EM P 03 Energy Prices Slides

8/9/2019 EM P 03 Energy Prices Slides

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Energy Management :: 2007/2008

Class # 3P

Energy Prices

João Parente

[email protected]


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Energy Management

Energy transformation

PROBLEM 01

Consider a Combined Heat and Power Plant (CHP), based on a micro gas turbine recuperated cycle. The main

components and thermodynamic processes are described below:

Compressor;

Combustion chamber;

Recuperator (heat exchanger gas-gas)Economizer (heat exchanger gas-liquid)

Alternator

Summary description of the process: Atmospheric air is compressed in the compressor and heated by the

exhausts gas at the recuperator, then the compressed air passes through the combustor where natural gas is

burned. Combustion products are then expanded in the turbine that feeds the compressor and the alternator., .

Finally, in the economizer, using the sensible heat of the exhausts water is heated for heat production.

Class # 3P :: Energy Prices Slide 2 of 53


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Energy Management

Energy systems representation

Gas Turbine – Recuperated cycle in CHP

Functional scheme Fuel

ElectricityC T ~

Air

REC

ECO

Cold water


o waer

x aus s


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Energy Management


PROBLEM 01

Depending on the outputs that are considered, the efficiency of the CHP system will be:

a) Electricity production only ~ 30% ~

Considering all of this information, estimate the Primary Energy saves obtained by the full use of the CHP

electricity supplied by the grid.

Assume the followingvalues:

LHVNG = 39,5 MJ/m3

Primaryenergy conversioncoefficients:

Electricity –0,29 kgoe/kWhe

Natural gas–0,82 kgoe/m3



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Energy Management


SOLUTION P01

Assuming100 kWhNG

in thecombustion chamber (CHP):

* =

Heat: 100 kWhNG * (ηCG -ηe) = 55 kWhth

rmar energy:

1) CHP:

Natural gas: 100 kWhNG * 3600 kJ/kWh / LHVNG * 0,82 kgoe/m3=

= 100 kWhNG * 3600 kJ/kWh / (39500 kJ/m3) * 0,82 kgoe/m3 = 7,5 kgoe

1 Boiler + rid:

Natural gas: (55 kWhth / 0,85) * 3600 kJ/kWh / LHVNG * 0,82 kgoe/m3= 4,8 kgoe

Electricity: 30 kWhe * 0,29 kgoe/kWhe = 8,7 kgoe


aura gas + eec rc y: , goe+ , goe= , goe a mos w ce e pr mary energy


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Energy Management

Tariff in the SEP

Normal Low voltage – BTN (< 41,4 kVA)

Active energy (Variable term - VT)

Contracted Power (Fixed Term - FT)

BTN = E kWh * VT €/kWh + FT €/month

Low, average, high and very high voltage – BTE (> 41,4 kVA), MT, AT and MAT

Energy –Active energy (Variable term - VTae)

-

Contracted Power (Variable term - TVcp)

Peak hours power (Variable term - TVpp)

Fixed term (FT)

BTE, MT, AT, MAT =


Ea (kWh) * TVea (€/kWh) + Er (kvarh) * TVer (€/kvarh) + Pc (kW) * VTcp (€/kW.month)

+ Pp (kW) * VTpp (€/kW.month) + FT (€/month)


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Energy Management

Tariffs in SEP - BTN

Normal Low Voltage Tariff (BTN): Power scale

TARRIFFS CONTRACTED POWER (kVA)

