design of photo voltaic systems

7/31/2019 Design of Photo Voltaic Systems

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Design of Photovoltaic Systems

A . K. Mukerjee

Chief Scientific Officer (Retired)

Centre for Energy Studies

Indian Institute of Technology, Delhi

New Delhi 110016.


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A Typical PV System Introduction

DC LOADPV ARRAY

DIODE

BATTERY

F

ANS,

L

AMPS

e

tc.

AC LOADINVERTER GRID

Power

Conditioner


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Construction of PV Array 1. PV Array consists of several Modules

2. Single, polycrystalline or amorphous

silicon

3. Packing Factor


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Losses1. The transmission of the radiation is

reduced because of reflection of the

protective glass sheet on top andabsorption in it, and

2. The packing factor. That is, the entirearea of the module is not covered by

the solar cells but there are large gapsbetween the adjacent solar cells.


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Space Wasted by RoundSolar Cells Suppose that the radius of the cells is r. Then

the total area required to place four cells is: At = (2r + 2r) X (2r+2r) = 4r X 4r = 16r2

..(1) However, the area covered by the four cells,

the cell area, is: 4X ( r2) = 4 r2. Therefore, the ratio of the cell area to the

total area At, is: 4 r2 /16r2 = /4 = 0.7854.


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Connection of Array

A1

A2

A3

B1

B2

B3

C1

C2

C3

D1

D2

D3

S1

R2 R4R3R1

S3

S2

Figure 4. A typical array of solar modules with bypass diodes.

T1

T2


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Bypass Diodes If a module in a string fails due to some

reason, or comes under the shadow of someobject then the current in that module will

reduce drastically and will limit the currentfrom the other two which pass through it. Inshort the current through a string will reduce.In such a case the bypass diode associatedwith that module will allow the current topass through itself. For example if module D2fails then S2 will bypass the currentgenerated by D1 and D3.


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Hot Spot Formation1. Hot spots in a module.

2. A module consists of a large number of

solar cells connected in series. If one cell isshaded and the module is either shortcircuited or connected to a heavy load thenthe current from the other cells will cause

i2R heat to be generated in it. The cellunder shadow will present a highresistance.


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Hot Spots (continued)1. The other nine cells will approach open

circuit voltage Voc

2. This Voc will then be applied across theshaded cell and force a current in thereverse direction

3. This will not only reverse bias the junction,which may cause a breakdown, but alsoforce the current through its combinedseries and shunt resistors, R = (Rs + Rsh)


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Hot Spot in Shaded Cell


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Equivalent Circuit of a Solar

Cell


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The power conditioner

The power conditioner has two parts:

1. The maximum power pointtracker, and

2. The battery charge/dischargecontroller


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The Maximum Power Point

Tracker (MPPT)


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Converters and Algorithms1. Buck Converter based MPPT

2. Boost Converter based MPPTCommon Algorithms for Converters

1. Perturb and Observe (P&O), and

2. Incremental Conductance


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Block Diagram of MPPT Microcontrollers and DSPs


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A Typical P & O Algorithm


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A 2.2kW MPPT ResponseCurve

The response time of a buck basedMPPT with P&O algorithm

VOC

VMPP


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Battery charge/dischargecontroller


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A Lead Acid Battery


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Chemical Equations

For charging: the cell the positive terminalof a DC voltage, higher than that of the cell,is applied to the anode with the negative endattached to its cathode. The governingchemical equations are:

1. PbSO4 + 2H2O PbO2 + 4H+ +SO42- + 2e-At the anode, and

2. PbSO4 + 2e- Pb + SO42-

At the Cathode


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Discharging:

The equations at the anode andthe cathode become:

PbO2 + 4H+ + SO4

2- + 2e- PbSO4 + 2H2O

And,

Pb + SO4

2- PbSO4

+ 2e-


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Charge Versus Rate ofDischarge


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Life Cycles Versus Discharge


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Design of a 1 kW Stand AlonePhotovoltaic Power Supply

1. Average power output = 1 kW into a DC load at a DCvoltage of 108 Volts

2. Duration of operation = 24 hours/day

3. Average time of sunlight available = 8 hours/day

4. Number of sunless days = 2/week

5. Peak value of insolation in Delhi = 900Watts/meters2

6. Maximum depth of discharge of battery = 50 %7. Array should have a fixed tilt of 28 Deg. For Delhi


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Block diagram of the 1 kW PVpower supply

For DC loads only

DC LOADPV ARRAY

DIODE

BATTERY

FAN

S,

LAM

PS

etc.

