Masters G.M. Renewable and Efficient Electric Power Systems

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578 PHOTOVOLTAIC SYSTEMS

The 85% efﬁcient inverter will deliver

Inverter output = 142 Ah/day × 24 V × 0.85 = 2900 Wh/day @ 120 Vac

So, the design is just a bit shy of its goal of 3000 Wh/day ac. The system diagram

for this cabin, including some of the energy ﬂows for December, is shown in

Fig. 9.51.

The month-by-month energy that the system just designed will deliver is shown

in Fig. 9.52. The system delivers far more energy in the summer than was needed

in the winter.

−

Charge

controller

2500-W

Inverter

0.85

158 Ah/d

142 Ah/d

120 V

−

24 V

900 Ah

−

2900 Wh/d

−

Figure 9.51 Design for the cabin in Salt Lake City (additional fuses, breakers, and

diodes not shown). Energy values are for the design month, December.

Jan Feb Mar Apr May Jun Jly Aug Sep Oct Nov Dec

1000

2000

3000

4000

5000

6000

Demand 3000 Wh/d

Supply

Energy delivered (Wh/d)

Figure 9.52 A PV–battery system sized to cover the worst month delivers much more

energy than is needed during the rest of the year. Data are for the system designed in

Example 9.20.

STAND-ALONE PV SYSTEMS 579

9.5.10 Hybrid PV Systems

A PV system designed to supply the entire load in the worst month (the “design

month”) usually delivers much more energy than is needed during the rest of the

year. Outside of the tropics, it is not at all unusual for the energy supplied during

the best month to be nearly double that of the design month. Figure 9.52 shows

that to be the case for the cabin in Salt Lake City designed in Example 9.20.

After ﬁguring out the cost of a system that has been designed to be completely

solar, a buyer may very well decide that a hybrid system with most of the load

covered by PVs and the remainder supplied by a generator is worth considering.

The key to that decision is estimating the relationship between shrinking the

PV system and increasing the fraction of the load carried by the generator. An

analysis based on month-by-month calculations of energy supplied by the solar

portion of the system while varying the area of PVs is quite straightforward.

Figure 9.53 shows a pretty typical result (drawn for Salt Lake City with south-

facing array at a tilt of L + 15

◦

). As can be seen, signiﬁcant reductions in the

size of the solar system may still cover a high fraction of the annual load. For

example, Fig. 9.53 suggests that a PV system designed to deliver only 50% of

the load in the design month will still cover about 80% of the annual load.

An approximation to Fig. 9.53 that is helpful for computer analysis is the

following:

Design mo. solar fraction = 0.625 × Annual solar fraction

(Annual ≤ 0.80) (9.36)

Design mo. solar fraction = 0.50 + 28(Annual solar fraction − 0.80)

2.5

(Annual > 0.80) (9.37)

1009080706050403020100

100

Design Month Solar Percentage

Annual Load From Solar (%)

Figure 9.53 The fraction of the annual load supplied by solar as a function of the percent

of the load covered in the worst month of the year. Derived for Salt Lake City.

580 PHOTOVOLTAIC SYSTEMS

TA BLE 9.16 Characteristics of Generators for Hybrid PV Systems

Size

Maintenance Intervals (hours)

Range

Cost

Type

(kW)

Applications

($/W)

Oil Change Tune-up Engine Rebuild

Gasoline

(3600 rpm)

1–20 Cabin Light

use

$0.50 25 300 2000–5000

Gasoline

(1800 rpm)

5–20 Residence

Heavy use

$0.75 50 300 2000–5000

Diesel 3–100 Industrial $1.00 125 –750 500–1500 6000

Source: Sandia National Laboratories (1995).

When a generator is included in the system, the most versatile inverter is an

inverter-charger capable of converting dc from the batteries into ac for the load, as

well as converting ac from the generator into dc to charge the batteries. Switching

from one mode to the other can be done manually or with an automatic transfer

switch in the unit itself. The generator can be sized just to charge the batteries,

which is the usual case, or it can be sized large enough to charge batteries and

simultaneously run the entire household.

With a hybrid system, the battery storage bank can be smaller since the gen-

erator can charge the batteries during prolonged periods of poor weather. One

constraint on how small storage can be is to check to be sure that the load can’t

discharge the batteries at too fast a rate—certainly no faster than C/5. A nominal

3-day storage system is often recommended since it will avoid discharging too

rapidly, while at the same time keeping the number of times the generator has to

be ﬁred up to a reasonable level. Finally, the generator should be sized so that it

doesn’t charge the batteries too rapidly—again, certainly no faster than C/5.

Generators are somewhat costly, depending on the quality of the machine.

