Lecture Notes 5 BU 9Dec2011

8/3/2019 Lecture Notes 5 BU 9Dec2011

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Grkan KumbaroluAssoc. Professor of Industrial Engineering

Energy Policy and PlanningIE 461

Lecture Notes II-1.

Part II. Energy & Environmental Policy Modeling


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Contents

1.

Introduction........................................................................................................................... 3

2. Energy Payback ..................................................................................................................... 43. Energy Efficiency................................................................................................................... 8

4. Technological Learning ........................................................................................................ 12


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1. Introduction

Primary energy is undergoing various transformations until it reaches the final consumer in theform of useful energy as illustrated in Figure II.1 below.

Figure II.1. Energy Flow Diagram from Primary to Useful Energy(http://www.thegreenhomeadvantage.com/Portals/0/LLNL_Energy_Chart300_1156px_transparent.png)

Primary energy refers to energy sources as found in their natural state. Secondary energy is

the result of the transformation of primary sources. Final energy is the energy supplied to thefinal consumer for all energy uses. The final energy is available to the consumer to be converted

into useful energy. Electricity becomes for instance light, mechanical energy or heat whenconverted into useful energy by the consumer according to her respective use.

An energy producing system is composed of many components whose production, transport

and installation requires the input of energy. Such a holistic view is embedded in a life cycle

energy assessment as shown in Figure II.2.


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Figure II.2. Life Cycle Energy Assessment

The ratio between the energy input requirement and the energy output over an energy systems

lifetime is defines the energy payback as explained in the following.

2.

EnergyPayback

For any given energy system, the Energy Payback Ratio (EPR) is the ratio of total final energy

produced during a systems normal lifespan to the energy required to manufacture, install,maintain and fuel the system. Accordingly,

Extraction&Processingof

RawMaterials

ManufacturingofSystem

Components

Packaging&Transportation

Installation

Use&Maintenance

Decomissioningand/or

Recycling

Energy Input

Energy

Output


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, , , (2.1)

where ,, is the final energy output of type i (i=electricity, heat etc.) that is produced attime t. Hence, all the energy poduced over the full lifetime n of the system is summed up in thenominator. The denominator on the other hand includes the energy input required by the system

over its full lifetime. , is the energy required by activityj (j= the systems construction,operation, decommissioning etc.) for setting up (t0) and running (t1) the system. It should be

noted that the energy needed to produce, process and transport a fuel is considered underas the energy required to fuel the system, but not the energy content (heating value) of the fuel

itself. Naturally, a highEPR indicates good performance in terms of the lifetime energyproduction. A lowEPR on the other hand implies that a significant amount of energy is required

to build and maintain the system. This is also an indication of poor environmental performance

since, if fossil fuels are used to supply that energy, higher energy requirements will lead tohigher pollutant emissions. At the lower extreme is anEPR close to 1 implying that the system

consumes nearly as much energy as it generates over its lifetime.

Exercise 2.1.

The energy input required for the production of a 1 MW wind turbine are given below. Thisincludes the energy associated with extraction, processing, transportation and assembly of

materials. The energy needs associated with the O&M phase of the turbine are estimated to be

300 kWh/year. The energy required for disposing the turbine amounts to 30 MWh.

Material Used Energy Required

(MWh)

Steel 650

Aluminium 20

Copper 15

Sand 3

Glass 4

Polyester and epoxy 35Reinforced Iron (foundation) 180

Conrete (foundation) 350

Miscelaneuous 10

Total energy required for one wind turbine 1,267 MWh


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Consider a wind farm that consists of 15 wind turbines with 1 MW installed capacity each. The

annual electricity production from the farm is projected to be 39,420 MWh. The economiclifetime is 20 years. Assuming that all energy inputs of the wind farm (including O&M and

disposal) are additive from the individual turbines input needs, compute the Energy Payback

Ratio of the wind farm.

Solution:

39,420201,267 15 0.3 15 20 30 15 40.3

Energy Payback Time (EPT) is the length of time (in years) that a system needs to produce the

amount of energy that is required to manufacture, install, maintain and fuel the system. That is,

, , , 1(2.2)

As opposed toEPR, a lowEPTindicates good performance.

Exercise 2.2.

Consider a solar PV system with 10 kW installed capacity, consisting of five solar panels. Thetotal energy required for a solar panel (including raw materials, manufacturing, transportation

and installation) is 578 kWh as detailed below. The rest of the 10 kW-PV system requires 75

kWh of energy in total up to the point of installation, and 50 kWh/year is required for operation& maintenance. The solar PV system will produce 162 kWh of electricity per year. What is the

Energy Payback Time of this system ?

