EEL303: Power Engineering I - Tutorial 5
1. A single circuit 50 Hz, 3-phase transmission line has the following parameters per km:
R=0.2 ohm, L=1.3mH and C=0.01µ F. The voltage at the receiving end is 132 KV. If
the line is open at the receiving end, find the rms value and phase angle of the following:
(a) The incident voltage to neutral at the receiving end (reference).
(b) The reflected voltage to neutral at the receiving end.
(c) The incident and reflected voltages to neutral at 120 km from the receiving end.
Solution: The series impedance per unit length of the line
z = r + jx = (0.2 + j1.3 × 314 × 10−3 ) = (0.2 + j0.408) = 0.454̸ 63.88
T he shunt admittance = jwc = j314 × 0.01 × 10−6 3.14 × 10−6 ̸ 90
√
√
z
0.454
T he characteristic impedance Zc =
=
× 105 ̸ (63.88 − 90) = 380̸ −13.06
y
0.314
√
√
γ = yz = 0.314 × 0.454 × 10−6 ̸ (90 + 63.88)/2 = (0.2714+j1.169)×10−3 = (α+jβ)
T he receiving end line to neutral voltage Vr =
132 × 1000
√
= 76200volts
3
The receiving end current under open circuited condition Ir = 0
(a) The incident voltage to neutral at the receiving end (x=0) is
Vr + Ir Zc
2
Since, it is no load condition, Ir = 0
Incident voltage =
Vr
76200
=
= 38100volts
2
2
(b) Similarly, the reflected voltage to neutral at the receiving end
Vr − Ir Zc
Vr
=
= 38100volts
2
2
(c) The incident voltage at a distance of 120 km from the receiving end
Vr+ = Vr exp(αx)exp(jβx)
Vr+ = 76.2exp(0.2712 × 120 × 10−3 exp(j1.69 × 120 × 10−3 ) = 78.7̸ 8.02
The incident voltage at a distance of 120 km from the receiving end
=
78.7
̸ 8.02 = 39.35̸ 8.02
2
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EEL303: Power Engineering I - Tutorial 5
Vr− = Vr exp(−αx)exp(−jβx)
Vr+ = 76.2exp(−0.0325)exp(−j0.140) = 73.76̸ − 8.02
The rreflected voltage at a distance of 120 km from the receiving end
=
73.76
̸ − 8.02 = 36.88̸ − 8.02
2
2. Determine the efficiency of the line in the above problem if the line is 120 km long and
delivers 40 MW at 132 KV and 0.8 p.f. lagging.
Solution:
40 × 1000
T he receiving end currentIr = √
= 218.7amps
3 × 132 × 0.8
From the above problem,
Zc = 380̸ − 13.06
For 120 km length of the line,
exp(αx)exp(jβx) = 1.033̸ − 8.02
exp(−αx)exp(−jβx) = 0.968̸ − 8.02
Taking Vr as reference,
Ir = 218.7̸ − 36.8
Vr + Ir Zc
exp(αx)exp(jβx)
2
76200 + 380 × 218.7̸ − 13.06̸ − 36.8
Vs+ =
× 1.033̸ 8.02 = 74.63̸ − 18
2
Vr − Ir Zc
Vs− =
exp(−αx)exp(−jβx)
2
76200 + 380 × 218.7̸ − 49.86
× 0.968̸ − 8.02 = 32.619̸ 62.37KV
Vs− =
2
Vs = Vs+ + Vs− = 74.63̸ − 18 + 32.619̸ 62.37 = 86077 + j5751 = 86.26̸ 3.82
Vs+ =
Now,
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EEL303: Power Engineering I - Tutorial 5
Vr
Vr
+ Ir
− Ir
Z
Z
Is = c
exp(αx)exp(jβx) − c
exp(−αx)exp(−jβx)
2
2
Vs+ Vs−
Is =
−
= 200.39̸ − 29.9
Zc
Zc
P ower at the sending end = 3×|Vs ||Is |cosϕs = 3×86.26×200.39cos33.72 = 43.132M W
% Ef f iciency =
40
× 100 = 92.7%
43.132
3. Determine the ABCD parameters of the line of above problem and verify the sending
end quantities found in the above problem.