They require periodic oil changes, tune-ups, and major overhauls. Home-size

generators burn fuel at varying rates, but a rough estimate is that they will

generate about 5 kWh per gallon of fuel. Table 9.16 summarizes some of the

cost and maintenance characteristics of generators.

9.5.11 Stand-Alone System Design Summary

While the above series of examples and sets of equations for stand-alone sizing of

PV systems may at ﬁrst appear daunting, it is actually quite straightforward and

can fairly easily be set up in a spreadsheet, which allows multiple design options

to be quickly explored. The following is a summary of the design approach for

the principal components of the system.

Load Analysis

1. Analyze the load to determine daily watt-hours of demand. See Table 9.10

for representative load power. Adjust ac loads using an inverter efﬁciency

STAND-ALONE PV SYSTEMS 581

(default 85%) to ﬁnd their dc equivalent and add it to the load that runs

directly on dc:

Total dc load (Wh/day) = dc load (Wh/day) +

ac load (Wh/day)

Inverter efﬁciency

(9.38)

2. Use Tables 9.10 and 9.13 to estimate the maximum steady-state power

demand (W) and, with the help of Table 9.11, pick an appropriate system

voltage (12 V, 24 V, or 48 V).

3. Convert total dc load to a load expressed as Ah @ the system voltage:

Total load (Ah/day @ System voltage) =

Total dc load (Wh/day)

System voltage (V)

(9.39)

4. Use Fig. 9.53 [or Eqs. (9.36) and (9.37)] to help decide what annual solar

fraction to design for and to determine from that the solar fraction in the

design month. The design-month load is then

Design-month load (Ah/day) = Design-month fraction

× Total load (Ah/day) (9.40)

PV Sizing

5. Using insolation data for the site (Appendix E), ﬁnd the insolation

(kWh/m

-day = hours @ 1-sun) in the worst month of the year (the

design month).

6. Pick a PV module and use its rated current I

along with an estimated

Coulomb efﬁciency (default 0.90), de-rating factor (default 0.90), and

the design month insolation to determine design-month Ah/day delivered

per string:

Ah/day-string = Insolation (h/day @ 1-sun) × I

(A) × Coulomb.

× De-rating (9.41)

7. Find an integer value for the number of parallel strings of modules

based on

Strings in parallel =

Design-month load (Ah/day)

Ah/day per module in design month

(9.42)

8. Find the number of modules in series from

Modules in series =

System voltage (V)

Nominal module voltage (V)

(9.43)

582 PHOTOVOLTAIC SYSTEMS

Battery Sizing

9. Decide the number of days of storage needed. For a totally solar system,

use design-month insolation in Fig. 9.46 as a guide. For a hybrid sys-

tem, the choice depends on how inconvenient running the generator is

perceived to be (3 days of storage is a reasonable default).

10. Find the usable storage needed from the Total load (Ah/day) found in

Step 3 and the days of storage found in step 9:

Usable storage (Ah) = Total load (Ah/day) × Days of storage (days)

(9.44)

11. Using the days of storage as an indicator of the discharge rate C/T , along

with anticipated minimum battery temperature, ﬁnd the temperature and

discharge rate factor (T , DR) from Fig. 9.42.

12. Pick the maximum depth of discharge (MDOD defaults: deep-cycle lead-

acid 0.80, auto SLI 0.25, NiCad 0.90) subject to the freeze constraint

given in Fig. 9.39.

13. Find the total storage capacity from

Total storage capacity

(Ah @ C/20, 25

◦

Usable storage capacity (Ah)

(MDOD) × (T , DR)

(9.45)

14. Check to be sure the battery capacity is sufﬁcient to avoid a too-rapid

discharge rate (default C/5 rate). Use the maximum power demand found

in step 2 along with a maximum discharge rate of C/5 to ﬁnd the minimum

storage capacity:

Minimum storage

capacity (Ah)

Maximum load power (W) × 5h

System voltage (V) × Max depth of discharge

(9.46)

Choose the tentative storage capacity to be the larger of the total and

minimum.

15. Choose a battery (voltage and Ah) and determine the number of batteries

in series in each string and the number of parallel strings:

Number of batteries in series =

System voltage

Nominal battery voltage

(9.47)

Number of strings of batteries

in parallel

Total storage capacity (Ah)

Capacity of a single battery (Ah)

(9.48)

Actual battery capacity is then

Actual battery capacity (Ah) = Ah/battery × No. of parallel strings

(9.49)

STAND-ALONE PV SYSTEMS 583

Generator Sizing

16. A generator should be able to replenish the batteries in a reasonable

amount of time, but no faster than at a C/5 rate. Including a charger

efﬁciency factor (default 0.80), the generator output to the battery charger

should be

Generator (W) =

Total storage capacity (Ah) × System voltage (V)

Charge time (h) × Charger efﬁciency

(9.50)

The actual generator size will be determined by what is available on

the market.

