use – new use] x energy cost
Example: a light bulb used for 2,000 hrs per year
– Now:100 W tungsten bulb
– Future?: 20 W compact fluorescent
Saving = [(100x2000) – (20x2000)] x (0.08/1000)
– £12.80 per year saved
– Lamp cost is £1
– Payback less than 1 month
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Energy management training 2009
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Operating time is critical …
In our example ……
If the operating time was 1 hr / week
• Saving would reduce to £0.36 year
• Payback would increase to 3 years
25
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Energy management training 2009
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Don’t always blame the biggest equipment
• 30 kW water transfer pump
– Used 1 hr / week
– 30 kW x 1 hour x 52 weeks x 8p /
kWh
= 1,560 kWh =124
• 500 W yard light
– Used 10 hr / day every day year
round
– 0.5 kW x 10 hours x 365 days x 8p
= 1,825 kWh = £146
• The larger piece of equipment
used less
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Energy management training 2009
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Worked example
Calculate the payback on a variable speed drive to
be fitted on a water circulation pump
• Pump is rated at 5 kW
• Savings will be 25%
• Pump operates for 6,000 hours per year
• Average cost of electricity is 7p / kWh
• Drive cost is £1,500
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Calculation is:
(old technique use – new technique use)
x cost of energy = saving
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Answer
[(5kW x 6000hrs) – (5kW x 0.75 x 6000)] x £0.07
= £525.00
Payback time = £1500/£525 = 2.85 years
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© FEC Services Ltd 2009
Energy management training 2009
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Session 1 summary
• You buy energy to do a job
– It is the cost of doing the job that matters
• All fuels have different costs and …
– They can be compared on the same basis by converting
the cost in p / kWh
• When comparing fuels
– Consider both fuel cost and efficiency
• To calculate savings:
(old technique use – new technique use) x cost of energy
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© FEC Services Ltd 2009
Energy management training 2009