Three Phase Load Flow Calculations

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jcbeck

Member
Location
UK
Working on a three phase problem (see below).

My questions:

1. Series reactance and resistance stated as per km per phase. Do I need to multiply by sqrt 3 to get line values?
2. I believe I should convert the system to pu - am I on the right lines here?
3. Subject to point 2 above, I am thinking of setting Vbase as 11 kV and VAbase as 1 MVA - is this a good choice or am I off the mark?
4. How do I handle the 1 Ohm resistor on the secondary transformer?
5. To calculate Vr, I am planning to estimate Vr0, then use iterative calculations until no material change in Vr

Thanks as always to the community, I'm learning a lot and appreciate the assistance. Any other guidance or advice gratefully received.

JCB

______________________________________________________________________________________________________________
An isolated three-phase load is supplied from a 33 kV/11 kV substation via a 5MVA transformer with an impedance of 8% (calculated on transformer rating) and a 28 km long three-phase 11 kV overhead line. The overhead line has the following series parameters:

Series inductive reactance = 0.264 ?/km/phase

Series resistance = 0.245 ?/km/phase

Line capacitance and shunt leakage effects may be neglected.

The line supplies a small industrial site fed from a 760 KVA, 11 kV/660 V delta/star transformer with an impedance of 6% (calculated on transformer rating) and negligible resistance. The transformer secondary is earthed via a 1? resistor.

The site has a load of 640 kVA at a power factor of 0.96 lagging. The voltage at the 33kV bus at the substation (
[FONT=Times New Roman,Times New Roman][FONT=Times New Roman,Times New Roman]V[/FONT][/FONT][FONT=Times New Roman,Times New Roman][FONT=Times New Roman,Times New Roman][/FONT][/FONT]s) is maintained at 33.3kV and the 33kV fault level at the substation is 580 MVA.

[FONT=Times New Roman,Times New Roman][FONT=Times New Roman,Times New Roman][/FONT][/FONT][FONT=Times New Roman,Times New Roman][FONT=Times New Roman,Times New Roman][/FONT][/FONT]
 

jcbeck

Member
Location
UK
Hi Bob

Sorry, should have been more specific. Looking to calculate the receiving voltage at the 660 kV load bus, the substation power and reactive power flows and the power loss in the 11 kV transmission line.

Any guidance you can give would be much appreciated.

JCB
 

bob

Senior Member
Location
Alabama
For 3phase AC system
VD= sqrt 3*Iph*(R*cos(theta) + X*sin(theta))*L
where,
Iph=phase current
R= ac resistance of the cable per unit length
X= reactance of the cable per unit length
cos(theta)= power factor
L=Length of the cable
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
Three Phases Load Flow Calculation.jpg First of all you have to add a few data as transformer resistance [if you have copper losses -pk[w] Rcu=pk/3/Iph^2 Iph=S/SQRT(3)/VL-L
You may employ a more accurate formula:
VS=sqrt3*(VL^2/3+Uqa^2+2*VL/SQRT(3)*Uqa+Uqb^2) where:
Uqa=R*I*COS(FI)+X*I*SIN(FI)
Uqb=-R*I*SIN(FI)+X*I*COS(FI)
sin(d)=uqb/VS*sqrt(3)
At 11 kV side of 11/0.76 kV transformer VS you have to multiply by 11/.76 and to replace angle Fi with Fi+d.
Some answer to your questions:
1)You may keep the impedance per phase and multiply by sqrt(3) the resulted voltages.
2-3) I think a Sbase of 100 MVA it is more convenient.
4) The 1 ohm resistance will be involved in a phase-to-ground short-circuit.

Xfrm Voltage Drop.jpg
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
33 kV Load and Short Calculation VoltDrop.jpg 33 kV Load and Short Calculation Short_Circuit.jpg
The first inserted jpg represents the voltage drop considering the load MVA constant and the current depending on voltage. The low voltage is reduced so in order to limit the substation voltage to 33.3 KV.
The second jpg represent the short-circuit calculation according IEC 60909-0 [I?k1 represent the phase-to-ground short-circuit rms and I?k3 all three phases in short-circuit rms.]The diagram shows the impedances ?positive, negative and zero sequence-for one single phase?to-ground short-circuit. For actual short-circuit current you have to multiply be 3[zero sequence is equal in all three phases].
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
I did a mistake, I am sorry! I changed between rated low voltage and 11/0.66 kV rated kVA [0.76]. Then actually the low voltage has to be 618.9 V [if the System voltage will be 33.3 kV]. Also the short-circuit I?k1 will be: 6.896 kA [at 660 V transformer terminal] and I?k3= 5.7287 kA.:ashamed1:
 
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