PW-C3 · Transmission and per-unit

Voltage drop and loss: raise the voltage, cut the loss

When current flows through the line impedance, the receiving-end voltage drops and the line resistance loses power as heat. Yet when the same power is sent at a higher voltage, the current falls in inverse proportion, so the voltage drop shrinks as 1/V and the loss as 1/V²; drag the voltage yourself and see why the loss falls so steeply.

Drag the transmission voltage and watch the loss

The same power is sent over the same line. Dragging the handle raises the transmission voltage: the current bar shrinks as 1/V, but the loss bar goes as 1/V² and shrivels much more steeply, and raising the voltage 2 times halves the current and cuts the loss to 1/4.

Drag the handle left and right to set the transmission voltage between 0.5 and 4 times the reference. The loss is taken as 10% at the reference voltage, so raising the voltage 2 times drops it to 2.5%, and 4 times to just over 0.6%.

Current and loss (vs reference voltage = 1)

P_loss ≈ 10.0%   η ≈ 90.9%

V × 1.0   I × 1.00   (P_loss × 1/V²)

Current flows, voltage drops

When current flows through the line Z = R + jX, the receiving-end voltage is lower than the sending end. The approximate per-phase voltage drop is e ≈ I(R cosφ + X sinφ), so it depends not only on the current but on the load's power factor, and because X is large on high-voltage lines, the X sinφ term enlarges the drop at a lagging power factor. If the receiving-end voltage falls too far, equipment does not run properly, so the voltage drop is a quantity always watched in transmission.

Loss goes as the current squared

The power lost in the line resistance is P_loss = 3 I²R in three-phase. The key is that it goes with the square of the current: a small cut in current cuts the loss by its square, and a rise in current makes the loss shoot up. So when sending power far, the way to hold down loss is above all to reduce the current itself.

ObserveP_loss = 3 I² R

Three-phase line loss is 3 I²R, the square of the current times the resistance.

ChooseP_loss → 1 / ?

For the same power, loss is inverse to voltage squared.

Fill ine ≈ I(R cosφ + X ?)

The reactive term of the drop is X times sinφ.

On your ownV × 2 → P_loss × ?

Voltage ×2 means loss × 1/4.

Raise the voltage, lower the current

When the same power P = √3 V_L I_L cosφ is sent, the current is I = P/(√3 V cosφ), inversely proportional to the voltage. So raising the voltage shrinks the voltage drop as 1/V, and the loss, which goes with the square of the current, as 1/V², so raising the voltage 2 times cuts the loss to 1/4. Measured as a fraction of the sending-end voltage, the percentage voltage drop also shrinks as 1/V². That is exactly why power is stepped up to extra-high voltage the farther it is sent, and power-factor correction (B2) lowers loss too by reducing the current for the same power.

Back to the first screen

As the voltage was dragged up, the current bar shrank as 1/V, but the loss bar went as 1/V² and collapsed much more steeply. Sending the same power while cutting only the current makes the loss follow by its square, and the voltage drop shrank along with the current. Raising transmission to extra-high voltage of hundreds of kilovolts and correcting the power factor to shed reactive current come down to the same thing: holding down loss by reducing the current flowing in the line.

Voltage drop and power loss: when current flows through a line Z = R + jX, the receiving-end voltage drops. Approximate voltage drop e ≈ I(R cosφ + X sinφ), line loss P_loss = 3 I²R. For the same power the current is I = P/(√3 V cosφ), so raising the voltage shrinks the voltage drop as 1/V and the loss as 1/V² (voltage ×2 → loss 1/4). Extra-high-voltage transmission and power-factor correction are both ways of holding down loss by reducing the current.

The next step

Voltage drop swings with the load. At light load late at night the receiving-end voltage is high, and at heavy load at midday it drops sharply, so the next unit (PW-C4) works out voltage regulation: how, by changing transformer taps or adjusting reactive power with compensating equipment such as capacitors and reactors, the receiving-end voltage is held within the prescribed range however the load changes.