Category: power transmission

Advert from Engineering. Vol: CXLVII, No: 3827…

Advert from Engineering.
Vol: CXLVII, No: 3827 – May 19th 1939
The Variable Speed Gear of London
Hydraulic variable speed transmission

Advert from Engineering. Vol: 211, No: 1 – Apr…

Advert from Engineering.
Vol: 211, No: 1 – April 1971
Teleflex Morse of Basildon
Motion, power, signal transfer consultancy and equipment

Advert from Engineering. Vol: CXLVII, No: 3827 – May 19th…

Advert from Engineering.
Vol: CXLVII, No: 3827 – May 19th 1939
The Morse Chain Co of Letchworth
Chain drives

More Than You Ever Wanted to Know About Electrical Engineering: Power Transmission Losses

We’ve been talking about step-down transformers
that can convert between a high line voltage and more moderate local
voltages. You might be asking yourself why we need to do such a thing in
the first place. Why can’t we just transmit utility power at 120 V or
240 V?

All right, let’s try it. Say we need to transmit 20 MW of
power over a 100 km. If we do it at 240 Vrms, we’ll have a current of
83.3 kArms flowing through the conductor.

image

Some
of the 20 MW we’re transmitting is going to be lost – that is,
dissipated as heat. We’d like to minimize losses, so let’s say we’re
aiming for 97% efficiency – no more than 3% of that 20 MW lost. Since
power dissipated is a function of current and resistance, we know we’ll
need a conductor with an overall resistance of 8.64 x 10^-5 Ohms over a
distance of 100 km.

image

The resistance of a conductor
is a function of its length, l, its cross-sectional area, A, and the
conductor’s resistivity,
ρ, which is a property of the material it’s made of.

We’ll assume a resistivity here of 8 x 10^-8 Ohm-meters. From here, we
can figure out what kind of cross-sectional area our conductor needs –
in other words, how large the cable’s diameter has to be.

image

To make this work, you’d need a conductor with a radius of 5.4 m. Clearly, this isn’t going to happen.

If
you want a conductor that’s actually practical, that means you need to
transmit power at high voltage – since a higher voltage at the same
power will result in a lower current, this means your losses will be
lower, allowing you to use a much smaller conductor. Here are the same
calculations done with a transmission voltage of 240 kVrms instead of
240 Vrms.

image

In this case, you need a conductor with a radius of about 0.5 cm, which is much more reasonable.