‐

Weekly cycle

Winter legal time Summer legal time

Monday ‐ Friday Monday ‐ Friday

09.30 12.00 h, ,

Simple Tariff 1,15 ‐ 2,3 ‐ 3,45 ‐ 4,6 ‐ 5,75 ‐

6,9 ‐ 10,35 ‐ 13,8 ‐ 17,25 ‐ 20,7

Dual‐Tariff 3,45 ‐ 4,6 ‐ 5,75 ‐ 6,9 ‐ 10,35 ‐13,8 ‐ 17,25 ‐ 20,7

Pea

18.30/21.00 h Pea 09.15 12.15

Full

07.00/09.30 h

12.00/18.30 h

21.00/24.00 h

Full 07.00/09.15 h

12.15/24.00 h

Off ‐peak 00.00/07.00 h Off ‐peak 00.00/07.00 h

Simple Tariff 27,6 ‐ 34,5 ‐ 41,4

Non‐Intensive Use Tariff 27,6 ‐ 34,5 ‐ 41,4

Saturday Saturday

Peak 09.30/13.00 h

18.30/22.00 h Peak

09.00/14.00 h

20.00/22.00 h

Off ‐peak

00.00/09.30 h

13.00/18.30 h Off ‐peak

00.00/09.00 h

14.00/20.00 h

Intensive Use Tariff 27,6 ‐ 34,5 ‐ 41,4

Sazonal Triple‐Tariff 27,6 ‐ 34,5 ‐ 41,4

Sazonal Simple Tariff 3,45 ‐ 4,6 ‐ 5,75 ‐ 6,9 ‐ 10,35 ‐

22.00 24.00 h 22.00 24.00 h

Sunday SundayOff ‐peak 00.00/24.00 h Off ‐peak 00.00/24.00 h

, ‐ , ‐ ,

Sazonal Dual‐Tariff 3,45 ‐ 4,6 ‐ 5,75 ‐ 6,9 ‐ 10,35 ‐

13,8 ‐ 17,25 ‐ 20,7

Sazonal Triple

‐Tariff

3,45 ‐ 4,6 ‐ 5,75 ‐ 6,9 ‐ 10,35 ‐

13,8 ‐

17,25 ‐

20,7

a y cyc e

Winter legal time Summer legal time

Monday ‐ Friday Monday ‐ Friday

Peak 09.30/11.30 h

19.00/21.00 h

Peak 10.30/12.30 h

20.00/22.00 h

Seasonal tariffs: these tariffs can only be applied to

entities with seasonal consumptions - characterized

by a period of at least 5 consecutive months per year

Full

08.00/09.30 h

11.30/19.30 h

21.00/22.00 h

Full

. .

20.30/19.30 h

12.30/20.00 h

22.00/23.00 h

Off ‐peak 00.00/08.00 h

22.00/24.00 h Off ‐peak

00.00/09.00 h

23.00/24.00 h


Selling tariffs to the final consumer (2008) www.erse.pt/

w ou any consump on. n s sense, ese ar s

do not apply to the residential sector.


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Energy Management

Tariffs in SEP - BTN

Exemple: Selling tariffs to final consumers (residential) in BTN (≤ 20,7 kVA e > 2,3 kVA)

Contracted

power

(kVA)

Fixed term for simple tariff

(€/month)

Fixed term for dual tariff

(€/month)

Dual tariff:

Peak (peak + full)

3.45 5,74 8,32

4.6 7,45 10,67

5.75 9,15 13,02

6.9 10,85 15,37

10.35 15,7 21,99

Of -peak (of -peak)13.8 20,61 28,7117.25 25,42 35,26

20.7 30,42 42,14

Selling tariffs to the final consumer (2008) www.erse.pt/

Energy cost

(variable term) €/kWh €/kWh

Peak hours0,1132

0,1143

Off ‐peak

hours 0,0614

Application: A house with a 6,9 kVAcontracted power has a monthly consumption of 200 kWh at peak hours and 100 kWh at off-

eak hours. What is the best o tion in the Re ulated Market ?



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Energy Management

Problems

EXAMPLE 1:

Consider a house with an electrical contract of 10,35 kVA. The present contract is in simple tariff. The

consumption power daily profile is represented in the next figure.

.

W, while thenewone isexpected toconsume100W.

Consider the followingsimplifications:

b) Constant power consumption of the fridge

Questions:

b) Isdual tariff, cycledaily, less expensive?

c) Considering that the contract has been optimized, what will be the annual saves fromthe fridge

substitution?



