Question { MEIL, 5778 }

For a 500 MVA 400/220/33KV transformer following data given
HV power = 500 MVA
IV power = 500 MVA
LV power = 5 MVA
If input power is 500 MVA then how output power is
(500+5)MVA.
Can anyone help

Answer

If the tertiary winding is connected in Delta only in order
to cancel the current unbalance in steady state no current
will flow through it, that means no power consuming. See
[for instance]:http://www.electrical4u.com/tertiary-winding-
of-transformer-three-winding-transformer/

Question { 6763 }

How to calculate incomer of a capacitor bank/feeder from its
KVAR,
AND how to calculate the copper busbar for rated incomer,like
for 400A we calculate the 40x10 of aluminum busbar.

Answer

If the system is three-phase: KVAR=sqrt(3)*Icap[A]
*Volt/1000 then:

Icap=1000*kVAR/Volt/sqrt(3)

If the ambient temperature will remain the same and if
permissible conductor temperature will be the same then
copper bus bar cross section required will be 70%[approx.]
from Aluminum bus bar.

Question { 3885 }

What will happen to transformer if we gave 5HZ input instead
of 50/60Hz??? explain with reason

Answer

Since V=K*B*frq [V= primary supply voltage -approx.
equal to E-emf]; B=magnetic flux density ; frq=frequency]
then if frq=5 Hz and V=constant B[5Hz]=10*B[50 Hz].The no-
load[excitation] current will be more than 300% rated
current.

You have to reduce the voltage to 10% from rated at 50 Hz
in order to get the same B.

Question { Bhel, 27528 }

how much current capacity of 300 sq. mm cable (4 core
aluminium) and 150 sq. mm cable ( 4 core aluminium)

Answer

It depends on lot of factors. The duration of the load:
short-time[as in short-circuit case],continuous load,
intermittent load. It depends on also on ambient type [air,
underground-direct laid or in concreted duct],on thermal
resistivity of the ambient [if it is in underground],on
ambient temperature, on insulation type[bare, pvc, xlpe,
epr etc.] and other factors.

Let's say in short-circuit case for 1 sec 300 square mm
aluminum conductor pvc insulated. Permissible temperature
[for PVC short-time]= 150oC.Ambient air 30oC.
Ipermissible=27 kA[27000 A].

For the same cable in open air 30oC[NEC Table 310.15(B)
(17)] continuous load [75 oC]600 MCM=300 sqr.mm permissible
545 A

Question { 3573 }

What happens instead of using 3 phase, if we give loop of
single phase to all three phases ??

Answer

If the supply system was 3 phase with neutral the single-
phase consumers [lighting, single-phase motors and so on]
could work [you have to check the other two phase are
disconnected at first].The current on the remaining phase
will rise since all load will stay on this single phase [it
depends on how was the load distributed per phase]. The
three-phase motor will not work at all-no start- no current
for ungrounded stator winding and high current for grounded
one].

If the supply system was 3 phases without neutral nothing
would work [you supply one-phase without return through
neutral].

Question { Rasandik, 4259 }

How to Select transformer for 100HP
motor plz... With calculation

Answer

See:

https://www.relayspec.com/catalogs/002050/2006/11-12.pdf

Question { 3244 }

in GEAFOL Cast-Resin trnaformer of 11/.72kv which has
tapping of 11550v,11250,11000,10246v in primary side. To
obtain 700v. which tapping should be used of HV side as we
have 11000v/720v tapping now .

Answer

If the primary voltage [supplied by Utility] it is NOW 11
kV and your transformer at no-load shows 720 V at its
secondary terminals the ratio primary/secondary voltage is
11/0.72=15.278. If you want to reduce the secondary to 700
V you have to change the ratio to 11/0.7=15.714.

Let's say the rated secondary voltage [at no-load] is 720 V.

That means the ratio 11250/720=15.625 it is the most
suitable and you'll get 11000/15.625=704 V at secondary
terminal [at no-load].

Question { Bhel, 10354 }

Why transformer rating is represent in KVA?

Answer

A power transformer contains two main part –the windings-
copper or aluminum conductors insulated –and the magnetic
core.

Magnetic core requires an excitation current in order to
maintain the required magnetic field.

The excitation current depends on supplied voltage and if
the supplied voltage is over a certain limit the excitation
current presents a threshold and start to be very high.
This voltage is the rated voltage.

On the other hand, the conductor insulation rated
temperature is limited-according to insulation class-since
above this limit the insulation will be damaged.

The temperature arise depends of the current passing
through conductor and producing heat losses. These losses
depend only on the absolute value of the current and it is
not connected to the active load [kw=volt*current*p.f. then
the maximum kw it is when p.f.=1 then kw=kVA].

The product of voltage with current is measured in VA or
kVA=1000 VA or MVA=1000000 VA.

