Masters G.M. Renewable and Efficient Electric Power Systems

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598 PHOTOVOLTAIC SYSTEMS

20181614121086420

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1.0

1.5

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4.0

VOLTAGE (V)

CURRENT (A)

NOON

11 am, 1 pm

10, 2

9, 3

Pump

−

Figure P9.3

9.4 Imagine a very simple, reliable solar water heater using PVs directly con-

nected to the electric resistance heating element of a small electric water

heater. The water heater is one that would deliver 2.88 kW if it were wired

to normal 120 V ac.

amps

volts

4 PV modules

Idealized

−

curve

for 1 module

1-sun

Electric

water

heater

rated at

2.88 kW

and 120V

Figure P9.4a

Suppose you have 4 identical PV modules, each with an idealized, 1-sun,

I –V curve as shown above. Three ways that you can wire up the PVs are

A, B, and C as shown.

−

(c)(b)(a)

Figure P9.4b

Plot the I –V curves for the three combinations (A, B, and C) and decide

which one delivers the most energy in a day’s time. Explain your answer.

9.5 A clean, 15-%-efﬁcient module (STC), 1 m

in size, has its own 90% efﬁ-

cient inverter. Its NOCT is 44

◦

C and its rated power degrades by 0.5%/

◦

above the 25

◦

C STC.

PROBLEMS 599

6 kWh/m

1 m

h = 15%

amb

= 15°C

90% eff

dc-to-ac

kWh?

Figure P9.5

a. What would its STC rated power be?

b. Find the kWh that it would be expected to deliver on a 15

◦

C day with

daylong insolation equal to 6 kWh/m

9.6 You have decided to enter the business of selling single-axis, polar mount,

trackers and want to know how much you can sell them for on a dollar-

per-square meter basis to make the resulting PV system no more expensive

than when using a ﬁxed mount system. Using simple dc, STC ratings along

with the following assumptions, how much can the trackers cost $/m

Fixed mount

Polar mount, 1-axis tracker

Figure P9.6

1. Trackers see one-third more insolation.

2. Modules cost $4/W and have conversion efﬁciencies of 10%.

3. Inverters cost $1 per W of module.

4. Installation cost is the same for ﬁxed mount and trackers.

9.7 You are to size a grid-connected south-facing PV system for Las Vegas to

deliver 4000 kWh/yr. Use a tilt equal to latitude minus 15

◦

a. What should be the ac-rated power of the PV/inverter system?

b. What temperature derating should you use if the modules have an NOCT

of 45

◦

C, the maximum power drops by 0.36%/

◦

C, and for ambient the

average annual daily high given in Appendix E is appropriate.

c. For 3% dirt, 3% mismatched modules, and a 92% efﬁcient inverter,

what should the DC, STC rated power of the modules?

d. If the modules are 13% efﬁcient, what area of PVs would be required?

e. If the installed system cost is $6 per dc, STC W, what is the total cost

of the system?

600 PHOTOVOLTAIC SYSTEMS

f. Suppose the system is paid for with a 6%, 30-year loan with interest

being tax deductible, The customer is married and has a taxable income

around $200,000/yr, which puts them in the 35.5% marginal tax bracket.

Nevada has no state income tax. There is a renewable energy credit

(Green Tag program) that will pay the owner 5¢/kWh generated. What

would be the ﬁrst year cost of PV electricity for this customer (¢/kWh)?

9.8 A grid-connected PV array consisting of sixteen Shell SP150 modules

(Table 9.4) can be arranged in a number of series and parallel combina-

tions: (16S, 1P), (8S, 2P), (4S, 4P), (2S, 8P), (1S, 16P). The array delivers

power to a Sunny Boy SB2500 inverter (Table 9.5). Using the input volt-

age range of the maximum power point tracker (MPPT) and the maximum

input voltage of the inverter as design constraints, what series/parallel com-

bination of modules would best match the PVs to the inverter?

9.9 A 1.5 V AA battery that costs $1 is rated at 1.8 Ah. What is its cost

per kWh?

9.10 Suppose a 12-V battery bank rated at 200 Ah under standard conditions

needs to deliver 600 Wh over a 12-h period each day. If they operate at

−10

◦

C, how many days of use would they be able to supply?