17. The generator output and run time can be estimated from the annual solar

fraction (step 4) along with the annual dc load and the charger efﬁciency

(default 0.80):

Generator (kWh/yr) =

Total dc load (Wh/day) × 365 day/yr

×(1 − Annual solar fraction)

Charger efﬁciency × 1000 (W/kW)

(9.51)

Generator run time (h/yr) =

Generator (kWh/yr)

Generator rated power (kW)

(9.52)

System Costs

18. In lieu of actual data on component costs, the following can be used for

preliminary estimates of total system cost:

Photovoltaics: $4.00/W

Batteries: Golf cart $0.07/Wh; True Deep cycle $0.10/Wh

Inverter-charger: $0.60/W

Generator: Light duty $0.50/W; heavy duty $0.75/W

Installation + BOS: 20% × (PV + batteries + inverter + generator $)

19. Annual fuel costs can be estimated based on unit cost and use rate (default:

$1.50/gal × 0.2 gal/kWh = $0.30/kWh)

Fuel ($/yr) = Generator (kWh/yr) × Fuel cost ($/gal)

× Fuel rate (gal/kWh) (9.53)

Maintenance costs depend on hours of use, maintenance intervals and unit

costs. (Defaults: $100 tune-ups @ 300 h, $400 overhauls @ 4000 h):

Maintenance ($/yr) =

Generator run time (h/yr) × Rate ($/service)

Maintenance interval (h/service)

(9.54)

584 PHOTOVOLTAIC SYSTEMS

14,000

13,000

12,000

11,000

10,000

9000

8000

7000

6000

Capital Cost of System ($)

30 40 50 60 70 80 90 100

Annual Percentage Solar

Figure 9.54 The capital cost of the Salt Lake City off-grid cabin drops quickly when it

includes a backup generator to supply even a small portion of the annual load.

While these 19 design steps may imply a linear ﬂow from top to bottom, a real

design will of course involve many iterations as different modules and batteries

are analyzed and as the economics of various solar fractions are determined.

Figure 9.54 is based on applying the above sequence of steps to the cabin

already analyzed in previous examples, and it does so for a system that is

100% solar for comparison with hybrid systems having annual solar fractions

of 80% 60% and 40%. Even though the cost estimates used are crude, the

conclusions drawn are quite robust. That is, the capital cost of a system drops

rather dramatically when a generator provides even a small fraction of the annual

energy.

9.6 PV-POWERED WATER PUMPING

One of the most economically viable photovoltaic applications today is for water

pumping in remote areas. For an off-grid home, a simple PV system can raise

water from a well or spring and store it in either a pressurized or an unpressur-

ized tank, or it can circulate water through a solar water heating system. Water

for irrigation, cattle watering, or village water supply—especially in developing

countries—can be critically important, and the value of a PV water-pumping

system in these circumstances can far exceed its costs.

The simplest PV water-pumping systems consist of just a PV array attached

to a dc pump. Water that is pumped when the sun is out may be used at that

time or stored in a tank for use later, so the disadvantages of battery storage

can be avoided. The result can be a system that combines simplicity, low cost,

and reliability. Matching photovoltaics and pumps in such directly coupled sys-

tems (without battery storage), along with predicting their daily performance, is

PV-POWERED WATER PUMPING 585

−

Pump

"Load"

ELECTRICAL SIDE

HYDRAULIC SIDE

dc motor

Figure 9.55 The electrical characteristics of the PV–motor combination need to be

matched to the hydraulic characteristics of the pump and its load.

actually a quite challenging task. More complex systems may include a battery

and inverter to run a conventional ac pump, along with a linear current booster

(LCB) to improve performance in low-light conditions.

As suggested in Fig. 9.55, a simple, directly coupled PV–pump system has

(a) an electrical side in which PVs create a voltage V that drives current I

through wires to a motor load and (b) a hydraulic side in which a pump creates

a pressure H (for head) that drives water at some ﬂow rate Q through pipes

to some destination. The ﬁgure suggests that the hydraulic side is a closed loop

with water circulating back to the pump, but it may also be an open system

in which water is raised from one level to the next and then released. On the

electrical side of the system, the voltage and current delivered at any instant

are determined by the intersection of the PV I –V curve and the motor I –V

curve. In the hydraulic system, H is analogous to voltage while Q is analogous

to current; as we shall see, the role of H –Q curves in determining a hydraulic

operating point is exactly analogous to the role of I –V curves that determine

the electrical operating point.

