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Lab No. 3

Generator:

Almost all the hydraulic turbine-driven generators used are synchronous alternatingcurrent machines, which produce electrical energy by the transformation of the rotational

mechanical energy. The electrical and mechanical design of each generator must confirm

to the electrical requirements of the power distribution system to which it will be

connected, and also to the hydraulic requirements of its specific plant. The electrical

characteristics of the generator used in our design are:

SN. Specifications Ratings

1 MVA 7.2

2 KV 11

3 Phase 3

4 Frequency 505 Power Factor 0.85

6 No. of Poles 18

7. RPM 333

Generator excitation system:

Brushless excitation system was used for the excitation of the generator as it is

the best excitation system for medium sized power plant due to the absence of carbon

brushes, brushless system requires less maintaince and its outage time is less.

Generator neutral grounding:We preferred to select the resistance grounding method . Its value is selected such

that the value of Xcg >> Rn or Xcg/Rn>>1. Since the value of resistance was very

high, the distribution transformer method for generator neutral grounding was done.

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Low voltage busbars:

We employed 11 KV medium voltage single bus bar arrangement and unit

Generator Transformer Scheme was used. Unit Generator scheme is a conventional

scheme preferred for medium sized power plants in a small power system where the plant

size is significant for the system. With the use of individual transformer for each

generator unit, unit protection of generator and transformer becomes easy. Overall

protection system becomes simple and easy to locate the faults.

High voltage busbars (66 kv):

We have chosen the Modified Main And Transfer Bus scheme considering its

reliability, flexibility, availability in operation and protection simplicity.

This type of scheme has many advantages as follows:

Flexible operation Load can be shifted to any bus during maintaince of other bus.

Periodic Maintence of breaker can be done without interrupting the powersupply.

Breakers:

The circuit breakers are used to protect the equipment during the fault condition.

The rated voltage and the breaking current are the parameter that calls for the desired

circuit breaker to be used in the protective scheme of an electrical system. We have used

the vaccum circuit breakers as the generator breakers (synchronizing breaker), SF6 circuit

breakers at the HV side and ACB at the station transformer side. A VCB breaker suffers

from the current chopping phenomenon. So to avoid this we have used the surge arrestor

in parallel with each of its phases.

Transformer:

We have used 3 phase delta / star transformer to step up the voltage from 11 kv to

transmission line voltage i.e. 66 KV. The MVA capacity of the transformer is 7 MVA.

Station transformer of 11/ 0.4 KV delta /star is used to supply power for the auxillaries.

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Calculation of resistance grounding

Generator : MVA rating =7.2 MVA, 11KV, 0.85 pf, 377.9 A

Capacitance of terminals of generator with respect to ground Co= 0.135 F for 11 KV

level.

Capacitance of surge capacitance = 0.25 F

Other capacitances are negligible.

Cg = 3(C0 + Cs) = 1.155 F

Xcg =#

$@/A= 2756 = Rn

The value of Rn is in range of high resistance , hence distribution transformer is selected.

A distribution transformer of ratio 11/ % is selected.

Transformation ratio = 13.23

A neutral resistor to be inserted in secondary is

Rn= Rn/n

2= 15.74

Rating of Distribution transformer

The fault current that flows in secondary resistor is

Isec(max)= Vsec(max)/ Rsec = 480/15.74 = 30.5 A

The thermal rating of the transformer should be

KVA = Esec rated * Isec max = 14.6 KVA

Hence,

Dry type transformer = 15 KVA, 11/ %

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Lab No. 4

Type Of Transmission Line Circuit

In our Scheme we used the double Transmission line circuit for the power evacuation.

From the standard table of ACSR conductors,The current carrying capacity of the line is 430 A at 20

0C. For reliable operation

and for maintenance purpose, we used parallel feeders and each feeder is capable of

carrying full load current which is 115 A. Also while using double circuit the capacity to

evacuate the power increases.

Line current =#&"

$#"

High Voltage Bus Bar Scheme Selection:We have chosen the Modified Main And Transfer Bus scheme considering its

reliability, flexibility, availability in operation and protection simplicity.

This type of scheme has many advantages as follows:

Flexible operation

Load can be shifted to any bus during maintaince of other bus.

Periodic Maintence of breaker can be done without interrupting the powersupply.

Fig: Modified Main and Transfer Bus

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Modes of Operation:Normal operation ( Main Bus Operation) :

For maintenance of circuit breaker 1 - :

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Fault occurring at Main Bus:

Calculation of Fault current and Fault MVA level

The fault occurs at various points as shown in the figure:

Fig: 1

For the generator capacity of 7.2 MVA and synchronous speed 333 RPM the per unit sub-

transient reactance of the generator selected from the graph below is 22.5%.

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Generator Sub-Transie

Similarly for the transforme

transformer impedance is sel

Transformer Impedanc

t Reactance

r capacity of 8MVA and high voltage windi

cted from the table below as 9.87%.

e

g 33KV the

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With reference to, ACSR conductor table, for Bear conductors, the line reactance is taken

to be 0.220 /km.