9.11 A small cabin needs 2.4-kWh per day, all of which is used in the evening

from 7 pm to 11 pm. If you decide to supply only enough 12-V battery

storage to cover each evening, what should the minimum rated A-hr of

batteries be if they shouldn’t be discharged more than 80 percent? Assume

battery temperature is 25

◦

9.12 Analyze this PV/battery system (ignore battery discharge rate and temper-

ature constraints):

1. PV Module characteristics: P

= 100 W, V

= 20.0 V, V

= 22 V,

= 5.0 A, I

= 6A

2. Battery characteristics: 6 V, 100 Ah (each)

3. Coulomb efﬁciency 90%; Maximum discharge 80%

4. Insolation: 6 kWh/m

day

5. Inverter efﬁciency: 80%;

6. Dirt, etc, efﬁciency: 90% (10% loss)

6V,

100 Ah

−

80%

efficient

Inverter

−

Output

6 kW/m

-day

Figure P9.12

PROBLEMS 601

a. Find the watt-hours/day that would reach the loads (assume all PV

current passes through the batteries):

b. If the load requires 600 Wh/day, for how many cloudy days in a row

can fully-charged batteries supply the load?

9.13 Consider the design of a small PV-powered light-emitting-diode (LED)

light. The PV array consists of 8 series cells, each with rated current 0.3 A

@ 0.6 V. Storage is provided by three series AA batteries that each store

2 Ah at 1.2 V when fully charged. The LED provides full brightness when

it draws 0.4 A @ 3.6 V.

3.6 V

light

LED

3 AA

batteries

0.4A

On/Off

−

Blocking diode

Figure P9.13

The batteries have a Coulomb efﬁciency of 90% and for maximum cycle

life can be discharged by up to 80%. Ignore any complications associated

with temperature, dirt, discharge rate, etc.

a. How many hours of light could this design provide each evening if the

batteries are fully charged during the day?

b. How many kWh/m

-day of insolation would be needed to provide the

amount of light found in (a)?

c. With 12-%-efﬁciency cells, what PV area would be required?

9.14 Consider a PV a rray charging batteries for an off-the-grid, stand-alone

system near Boulder, Colorado You plan to run everything in the house on

120 V ac.

a. Suppose the load is estimated to consist of the following 120 V ac appli-

ances. Supplement load data given below with values from Table 9.10.

A 19-cu. ft. refrigerator/freezer.

A 1000-W microwave used 15 min./day.

Five 20-W compact ﬂuorescent lamps, each used 8 hrs/day.

A horizontal-axis washing machine used 3 hrs/week

A 24-V well pump that delivers 288 gal/day at 1.6 gpm of water from

100-ft depth

602 PHOTOVOLTAIC SYSTEMS

A 19-inch color TV used 2 hrs/day and drawing standby power the other

22 hrs/day

A satellite receiver system for the TV, used 2 hours/day and in standby

mode 22 hrs.

A laptop computer used 4 hrs/day,

Six 3-W transformer units for chargers that run all day long

Find the total watt-hrs per day needed by these appliances.

b. Pick an appropriate system voltage.

c. How many amp-hours/day would the battery bank have to deliver if

the loads are all ac provided by an 85% efﬁcient inverter?

d. How many Ah/d would have to be delivered to the batteries if they

have a Coulomb efﬁciency of 90%?

e. Suppose 5% of the PV output is lost due to dirt, etc. How many Ah/d

should the PVs provide before that derating?

f. Using the worst month in Boulder with south-facing panels tilted at

L + 15, how many AstroPower APex 90-W modules with DC, STC

rated current 5.3 A and rated voltage 17.1 volts would be needed in

series and parallel? If they cost $400 each, what is the cost of PVs?

g. What is the maximum current that you might expect in the wires con-

necting the array to battery system (just use the rated current)? What

gage wire would you suggest using (Table 1.3)?

h. If your design goal is to provide needed electricity 95% of the time,

about how many days of usable battery storage would you need?

i. If the coldest temperature the batteries might experience is −20

◦

what is the maximum depth of discharge that lead-acid batteries could

tolerate without freezing (Figure 9.39)?

j. If a C/72 discharge rate is assumed along with −20

◦

C and the results

from (c), what should be the rated (nominal) Ah capacity of the battery

bank? (use Figure 9.42 and Eq. 9.32)?

k. Suppose you use Concorde PVX 1080 batteries (Table 9.15). How

many batteries in series and how many in parallel would you rec-

ommend (round up if necessary). At $160 each, how much would the

batteries cost?

l. Assume a power control unit with inverter costs $1/watt and they come

in 500 W increments (1 kW, 1.5 kW, 2 kW, 2.5 kW, etc). You want

one big enough to cover all your appliances on at once. Pick an inverter

and how much would it cost?

m. Draw the system “wiring” diagram showing series/parallel combina-

tions of the PV modules and batteries, similar to Figure 9.51.

n. What would the total system cost be?

o. Using the annual insolation in Boulder, estimate the average Amp-

hours/day (at the system voltage) that the PVs could deliver all year

PROBLEMS 603

long. If all of that electricity is run through the batteries and inverter

and ends up being used in the house, how many kWh/year would be

delivered (i.e., in the better months you use all of the kWh not just the

design load speciﬁed in (a))?

p. Paying for the system with a 15-year, 5% loan, ﬁnd the ﬁrst-year cost

of electricity ($/kWh) for someone in the 36% marginal tax bracket if

the loan interest is tax deductible (see Section 5.3.8).

