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Advanced Power Systems

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Title: Advanced Power Systems

1
Advanced Power Systems

ECE 0909.402-02, 0909.504-02
Lecture 7 Distributed Generation Energy
Economics
5 March 2007
Dr. Peter Mark Jansson PP PE
Associate Professor Electrical and Computer
Engineering

2
Aims of Todays Lecture

Cover other Current Topics
Upcoming FE Exam registration deadlines
Mid Term Exam 19 Mar
APS Final Project Options
Overview of Chapter 4 5 concepts
Finish Up Distributed Generation Technologies
Introduce Economics of Energy Systems

3
Aims of Todays Lecture (cont)

Stretch Break
Finish Chapter 5 and Engineering Economics of
Investments
Next Week
Have a Safe and Restoring Spring Break

4
FE Examination Announcement

The College of Engineering strongly recommends
students sitting for the Fundamentals of
Engineering (FE) exam. There is a July 3rd
application deadline for sitting in the FE exam
given on October 27. However, we urge you to get
your application completed much earlier than the
July 3rd deadline, and preferably in late March
or early April. It will be easier to locate
faculty members for references before summer
begins, and there will be more leeway to fix
issues that may arise.
Please contact the persons above if you have any
questions. More information on application fees
and exam dates can also be found on the website
http//www.state.nj.us/lps/ca/pels/pelsfees.htm
The application form and instructions can be
found at the following website
http//www.state.nj.us/lps/ca/pels/pelsinfo.htm

5
APS Mid Term Exam

Last half of period Monday 19 March 2007
Exam Topics Covered
Lectures 1-7 (including AIC and last weeks
Teleconference Lecture and Meeting)
Chapters 1-5
HW 1-6
Articles Researched and Reviewed
Recommended Study above and LCs

6
APS Final Project (30)

Select Teams (2 max per team for UGs Grad Only
1 pp)
To get an A select a APS and a specific site /
get approval for scope by 19 March 2007 (HW 6)
Sample Projects
Wind Farm Potential for 1200 acres in Cumberland
County
PV system design for Dr. Janssons Pitman
Home/Garage
Wind turbines for Berkeley Township, Ocean County
Offshore Wind System design (Del Bay or Atl Ocean
off AC)
Fuel cell for Rowan campus building
Microturbine/Cogen system for local industry
Micro-Hydroelectric facility for Farm in Pa.
Solar Concentrator in desert southwest

7
Final Project Content

All APS projects must include
Specific site collect data on energy use, and
other key site information critical for effective
design of your advanced power system.
Detailed advanced power system design
specifications, size, model, equipment, plans,
interconnection, alternatives considered, etc.
Cost benefit analysis SPB, SIRR, NPV, IRR,
NPV/IRR w/ fuel escalation and cash flow.

8
Final Project Format

(Team) Technical Paper (10)
6 pages maximum (4,000 - 6,500 words)
2 column technical paper IEEE format
Due 16 April 2007 (A max), 23 Apr (B max), 30 Apr
(C max)
25 - minute (Team) Presentation (20)
22 minutes of Slides/Demo etc. (min 30 slides)
3 minutes QA
Random date selection

9
Microturbines

Very small gas turbines (NG or waste gas)
Typically 500 W to 300 kW
Typical Microturbine Components
Compressor
Turbine
P-M generator
Combustion chamber
Heat exchanger (recuperator)

Often all on one shaft
10
Leading Manufacturers

Capstone Turbine Corporation
One moving part common shaft 96,000 rpm
C30 - 30 kW unit
? LHV 26, Heat Rate 13,100 Btu/kWh
C60 - 60 kW unit
? LHV 28, Heat Rate 12,200 Btu/kWh
Elliot Microturbines
TA100R - 105 kW unit
? LHV 29, Heat Rate 11,770 Btu/kWh
172 kW thermal potential for hot water ? TTE gt
75

11
Leading Mfgrs. (continued)

Ingersoll Rand
Design Life 80,000 hours with overhauls
MT250 - 250 kW unit
? LHV 30, Heat Rate 11,380 Btu/kWh
MT70 - 70 kW unit

12
Elliot Microturbine Application

The Elliot TA 100A produces its full output of
105 kW when burning 1.24 x 106 Btu/hr of natural
gas. Its waste heat is used to supplement an
existing boiler by raising its temperature from
120 -140 oF. It operates for 8000 hours/year.
A) If 47 of the fuel is transferred to boiler
what should the waters flow rate be?
B) If boiler is 75 efficient and NG is 6 per
MMBtu how much money will microturbine save in
displaced fuel?
C) If electric costs 8 / kWh what is annual
savings?
D) If OM is 1,500 per year what are net
savings?
E) If microturbine costs 220,000 what is Initial
RR and SPB?