MPPT

CHARGE/

DISCHARGE

CONTROLLER


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Assumptions

The following assumptions have been made:

The electrical efficiency of the circuit of the MPPT =90 %

The charge/discharge cycle efficiency of the battery(assuming new ones) = 90 %

The diode is usually a built-in part of the MPPT andtherefore neglected. However it is necessary to save

the circuit from accidental input voltage inversion. Wiring and cabling will introduce another 5 % loss.


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Calculations

The energy requirement for 7 days will becalculated below.

Power required = 1000 Watts Therefore, energy needed for 7 days = 1000

W X 24 Hr X 7 days = 168,000 Watt hours. For an 8 hour sunlit day the energy given

directly to the load is: 1000 W X 8 Hr X 5 Days = 40,000 W Hr ----

---------------- A1 Since sunlight is available for only 5 days.


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Calculations (Continued)

Hence, the rest of the energy must be storedin and supplied by the battery bank.

This energy is:

168,000 40,000 = 128,000 W Hr --------A2

Again, the charge discharge efficiency ofthe battery bank is 90 %. Therefore, the

energy supplied to the battery is: 128,000/0.9 = 142,222 W Hr -----------------

--A3


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The size of the battery bank

If the voltage of the battery bank is 108Volts, as desired, then its charge is:

1, 42,222 W - Hr/108 V = 1316.87 Ampere

hours.-------- B1 Since it is assumed that the batteries must

retain 50 % of the charge after discharge,their charge holding capacity must be twice

this value. That is: Total charge = 1,316.87 X 2 = 2,633.75 A

Hr------------B2


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Battery Sizing (Continued)

1. For 108 volts a string of 9 batteries, of 12volts each, must be used.

2. The charge capacity of each battery must be:3. 2633.75 A - Hr/9 = 292.6 A Hr.-----------B3

4. In case 300 A - Hr batteries, which are rare,

are not available then two strings of 9, 150 A Hr ones may be connected in parallel


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Series Parallel Connection


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Diodes D1 , D2 ,D3 and D4

1. At 108 volts the load current is:

2. 1000 W/108 V = 9.26 Amperes -----B4

3. Nominal voltage, during conduction,across them is 0.7 Volt.

4. Therefore both the diode will dissipate

9.26 A X 0.7 V = 6.8 Watts -----B5


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Dissipation in Diodes

1. Each diode will conduct half thecurrent of 9.26 Amperes, that is, 4.63

Amperes2. Each diode, with 100 % overrating,

should be 10 amperes, 200 volts

3. Energy Consumed by Diodes:

6.8 W X 24 hrs X 7 days = 1,142.4 WHr -------------- B6


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Dissipation in D3 & D4

Energy passing through D3 & D4 is:

1. 40,000 W Hr +1, 42,222 W Hr = 1,

82,222 W Hr --------- B72. This energy is passed in:

8 hours X 5 days = 40 hours -------------- B8

3. Hence the power is: 1, 82,222 W Hr/40 Hr

= 4,555.55 Watts -------------- B94. This amounts to: 4,555.55 W/108 V = 42

Amperes --------------- B10


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Power loss in D3 & D4(Continued)

1. The voltage drop across the diodes is0.7 Volts.

2. Therefore the power dissipated in D3and D4 is: 42 A X 0.7 V = 29.53 Watts

3. Therefore energy consumed is:

29.53 W X (8 Hours X 5 days) = 1181W Hr --------------- B11


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Total Array Energy

1. Thus the energy consumed by the fourdiodes is:

1,142.4 W Hr + 1,181 W Hr = 2,323.4 W Hr ------------------ B12

2. This must be supplied by the PV array.Therefore the total array energy rises to:

40,000 W Hr +1, 42,222 W Hr + 2,323.4W - Hr = 1, 84,545.4 Watt hours -----B13


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Total Array Energy (Contd.)

