Transmission Line Calculator: ABCD Constants, Voltage Regulation and Efficiency

Model a three-phase transmission line from its resistance, reactance, susceptance and length, and get the sending end values, the regulation and the efficiency.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

Choose the model by line length: a short line (under about 80 km) uses the series impedance alone, a medium line (80 to 250 km) adds the capacitance as a nominal pi or nominal T, and a long line (over about 250 km) uses the exact distributed formulas. Enter the length and the resistance, reactance and susceptance per km.

Enter the receiving end line voltage, the load power and the power factor, and whether it is lagging or leading. The calculator works per phase and finds the ABCD constants, then the sending end voltage V_S = A V_R + B I_R and the sending end current I_S = C V_R + D I_R.

It gives the voltage regulation, the line losses, the efficiency, the angle between the two end voltages and the surge impedance loading. A leading load can give a negative regulation, which is the Ferranti effect.

Formulas used

  • V_S = A V_R + B I_R
  • I_S = C V_R + D I_R
  • A = D = 1 + \frac{ZY}{2}\ \ (\text{nominal } \pi)
  • A = D = \cosh\gamma\ell,\ \ B = Z_c\sinh\gamma\ell
  • \text{VR} = \frac{|V_S|/|A| - |V_R|}{|V_R|}

Worked example

With the default numbers in the calculator above:

  1. Series impedance of the whole line: 82.46 ohm at 75.96 degrees
  2. Shunt admittance of the whole line: 560.00 uS
  3. ABCD constants: A = 0.98 at 0.33 degrees; B = 82.46 ohm at 75.96 degrees; C = 553.73 uS at 90.16 degrees
  4. Receiving end current (taken as the reference, the voltage at 0 degrees): 291.59 A at -25.84 degrees
  5. Sending end voltage (phase): 140897.98 V at 7.82 degrees. Line voltage 244.04 kV
  6. Sending end current: 262.36 A at -11.53 degrees
  7. Angle between the sending and receiving end voltages: 7.82 degrees
  8. Voltage regulation: 13.47 %
  9. Sending end power: 104.63 MW
  10. Power lost in the line: 4.63 MW
  11. Transmission efficiency: 95.57 %
  12. Surge impedance loading (lossless line): 128.05 MW

Change the numbers in the calculator to see the full working, every step, for your own case.

A second worked example

A different case, with these numbers: Line model Short line (up to about 80 km): series impedance only.

  1. Series impedance of the whole line: 82.46 ohm at 75.96 degrees
  2. ABCD constants: A = 1.00 at 0.00 degrees; B = 82.46 ohm at 75.96 degrees; C = 0 uS at 0.00 degrees
  3. Receiving end current (taken as the reference, the voltage at 0 degrees): 291.59 A at -25.84 degrees
  4. Sending end voltage (phase): 143624.13 V at 7.38 degrees. Line voltage 248.76 kV
  5. Sending end current: 291.59 A at -25.84 degrees
  6. Angle between the sending and receiving end voltages: 7.38 degrees
  7. Voltage regulation: 13.07 %
  8. Sending end power: 105.10 MW
  9. Power lost in the line: 5.10 MW
  10. Transmission efficiency: 95.15 %
  11. Surge impedance loading (lossless line): 128.05 MW

Common questions

When should I use the long line model?

Above about 250 km, where the line cannot be treated as lumped. It uses hyperbolic functions of the propagation constant. For shorter lines the nominal pi model is close enough.

What is surge impedance loading?

The load at which the reactive power made by the line capacitance equals the reactive power used by the line inductance. Below it the line makes net reactive power, above it the line uses it.

Why is the regulation negative for a leading load?

A leading (capacitive) load pushes current that raises the voltage along the line, so the receiving end voltage is higher on load than it would be on no load.

Which voltage does it use?

You give the line-to-line receiving end voltage. The calculation is per phase, so the phase voltage is that divided by root three, and the answers are converted back to line values.

Related calculators

Got a question about this? Ask an engineering question and get a plain-language answer with references.

Further reading: Grainger and Stevenson, Power System Analysis (transmission line models); Kothari and Nagrath, Modern Power System Analysis.

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