Activity Energy Required

(kWh)

Production of metallurgical grade silicon 2

Production of electronic grade silicon 7

Production of ingot 15

Cell Fabrication 81


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Balance of System 473

Total energy required for one solar panel 578 kWh

Solution:

578 75 50 162

The result can easily be identified by comparing the cumulative energy requirement and

production year by year. That is,

End of Year,

tEnergy Input,

Cumulative

Enery Output,

Cumulative

0 653 01 703 1622 753 3243 803 4864 853 6485 903 8106 953 972

Obviously, cumulative energy output exceeds the cumulative energy input at the end of year 6.Assuming that energy requirements and production occur uniformly throughout the year, the

exactEPTcan be computed by linear interpolation as

x=0.9 EPT =6.9 years


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3. EnergyEfficiencyIn an energy using/producing process, system or machinery, the ratio of all useful energy output

to all energy input defines the energy conversion efficiency energy conversion

, i.e.

(3.1)

Typically, is a dimensionless number between 0 and 1, but generally expressed as apercentage). The efficiency of Combined Heat and Power (CHP) generation plants, for example,

is about 70-80%. Conventional heat & power generation, on the other hand, has a much lowerenergy efficiency as the heat obtained as a by-product during electricity generation is wasted.

Figure 3.1 provides an illustration of the energy efficiency in conventional and CHP systems. In

this example of a typical CHP system, to produce 75 units of electricity & heat, the conventional

generation or separate heat and power systems use 154 units of energy (98 for electricityproduction and 56 to produce heat) resulting in an overall efficiency of 48.7 %. However, for

producing the same 75 units of electricity & heat, the CHP system needs only 100 units of

energy input. Accordingly, the overall energy conversion efficiency of the CHP system is 75%.

Figure 3.1. Energy Conversion Efficiency for Conventional and Combined Heat & PowerGenerationSource: http://www.epa.gov/chp/basic/methods.html

Energy efficiency can be described as the amount of a specified activity per unit of energy used.

In the above CHP system, for example, 1 unit of energy used produces 0.75 units of electricity &

68 Units Loss

11 UnitsLoss 25 Units Loss


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heat. Hence the overall energy conversion efficiency is 0.75, or 75%. In this example, activity is

the production of energy and it is measured in energy units. However, in a wider context, activitycould refer to the production of any good or service. In that case, the understanding of energy

efficiency in general (as opposed to the energy conversion efficiency, which is used for energy

conversion systems in particular) applies and is measured by the Specific Energy Consumption

as defined in the following..

The amount of energy required to produce a certain amount of good or service is called Specific

Energy Consumption SEC, i.e.

(3.2)

wherePkstands for the amount of activity (good or service) kproduced, and

denotes the

total net energy input1 that is required for its production. Reducing the amount of energy requiredto provide the same product or service, that is reducing SECk, is referred to as energy efficiency

improvement. Hence, SECkcan be considered as an indicator of energy efficiency. For the

computation of SEC, the amount of activity is measured in physical terms. Therefore, SEC is aphysical energy efficiency indicator. However, the production of goods or services can also be

measured in economic terms (based on the value of products). In that case, the Energy Intensity

EIis computed as

(3.3)

where Vkis the value of good or service kproduced. EI is an economic energy efficiency

indicator.

Physical production data is often not available and sectoral value added is published undernational statistics at a rather aggregate level where the output of different products and

subsectors is combined. Therefore, most aggregate level energy efficiency analyses are

inevitably based on the energy intensity.

Whatsoever, physical energy efficiency indicators should be preferred to economic ones as thelatter includes price information that may cause misleading energy efficiency conclusions. For

example, if the value of a product rises due to market-based reasons that are not related to the

1Energy input can be based on the Lower Heating Value (LHV) or the Higher Heating Value (HHV) of fuels. HHV is a better

measure of the energy inefficiency of processes and is therefore a preferable basis for energy efficiency analysis. However, it

should be noted that most national and international energy statistics are based on LHV.


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technology/process of production (e.g. change in the cost of raw materials, shortage of supply

etc.) thenEImay decrease although the actual energy efficicency does not change.

Both EI and SEC are influenced by the structure of a sector, which can be defined by eithermixof products ormix of activities. Different energy efficiency conclusions may emerge depending

on which definition is used. The production of ammonia, for example, can be done by at leasttwo alternative processes:

i. Steam reforming, i.e. conversion of natural gas, LPG or petroleum naphtha into gaseoushydrogen, and catalytically reacting the hydrogen with nitrogen.

ii. Partial oxidation of high viscosity, sulfur-rich crude oil residuesThe former process, steam reforming, has a low SEC whereas the latter, partial oxidation of

residues, has a high SEC. Ifmix of products defines sector structure, than the higher SEC ofpartial oxidation is considered to be a matter of energy efficiency. If, on the other hand, mix of

activities is used to define sector structure, than it is considered to be a matter of differences insector structure. It should be noted, however, thatproduct mix is ae commonly used indicator of

sector structure.

The sector structure is an important factor for international comparisons of energy efficiency.