Length of the line =50km

Total reactance of the line= 0.220 /km * 50km = 11

Now, we take,

Base KV (on LV side of transformer) =11 KV

Base KV (on HV side of transformer) =72 KV

Base MVA = 7.2 MVA

Base reactance (on LV side) = (base kv)2/ base MVA = (11 2 / 7.2)

= 16.8

Base reactance (on HV side) = (base kv) 2/ base MVA = (72 2 / 7.2)

= 720

Base Current =$#"

###"= 377.9 A

PU reactance of Transmission line =##

$"= 0.01563 pu

PU reactance of Transformer (PU new ) = PU old *(Base KV old /Base KV new )2

*(Base MVA new /Base MVA

old )

=0.096*(11/11) 2 *(7.2/7) =0.0987 pu

Now,

Xpu (generator) =0.225pu

Xpu (Transformer) =0.0987pu

Xpu (line) =0.01563pu

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Calculations of Fault at different positions:

Fault at position 1:

1

IJ { - % -

-

- %

=0.10633

I fault (pu) =#"

"#"= 9.403 pu

Ifault

(actual) = Ifault

(pu) * Base Current = 9.403 * 377.9 A

= 3554 A

Fault MVA = Base MVA / (X pu ) eq =67.713 MVA

Fault at position 2:

L VH V

(Xpu

)eq

= 0.225 pu

I fault (pu) = 1

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Fault at Position 3:

3

IJ

{ - % -

-

=0.00763

Ipu =#

"""= 131.04A

Iact = 131.04 * 57.73

= 7565.40 A

Fault MVA = Base MVA / (Xpu

)eq

=943.64 MVA


L V H V

4

Xpu = 0.225 + 0.0.0987 = 0.3237 pu

Ipu =#

"$

% pu

Iact = 3.089 * 57.73 A =178.34 A

Fault MVA = Base MVA / (X pu ) eq=22.24 MVA

0.225 0.09870.021

0.021

0.225 0.0987

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c b 55

Xpu = 0.01563 pu

Ipu =#

""#' pu

Iact = 63.97 * 57.73 = 3693.5 A

Fault MVA = Base MVA / (X pu ) eq=46.065 MVA


c b 56

Xpu =#

!

!

\a

Ipu =#

""#&$' \a

Iact = 70.15 * 57.73 = 4049.76 A

Fault MVA = Base MVA / (Xpu

)eq

=505.263 MVA

0.021

0.0210.0987

0.0987

0.225

0.225

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Ratings Of Circuit Breakers:

CB Rate

Volt

KV

Rated

insulaton

level KV

Freq

uency

Normal

Current

A

Breaking

Current

KA

Making

Current

KA

Short time

Rating

Current

Rated

Impulse

Current

1 12 60 50 400 8 20.4 8 20.4

2 72.5 325 50 800 12.5 31.875 12.5 31.875

3 72.5 325 50 800 12.5 31.875 12.5 31.875

4 12 60 50 400 8 20.4 8 20.4

5 72.5 325 50 800 12.5 31.875 12.5 31.875

6 72.5 325 50 800 12.5 31.875 12.5 31.875

7 72.5 325 50 800 12.5 31.875 12.5 31.875

8 0.6 325 50 800 12.5 31.875 12.5 31.875

9 0.6 325 50 800 12.5 31.875 12.5 31.875

10 72.5 325 50 800 12.5 31.875 12.5 31.875

Conclusion

Hence by the end of the report we designed the 12 MW DhuncheHydropower Project on the basis of Medium Hydropower Project (NEA 1997)Method. In this report we also calculated the catchments area, head, thedischarge and output of the plant. This report also include turbine selection on thebasis of various criteria ,SLD using different bus bar ,neural grounding , excitationschemes used in various power plants existing in Nepal. Report include faultcurrent calculation at different bus bar and circuit breaker ratings.

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The Table for the proper selection of breaker is shown below

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ACSR CONDUCTOR IS : 398 [PART-II

Code Name

Resistance

At 20oC

Tensile

StrengthOverall

Diameter

Current Rating Inductive Reactance

Calculated

Breaking LoadIn Still Air With Wind 30 MM 50 MM

Ohm/km N/mm2 Mm A A Spacing Spacing

MOLE 2.718 407 4.50 40 70 0.352 0.374 3.97

SQUIRREL 1.374 771 6.33 76 120 0.325 0.355 7.61

GOPHER 1.098 952 7.09 85 130 0.318 0.349

WEASEL 0.9116 1136 7.77 95 150 0.314 0.345 11.12

FERRET 0.6795 1503 9.00 115 175 0.308 0.339

RABBIT 0.5449 1860 10.05 135 200 0.305 0.335 18.25

MINK 0.4565 2207 11.00 165 250 0.302 0.353

HORSE 0.3977 6108 13.95 185 270 0.296 0.327

BEAVER 0.3841 2613 12.00 176 257 0.299 0.327

RACOON 0.3656 2746 12.30 180 260 0.298 0.329 26.91

OTTER 0.3434 2923 12.60 185 270 0.297 0.328

CAT 0.3020 3324 13.50 195 290 0.296 0.327

DOG 0.2745 3299 14.20 205 305 0.283 0.315 32.41

LEOPARD 0.2193 4137 15.85 275 395 0.259 0.282

COYOTE 0.2214 4638 16.86 260 380 0.238 0.268

TIGER 0.2221 5758 16.50 265 385 0.240 0.271

WOLF 0.1844 6880 18.10 305 425 0.235 0.266 67.34

LYNX 0.1589 7950 19.60 335 470 0.230 0.261

PANTHER 0.1375 9127 21.00 370 510 0.225 0.256 89.67

LION 0.1223 10210 22.30 405 560 0.222 0.252

BEAR 0.1102 11310 22.90 430 590 0.220 0.250

GOAT 0.08989 13780 26.00 495 665 0.213 0.224

SHEEP 0.07771 15910 28.00 554 745 0.210 0.240

KUNDAH 0.07213 9002 26.82 575 775 0.211 0.242 88.79

DEER 0.06786 18230 29.90 590 800 0.207 0.237

ZEBRA 0.06915 13245 28.62 610 812 0.205 0.237 130.32

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