Lecture No 2
Advance Power Systems - Lecture Plan.
1) Power Transfer Between Two Nodes
2) Line Design Problem – Selection of Conductor Dia.
3) Transmission Line Modelling.
a) Medium Line – Approximate Solution
b) Long Line – Exact Solution
4) Limitations of EHV Transmission Lines.
University of the Punjab, L
Power Transfer Between Two Nodes
In an AC supply system, the power transfer between two
nodes depends upon 1) Power Angle and 2) line reactance (X)
As resistance (R) is very small as compared to reactance
2 Line Design Problem
A 3-phase 132 KV transmission line is to be
developed to deliver 50 MW at P.F. 0.8
lagging, over a distance of 40 KM. Determine
conductor size and efficiency of transmission
line.
Maximum operating temperature of line =55
C.
b) Also determine conductor size if the line
efficiency is to be 95%.
TRANSMISSION LINES MODELLING
Medium line
The Long Line Model= Exact Solution
Exercise for homework
1) Determine the efficiency and regulation of a 3phase, 100km 50 Hz
line delivering a load of 20MWat 0.8 p.f. lagging and 66KV. The line
parameters per phase are R= 0.1 ohms/km, inductive reactance =
0.35 ohms/km and capacitance =0.01 micro farad/km. Neglect
leakage resistance and use nominal T method for modelling the line.
2) A 3 phase line has resistance, inductance and capacitance per phase
of 10 ohms, 0.1 H and 0.9 microfarad respectively. Using exact
solution calculate the receiving end voltage if the voltage at sending
end is 140 KV.
Limitations of EHV a.c. Transmission Lines
Limitations of EHV Transmission
• Corona loss and radio Interference.
• Heavy supporting structures and erection difficulties.
•
Insulation requirements.
•
Stability considerations.
•
The cost of transformers, switchgear equipment and
protective equipment increases with increase in
transmission line voltage.
•
Current carrying capacity.
•
Ferranti effect.
• Environmental and biological aspects
Limitations – Contd.
Heavy supporting structures and erection difficulties
• E.H.V. lines carry large mechanical loadings on towers due to
bundle conductors, large air and ground clearances, considerable
dynamic forces due to broken conductors etc.
• The towers with fabricated steel member, are usually employed in
E HV. transmission. Since the cost of steel towers varies from 30 to
50% of the total cost of line for voltages upto 500 kV, therefore,
better and cheaper designs must be evolved to effect economy.
•
For the erection of E.H.V. lines, problems of transportation and
erection arise as the supporting structures are to be transported over
long distances, moreover a high standard workmanship is required
. Audible Noise:
EHV transmission lines and substations also produce audible noise. The sources of
noise generation are corona, humming of transformers, cooling systems and
mechanical and electrical auxiliaries.
The audible noise (AN) generated by a transmission line is a function of the
following:
(i) Voltage gradient on the surface of a conductor.
(ii) Number of sub-conductors in a bundle.
(iii) Conductor diameter.
(iv) Atmospheric conditions.
(v) Lateral distance between the line and the point of measurement of noise.
The audible noise is caused by vibrations produced in air due to change in the air
pressure.
Below 500 kV, audible noise (AN) levels do not exceed the permissible noise but the
lines of 500 kV and above must be designed for a satisfactory audible noise leve
Environment and Biological Aspects:
The biological effect of electric field of EHV lines and EHV substations has
been studied extensively during 1970s. The electrostatic and electromagnetic
fields produced by EHV/UHV transmission lines induce currents and voltages
in animals, birds and human beings. EHV/UHV lines are designed such that
maximum electrostatic field gradient does not exceed 9 kV/m at mid span
under the line near the ground level.
It has been found that very low frequency magnetic fields even of weak
intensities can affect certain cellular processes. Induced current densities
below 10 mA/m2 have no significant biological effect. Currents 10-100
mA/m2 cause minor biological effects. Current densities exceeding 100
mA/m2 cause health hazards.