13
LM 1

A) If we can not use the waste heat is it
economical to install a microturbine in this
application?
(assume you require lt 10 yr Simple Payback)
B) What if your electric costs are 11 / kWh?
C) What is lowest price electricity can be to
still meet your 10 yr simple payback requirement?

14
Microturbines in CHP applications rules of
thumb

Electric Production ? E 29
Waste-heat Recovery ? WH 47
Overall LHV conversion efficiencies can be gt 75

15
Reciprocating IC Engines

Very small piston-driven, 4 stroke ICEs
Typically 500 W to 6,500 kW
Typical Operation
Intake, Compression, Power, Exhaust
Spark ignited (Otto cycle)
Compression ignition (Diesel cycle)
Multi-fuel gasoline, natural gas, kerosene,
propane, fuel oil, alcohol, waste gas

16
Advanced Reciprocating Engines

Current design is cheapest of all DGs
Efficiencies are good
Today electrical ? 37-40
Turbocharged Thermal fuel ? horsepower
Otto cycle ? LHV 41, ? HHV 38
Diesel cycle ? LHV 46, ? HHV 44
ARES Targets ? ELECTRIC 50, ? CHP 80
HRSG can increase overall ? CHP 85

17
Stirling Engines

Energy is supplied from outside the system
External combustion, can run on any heat source
Invented in Scotland and patented in 1816
Used quite extensively until early 1900s
Eliminated from market by efficient technologies
Current efficiencies relatively low lt 30
Size ranges from 1 25 kW
No explosions, relatively quiet devices
Good match with solar dishes
Four states of Transition in Stirling Cycle

18
Concentrating Solar Power

Solar Dish / Stirling Power Systems
Parabolic Trough Systems
Solar Central Receiver
Disadvantages All these technologies require
direct solar beam radiation. Many regions of the
US have a large fraction of diffuse solar
radiation

19
Solar Dish / Stirling Power System

Multiple mirrors concentrate sun to one pt
Thermal receiver converts sunlight to heat
Heat is delivered to Stirling engine
Stirling engine drives electrical generator
? 20 typical, ? 30 peak
Sterling engine in this 725 oC application gt 36
Potential stand-alone power plants
Still have very high capital costs and electricity

20
Solar Dish Sterling
21
SES site

http//www.stirlingenergy.com/whatisastirlingengin
e.htm

22
Solar Parabolic Troughs

Parabolic reflectors concentrate sun to one tube
Stainless steel tube (g-ins) converts light to
heat
Heat transfer fluid delivered to heat exchangers
Steam turbine drives electrical generator
? 10 typical, costs 12 / kWh
Advanced applications lt 5 / kWh
Development ongoing in Spain, Greece, Italy,
India and Mexico

23
Solar Parabolic Trough
24
Solar Parabolic Trough
25
Solar Central Receivers

Multiple computer controlled mirrors concentrate
sun to one central receiver tower
Central receiver (Power Tower) converts sunlight
to heat
Heat is delivered to Steam Generator
Solar One Water/steam system not well matched
Solar Two Molten nitrate salts (97 match)
Extremely high capital costs to prove economics

26
Solar Two
27
Solar Two
28
Concentrating Solar Power

Technology Concentration Ann Effic.
Solar Dish Stirling 3000 x 21
Solar Power Tower 1000 x 16
Parabolic Trough 100 x 14

29
Biomass

Current global installed capacity 14 GW
Low efficiencies lt 20
Wet fuels, low HV, small plants located near fuel
High costs 9 / kWh
Future Options to reduce costs
Co-firing with other fuels (coal)
Syngas production via pyrolysis
Anaerobic digesters common in wastewater TP

30
LM 2

Order these concentrating solar power
technologies from least to most efficient
A) Solar power tower
B) Parabolic trough
C) Solar dish Stirling

31
Micro-Hydropower

19 of worlds electricity today is hydro
Many countries in S. America and Africa are gt90
hydro
Typically Hydro plants are above 30 MW
Micro-hydro is lt 100 kW
Run-of-the-river
No expensive dam construction
Penstock (pipeline) delivers water under pressure
to a turbine generator lower down

32
Micro-hydropower
33
Micro-hydropower potential

Three types of energy
Potential z (height m or feet)
Pressure p / ? (pressure / specific weight)
Kinetic v2 / 2g (velocity / grav. constant)
THEORETICAL POWER (rules of thumb)
Power (W) 9810 Q (m3/sec) H (m)
Power (W) Q (gpm) H (ft) / 5.3

34
Net Head

Net Head (HN) Gross Head Pipe Loss
Gross Head elevation
Pipe Loss depends on Pipe size and Flow rate (
see Figure 4.19 p. 199)
Assuming a 50 efficient turbine generator
P (W) Q (gpm) HN (ft) / 10
P (kW) 5 Q (m3/s) HN (m)

35
Example of micro-hydro

Assuming a 200 gpm flow rate and a 3-in PVC
penstock, what is the site potential if it is a
200 foot run to an elevation 100 feet lower?
HG 100 feet
Pipe loss 6 ft of head per 100 feet of run (p.
199)
HN 100 (6 x 2) 88
P (W) Q x HN / 10 200 x 88 / 10 1.76 kW

36
LM 3 Assess your own site

A) How much water must you divert from your
stream (gpm) to achieve a 2 kW output (assume
rule of thumb calcs) if the elevation drop on
your farm is 160 feet and you assume a penstock
of 300 feet of 3 polypropylene (see page 199)?
B) How much would you be saving annually if your
electric costs are 13 / kWh?

