Question { 3134 }

In 6600/√3/110/√3v voltage transformer,how much output
voltage in open delta transformer,after removing 1-phase of
primary side on transformer?

Answer

The open-delta connection it is actually 2 single phase transformer in series.
The voltage between two high voltage terminals it is 6600 V and each secondary
phase presents 110/√3 V.
If all three H.V.phases are connected the voltage between secondary terminals[in series] is 110/√3<0o -110/√3<120o=110 V<30o .
If the removed phase it is an extreme one then only one transformer stays connected [two voltages at ends] and the voltage is 110/√3 V.
If the removed phase it is the middle then both transformers stays connected with only one 6600 V and the voltage is 2*110/√3/2<0o=110/√3 V.

Question { 3134 }

In 6600/√3/110/√3v voltage transformer,how much output
voltage in open delta transformer,after removing 1-phase of
primary side on transformer?

Answer

The open-delta connection it is actually 2 single phase transformer in series.
The voltage between two high voltage terminals it is 6600 V and each secondary
phase presents 110/√3 V[ some time even 100 or 110V].
If all three H.V.phases are connected the voltage between secondary terminals[in direct series] is 110/√3<0o +110/√3<120o=-110/√3 V<240o or oppositely connected 110/√3<0o -110/√3<120o=110 V<30o .
If the removed phase it is an extreme one then only one transformer stays connected [two voltages at ends] and the voltage is 110/√3 V.
If the removed phase it is the middle then both transformers stays connected with only one 6600 V and the voltage is 2*110/√3/2<0o=110/√3 V[or 110 V] for direct connection or 0 if they are oppositely connected.

Question { 5110 }

difference between diode and scr?

Answer

See:
http://www.onsemi.com/pub_link/Collateral/HBD855-D.PDF
and:
http://people.ee.duke.edu/~cec/final/node85.html

Question { ABB, 7236 }

what effect on electrical equipment if PF is gone over 0.996
in leading when provider say that its keep in minimum
0.996lagging

Answer

The interruption of a capacitive current can cause
dielectric problems for the switching device. The high
inrush currents can cause damage to the capacitors of the
capacitors bank and to weld the contacts of the switch
together.See[for instance]:
http://www.icrepq.com/ICREPQ'09/480-kulas.pdf

Question { PSU, 6499 }

why current is leading in capacitor?

Answer

In a close circuit [in a loop] the current flowing in the
conductor produces a magnetic flux [F]. F=L*i [L=circuit
inductance, i=instantaneous current intensity].
In a.c. the flux will change from maximum positive to
maximum negative and again to maximum positive-ideal sine
wave form-in a full wave time-and in turn will produce a FEM
[equal to supply voltage-if we neglect the conductor
resistance].
v=L*di/dt -current derivate in time.
If i=Imax*sin(w*t)[ considering i=0 when t=0] w=2*pi()
*f f=frequency[50 or 60 Hz]
Simplifying: d(sin(wt))/dt=w*cos(wt) and d(cos(wt))/dt=-
w*sin(wt).
v=L*w*Imax*cos(w*t) if L*w*Imax=Vm then:
v=Vm*cos(w*t)
Since at time 0 voltage v=Vm but i=0 and i=Im will be
only after 1/4 of a wave , voltage leads the current.
In a condenser -capacitor-an electric field is installed
[if in the circuit is a voltage supplied ] and the plates
will be charged in short time accumulating the current. In
d.c. after the charge is finited no current will flow
further.
In a.c. a steady state current will flow following the
voltage variation.
From the relation :Q=C*v Q=electric charge C=Capacity
V=voltage
By definition i=dQ/dt then i=C*dv/dt
since v=Vm*cos(w*t)
dv/dt=-w*Vm*sin(w*t) i=-C*w*Vm*sin(wt) .
At t=0 i=0 but before this at wt=-pi/2 rad[-90 dgr] i=Im
that means current leads[the voltage].
90drg. lead=360-90=270 dgr. lag., of course. Conventional
we take only up to 180 dgr. for the angle between voltage
and current of the same phase.

Question { 6402 }

how to calculate Transformer loss?

Answer

Transformer losses are winding losses[copper or aluminum
conductor losses] Pcu and steel

laminate core losses Pfe Ptot=Pcu+Pfe

If you would know the resistance per phase at working
temperature[R] you could calculate winding losses
Pcu=n*Irat^2*R [n=no.of phases;Irat=transformer rated
current].

In order to calculate laminate core losses you need to know:

unit losses of the steel laminate[w/lb or w/kg],total core
weight [lb or kg].

Pfe=[w/lb]*weight[lb].

In order to state the unit losses[from Manufacturer
Catalogue] you'll need magnetic flux density[B[wb/m^2 or
gauss/inch^2] and the current frequency[50 Hz or 60 c/sec].

Question { 10728 }

what is the major difference between dc generator and
synchronous generator?

Answer

Both transform kinetic energy[mechanical energy]of a primer
[D.G. or turbine] into electrical energy. But d.c.generator
will generate direct current and the synchronous generator
generates alternative current.

Both have d.c. supplied poles-d.c.gen. on stator synchron.
on the rotor.