Safe line to ground clearance of 20 m at mid span is recommended for 400 kV
lines and 24 m for 1,100 kV lines. This permits movement of vehicles safely
Lecture No 3-Recap of the line analysis . Date. 9.1.23
• Problem No 1----- (Short Line Representation)
• A 69-kV, three-phase short transmission line is 16 km long. The line
has a per phase series impedance of 0.125 + j0.4375 Ω per km.
Determine the sending end voltage, voltage regulation, the sending
end power, and the transmission efficiency when the line delivers
• (a) 70 MVA, 0.8 lagging power factor at 64 kV.
• (b) 120 MW, unity power factor at 64 kV
• (Solution to be provided on Board Part a)
• Part b is assignment to be submitted in next week
Line Compensation (Problem)
• Shunt capacitors are installed at the receiving end to improve the line
.
• Problem No 2-----The line of the Problem solved delivers 70 MVA, 0.8
lagging power factor at 64 kV.
• Determine the total MVARs required and the capacitance per phase
of the Y-connected capacitors when the power factor is improved to
0.9.
• Series compensation can also be applied to improve the line
performance from stability point of view.
Line compensation for improvement of performance
Solution to shunt capacitor problem
Solution to shunt capacitor problem
Series Compensation
Medium Line Modelling
• Problem No 3----A 230-kV, three-phase transmission line has a per phase
series impedance of z = 0.05 + j0.45 Ω per Km and a per phase shunt
admittance of y = j3.4 × 10−6 siemens per km. The line is 80 km long. Using
the nominal π model, determine
• (a) The transmission line ABCD constants.
• Find the sending end voltage and current, voltage regulation, the sending
end power and the transmission efficiency when the line delivers
• (a) 200 MVA, 0.8 lagging power factor at 220 kV.
• (b) 306 MW, unity power factor at 220 kV
• Part B is assignment to be submitted next week.
Medium line analysis by pi method
Medium line contd.
Pi Model For Medium Lines
• The line impedance and shunt admittance are
• Z = (0.05 + j0.45)(80) = 4 + j36
• Y = (j3.4 × 10−6)(80) = j0.272 × 10−3 siemens
• The ABCD constants of the nominal π model are
• A = (1 +ZY/2)
• = (1 +j(4 + j36)(j0.272 × 10−3)/2)
• = 0.9951 + j0.000544
• B = Z = 4 + j36
• C = Y (1 +ZY/4) = j0.0002713
• The receiving end voltage per phase is
• VR =220/ √3 KV
• = 127/ 0◦ kV
Pi Model For Medium Lines
• The current per phase is given by
• IR = 200000/ − 36.87◦/(3 × 127/ 0◦) A
• = 524.864/ − 36.87◦ A
• The sending end voltage is
• VS = A(VR) + B(IR) = (0.9951 + j0.000544)(127/ 0◦) +
• (4 + j36)(524.864 × 10−3/ −36.87◦)
• = (140.1/ 5.704◦) kV
• The sending end line-to-line voltage magnitude is
• |VS(L−L)| = √3 |VS| = 242.67 kV
Pi Model For Medium Lines
• The sending end current is
• IS = CVR + DIR = (j0.0002713)(1270006 0◦) + (0.9951 +
j0.000544)(524.8646 −36.87◦) = 502.386 −33.69◦ A
• The sending end power is
• = 163.179 MW+ j134.018 Mvar
• = 211.16/ 39.396◦ MVA
• Voltage regulation in Percent V R =(242.67 − 220)/(220 × 100) =
10.847%
• Transmission line efficiency is
• η ={160/(163.179)} × 100 = 98.052%
Assignment to be submitted after one week
• Problem No 1--Part B
• Problem No 2– Determine the MVARs required when the sendinf end
voltage= 69KV. All other conditions and parameters are the same as
mentioned in the problem no 2 above in slide.
• Problem No3 Part B. Solve it using the T model of Transmission line.
• Which of the two models (T and Pi) for transmission lines would
provide a result with less error. Discuss why.
Long line model
Long line solution
Long Line Solution