ELE2704: Electricity Supply Systems - Conductor Materials Solution

This document presents a comprehensive solution to Question 1.1 of the ELE2704 Electricity Supply Systems assignment, focusing on conductor materials and cable design. The solution begins with a schematic diagram of the source, cables, and load, followed by detailed calculations for load current distribution, cable resistance, and power loss in both copper and aluminum cables connected in parallel. The analysis includes determining the sending end voltage under a 5% voltage drop at the load, exploring the advantages and disadvantages of paralleling copper and aluminum cables. Additionally, the solution addresses power factor correction using a capacitor bank, calculating the line current, generator voltage, and complex output power. It also determines voltage regulation and transmission efficiency, providing a complete understanding of the electrical system's performance. The document provides a detailed and step-by-step solution to the problem.

Response to question 1.1:
a) The schematic diagram for the source, cables, load and the equivalent circuit is given
below.
b) Given that,
i) Load current IL = 125 A.
Resistivity of copper ρCu = 17.5 nΩ.m
Resistivity of aluminium ρAl = 31.8 nΩ.m
Length of both cables = L = 175 m.
Cross-section of the aluminium core AAl = 50 mm2=50∗10−6 m2 .
Cross-section of the copper core ACu = 35 m m2=35∗10−6 m2 .
2 core Copper
cable
2 core Aluminium
cable Lumped
resistance
ICu
IAl
Vs

Now, the resistance of Cu cable RCu = (ρCu*L)/ACu = 17.5∗10−9∗175
35∗10−6 = 0.0875 Ω.
Now, as connected effectively in series the total resistance of the 2-core copper cable RCu2 core
= 2*0.0875 = 0.175 Ω.
Similarly, the resistance of the Al cable = RAl = (ρAl*L)/AAl = 31.8∗10−9∗175
50∗10−6 = 0.1113 Ω.
Now, the resistance of the 2-core aluminium cable RAl 2 core = 2*0.1113 = 0.2226 Ω.
Now, as the connection is parallel the load current is subdivided into parts based on the
resistance of the copper and the aluminium cable.
Hence, ICu = (IL*RAl)/(RCu + RAl) = (125*0.2226)/(0.2226+0.175) = 69.98 Amps.
Similarly, IAl = (IL*RCu)/(RCu + RAl) =(125*0.175)/(0.2226+0.175) = 55.018 Amps.
ii) The cable power loss PCu = ICu
2 RCu2 core = 69.982*0.175 = 857.0687 Watts.
The cable power loss PAl = I Al
2 RAl 2 core = 55.018^2 *0.2226 = 673.8058 Watts.
iii) When 5% voltage is dropped at load occurs then the sending end voltage will be equal to
sum of the voltage drop at either of the aluminium or copper line added with 5% drop at load.
VS = I Cu RCu + 0.05*I Cu RCu = 69.98*0.175 + 0.05*I Cu RCu = 12.247 + 0.05*12.247 = 12.247 +
0.6123 = 12.859 Volts.
c) The advantages of paralleling the copper and aluminium cable is that, in case one of the
material starts to have erosion the other cable can be used. The other advantage is that the
same voltage is applied across the two conductors and hence the loads connected with those
will have the same illumination.

The disadvantages of using the parallel connection of these two particular cable is that the
current do not remains the same as in parallel circuit the current is divided and the current
tends to increase as more branches are connected. Hence, the supply voltage should be
capable of delivering the extra current, otherwise, the whole circuit will not work due to less
current. In parallel circuit operation the circuit loading become complex.
Response to Question 1.2:
a) The equivalent single-phase circuit of a three-phase line is given by,
The load is star connected. Impedance of the line Z = (R + jX) = (1 + j4) Ω/phase.
The sending current IS = IR = 250 A.
Receiving end voltage V R= 11∗103
√ 3 ∠30°
i) Now, sending end phase voltage = VR + Z*IR = 11∗103 / √3(cos(30) + jsin(30)) + (1+j4)250
= 6350.853(0.866 + 0.5j) + 250 + 1000j = 5500 + 3175.426j + 250 + 1000j = 5750 + 4175j =
7105.9∠35.983°
|VR| = 11∗103
√3 = 6350.853 Volts.