1. This energy is given by the MPPTwhich itself has an efficiency of 90 %.

Hence the energy delivered at the inputof the MPPT from the array is:

1,84,545.4/0.9 = 2, 05,050.44 Watt

hours -------------- B14


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Energy given to MPPT Input

1. As assumed earlier there is a 5 % lossin wiring and cabling, hence the output

of the array should be:2, 05,050.44 + (5 X 2, 05,050.44)/100= 2, 05,050.44 + 10,252.52 = 2,

15,302.96 W Hr----B15


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. Array Size

This energy of 2, 15,302.96 W Hr isto be generated by the array in 5 days

with 8 hours of sunlight on each day.Hence the power of the array becomes:

2, 05,050.44 W Hr/ (8 hr X 5 days) =

5126.25 Watts ------------------ C1


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Daily Variation of Insolation

Modules are Rated at 1000 W/m2


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Array Size

Average works out to:

0.635 X 900 W = 571 Watts ------- C2

Therefore the total size of the arraybecomes from C1:

5126.25 Watts X 1000/571 = 8977.67

Watts ---------- C3


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Electrical Parameters PM 150

1. Maximum Power Rating Pmax. (Wp)* 150.0

2. Minimum Power Rating Pmin (Wp)* 180.03. Rated Current IMPP (A) 4.80

4. Rated Voltage VMPP (V) 34.0

5. Short Circuit Current Isc (A) 5.06. Open Circuit Voltage Voc (V) 42.8


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Number of Modules

1. Since these modules are calibrated at 1000W/m2 their MPP value will reduce at 900W/m2. The real MPP voltage will be then:

34 X 9/10 = 30.6 Volts -------------------- C72. Therefore 3 modules in series will yield 91.8

Volts and the total number of modulesrequired for the array:

8977.67 Watts/150 Watts (Wp) = 59.85 = 60----- C83. Number of strings, with each strings

containing 3 modules is 60/3 = 20


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Costing

The cost of the components can be tabulatedbelow:

Solar module @ Rs. 200.00 per Watt =

8977.67 Watts X 200 = Rs. 17,95,534.00 18 batteries, 150 A Hr, 12 Volts each @ Rs.

10,000.00 each = Rs. 1,80,000.00 MPPT and Charge/Discharge controller

= Rs. 50,000.00_

Total = Rs.20,25,534.00


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Comparision

If a life time of 10 years is taken for the arrayand 5 years for the battery bank then thecost amounts to: Rs. 22, 05,534.00. In tenyears the electricity produced is equal to:

1 X 24 hours X 365 days X 10 years = 87,600kW-Hr. ------------- D1

Therefore the cost of this energy is: Rs. 22, 05,534/87,600 = Rs. 25.17 per kW

Hr --------------D2


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Comparison (Continued)

1. At present the cost of domestic electricityfrom the grid is Rs. 4.60 per kW Hr

2. Cost of electricity from Diesel = Rs.12.50/unit

3. If the life span of the PV array is taken tobe 20 years, the PV generated power will

compete with diesel generated power


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The Additional Benefits of PVPower

1. Carbon credits, and

2. Lack of emission of polluting gases.


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Design of Solar Pump

1. The pump horsepowerHP = (4.19 X 10-6) (GPD)(h) ---------- E1

(PT)(PTF)()

GPD is the gallons per day to be pumped,PT is the pumping time,PTE is the pumping time factor,h is the effective height and is the wire-to-water efficiency of the pump-motor combination.


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Horsepower in MKS Units

In MKS, the horsepower is given by

HP = (3.658 X 10-6) (LPD)(h) ----- E2

(PT)(PTF)()

Where now LPD is the pumpingrequirement in liters per day

h is the effective pumping height inmeters.


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Pumping Time Factor

Use of an MPPT in the system normallyincreases the daily volume pumped by

an additional 20%. Hence, a reasonabledefault value for PTF when a MPPT isused is 1.2 if the pump is connected

directly to the PV array, then the PTFwill be 1.0.


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Pump Efficiency

The wire-to-water efficiency, , will bespecified by the pump manufacturer.

For fractional horsepower pumps, it istypically about 25% while larger pumpswill be more efficient.


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Numerical

Numerical:

Specification for pumping system:

Volume of water to be lifted = 2000 gallons/day.

Water reservoir = 200 ft. underground

Worst case peak Sun day = 6 hrs.

PTF = 1

Peak Sun = 6 hrs.

Assume pump efficiency = 25%

Piping friction losses = 5%


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Calculations

Therefore effective height = 200X 1.05 = 210ft.

Substituting in equation E1, pump HP = 1.17.However, the service factor is 25% for a 1HPmotor which means that a 1 HP motor canoperate at 1.25 HP without any damage toitself.

1.17 HP = 1.17 X 746W = 872.82 W.

Pump Operating DC Voltage = 96 V


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Solar Array Calculations

Since this is the load, the solar arraywattage can be calculated as given in

section C above. It is important tonotice that the use of an MPPT in thesystem normally increases the daily

volume pumped by an additional 20%.Therefore, the final array size will beless by 20%.


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Design of a PV operated Pump

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design of photo voltaic systems

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