Similarly, structural indicators need to be considered for intercompany comparisons of energyefficiency. Comparing the sectoral EIs or SECs of two different countries/plants at a given

structure provides information on relative energy efficiency (relative to the country/plant

compared) but not on actualenergy efficiency (relative to what can be achieved). Therefore, theEIs orSECs are compared with a reference value, which is based on the lowest energy

consumption that can be realized using the most energy efficient technology available. That is,

(3.4)

and

(3.5)

The difference between the actual and referenceEIs orSECs can be used as a measure of energyefficiency. It indicates the energy efficiency improvement potential: what reduction in energy

input (per unit or value produced) can be achieved at a particular sector structure. This difference


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can then be used for international or intercompany energy efficiency comparisons at different

sector structures.

Exercise 3.1.

Cement clinkers are formed by the heat processing of cement elements in a kiln, which is the

most energy intensive part of cement production. Therefore, the clinker content of cement isconsidered to be a structural indicator in energy efficiency analysis. According to the type of

cement, the clinker content may vary significantly.

Assume that the Reference technology SECof cement can be computed as a function of the

clinker contentas

2 + 3r

where SECis the specific energy consumption of cement, measured in GJ/t, and rdenotes the

clinker to cement ratio.

Five plants have the following energy consumption, cement production and clinker to cement

ratio values:

Plant # Energy Use (GJ) Cement Production (t) Clinker to Cement Ratio

1 46,000 10,000 0.602 61,100 13,000 0.553 90,000 20,000 0.624

23,000 5,000 0.68

5 42,300 9000 0.58

Compare the energy efficiencies of these plants and rank them according to their energy

efficiency improvement potentials

Solution:

The actual SECs of the plants and reference SECs corresponding to the clinker to cement ratiosare computed as

Plant # Actual SEC (GJ/t) Clinker to Cement Ratio Reference SEC (GJ/t)

1 4.6 0.60 3.80

2 4.7 0.55 3.65

3 4.5 0.62 3.86

4 4.6 0.68 4.04

5 4.7 0.58 3.74


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The energy efficiency improvement potentials SECare then computed as the difference

between actual and reference SECs:

Plant # SEC (GJ/t)1 0.80

2 1.05

3 0.64

4 0.56

5 0.96

The solution can also be depicted graphically.

4. TechnologicalLearningEnergy and environmental policy supports the deployment of new, more efficient and/orrenewable energy technologies. The diffusion of new technologies, however, is subject to costcompetitiveness. The technical performance of a technology increases and its production cost

decreases substantially as producers gain experience with the technology. Labor productivity

improves physically unchanged equipment due to improvements in work methods, plant layout,

material handling and organization of production. Empirical evidence indicates that the cost ofnew technologies decays exponentially as a result of the learning effect. The most common form

of the relationship between cost and production level is given as

(4.1)where C0 is the Cost of the first unit produced, and CCum the cost per unit capacity at Cumulative

production capacity level Cum; b is an experience parameter. The percentage change in cost ()from cumulative production level Cum1 to Cum2 can be computed as

1 (4.2)

The cost associated with the cumulative production levels are computed from equation (7.1) as


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1(4.3)

2 (4.4)

Substituting (7.3) and (7.4) into (7.2) yields

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(4.5)

The ratiorepresents the increase in cumulative capacity. Letting , equation (4.5)

is rewritten as

1 (4.6)

expresses the rate at which costs decline when cumulative capacity increases x times. This isknown as the Learning RateLR. Its complement, 1-LR, is known as the Progress Ratio PR and

expresses obviously the remaining fraction of cost (per unit of capacity) after cumulative

production capacity increasesx times. Hence,

1 (4.7)

It has become standard practice to refer to LR as the rate of cost decline for each doublingof

cumulative production capacity.

Figure 4.1 plots the learning curve for a doubling of cumulative capacity under variousPR

assumptions. As can be observed from the figure, the cost reduction rate becomes less as

cumulative production capacity increases. It should also be noted that the learning curve is time-

independent.


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Figure 4.1. The Learning Curve

Exercise 4.1.

The annual production capacity and cost (in real values) of solar PV modules for the period2005-2011 in Europe is given below.

Year Production

Capacity(GW)

Module Cost

(/W)

2005 0.6 5.832006 0.8 5.452007 1.1 5.302008 1.9 3.752009 2.0 3.402010 2.4 3.052011 3.5 2.84

Calculate the learning rate for each year

Solution:

First, the cumulative capacity is computed and then the experience index b from

0

2

4

6

8

10

12

100 200 400 800 1600 3200 6400

Cost

PR=80%

PR=85%

PR=90%

PR=95%

CumulativeCapacity


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which is obtained by combining equations (4.2) and (4.5). The learning rate can then be

computed for each year by using formula (4.6). Results are as follows:

Year Production

Capacity

(GW)

Cumulative

Capacity

(GW)

b LR

2005 0.6 0.6 0,08 0,072006 0.8 1.4 0,05 0,032007 1.1 2.5 0,61 0,292008 1.9 4.4 0,26 0,092009 2.0 6.4 0,34 0,10

2010 2.4 8.8

0,21 0,072011 3.5 12.3

Lecture Notes 5 BU 9Dec2011

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