9.6.1 Hydraulic System Curves

Figure 9.56a shows an open system in which water is to be raised from one

level to the next. The vertical distance between the lower water surface and the

elevation of the discharge point is referred to as the static head (or gravity head),

and in the United States it is usually given in “feet of water.” Head can also be

measured in units of pressure, such a s pounds per square inch (psi) or pascals

(1 psi = 6895 Pa). To convert between these two equivalent approaches to units,

just picture the pressure that a cube of water exerts on its base. For example, the

pressure that a 1-ft cube weighing 62.4 lb would exert on its 144 square inches

of base would be

1 ft of head = 62.4lb/144 in.

= 0.433 psi (9.55)

Conversely, 1 psi = 2.31 ft of water. Typical city water pressure is about 60 psi

which corresponds to a column of water roughly 140 ft high.

586 PHOTOVOLTAIC SYSTEMS

(a)

Pump

Total Dynamic Head (ft)

Discharge,

(gpm)

(b)

Static

Head

Static Head

Friction Head

Figure 9.56 An “open” system and the resulting “system curve” showing the static and

friction head components.

If the pump is capable of supplying only enough pressure to the column of

water to overcome the static head, the water would rise in the pipe and just

make it to the discharge point and then stop. In order to create ﬂow, the pump

must provide an extra amount of head to overcome friction losses in the piping

system. These friction losses rise roughly as the square of the ﬂow velocity (as is

suggested in Figure 9.56); they depend on roughness of the inside of the pipe and

on the numbers of bends and valves in the system. For example, the pressure drop

per 100 ft of plastic water pipe for various ﬂow rates and diameters is presented

in Table 9.17. In keeping with U.S. tradition, ﬂow rates are given in gallons per

minute (gpm), pipe diameters in inches, and head in feet of water.

Table 9.18 gives pressure drop of various plumbing ﬁttings expressed as equiv-

alent lengths of pipe. For example, each 3/4



◦

elbow in a plumbing run adds

to the pressure drop the same amount as would 2.0 ft of straight pipe. So we can

TABLE 9.17 Pressure Loss Due to Friction in

Plastic Pipe, Feet of Water per 100 ft of Tube for

Various Nominal Tube Diameters

gpm 0.5 in. 0.75 in. 1 in. 1.5 in. 2 in. 3 in.

1 1.4 0.4 0.1 0.0 0.0 0.0

2 4.8 1.2 0.4 0.0 0.0 0.0

3 10.0 2.5 0.8 0.1 0.0 0.0

4 17.1 4.2 1.3 0.2 0.0 0.0

5 25.8 6.3 1.9 0.2 0.0 0.0

6 36.3 8.8 2.7 0.3 0.1 0.0

8 63.7 15.2 4.6 0.6 0.2 0.0

10 97.5 26.0 6.9 0.8 0.3 0.0

15 49.7 14.6 1.7 0.5 0.0

20 86.9 25.1 2.9 0.9 0.1

PV-POWERED WATER PUMPING 587

TA BLE 9.18 Friction Loss in Valves and Elbows Expressed as Equivalent Lengths

of Tube

Fitting 0.5 in. 0.75 in. 1 in. 1.5 in. 2 in. 3 in.

90-degree ell 1.5 2.0 2.7 4.3 5.5 8.0

45-degree ell 0.8 1.0 1.3 2.0 2.5 3.8

Long sweep ell 1.0 1.4 1.7 2.7 3.5 5.2

Close return bend 3.6 5.0 6.0 10.0 13.0 18.0

Tee—straight run 1.0 2.0 2.0 3.0 4.0

Tee— side inlet or outlet 3.3 4.5 5.7 9.0 12.0 17.0

Globe valve, open 17.0 22.0 27.0 43.0 55.0 82.0

Gate valve, open 0.4 0.5 0.6 1.0 1.2 1.7

Check valve, swing 4.0 5.0 7.0 11.0 13.0 20.0

Units are feet of pipe for various nominal pipe diameters.

add up all the bends and valves in a pipe run and ﬁnd what equivalent length of

straight pipe would have the same pressure drop.

The sum of the friction head and the static head is known as the total dynamic

head (H).

Example 9.21 Total Dynamic Head for a Well. What pumping head would

be required to deliver 4 gpm from a depth of 150 ft. The well is 80 ft from

the storage tank, and the delivery pipe rises another 10 ft. The piping is 3/4-in.

diameter plastic, and there are three 90

◦

elbows, one swing-type check valve,

and one gate valve in the line.

150 ft

80 ft 10′

4 gpm

Valve

check

3/4″

PVC

Pump

Well

Tank

Solution. The total length of pipe is 150 + 80 + 10 = 240 ft. From Table 9.18,

the three ells add the equivalent of 3 × 2.0 = 6 ft of pipe; the check valve adds