ece 333 renewable energy systems lecture 21: photovoltaic systems, water pumping prof. tom overbye...

ECE 333 Renewable Energy Systems

Lecture 21: Photovoltaic Systems, Water Pumping

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Exam 2 Average: 78 • Read Chapter 8

2

Amortizing PV Costs

• Simple payback is the easiest analysis, which assumes there is no cost for money and no inflation. Annual cost is just total cost divided by lifetime– This can give a quick ballpark figure

• Example: Assume 5 kW system with a capacity factor of 18%, an installed cost of $ 5/W (after tax credits), and a lifetime of 20 years with no maintenance costs. What is $/kWh?

3

$5/W×5000 W20 yrs $1250/yr

= =$0.158/ kWh5 kW×0.18×8760 7884 kWh/yr

Amortizing PV Costs

• More detailed analysis uses the capital recovery factor using an assumed discount rate

• Redo the previous example using a discount rate of 5% per year

4

20

20

(1 ) 0.05 1.0525,000

(1 ) 1 1.05 1

0.132725,000 2006.6

1.6533

n

n

d dA P

d

A

$2006.6/yr=$0.2545/ kWh

7884 kWh/yr

These values varylinearly with theassumed PV installed cost

Complications

• "It's tough to make predictions, especially about the future", Yogi Berra (a baseball player/coach appearing as a player or coach in 21 world series)

• There is uncertainly about the rate of electric rate inflation, and the decreasing costs of solar panels

• Also, how longwill you own thehouse, how is PV included inhome's value

5https://qzprod.files.wordpress.com/2014/11/us-consumer-price-indexes-year-on-year-change-core-cpi-headline-cpi_chartbuilder.png?w=1280

Residential Solar Power Purchase Agreements

• Solar PPAs are set so a customer has a developer design, permit, finance and install a PV system on the customer's property

• Customer then buys the solar power from the developer at a fixed price (perhaps slowing inflating), typically below the utility rate for a fixed time period (say 20 years)

• The PPA requires little up front cost from the customer, and the developer maintains the system

• Can be transferred to a new owner if house is sold

6

Utility Scale Solar Prices

7Image Source: Steven Chu talk at NRC Next Generation Electric Grid Workshop, Irvine, CA, Feb 11, 2015

Solar PV isusuallypurchasedon long-term powerpurchaseagreements

(PPAs)

In the News: Minnesota and Renewable Energy

• This month Minnesota is considering changing a requirement that the state get 1.5% of energy from solar by 2020– Proposed bill would allow utilities to meet requirements by

wind, hydro or biomass if cheaper than solar– It would also create rebates for geothermal heat pumps, wind,

solar thermal or storage; would also allow new nuclear– Also impacts net metering

• Bill is getting a negative reception from some, " "Never have I seen some people so upset over a bill that reduces energy pollution and lowers energy costs." Minnesota State Rep. Pat Garofalo

8

Stand-Alone Configurations

9

Key Issue is Intermittency of Renewable Sources

• Unlike hydro, nuclear, and fossil, renewable energy from wind or solar PV cannot be stored in fuel form

• In designing systems powered exclusively by such systems some storage is usually needed

• Image on right shows that wind stopped in BPA areafor 11 days

• In Dec 2014 Chicago had ten days with only 30 minutesof total sun, and only got 16%of available sun up to Dec 22 10Image Source: Steven Chu talk at NRC Next Generation Electric Grid Workshop, Irvine, CA, Feb 11, 2015

Battery I-V Curves

• Energy is stored in batteries for most off-grid applications

• An ideal battery is a voltage source VB

• A real battery has internal resistance Ri B iV V R I

11

Battery I-V Curves

• Charging– I-V line tilts right with a slope of 1/Ri, applied voltage must be greater than VB

• Discharging battery- I-V line tilts to the left with slope 1/Ri, terminal voltage is less than VB

• A blocking diodecan be usedto preventthe battery fromdischarging into the PV atnight.

12

Hourly I-V Curves

• Current at any voltage is proportional to insolation

• VOC drops as insolation decreases

• Can just adjust the 1-sun I-V curve by shifting it up or down 13

Batteries and PV Systems

• Batteries in PV systems provide storage, help meet surge current requirements, and provide a constant output voltage.

• Lots of interest in battery research, primarily driven by the potential of pluggable hybrid electric vehicles– $2.4 billion awarded in August 2009

• There are many different types of batteries, and which one is best is very much dependent on the situation– Cost, weight, number and depth of discharges, efficiency,

temperature performance, discharge rate, recharging rates

14

Lead Acid Batteries

• Most common battery for larger-scale storage applications

• Invented in 1859• There are three main types: 1) SLI (Starting, Lighting

and Ignition) : optimized for starting cars in which they are practically always close to fully charged, 2) golf cart : used for running golf carts with fuller discharge, and 3) deep-cycle, allow much more repeated charge/discharge such as in a solar application

15

Basics of Lead-Acid Batteries

+ 22 4 4 2Positive Plate: PbO + 4H + SO + 2 PbSO 2H O (9.21)e

22 4 4Negative Plate: PbO + SO PbSO 2e (9.22)

