Design Of Photo Voltaic Systems

Design of Photovoltaic Systems A . K. Mukerjee Chief Scientific Officer (Retired) Centre for Energy Studies Indian Institute of Technology, Delhi New Delhi – 110016.

A Typical PV System PV ARRAY

DC LOAD

Conditioner

DIODE BATTERY

AC LOAD

GRID

etc.

Power

FANS, LAMPS

Introduction

IN V E R T E R

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Construction of PV Array 

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1. PV Array consists of several Modules 2. Single, polycrystalline or amorphous silicon 3. Packing Factor

Losses 1.

2.

The transmission of the radiation is reduced because of reflection of the protective glass sheet on top and absorption in it, and The packing factor. That is, the entire area of the module is not covered by the solar cells but there are large gaps between the adjacent solar cells.

Space Wasted by Round Solar Cells 

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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 (π r2) = 4 πr2. Therefore, the ratio of the cell area to the total area At, is: 4 πr2 /16r2 = π/4 = 0.7854.

Connection of Array T1 R1

R2

A1

R3

B1

R4

C1

D1

A2

B2

C2

D2

A3

B3

C3

D3

S1

S2

S3

T2

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

Bypass Diodes 

If a module in a string fails due to some reason, or comes under the shadow of some object then the current in that module will reduce drastically and will limit the current from the other two which pass through it. In short the current through a string will reduce. In such a case the bypass diode associated with that module will allow the current to pass through itself. For example if module D2 fails then S2 will bypass the current generated by D1 and D3.

Hot Spot Formation 1. 2.

Hot spots in a module. A module consists of a large number of solar cells connected in series. If one cell is shaded and the module is either short circuited or connected to a heavy load then the current from the other cells will cause i2R heat to be generated in it. The cell under shadow will present a high resistance.

Hot Spots (continued) 1.

2.

3.

The other nine cells will approach open circuit voltage Voc This Voc will then be applied across the shaded cell and force a current in the reverse direction This will not only reverse bias the junction, which may cause a breakdown, but also force the current through its combined series and shunt resistors, R = (Rs + Rsh)

Hot Spot in Shaded Cell

Equivalent Circuit of a Solar Cell

The power conditioner The power conditioner has two parts: 1. The maximum power point tracker, and 2. The battery charge/discharge controller

The Maximum Power Point Tracker (MPPT)

Converters and Algorithms 1. Buck Converter based MPPT 2. Boost Converter based MPPT Common Algorithms for Converters 1. Perturb and Observe (P&O), and 2. Incremental Conductance

Block Diagram of MPPT 

Microcontrollers and DSPs

A Typical P & O Algorithm

A 2.2kW MPPT Response Curve 

The response time of a buck based MPPT with P&O algorithm VOC

VMPP

Battery charge/discharge controller Note:

Temp. Comp. at B+ = 14.56 mV/C (16.13mV/C) Float Charging = 14.5 V Load disconn. = 11.5 V Load Re-conn. = 12 V

1N 5822 F1

D1 +

2

B+

D3

SW 1A

100 25V

1

1N 5822

To L2

C1

D2

PV+

1

47K

(1N 4148)

1N 4007

1

22K 2K2)

U1A LM324

OUT

D8 -

4 5

2

2

R11

39K 1%

R5

1M

R10

270

8K2

2K2

5K6

R15

R7

4K7

R13

U2B LM324

D10 OUT

7

3V3 D11

-

D9

39K 1%

1N 4148

R17

Green

C2

LED

(Charging) R14

0.01

1K

R8

6

+

V+

5K6

1

B-

R12

2K2

11

R3

R4

3

2M2

8K2 1%

3

R2

V+

D5

V-

D4

R1

+

R18

R16

R6

(8K2) 10K

11

2

4

3

8K2 1%

R9

D7 to D14 8 * 1N 4148

1N 4148

MTP 2955 E

V-

1N 5822

1K8

PVC3

0.1 R22

22K

U3C LM324

OUT

R26

2

4K7

R25

2K7

C4

11 13

5V

12

U4D LM324 +

D14

RED

1K8

0.01 OUT

3V3

LED

47K

0.1

R29

C5

R28 14

1

SW 1B

2

R30

2K2

1

R21

18K/1% R19

2K7

R20

18K/1%

R23

10K

R27

270K

BC107A 2 D15

3 Q1 1

4V7

C7

0.01

4

4

+

V+

10

8

3

-

V-

R24

V-

11 9

D13

Battery Low

12K

D12

V+

LM 385 - 2.5V

Charge Controller for CFL 5W/7W based Lantern

TO CT OF TRANSFORMER BASE WINDING

A Lead Acid Battery

Chemical Equations 

For charging: the cell the positive terminal of a DC voltage, higher than that of the cell, is applied to the anode with the negative end attached to its cathode. The governing chemical equations are:

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

2. PbSO4 + 2eAt the Cathode

Pb + SO42-

Discharging: The equations at the anode and the cathode become: 

PbO2 + 4H+ + SO42- + 2e2H2O

PbSO4 +

And, 

Pb + SO42-

PbSO4 + 2e-

Charge Versus Rate of Discharge

Life Cycles Versus Discharge

Design of a 1 kW Stand Alone Photovoltaic Power Supply 1. Average power output = 1 kW into a DC load at a DC voltage 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 = 900 Watts/meters2 6. Maximum depth of discharge of battery = 50 % 7. Array should have a fixed tilt of 28 Deg. For Delhi

Block diagram of the 1 kW PV power supply 

For DC loads only MPPT DC LOAD

PV ARRAY

CHARGE/ DISCHARGE CONTROLLER

BATTERY

FANS, LAMPS etc.