Solution:
γl = (0.2712 + j1.169)120 × 10−3 = 0.03254 + j0.1402
A = coshγl = cosh(0.03245 + j0.1402)
A = cosh0.03254cos0.1402 + jsinh0.03254sin0.1402 = 0.99 + j0.004435 = 0.99̸ 0.26
B = Zc sinhγl
sinhγl = sinhαlcosβl + jcoshαlsinβl
sinhγl = sinh0.03254cos0.1402+jcosh0.03254sin0.1402 = 0.031958+j0.1386 = 0.1422̸ 77
B = Zc sinhγl = 380̸ − 13.06 × 0.1422̸ 77 = 54.03̸ 64
Vs = AVr + BIr = 0.99̸ 0.26 × 76200 + 54.03̸ 64 × 218.7̸ − 36.8
Vs = 85908 + j5380 = 86.07̸ 3.588
4. A three phase 50 Hz transmission line is 400 km long. The voltage at the sending end
is 220 KV. The line parameters are r=0.0125 ohm/km, x=0.4 ohm/km and y = 2.8 ×
10−6 mho/km. Find the following:
(a) The maximum permissible line length if the receiving end no-load voltage is not to
exceed 235 KV.
(b) For part (a), the maximum permissible line frequency, if the no-load voltage is not
to exceed 250 KV.
Solution: The total line parameters are:
R=0.125 × 400 = 50 ohms
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EEL303: Power Engineering I - Tutorial 5
X = 0.4 × 400 = 160 ohm
Y = 2.8 × 10−6 × 400 ̸ 90 = 1.12 × 10−3 ̸ 90 mho
Z = R+jX = (50+j160) 168 ̸ 72.6 mho
YZ = 1.12 × 10−3 ̸ 90 ×168 ̸ 72.6 = 0.188 ̸ 162.6
(a) Maximum permissible no-load receiving end voltage = 235 KV
|A| = |
220
Vs
|=
= 0.936
Vr
235
Now,
|A| = 1 +
YZ
2
l2 × j2.8 × 10−6 × (0.125 + j0.4)
= (1 − 0.56 × 10−6 l2 ) + j0.175 × 10−6 l2
2
Since, the imaginary part will be less than 1/10th of real part, |A| can be approximated as
|A| = (1 − 0.56 × 10−6 l2 ) = 0.936
|A| = 1 +
l2 =
1 − 0.936
0.56 × 10−6
l=338 km.
(b)
Vs
220
|=
= 0.88
Vr
250
1
f
f
|A| = 1 + × j1.12 × 10−3 × (50 + j160 × )
2
50
50
Neglecting the imaginary part, we can write
|A| = |
|A| = 1 −
1
f2
× 1.12 × 1.12 × 10−3 × 160 2 = 0.88
2
50
Simplifying, we obtain the maximum permissible frequency as f=57.9 Hz
5. A 275 KV transmission line has the following line constraints: A=0.85̸ 5; B=200̸ 75
(a) Determine the power at unity power factor that can be received if the voltage profile
at each end is to be maintained at 275 KV.
(b) What type and rating of compensation equipment would be required if the load is
150 MW at unity power factor with the same voltage profile as in part (a).
(c) With the load as in part (b), what would be the receiving end voltage if the compensation equipment is not installed.
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EEL303: Power Engineering I - Tutorial 5
Figure 1:
Solution: (a) Given, |Vs | = |Vr | = 275 KV;
α = 5, β = 75.
Since, the power is received at unity power factor,
Qr = 0 =
275 × 275
0.85
sin(75 − δ) −
× (275)2 sin(75 − 5)
200
200
0 = 378sin(75 − δ) − 302
which gives, δ = 22.
