power sysetm
-
Upload
femi-prince -
Category
Documents
-
view
216 -
download
3
Transcript of power sysetm
![Page 1: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/1.jpg)
![Page 2: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/2.jpg)
![Page 3: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/3.jpg)
![Page 4: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/4.jpg)
![Page 5: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/5.jpg)
![Page 6: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/6.jpg)
Per Unit CalculationsA key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually refer
impedances to the different sides of the transformers
This problem is avoided by a normalization of all variables.
This normalization is known as per unit analysis.
actual quantityquantity in per unit base value of quantity
![Page 7: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/7.jpg)
Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB
2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.
3. Calculate the impedance base, ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB 5. Convert actual values to per unit
Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
![Page 8: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/8.jpg)
Three Phase Per Unit
1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different
voltage level, VB. Voltages are line to line. 3. Calculate the impedance base
Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases
3BS
2 2 2, , ,3 1 1
( 3 )3
B LL B LN B LNB
B B B
V V VZ
S S S
Exactly the same impedance bases as with single phase!
![Page 9: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/9.jpg)
Three Phase Per Unit, cont'd4. Calculate the current base, IB
5. Convert actual values to per unit
3 1 13 1B B
, , ,
3I I3 3 3B B B
B LL B LN B LN
S S SV V V
Exactly the same current bases as with single phase!
![Page 10: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/10.jpg)
![Page 11: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/11.jpg)
![Page 12: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/12.jpg)
![Page 13: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/13.jpg)
![Page 14: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/14.jpg)
![Page 15: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/15.jpg)
![Page 16: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/16.jpg)
![Page 17: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/17.jpg)
![Page 18: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/18.jpg)
![Page 19: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/19.jpg)
![Page 20: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/20.jpg)
![Page 21: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/21.jpg)
![Page 22: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/22.jpg)
![Page 23: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/23.jpg)
Load Representation
![Page 24: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/24.jpg)
![Page 25: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/25.jpg)
![Page 26: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/26.jpg)
Transmission Circuit Calculations
![Page 27: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/27.jpg)
Short Transmission lineIn the case of a short transmission line the capacitance and
conductance to earth may be neglected. Leaving only the series resistance and inductance to be taken
into consideration. The current entering the line at the sending-end termination
is equal to the current leaving at the receiving-end, and this same current flows through all the line sections.
The R and L parameters may therefore be regarded as ' lumped ' .
![Page 28: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/28.jpg)
The equivalent circuit diagram and the vector diagram for a short line are shown below in which:
Equivalent circuit for a short transmission line
![Page 29: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/29.jpg)
Vector diagram for a short transmission line .
![Page 30: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/30.jpg)
The currents IS and IR will be equal in magnitude but not in phase.
Since there is a phase-shift of voltage along the line. R is obtained from a knowledge of the line length ,the size of
conductor and the specifics resistance of the conductor material ,
while XL is calculated from the conductor spacing and radius using the formula derived in Chapter 3.
![Page 31: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/31.jpg)
Referring to the equivalent circuit :
Hence, if the receiving-end conditions are known the necessary sending-end voltage may be calculated .
(6.1 )S RI I a
( ) (6.1 )
S R L R
R R
V V R jX I bV Z I
![Page 32: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/32.jpg)
It will be noted that previous are phasor equations , a more approximate method involving scalar quantities is as follows: Referring to the vector diagram,
cos sinSX R R R R L RV V I R I X
cos sinSY R L R R RV I X I R
2
2 1 2
=[ ( cos sin )
+( cos sin ) ] S R R R R L R
R L R R R
V V I R I X
I X I R
![Page 33: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/33.jpg)
However (IR XL) and (IR R) are very much less than VR and the small voltage is in quadrate with the much larger VSX ,
The voltage regulation of the line is given by the rise in voltage when full loads is removed , or :
cos sinS SX R R R R L RV V V I R I X
( cos sin )% S R R L R
RR R
V V R Xage voltage regulation I
V V
![Page 34: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/34.jpg)
Example A three-phase line delivers 3 MW at 11 KV for a distance of 15 Km . Line loss is 10 % of power delivered , load power factor is 0.8 lagging . frequency is 50 Hz , 1.7 m equilateral spacing of conductors . Calculate the sending-end voltage and regulation .
Solution
11,000Receiving-end phase voltage = 6.3603 RV
3
3
Line current = phase current ( assuming a star connection ) 3,000 10 = 197 A
3 11 10 0.8
![Page 35: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/35.jpg)
Assuming that the conductors are manufactured from copper having a resistance of 0.0137 ohms per meter for a cross-sectional area of 1 mm2 , the conductor cross-section is 80 mm2 corresponding to a radius of 5 mm .
2
3
Total line loss =3 (in three conductors)10 = 3,000 10
100
I R
3
2
300 10 3 197
2.58 ohms
R
![Page 36: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/36.jpg)
71Inductance = (1 4 log ) 10 /2 c
dL H metre
r
37 3
length
1 1.7 10 = 314 (1 4 log ) 10 15 102 5
=5.75 ohms
L
c
X L
![Page 37: power sysetm](https://reader036.fdocuments.in/reader036/viewer/2022070601/577cce0d1a28ab9e788d2d27/html5/thumbnails/37.jpg)
cos sin = 6,350 + ( 197 2.58 0.8) + ( 197 5.75 0.6) = 6,350 + 1057 = 7,407 V per phase = 12,780 V line
S R R R R L RV V I R I X
( cos sin )Regulation = =
1,057 = 16.7 %6,350
S RR L RR
R R
V VR XI
V V