Solving Complex ABCD Equations: 50 Hz, 3-Phase, High-Voltage Line

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FIGURE 3(a) represents a 50 Hz, three-phase, high-voltage, transmission
line. For one phase, the relationships between the sending end voltage
and current and the receiving end voltage are given by the complex
ABCD equations:

Vsp = Vrp(A1 + jA2) + Irp(B1 + jB2)

Isp = Vrp(C1 + jC2) + Irp(D1 + jD2)


where VSP is the sending-end phase voltage, ISP the sending-end phase
current and VRP is the magnitude of the open-circuit receiving end phase
voltage.


(a) Given the parameter values in TABLE A and if the magnitude of the
receiving-end line voltage VRL is measured as 154 kV when feeding
a balanced load of 40 MVA at a power factor of 0.9, calculate the
value of the sending-end phase voltage VSP and sending-end phase
current ISP.

[N.B. VSL = √3 × VSP and the total power in a three-phase load is
given by P = √3VI cos θ.]

(b) Hence or otherwise calculate the sending-end power and thus the
power lost in the cable.

(c) If the line is modeled by the Π-circuit of FIGURE 4(b), see if you
can estimate the primary line coefficients R, L, G and C. The line is
50 km long.


A1= 0.8698
A2= 0.03542
B1= 47.94 Ω
B2= 180.8 Ω
C1= 0 S
C2= 0.001349 S
D1= 0.8698
D2= 0.03542

so

a) P=sqrt(3)VI
I=40000000/(sqrt(3)*154000) = 149.961 A
Vrp=Vrl/sqrt(3) = 88912 V

Using formulas given for ABCD.
Vsp =84524.79 + j30262.16 = 89778.8∠19.70
Isp = 130.44 +j125.252 = 180.83∠43.84

b) p= sqrt(3)VI cos θ = sqrt(3)*154000*149.961*0.9 =36 MW

Vsl = sqrt(3)*89778.8 = 155501.4*149.961*0.9 = 36.3MW

350960.2 W lost in the cable.

c) I don't have a clue?
 

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Hi mate, Did you ever find the solution to part (C)? I cannot find for the life of me find any info on how to solve it. I have very similar answers to you for the rest of 4 but (c) has become a road block. Any help would be much appreciated, it almost feels like something is missing from the question!
 
I think the OP has come & gone. I'd like to look at this but can't open the .docx with my office 2003. Could you make a pdf file out of it? Or describe fig. 4(b) in words?I do know how to produce the ABCD parameters for a transmission line given R,L,G and C per-unit-length quantities.
 
Abcd

Hi mate, I am out of the country at the moment. But basiacally i have a table of values from A1,A2 through to D1, D2 but I can't find the formulas i need to find R,L,G,C.. The values in the table do have the relevant units ie ohms, S, etc with their respective imaginary part. The tranmission line is contructed as a pi circuit with Y being the 2 parallel resistors and z being the top one. It is a 50 hz 3 phase line with a 50 km length..

This probably isn't much use but i didn't want you to think i was ignoring you, I appreciate the offer to help.
 

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Thanks mate,

I do want to stress however that i am not looking for the answer here. I just need a nudge in the right direction or better still a good resource to look at. I am not finding my books very useful...
 
Hello everyone,
I'm struggling with the same here, I don't know how to find R, L, G, C from given data. Has anyone got an idea?