mid term problem 3 solved

8/2/2019 Mid Term Problem 3 Solved

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Example 5.1

Design a PV system to process 10 KW of power at 230 V, 60 HZ, single-phase. Determine the following:

i) Number of modules in a string and number of strings in an arrayii) Inverter specification and one-line diagramThe PV module data is given below.

max (PMPP): nominal power generated at the MPP traum voltage, VMPP: Voltage at maximum power point (

t at MPP, IMPP

en-circuit voltage)

rt-circuit current)

Solution

To acquire maximum power from the PV array, in the inverter design, we select a modulation index (in amplitude) of Ma=

Where Vac is the output ac RMS voltage and V idc is the input DC voltage

The inverter is designed to operate at the MPP of the PV array. Therefore the number of modules to be connected in ser

string is given by:

NM=Vidc/VMPP where VMPP is the voltage at the MPP of PV of the module.

NM = 361.4/50.6 = 7.14 7 modules

The string voltage is given as

SV = NM x VMPP

SV = 7 x 50.6 = 354.2 V

i) The power generated by one string is given by:SP = NM x PMPP where PMPP is the nominal power generated at the MPP tracking.

SP = 7 x 300 = 2100 W

To calculate the number of strings for a 10 KW PV system, we divide the PV power rating by power per string:

NS: number of strings

AP: array power

SP: string power.

NS = (10 x 103)/2100 = 5

Therefore we have five strings and one array to generate 10 KW of power.

ii) In the final design, the inverter should be rated such that it is able to process generation of 10 KW and supply the230 VAC from its array at its MPPT.

The string voltage is specified as Vidc = 354.2 V

The modulation index is given as follow

Let us select the switching frequency of 6 KHz. Therefore the frequency modulation index is given by

Mf= fs/fe = 6000/60 = 100


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See the one-line diagram.

PV specifications for 10 KW generation.

s per string per array r of arrays oltage (V)

Inverter Specifications

oltage

)

rating voltage, Vac (V) de modulation index, M ncy modulation index, M

Example 2.2

Design a PV system to process 500 KW of power at 460 V, 60 Hz, three-phase AC, and using PV data of Example 2.1. Dete

the following:

i) Number of modules in a string and number of strings in an arrayii) Inverter and boost specificationiii) The output voltage as a function of time and total harmonic distortioniv) The one-line diagram of this systemSolution

Modulation index = 0.9

Where VLL = 460 V

Vidc = 835 VWe will limit the maximum string voltage to 600 VDC. Therefore we can use a boost converter to boost the string voltage

V.

Let us select an approximate string voltage of 550 V, we have:

i) The number of modules in a string is given by:Vstring/VMPP = 550/50.6 11

where VMPP is the voltage of a module at MPPT

The string power SP = NM x PMPP

Using a module rated at 300 W

SP = 11 x 300 3300 W

The string voltage, SV is given by

SV = 11 x 50.6 = 556.6 V

PV specifications

s per string per array r of arrays oltage (V)

If we design each array to generate a power of 20 KW, then the number os strings, NS, in an array is given by:

NS = power of one array/power of one string

NS = 20/3.3 = 6


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The number of array, NA, for total power generation is

NA = PV generation/power of one array = 500 KW/20 KW = 25.

ii) The inverters should be rated to withstand the output voltage of a boost converter and should be able to supply required power. The inverter is rated at 100 kW with input voltage of 835 VDC and the amplitude modulation index of 0.

output voltage of inverter is 460 VAC.

The number of inverters, NI, needed to process a generation of 500 KW is given by

NI = PV generation/power of one inverter = 500/100 = 5

Hence we need to connect five inverters in parallel to supply the load of 500 KW.

If a switching frequency is set at 5.04 kHZ, the frequency modulation index is then:

Mf= fs/fe = 5040/60 = 84

The number of boost converters needed is the same as the number of arrays, which is 25, and the power rating of each b

converter is 20 KW.

The boost converter input voltage is equal to the string voltage:

Vi = 556.5 V

The output voltage of the boost converter is equal to the inverter input voltage

Vidc = Vo = 835 V

The duty ration of the boost converter is given by:

D = 1 - Vi/Vo = 1 556.6/835 = 0.33

iii) With a frequency modulation of 84, the harmonic content of the output voltage was computed as follows:

()

( )

( ) ( )

( )

() () ()The total harmonic distortion is given by

()

See one-line diagram

Example 3.2:

We must use 10 hp, 1760 r/min, 440 V, 3 phase induction motor as an asynchronous generator. Full-load current of the m

10 A and full-load power factor is 0.8.

Find:

1) Required capacitance per phase if capacitors are connected in delta.2) Required prime mover speed in rpm.


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Solution

1) Apparent power S = 3 E I = 1.73 * 440 * 10 = 7612 VAActive power P = S cos = 7612 * 0.8 = 6090 W

Reactive power Q = = 4567 VAR

For machine to run as an asynchronous generator, Capacitor bank must supply minimum 4567 / 3 phases = 1523 VAR pe

Voltage per capacitor is 440 V because capacitors are connected in delta.Capacitive current Ic = Q/E = 1523/440 = 3.46 A

Capacitive reactance per phase Xc = E/I = 127

Minimum capacitance per phase: C = 1 / (2**f*Xc) = 1 / (2 * 3.141 * 60 * 127) = 21 microfarads. If load also absorbs rea

power, capacitor bank must be increased in size to compensate.

2) Prime mover speed should be used to generate frequency of 60 Hz:Typically, slip should be similar to full-load value when machine is running as motor, but negative (generator operation):

Slip = 1800 - 1760 = 40 rpm

Required prime mover speed N = 1800 + Slip = 1840 rpm.

Example 1.

Consider a mountain stream with an effective head of 25 meters (m) and a flow rate of 600 liters ()per minute. How mu

power could a hydro plant generate? Assume plant efficiency() of 83%.

H = 25 m Q = 600 /min 1 m3/1000 1 min/60secQ = 0.01 m3/sec

= 0.83 P 10QH = 10(0.83)(0.01)(25) = 2.075P 2.1 kW

How much energy (E) will the hydro plant generate each year?

E= PtE= 2.1 kW 24 hrs/day 365 days/yrE= 18,396 kWh annually

About how many people will this energy support (assume approximately 3,000 kWh / person)?

People = E3000 = 18396/3000 = 6.13 About 6 peopleExample 2.

Consider a second site with an effective head of 100 m and a flow rate of 6,000 cubic meters per second (about that of Ni

Falls). Answer the same questions.

P 10QH = 10(0.83)(6000)(100)P 4.98 million kW = 4.98 GW (gigawatts)

E = Pt = 4.98GW 24 hrs/day 365 days/yrE= 43,625 GWh = 43.6 TWh (terrawatt hours) People = E3000 = 43.6 TWh / 3,000 kWhPeople = 1.45 million people

(This assumes maximum power production 24x7)


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mid term problem 3 solved

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