Pr =
275 × 275
0.85
cos(75 − 22) −
× (275)2 cos(75 − 5) = 227.6 − 109.9 = 117.7M W
200
200
(b) Now |Vs | = |Vr | = 275 KV
Power demand by load = 150 MW at UPF
PD = Pr =150 MW; QD =0
150 =
275 × 275
0.85
cos(75 − δ) −
× (275)2 cos(75 − 5) = 227.6 − 109.9 = 117.7M W
200
200
150 = 378cos(75 − δ) − 110
δ = 28.46
Qr =
275 × 275
0.85
cos(75−28.46)−
×(275)2 cos(75−5) = 274.46−302 = −27.56M V AR
200
200
Thus, in order to maintain 275 KV at the receiving end, Qr = -27.56 MVAR must be
drawn along with the real power of Pr = 150MW. The load being 150 MW at unity
power factor, i.e., QD =0, compensation equipment must be installed at the receiving
end.
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EEL303: Power Engineering I - Tutorial 5
-27.56 + QC = 0
QC = 27.56 MVAR
i.e., the compensation equipment must feed positive VARs into the line.
(c) Since, no compensation equipment is provided
Pr = 150MW; Qr = 0
Now, |Vs | = 275 KV, |Vr | = ?
Figure 5
Substituting this data in Pr and Qr , we have
275 × |Vr |
0.85
cos(75 − δ) −
× |Vr |2 cos70
200
200
275 × |Vr |
0.85
0=
sin(75 − δ) −
× |Vr |2 sin70
200
200
150 =
(1)
(2)
From, Eq. (2), we get
sin(75 − δ) = 0.0029|Vr |
√
cos(75 − δ) = (1 − 0.00292 |Vr |2 )
Substituting in Eq. (1), we obtain
√
150 = 1.375|Vr | (1 − 0.00292 |Vr |2 ) − 0.00145|Vr |2
(3)
(4)
(5)
Solving the quadratic and retaining the higher value of |Vr |, we obtain
|Vr | = 244.9 KV.
6. Input to a single-phase short line shown in below Figure is 2000 KW at 0.8 lagging power
factor. The line has a series impedance of (0.4+j0.4) ohms. If the load voltage is 3KV,
find the load and receiving end power factor. Also find the supply voltage.
Figure 2:
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EEL303: Power Engineering I - Tutorial 5
Solution: Sending end active/reactive power = receiving end active/reactive power
+ active/reactive line losses.
For active power,
|Vs ||I|cosϕs = |Vr ||I|cosϕr + |I|2 R
(6)
|Vs ||I|sinϕs = |Vr ||I|sinϕr + |I|2 X
(7)
For reactive power,
Squaring eq. (6) and (7), adding and simplifying, we get
|Vs |2 |I|2 = |Vr |2 |I|2 + 2|Vr ||I|2 (|I|Rcosϕr + |I|Xsinϕr ) + |I|4 (R2 + X 2 )
(8)
From the numerical values given,
|Z|2 = (R2 + X 2 ) = 0.32
2000 × 103
= 2500 × 103
0.8
|Vs ||I|cosϕs = 2000 × 103
|Vs ||I| =
|Vs ||I|sinϕs = 2000 × 103 × 0.6 = 1500 × 103
(9)
(10)
(11)
(12)
From eq. (6) and (7), we get
|I|cosϕr =
2000 × 103 − 0.4|I|2
3000
(13)
|I|sinϕr =
1500 × 103 − 0.4|I|2
3000
(14)
Substituting all the known values in Eq. (8), we have
0.32|I|4 − 11.8 × 106 |I|2 + 6.25 × 1012 = 0
(15)
|I| = 725amps
(16)
Substituting for |I| in Eq. (13), we get
cosϕr = 0.82
(17)
LoadPr = |Vr ||I|cosϕr = 3000 × 725 × 0.82 = 1790KW
(18)
|Vs ||I|cosϕs = 2000
2000
= 3.44KV
|Vs | =
725 × 0.8
(19)
Now,
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(20)
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