37
Aims of Todays Lecture

Part Two Chapter 5 concepts
Economics of DG resources
Energy Economic Methods SP, ISROR, NPV, IRR, NPV
and IRR w/ fuel escalation, AI and Cash Flow

38
Economics of Distributed Resources

Simple Payback
Initial (Simple) Rate of Return
Net Present Value (NPV)
Internal Rate of Return (IRR)
NPV and IRR with Fuel Escalation
Annualizing the Investment
Cash Flow

39
Simple Payback

Extra first cost () / annual savings (/yr)
Initial cost () / annual savings (/yr)
Units Years, months
Example CFLs cost 15 more than their 100 watt
counterparts but will save 8 per year in the
application you are considering. What is their
simple payback?
15 / 8/yr 1.875 years

40
CFL real data

Dr Jansson bought the following CFLs at Home
Depot and Kmart
23 watt CFL (7000 hours) at 4.94 (Kmart)
23 watt CFL (10,000 hours) at 4.22 (Home Depot)
vs.
100 watt incandescent (1000 hours) at 0.40
Assuming he pays 12 cents per kWh and has a
light in his home that needs these lumens for 3
hours each day, what is his simple payback?

41
CFL example

Initial Cost difference
Kmart - 4.94- 0.4 4.54
Home Depot - 4.22- 0.4 3.82
Lifetime of analysis (assume 10 years _at_ 1096
hrs/year 10,960 hours)
Kmart (10,960/7000) 1.57 lamps
Home Depot (10,960/10,000) 1.10 lamps
Incandescent (10,960/1,000) 10.96 lamps

42
CFL (cont)

Simple Payback (capital only)
Kmart CFL ? 1.57 x 4.94 7.76
Home Depot CFL ? 1.10 x 4.22 4.62
Incandescent ? 10.96x 0.40 4.38
Simple Payback (energy savings)
Kmart CFL ? 1.096 khrs x 23w 25.2 kWh
Home Depot CFL ? 1.096 x 23w 25.2 kWh
Incandescent ? 1.096 khrs x 100w 109.6kWh

43
CFL Simple Payback

Annual electric savings
Either CFL saves 109.6 25.2 kWh /year
84.4 kWh/year _at_ 0.12/kWh 10.13 / year
The Kmart Light has a simple payback of
4.54/10.13 0.448 years or 5.3 months
The Home Depot CFL has a simple payback of
3.82 / 10.13 0.377 years or 4.5 months

NOTE In neither example did we take into
consideration the lifetime savings of either CFL
to do that we need to consider how long each
will last
44
CFL Lifetime savings

Kmart CFL savings compared to Incandescent
7000/1000 hour life saves 7 lamps _at_ 0.4
2.80
Energy savings 7000hrs x (100-23 watts saved)
539 kWh over life _at_ 0.12/kWh 64.68
Total Savings 67.48 - 4.94 62.54
Home Depot CFL savings compared to Incandescent
10000/1000 hour life saves 10 lamps _at_ 0.4
4.00
Energy savings 10000hrs x (100-23 watts saved)
770 kWh over life _at_ 0.12/kWh 92.40
Total Savings 96.40 4.22 92.18

45
LM 4 your next PC

You are considering a new desktop PC and have all
the options determined, but there is one energy
efficient option you want to look into. The
feature says your PC will save 26 watts (on
average) if it is installed. The option costs
19.95. Will you take it if you only keep your
PCs for two years and your cost of electricity is
12/kWh? (Assume your PC is used continually and
you are an economically rational person) Show
your simple payback calculation and then explain
why, or why not.