16

Basics of Lead-Acid Batteries

• During discharge, voltage and specific gravity drops• Sulfate adheres to the plates during discharge and comes

back off when charging, but some of it becomes permanently attached

17

Battery Storage

• Battery capacity has tended to be specified in amp-hours (Ah) as opposed to an energy value; multiply by average voltage to get watt-hours– Value tells how many amps battery can deliver over a

specified period of time.– Amount of Ah a battery can delivery depends on its

discharge rate; slower is better

18

Figure showshow capacitydegrades withtemperatureand rate

Battery Costs have Been Decreasing

19Image Source: Steven Chu talk at NRC Next Generation Electric Grid Workshop, Irvine, CA, Feb 11, 2015

Battery Technologies

Type Density, Wh/kG

Cost$/kWh

Cycles Charge time, hours

PowerW/kg

Lead-acid,deep cycle

35 50-100 1000 12 180

Nickel-metal hydride

50 350 800 3 625

Lithium Ion 170 500-100 2000 2 2500

The above values are just approximate; battery technology is rapidly changing, and there are many different types within each category. For stationary applications lead-acid is hard to beat because of its low cost. It has about a 75% efficiency. For electric cars lithium ion batteries appear to be the current front runner

20

Number of Cycles Depends on Depth of Discharge

• The below graph shows results for a lead acid battery

21Image Source: http://www.mpoweruk.com/life.htm

Ballpark wouldbe 1000 cyclesat 50% discharge;if cost is $100per kWh, or $200per useable kWh, then thecapital cost is$0.20/kWhThis does notinclude energy cost;ballpark roundtrip efficiency is 80%

Stand-Alone System Energy Needs

• In many locations clouds can greatly reduce the available peak sun hours, sometimes for days at a time

• As a minimum the average peak sun hours must at least meet the average load– Peak sun hours and perhaps the load have a seasonal

dependence

• Sufficient storage is needed to supply full load at times when the sun isn’t available

• Probabilistic depending on location – how likely is a string of low sun days

• Inverter and battery efficiencies need to be considered 22

Estimating Storage Needs

23

Common PV System Usage: Pumping Water

PV systems are widely used for pumping water, particularly in developing countries; if water is stored, when it is pumped does not much matter

24

PV Powered Water Pumping

http://www.rajkuntwar.com/html/Solar.html

http://www.oksolar.com/pumps/

http://solar-investment.us/solar-pv-surface-and-bore-water-pumping/

25

DC Motors

• DC motors have a magnetic field produced in the stator, and then some mechanism to change the current flow in the rotor (armature) (either brushes or electronic)

• Advantages include high starting torque and speed control over a wide of values

• Disadvantages include higher initial cost and the need for a dc source– A luxury car may have more than 100 dc motors

26

DC Motors

• Main types are based on how the stator is powered:– Separately excited (separate windings) – Permanent magnet – Shunt connection (field winding is in parallel with the

armature); these motors have near constant speed regardless of load

– Series connection (field winding is in series with armature); these motors have high torque at low speed, but can have high speed with low torque

27

Separately Excited DC Motor

28Image Source: M.A. Pai, Power Circuits and Electromechanics, Stipes Publishing, 2007

• The equations are:

ff f f f

aa a a a m f

div i R L

dtdi

v i R L G idt

In steady-state (our concern) the derivativesare zero; if a permanentmagnet then Gif is a replaced by a constant k

With a shunt

configuration,vf = va

Torque, Speed and Voltage Relationships

• For a permanent magnetic dc machine in steady-state we get

• If Ra were zero then speed only depends on voltage

• Power to motor is

• Motor torque is

29

a a a mv i R k

a mmotor m a a m

a

v kP k i k

R

motor a mmotor a

m a

P v kT k k i

R

Motor will not start until solar PV has enough current sotorque is high enough to overcome static friction

Permanent Magnet DC Motor

30

DC Motor I-V Curve

Linear Current Booster (LCB) helps the motor be able to start in low sunlight

31

Hydraulic Pumping Curves

• For pumping, the two key values are head, H (height water is pumped), and flow rate, Q (rate at which water is pumped)

• Mechanical power is then

• Required electrical power depends on the efficiency

32

( ) . ( ) ( / min)

( ) . ( ) ( / )

mech

mech

mech

P HQ

P W 0 1885 H ft Q gal

P W 9 81 H m Q L s

. ( ) ( / min)( )elec

0 1885 H ft Q galP W

Efficiency

Example: Energy to Pump Water from Shallow Well

• How many kWh/day are required to pump 250 gallons/day with a 66 ft head, assuming 35% efficiency?

33

With efficiency given, the below equation works for any rate, so

calculate power required if the 250 gallons is pumped in one hour

. ( ) ( / min)( )

/( / min)

min/

elec

0 1885 H ft Q galP W

Efficiency

250gal dayQ gal

60 d

. / min

. .( ) W (for one hour)

.. kWh

elec

4 167galay

0 1885 66 4 167P W 148

0 35Answer 0 148

Cost is about $0.02 or$0.08 per 1000 gallons

Cost is higher whenpumpinginto a pressuretank (1 psi= 2.31 ft)