9.15 A PV module is directly connected to a water pump that needs to raise

water 10-ft high through 70 ft of 1/2



plastic tubing with twenty 90

◦

ells

(Figure P9.15a).

20 90° ells

70 ft of 1/2" pipe

pump

10 ft

−

Pump

Figure P9.15a

The pump curves of head H versus ﬂow Q as a function of input voltage

are shown in Figure P9.15b.

4.03.53.02.52.01.51.00.50.0

FLOW RATE (gpm)

HEAD (ft of water)

8 V

10 V

12 V

14V

16 V

18 V

20 V

Figure P9.15b Pump curves.

604 PHOTOVOLTAIC SYSTEMS

The hour-by-hour I –V curves for the PV module are shown in

Figure P9.19c along with the pump I –V curve:

20181614121086420

(volts)

Current (amps)

1-sun = 1000 w/m

Noon

11am and 1 pm

10 am and 2 pm

9 am and 3 pm

8 am and 4 pm

Figure P9.15c

Using Table 9.19, ﬁnd the equivalent length of the pipe and ells to create a

hydraulic system curve similar to that in Figure 9.56. Remember to include

the 10 feet of static head. Plot that system curve on top of the pump curves

in Figure P9.15b. Now ﬁnd the hour-by-hour pumping rate Q (gpm) that

will result.

9.16 Your client wants a quick design for a system to pump on the order of

1800 gallons per day from a depth of 200 ft. You decide to use a 45-V

submersible pump with an estimated efﬁciency of 35% along with PV mod-

ules having rated current of 5 A at a rated voltage of 16 V. The insolation

is estimated to be 6 kWh/m

day in the design month. Since this is only a

quick estimate, piping friction losses will be estimated at 5% of the static

head. Design the PV system.

APPENDICES

Appendix A: Useful Conversion Factors

Appendix B: Sun-Path Diagrams

Appendix C: Hourly Clear-Sky Insolation Tables

Appendix D: Monthly Clear-Sky Insolation Tables

Appendix E: Solar Insolation Tables by City

Appendix F: Maps of Solar Insolation

Renewable and Efﬁcient Electric Power Systems. By Gilbert M. Masters

ISBN 0-471-28060-7

 2004 John W iley & Sons, Inc.

605

606 USEFUL CONVERSION FACTORS

APPENDIX A

USEFUL CONVERSION FACTORS

LENGTH

1inch = 2.540 cm

1 foot = 0.3048 m

1yard = 0.9144 m

1 mile = 1.6093 km

1meter = 3.2808 ft

= 39.37 in.

1 kilometer = 0.6214 mile

AREA

1 square inch = 6.452 cm

= 0.0006452 m

1 square foot = 0.0929 m

1acre = 43,560 ft

= 0.0015625 sq mile

= 4046.85 m

= 0.404685 ha

1 square mile = 640 acre

= 2.604 km

= 259 ha

Renewable and Efﬁcient Electric Power Systems. By Gilbert M. Masters

ISBN 0-471-28060-7

 2004 John W iley & Sons, Inc.

USEFUL CONVERSION FACTORS 607

1 square meter = 10.764 ft

1hectare = 2.471 acre

= 0.00386 sq mile

= 10,000 m

VOLUME

1 cubic foot = 0.03704 cu yd

= 7.4805 gal (U.S.)

= 0.02832 m

= 28.32 L

1 acre foot = 43,560 ft

= 1233.49 m

= 325,851 gal (U.S.)

1gallon(U.S.) = 0.134 ft

= 0.003785 m

= 3.785 L

1 cubic meter = 8.11 × 10

−4

Ac ft

= 35.3147 ft

= 264.172 gal (U.S.)

= 1000 L

= 10

LINEAR VELOCITY

1 foot per second = 0.6818 mph

= 0.3048 m/s

1 mile per hour = 1.467 ft/s

= 0.4470 m/s

= 1.609 km/h

1 meter per second = 3.280 ft/s

= 2.237 mph

MASS

1 pound (avoirdupois) = 0.453592 kg

1 kilogram = 2.205 lb (avoirdupois)

= 35.27396 oz (avoirdupois)

1 ton (short) = 2000 lb (avoirdupois)

= 907.2 kg

= 0.9072 ton (metric)