Now, the sending end 3 phase voltage VS = √3 ( 5750+4175 j ) =9959.29+7231.31j
ii) The power angle can be calculated from the following equation,
tanδ= IXcos ϕR −IRsin ϕR
V R +IRcos ϕR + IXsin ϕ R
Now, for pf = 0 = cos ϕR
tanδ= −250
6350.853+250∗4 =−0.03401
Hence, δ =tan−1 (−0.03401 )=−1.0967 °
Now, similarly when pf = 0.8 = cos ϕR
tanδ= −250∗0.8−250∗4∗0.6
6350.853+250∗0.8+250∗0.6∗4 =0.0559
Hence, δ =tan−1 ( 0.0559 )=1.268 °
When, pf = 1 = cos ϕR
tanδ= −250∗4
6350.853+250 =−0.1515
So, δ=tan−1 ( −0.1515 ) =−1.456 °
iii) The voltage regulation for lagging power factor load is given by,
Voltage regulation = IRcos ϕR + IX sin ϕR
V S
Now, for pf = 0 = cos ϕR
Voltage regulation = 250*4/7105.9 = 14.07%
Now, when pf = 0.8 = cos ϕR

Voltage regulation = (250*0.8 + 250*4*0.6)/7105.9 = 11.258%
Now, when pf = 1 = cos ϕR
Voltage regulation = 250/7105.9 = 3.518%.
b)
iv) Now, the approximate % voltage regulation = RIcos ϕR −XI sin ϕR
V S
Now, by the above formula the percentage voltage regulation when pf 0.8 leading =
250∗0.8−4∗250∗0.6
7105.9 = -5.63%.
Response to question 1.3:
The single line diagram of the three-phase system is given below.
a) i) For the 50 MVA load at 0.8 lagging pf the real power = 50*10^6*0.8 = 40 MW.
Now, at unity pf the real power = 50 MW.
Hence, 50-40 = 10 MW of power is improved after capacitor bank is connected in the
secondary side of the 11-kV side of transformer.
Hence, the KVA rating of the 11-kV delta connected capacitor bank is 10*10^3 KVA.
ii) Now, the power factor is changed from 0.8 lagging to unity due to the capacitor bank.

Hence, the capacitive reactance per phase Xc = (KV^2)/(KVA) = ( 11
√ 3 )
2
10
= 121/(3*10) =
4.033Ω.
Now, Xc = 1/(2πfC) => 4.033 => C = 1/(2π*50*4.033) = 0.789 mF.
b) Now, the line current in the 33 kV side is IL = Transformer MVA / Transformer voltage =
100*10^6/33*10^3 = 3.03 kA.
c) Now, the line voltage of the generator = Feeder line drop + transformer voltage +
transformer drop = 3.03*10^3*(0.3+j1.5) + 33*10^3 + 3.03*10^3*(0.4+ j0.8) = 35121 +
6969j.
So, the line voltage VgL = 35121 + 6969j
d) The generator complex output power = line current*generator line voltage = (3.03*10^3)*(
35121 + 6969j) = 106416630 + 21116070j
e) Now, |Vs| = |VgL| = 35806 Volts.
Vs = VR + IZ.
Where, Z is the combined impedance of feeder and the transformer.
Z = 0.3 + j1.5 + 0.4 + j0.8 = 0.7 + 2.3j
VR = VS – IZ = 35121 + 6969j – 3.03*10^3(0.7 + 2.3j) = 33000
Hence, the voltage regulation = (|Vs| - |VR|)/|VR| = (35806 – 33000)/33000 = 8.5%
f) Transmission efficiency = (Power delivered)/(Power delivered + losses)
= (50*10^6*0.8)/(50*10^6*0.8 + (3.03*10^3)^2*(0.7) (As losses occur in resistance part
only)

= (40000000)/(40000000 + 6426630)) = 86.16%.
Response to question 1.4:
In the given figure the earth node F is not specified and hence the question has insufficient
information. Please specify the position of the earth node F to do the calculations of voltage
drops.