In Ferranti effect,Receiving end voltage is greater than
sending voltage,why? can anybody tell me the answer
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Answer Posted / larry
An answer to this one is not easy. First, there is a
difference between active power (the power that actually
does work) and reactive power (VARs, for Volts-Amperes
Reactive), which is power that does not itself produce
effective work, yet at the same time it is absolutely
necessary in power systems. Sounds like a contradiction, I
know. Alternating current (and voltage for that matter)
swings from positive to negative values, and the phase
relationship between voltage and current gets shifted
slightly due to inductance (i.e., magnetic field issues)
and capacitance (essentially a charge buildup that in its
most basic form is somewhat similar to battery effect).
Reactive power (VARs) that is required is affected by the
combined inductive reactance and capacitive reactance for
any given system. Long transmission lines therefore
amplify these factors, since there is logically more
inductance and more capacitance over longer distances.
Heavy motor loads (and transformer loads) on power systems
introduce large inductive loads (loads that require
generation of magnetic fields to operate). Consequently,
power systems must compensate for this inductance which
forces a phase shift between current and voltage. This is
accomplished by adding extra capacitance to bring things
closer to being phase for a more efficient power factor
(ratio of active power to apparent power). Inductive loads
cause the current to lag and voltage to lead, while
capacitive loads cause the current to lead and the voltage
to lag. So, the two are “combined” to get the most
efficient results and the targeted power factor.
When a power grid experiences interruption by distant loads
dropping off line, if the transmission line is still
charged the capacitance in the system that is always being
generated by the potential between transmission lines and
ground (or even with other lines) tends to build up much
like when you walk across a carpet and develop voltage that
is not being used. Then, when you touch a grounded object
the unusually high voltage discharges. This might be a
poor example, but the principle proves that devices—
including wires—can “float” up to charges even when a
generator is not directly causing this. This situation is
compounded in major transmission lines where separate
installations by the power company and even certain
customers’ installations specifically add capacitance to
counter what is normally heavier inductive loading.
Why is reactive power necessary and yet it technically
performs no effective work? It is what creates the
magnetic field that allows motors and transformers to do
their work. The magnetic fields are not collapsing and
consuming constant power to re-energize them—instead, they
use reactive power in what is essentially a closed circuit,
two-way street, give-and-take, with the power plant. A
full study of reactive power would be required to
understand this. As a powerplant operator (a new hire), I
have to adjust reactive power on my generators to
compensate for situations electrically downstream. On one
hand the system produces megawatts which is billable
electric power, and on the other hand we provide megavars
(billed to no one) which is something akin to priming a
pump but it is not the work produced for or by the pump.
By the way, buried cables have much higher capacitance than
overhead lines, and these systems introduce special
reactive power considerations. Consequently, the Ferranti
Effect is much more an issue with lengthy underground
transmission lines.
| Is This Answer Correct ? | 23 Yes | 9 No |
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