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Energy Management

Problems

PROBLEM 1:



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Energy Management

Problems

SOLUTION PROBLEM 1:

a)

Daily consumption is

Daily consumption = 6x4 + 5 + 4x5+3x9 + 5x1 = 81 kWh/day

Monthly consumption = 81 x 30 = 2430 kWh/month

Present electricity monthly bill:

Monthly bill = Fixed term + Variable term = 15,70 + 2430 x 0,1143 = 293 €/month

b)For the daily cycle:

Peak hours (8h –22h) = 6x3 + 5 + 4 + 3 x 9 = 54 kWh/day or 1620 kWh/month

= =

Present electricity monthly bill:

Monthly bill = 21,99 + 1620x0,1132 + 810x0,0614= 255 €/month

There is a benefit with the dual tariff, daily cycle.

c)

Supposing that, the contract has been changed for the dual tariff, daily cycle. The daily cost saving will be:

(considering that there are 14 peak hours and 10 off peak hours)

= =


, , , , ,

The cost saving in one year will be:

Monthly cost saving = 0,21x365 = 80 €/year


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Energy Management


PROBLEM 02

Imagine you are building yourself a new house and still haven’t decided on the heating system you will install.

After a brief market research you have narrowed down your options to only two: a heating system based on

electrical resistances and a heating system based on a hot water circuit heated by a natural gas boiler.

Based on the records you kept form your previous house, you got to the conclusion that you will use the

heating system 16 hours per day (between 8:00 to and 24:00), only on week days and only during 5 winter

months, with an average power of 2kW.

a) Considering that you have an electrical contract of 10.35kVA with dual tariff on a daily cycle, determinewhich of these two solutions is best (assume an 80% efficiency for the hot water boiler, a LHV of 40MJ/m3

and an average price for natural gas of 0.596778 €/m3 ).

b) Comment on the above solution bearing in mind what you have learned about Useful, Final and Primary

Energy.



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Energy Management


SOLUTION P02

The heating system runs 16 hours per day. According to the existing tariff plans, this corresponds to 14 hours on peak and 2

hoursoff-peak (dualelectricity tariff).

Theannual cost of the first solution is thereforegivenby:

(14h* 2kW* 0.1132 €/kWh+2h* 2kW* .0614 €/kWh) * 5days/week* 4weeks/month * 5months/year =314.52 €/year

s or e a ura as sou on, e annua amoun o na energy consume s gven y e c ency :

16h* 2kW* 5days/week* 4weeks/month * 5months/year / 0.8=4000 kWh/year

Consideringa lower heatingvalueof9054kcal/m3, thisyields:

4000 kWh/year * 860kcal/kWh/ 9054kcal/m3 =380m3/year

Which, considering thecurrent tariff results in:

* . .

Answer: The best solution is the natural gas boiler . The fixed terms are not taken into account, as it would be necessary to

allocate a certain percentage of its value to the heating use, which in this case cannot be done. However, considering that the

total fixed value is assigned toheating the final result remains the same, as the electricit fixed costsarehigher than the ones for


naturalgas (4.27€/month).


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Energy Management


SOLUTION P02

Comments: In this problemthe natural gas solution proves to be the best. Additionally, if we calculate the price

of natural gas per kWh of final energy, we obtaina value of 0.056684 €/kWh, which is lower than the electricity

prices. However, considering that a typical natural gas heating systemwill have an efficiencyof approximately

80%, the priceper kWh of useful energy will goup to 0.070857€/kWh, surpassing the off-peak value for adual

tariff electricitycontract (0.0614 €/kWh), wherewewouldconsider a100%efficient electric resistancesystem.

This means that, assuming the maximumpower usage (10.35 kW) during all the off-peak period (10h per day),

themonthly fee usingelectricitywouldbecome lower than the oneusingnatural gas:

Electricity: 10.35kW* 10h* 30 days/month * .0614 €/kWh+21.99 €/month=212.64 €/month

* *. . . .

This scenario, although theoretical, alerts to fact that an energy sourcethat is morePrimary Energy consuming

shouldalways be moreexpensive thanothers (whichdoes not happen in thiscase), as the failure todosomay

. ,

the fixed terms are now considered, as it is assumed that all power is allocated to the heating system.

Otherwisethepricedifferencewould increasefurthermore, emphasizing theaboveremarks.


EM P 03 Energy Prices Slides

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