Actually, by rotating the winding in the static magnetic
field as in d.c. gen. or by rotating the poles[that means
the magnetic field] as in synchron. we get alternative
electro-magnetic force[voltage],but in d.c. generator a
collector-or a semiconductor bridge-rectifies the current
and we get d.c. current

Question { 10728 }

what is the major difference between dc generator and
synchronous generator?

Answer

In three phase system the 3 phases windings are even
distributed along the stator periphery and since the 3
phase currents are 120 electrical degrees in time
distanced one from each other ,then an apparent rotating
magnetic field is installed[as an armature reaction].

The synchronous generator rotor rotates in the same time
with this field [stator itself does not move]. [differently
from the asynchronous[induction] generator in which rotor
velocity is more elevated than this field velocity and the
difference depends on the load.]

In d.c. generator no such rotating field is generated as
the armature reaction is always in d.c. field and it is
rotates with the rotor together[one could say it is
also "synchronous" rotation].

However this field disturbs the collector operation in the
commutation process.

In order to avoid this, commutation auxiliary poles are
located between excitation poles and are supplied by
armature current.

Question { 4400 }

Does the
electricity supply
authority send its
customers power
(kW) or voltape
(V)? Asked by Ivan
Muliro

Answer

This question is like this one:

"I get 100 cubic meter water or I get 1 kg/cm^2 [1 atm.
approx.] water pressure?"

I buy kwh but the Utility has to supply it at 415 V[+/-
5%,for instance] in order to use it.

Question { 7621 }

the inertia constants of two machines are respectively 30and
42 MJ per MVA.when the change in frequency due to change in
load in first machine is 12 Hz,the change in second machine
frequency will be?

Answer

If the change in load will be the same Pa1=Pa2 and the
rated frequency fo1=fo2

for the same time interval dt1=dt2

IF Pa=H*MVA/180/fo*d2R/dt2 [See IEEE-399 ch. 8.6.2 How
stability programs work]

and d2R/dt2=2*pi()*df/dt

then (df/dt)1=Pa/2/pi()*180*fo/MVA1/H1 and

(df/dt)2=Pa/2/pi()*180*fo/MVA2/H2

(df/dt)2/(df/dt)1=MVA1*H1/MVA2*H2

If MVA1=MVA2 then (df/dt)2=(df/dt)1*H1/H2

df2=12*30/42=8.57 Hz

Question { 18199 }

Is AC Current Scalar or Vector Component?

Answer

I forgot to mention that a vector field produces a "flux"
an integration of the vectors in a limited cross section
area and this result it is a scalar, indeed. This is the
case of magnetic flux[F-Wb] as an integration of magnetic
flux density[B-Wb/m^2].In this case you may consider the
current as integration of current density field in a
conductor cross section area. Then the "current" is
physically a "scalar" but ac current can be represented on
a paper as "phase-vector" as explained.

Question { 18199 }

Is AC Current Scalar or Vector Component?

Answer

In my opinion, if someone is speaking about "vector"
and "scalar" in the same time he is referring to the
physical definition and not to mathematical-symbolic one.

If "current" is a vector what could be "scalar" in symbolic
[complex number] definition?.

Question { 18199 }

Is AC Current Scalar or Vector Component?

Answer

Vector is a physical notion which means

a quantity that are fully described by both a magnitude and
a direction. Speed, force, electric field intensity,
magnetic field intensity, current density and other are
vector defined by a magnitude and a direction.

With a field of quantities define as vectors you may use
function like gradient, curl and divergence and product as
vectorial or scalar.

Scalar it is a quantity defined complete by the magnitude
as current intensity, magnetic flux ,electrical potential
and other.

In electrical calculation- in alternative current system-
it is employed a quantity defined as phasor or phasor-
vector formed of two parts: one real and one imaginary
quantity.

For instance, the real part of a current could be I*cos(fi)
where fi it is the angle between supplied voltage
[considered as fi=0] and I it is the measured current
[intensity].An imaginary part is annexed in order to
complete the image I*sin(fi).

Then we have three possibilities: only real part-
considering fi=0, only imaginary part considering fi=90
degree or a complex number I*cos(fi)+I*sin(fi)*i i=sqrt(-
1) an imaginary unit.

So, the ac current[intensity] is not a vector and not a
scalar but could be a complex number or a real quantity-
depends on how you'll take the "fi".

Question { 9414 }

why do we connect voltmeter in parallel,why not in
series,and why do we connect ammeter in series why not in
paprallel...

Answer

A voltmeter measures the potential difference between 2
points of an electric circuit and completes the circuit
which is open between this to points that means
is "parallel" with remaining elements in circuit.

An ammeter measures the current flowing through the circuit
so you have to break the circuit and put the ammeter "in
series" with the circuit elements.

Question { 11627 }

What is basic impulse insulation level? And how it is calculated for sub-
station?

Answer

IEEE Std 141/1993 ch. 6.3.2 "Insulation tests and ratings"
specifies:

"Electrical power and distribution apparatus assigned a
given insulation class should be capable of withstanding,
without flashover or apparent damage, a 1.2/50 full-wave
impulse test of specified crest kV. This specified crest
voltage is the basic impulse insulation level (BIL) of the
equipment."

The BIL is not calculated but is stated by standards.