46
Initial (Simple) Rate of Return

This calculation is the inverse of simple payback
Annual savings (/yr) / extra first cost ()
Annual savings (/yr) / initial cost ()
Units per year
Example 1-kW of PV costs 1800 after rebates but
makes 1350 kWh per year and earns more than 30
cents per kWh due to SRECs and electric savings.
What is the initial (simple) rate of return for
the system?
405/yr / 1800 22.5 per year

47
Net Present Value (NPV)

Considers the fact that money has value today
which is higher than it will likely be in the
future. (Inflation is not zero)
Consider what an investment P in an account
earning interest i will be worth
P Pi P (1i) next year
P Pi (PPi)i P(1i)(1i) the year after
that
In Fact Its Future value F P (1i)n

48
Present and Future Value

Future Value Current Sum and interest rate
If I give you 700 today how much is it worth in
five years? (assume you could earn interest of
4.5)
F P (1i)n ? 700(10.045)5 700(1.24618)
872.33

49
Present and Future Value

Present Value Future Sum and discount rate
If I give you 1000 in 7 years how much is it
worth today? (assume inflation or the discount
rate is 5.5)
P F / (1i)n ? 1000/(10.055)7
1000/(1.454679)
687.43

50
LM 5

Determine the following quantities and say which
is larger PV of A or FV of B
A) 15,000 8 years from now with inflation
running (or a discount rate of) at 7.0
Determine its Present Value
B) 6,500 deposited in a CD today that earns 6.1
interest annually for 5 years
Determine its Future Value

51
Net Present Value (NPV)

Considers a stream of benefits (/yr) that
persist for n years (life of device or
improvement) but will need to be discounted back
to today.
A benefit (/yr), d discount rate, n years

52
NPV example

Two 200-HP motors are being considered
High-efficiency 155 kW - 5800
Normal Efficiency 158 kW - 4800
Power costs are 8/kWh, motor runs 1800 hours per
year. Assume 20-yr life, 10 dr.
Which should you buy?

53
NPV example

The two motors annual costs
High-efficiency 155 kW x 1800 x 0.08 22,320
Normal Efficiency 158 kW x 1800 x 0.08
22,752
With a 20-yr life, 10 discount rate
PVF (1.1)20 1 / (0.1)(1.1)20
5.7275/0.67275 8.51
Which should you buy? - Compare NPVs
High-efficiency NPV 8.51 x 22,320 189,943
Normal Efficiency NPV 8.51 x 22,752 193,620

54
NPV example

Which should you buy? - Compare NPVs
High-efficiency NPV 8.51 x 22,320 189,943
Normal Efficiency NPV 8.51 x 22,752
193,620
Incremental First Cost 1,000 for high
efficiency, NPV of future savings 3,677
NET BENEFIT to your company
2,677 at your companys 10 discount rate

55
LM 6

Determine which of these two investments are best
for your company on a net present value (NPV)
basis, assume a discount rate of 7
A) A 13,000 hi-efficiency cooling system upgrade
that lasts 11 years and saves 1,800 per year
B) A 21,500 PV system that lasts for 20 years
and saves 2,200 / year

56
Internal Rate of Return (IRR)

Rate of return analysis is probably the most
frequently used exact engineering cost analysis
technique used in industry. Its major advantage
is that we can calculate a single figure of merit
that is readily understood and it requires no
interest rate, bond rate or discount rate to be
determined.

57
IRR

What is it?
The internal rate of return for any investment is
the discount rate that makes the present value of
the investments income stream total to zero

58
IRR applet

http//hadm.sph.sc.edu/COURSES/ECON/irr/irr.html

59
Another IRR approach
60
IRR example

A premium air conditioner costs an extra 1000
and saves 200 per year in electricity. It has a
lifetime of 10 years. What is its IRR?
SPP - 1000/200/yr 5 Years
Lifetime is 10 years look in Table 5.4

61
IRR

IRR is 15

62
LM 7

Determine which of these two investments has best
IRR for your company -
A) A 13,000 hi-efficiency cooling system upgrade
that lasts 15 years and saves 1,808 per year
B) A 19,500 PV system that lasts for 25 years
and saves 2,660 / year
Write IRR for each and then make your selection

63
NPV with fuel escalation
NPV We calculate an equivalent discount rate
64
IRR with fuel escalation
IRR We calculate an equivalent rate of return
65
Annualizing the Investment

Since money may be borrowed internally or
externally considering the capital investment as
a loan is always a helpful economic analysis
method
A P x CRF (i,n)
CRF 1/PVF

66
Annual Cost example

A 4-kW PV system in NJ generates 5,200 kWh per
year, it costs 8,500 after rebates. Assume a
6, 20-year loan. Assume your electric price is
12 cents/kWh and the SREC is worth 20 cents/kWh.
What is your annual cost and what is your
benefit-cost ratio?
Annual Cost A P x CRF ( 0.06, 20)
Annual Cost 8,500 x 0.0872 741 / year
Annual Benefits 1,664 / year
Benefit-Cost Ratio 1,664/741 2.25

67
LM 8

Determine which of these two APS investments has
best COST - BENEFIT ratio for your company.
(assume 8 loan)
A) A 12,000 wind turbine that saves 1,800 per
year and lasts 20 years.
B) A 18,500 PV system that lasts for 30 years
and saves 2,250 / year
Write Annual Cost for each and then make your
selection based on B-C ratio

68
Cash Flow