DIODE

Assumptions  

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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 and therefore neglected. However it is necessary to save the circuit from accidental input voltage inversion. Wiring and cabling will introduce another 5 % loss.

Calculations 

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The energy requirement for 7 days will be calculated 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.

Calculations (Continued) 

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Hence, the rest of the energy must be stored in and supplied by the battery bank. This energy is: 168,000 – 40,000 = 128,000 W – Hr --------A2 Again, the charge – discharge efficiency of the battery bank is 90 %. Therefore, the energy supplied to the battery is: 128,000/0.9 = 142,222 W – Hr -------------------A3

The size of the battery bank 

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If the voltage of the battery bank is 108 Volts, 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

Battery Sizing (Continued) 1. For 108 volts a string of 9 batteries, of 12 volts 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

Series – Parallel Connection CHARGE CONTROLLER

D 1

D 2

B T5

12

B T1

12

D 3

1

1

D 4

B T6

12

12

B T2

12

B T7

12

B T3

From MPPT

B T8

2

2

B T4

DISCHARGE CONTROLLER TO LOAD

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

Dissipation in Diodes 1. Each diode will conduct half the current of 9.26 Amperes, that is, 4.63 Amperes 2. 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 W – Hr -------------- B6

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 --------- B7 2. 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 -------------- B9 4. This amounts to: 4,555.55 W/108 V = 42 Amperes --------------- B10

Power loss in D3 & D4 (Continued) 1. The voltage drop across the diodes is 0.7 Volts. 2. Therefore the power dissipated in D3 and 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) = 1181 W – Hr --------------- B11

Total Array Energy 1. Thus the energy consumed by the four diodes 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 + 2,323.4 W - Hr = 1, 84,545.4 Watt – hours -----B13

Total Array Energy (Contd.) 1. This energy is given by the MPPT which itself has an efficiency of 90 %. Hence the energy delivered at the input of the MPPT from the array is:  1,84,545.4/0.9 = 2, 05,050.44 Watt – hours -------------- B14

Energy given to MPPT Input 1. As assumed earlier there is a 5 % loss in 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

. Array Size 

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This energy of 2, 15,302.96 W – Hr is to 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

Daily Variation of Insolation 

Modules are Rated at 1000 W/m2

Array Size  

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Average works out to: 0.635 X 900 W = 571 Watts ------C2 Therefore the total size of the array becomes from C1: 5126.25 Watts X 1000/571 = 8977.67 Watts ---------- C3

Electrical Parameters PM 150 2. Maximum Power Rating Pmax. (Wp)* 3. 4. 5. 6. 7.

150.0 Minimum Power Rating Pmin (Wp)* 180.0 Rated Current IMPP (A) 4.80 Rated Voltage VMPP (V) 34.0 Short Circuit Current Isc (A) 5.0 Open Circuit Voltage Voc (V) 42.8

Number of Modules 1. Since these modules are calibrated at 1000 W/m2 their MPP value will reduce at 900 W/m2. The real MPP voltage will be then: 34 X 9/10 = 30.6 Volts -------------------C7 2. Therefore 3 modules in series will yield 91.8 Volts and the total number of modules required for the array: 8977.67 Watts/150 Watts (Wp) = 59.85 = 60 ----- C8 3. Number of strings, with each strings containing 3 modules is 60/3 = 20

Costing The cost of the components can be tabulated below:  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

=

Comparision 

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If a life time of 10 years is taken for the array and 5 years for the battery bank then the cost amounts to: Rs. 22, 05,534.00. In ten years the electricity produced is equal to: 1 X 24 hours X 365 days X 10 years = 87,600 kW-Hr. ------------- D1 Therefore the cost of this energy is: Rs. 22, 05,534/87,600 = Rs. 25.17 per kW – Hr --------------D2

Comparison (Continued) 1.

2.

3.

At present the cost of domestic electricity from the grid is Rs. 4.60 per kW – Hr Cost of electricity from Diesel = Rs. 12.50/unit If the life span of the PV array is taken to be 20 years, the PV generated power will compete with diesel generated power

The Additional Benefits of PV Power 1. Carbon credits, and 2. Lack of emission of polluting gases.

Design of Solar Pump 1.

The pump horsepower HP = (4.19 X 10-6) (GPD)(h) ---------- E1 (PT)(PTF)(η) 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 η is the wire-to-water efficiency of the pump-motor combination.

Horsepower in MKS Units In MKS, the horsepower is given by  HP = (3.658 X 10-6) (LPD)(h) ----E2 (PT)(PTF)(η)  Where now LPD is the pumping requirement in liters per day  h is the effective pumping height in meters.

Pumping Time Factor 

Use of an MPPT in the system normally increases the daily volume pumped by an additional 20%. Hence, a reasonable default value for PTF when a MPPT is used is 1.2 if the pump is connected directly to the PV array, then the PTF will be 1.0.

Pump Efficiency 

The wire-to-water efficiency, η, will be specified by the pump manufacturer. For fractional horsepower pumps, it is typically about 25% while larger pumps will be more efficient.

Numerical   

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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%

Calculations 

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Therefore effective height = 200X 1.05 = 210 ft. Substituting in equation E1, pump HP = 1.17. However, the service factor is 25% for a 1HP motor which means that a 1 HP motor can operate at 1.25 HP without any damage to itself. 1.17 HP = 1.17 X 746W = 872.82 W. Pump Operating DC Voltage = 96 V

Solar Array Calculations 

Since this is the load, the solar array wattage can be calculated as given in section C above. It is important to notice that the use of an MPPT in the system normally increases the daily volume pumped by an additional 20%. Therefore, the final array size will be less by 20%.

Design of a PV operated Pump

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