Question { 3363 }

I know the Capacitor value- 12.5KVAR, voltage 440V gas
filled, then what is the formula for calculate the
microfarad value for this 12.5KVAR capacitor

Answer

Q=U^2*w*Cap w=2*pi*frq pi=3.14159 [approx.]
w=2*3.14159*50

kVA*1000=440^2*2*3.14159*50*Cap

Cap=12.5*1000/440^2/2/3.14159/50=0.000205.52 F=205.52 microF

Question { 3363 }

I know the Capacitor value- 12.5KVAR, voltage 440V gas
filled, then what is the formula for calculate the
microfarad value for this 12.5KVAR capacitor

Answer

Explanation:

Q=Reactive Power in VAR

U-voltage at capacitor terminals[single phase]in Volt

frq=frequency[50 or 60 HZ=cycles/second]

Q=U^2/Xcap Xcap=Capacity reactance [in ohms]=1/(w*Cap)
w=2*pi*frequency

Then Q=U^2*w*Cap or Cap=10^9*Q/(U^2*2*pi*frequency) if Q
is in kVAR and capacity in microFarad.

If U is the line-to-line voltage the same formula is good
for three-phases in Y but Q=3*U^2/Xcap for delta connection.

Question { 4061 }

Kindly give the brief explanations for checking 12.5 kvar
440V, 25 KVAR 440V, 12.5KVAR 690V, 25 KVAR 690V capacitors
microfarad values. what is the microfarad values for the
above four type of capacitors, and give the formula
explanations briefly. I know F-50HZ, voltage, and capacitor.

Answer

1)If this is only 1 capacitor and this is the voltage
between capacitor terminals:

Cap=10^9*Q/(Voltage^2*2*pi*frequency) if Q is in kVAR and
capacity in microFarad.

pi=3.14159

Cap=10^9*12.5/(440^2*2*pi*50)=205.5 mF

2)If this is reactive power of 3 capacitors in Y connection:

Voltage at a single capacitor terminals will be 440/sqrt(3)
=440/1.73=254 V

Reactive power of a single capacitor will be 12.5/3=4.167
kVAR

Then the capacity of a single capacitor will be :

Cap=10^9*4.167/(254^2*2*pi*50)=205.5 mF

3)If this is reactive power of 3 capacitors in Delta
[Triangle] connection:

Voltage at a single capacitor terminals will be 440 V

Then the capacity of a single capacitor will be :

Cap=10^9*4.167/(440^2*2*pi*50)=68.5 mF

If Q=25 kVAR Voltage=440 V then the capacity of a single
capacitor will be :

single and Y=411 microF ; Delta=137

If Q=12.5 kVAR Voltage=690 V then the capacity of a single
capacitor will be :

single and Y= 83.6 microF ; Delta=27.85

If Q=25 kVAR Voltage=690 V then the capacity of a single
capacitor will be :

single and Y= 167 microF ; Delta=55.7 microF

Question { 3025 }

what is the principle of switchgear and its function

Answer

In an electric power system, a switchgear is the
combination of electrical disconnect switches, fuses or
circuit breakers used to control, protect and isolate
electrical equipment.

See:

http://en.wikipedia.org/wiki/Switchgear

Question { 14187 }

Is it possible to connect a dYN1 transformer in
parallel with a YNd11 transformer.

Answer

It is not possible since there are not the same phase angle
shift [they have to be in the same group of connections]
See:

http://electrical-engineering-portal.com/principles-of-
transformers-in-parallel-connection-1

Question { 14187 }

Is it possible to connect a dYN1 transformer in
parallel with a YNd11 transformer.

Answer

You can turn DY1 into DY11 by swapping phase B with W.See:

http://files.engineering.com/getfile.aspx?folder=df1b9784-
7409-4333-a537-
780523d888a8&file=How_to_turn_a_Dy11_transformer_into_a_Dy1.
pdf

Question { 3131 }

Hi,

I've just got into my electrical engineering course and I need some help with a question of mine here.

During a site visit, we were shown a main LT panel which has switchgears, ammeters and different kinds of relays and the distribution network from that panel. In the main LT panel I've noted down the current readings as Ir - 163A, Iy - 150A and Ib - 147A. The Line to Line voltage is 415 V at 50 Hz and Line to Neutral voltage is 230V at 50 Hz.

From the above details, how would I calculate the kVA rating of transformer? The primary side voltage of the transformer is 11 kV and the secondary side voltage is 415V and the frequency is 50 Hz.

Answer

The transformer actual load [apparent power] S=0.23*
(163+150+147)=105.8 kVA.

You may calculate approximately taken the average current
(163+150+147)/3=153.33 A and the formula sqr(3)
*0.415*153.33=110.2 kVA [4% error].

But this is not the rated power. The rated power and the
rated current have to be written on manufacturer rating
plates.