Question { 3895 }

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

Something is not correct. If VL_L=415 V then VL_N=240 V
[not 230 V] and then

both calculation results will be the same [no error].

Question { 5972 }

can anyone pls expain me about how can we design a cable
tray??suppose if have to draw a 3 nos of 50 sqmm AYFY and 2
nos of 35 SQmm of AYFY cable through the tray..how can we
design the tray size?? regards...

Answer

For open top solid bottom cable tray and without any
clearance between cables the cable carrying capacity
has to be reduced to 73%. However the tray width will be
only:
0.3*34+34+34+34+30+30+0.3*30=181.2 mm
3*50+ 117*.73= 85.4 A ; 3*35+ 96*.73=70 A

Question { 5972 }

can anyone pls expain me about how can we design a cable
tray??suppose if have to draw a 3 nos of 50 sqmm AYFY and 2
nos of 35 SQmm of AYFY cable through the tray..how can we
design the tray size?? regards...

Answer

The cable tray is an assemble of tens or hundreds parts.
There are a big number of types and materials involved. You
may get a look in IEC 61537 or NEMA VE-1 or VE-2.

The simple way is to take the construction drawings and to
make a sketch of the cable tray run including the all
derivation on this drawing and then to send this to some
tray manufacturer as B-Line [for instance, you can find
many of them in web or Google].

Question { 5972 }

can anyone pls expain me about how can we design a cable
tray??suppose if have to draw a 3 nos of 50 sqmm AYFY and 2
nos of 35 SQmm of AYFY cable through the tray..how can we
design the tray size?? regards...

Answer

You intend to run 3 cables of 3*50 mm^2 and 2 cables of
3*35 mm^2 type AYFY, I guess.
It depends on type and width of the cable trays. Let’s say
you want to use an open top and ventilated tray [ladder or
well perforated plate] and you want to carry the maximum
possible currents. Then, according to IEC 60364-5-52 Table
A.52-17 Note 2-you have to keep two cable diameters
clearance between cables and at least 0.3 time cable
overall diameter from the wall-both ends.
A 3*50+25 0.6/1 kV [according to IEC 60502-1] aluminum
conductor [compact], pvc insulated, steel tape armored and
pvc serving overall diameter 34 mm and for 3*35+16 overall
dia= 30 mm.
Here is the width calculation:
|0.3*34|O|2*34|O|2*34|O|2*34|o|2*30|o|0.3*30| where O =one
3*50+ cable diameter and o=one 3*35+ cable dia.
That means:
0.3*34+34+2*34+34+2*34+34+2*34+30+2*30+30+0.3*30=445.2
Then the tray width will be more than: 445.2 mm.
In 30 dgr.C air permissible current-carrying capacity will
be:
[See Table A.52-11 – Current-carrying capacity in amperes
system E aluminum]
[Or Tableau B.52-1 for aluminum col.7]
3*50 117 A
3*35 96 A
You may reduce the cable tray width but you have to reduce
the current in order to limit the conductor temperature to
70 dgr.C.

Question { 2960 }

prove that inegration of { e^z / z^2 + pi^2 }dz is equal to
i/pi by Cauchy Residue Theorem.

Answer

Then pi=3.14159... and i=sqrt(-1), I guess.I don't know if z
is a real or complex number and if there is a pole
somewhere.

Question { 4294 }

I have 2 transformers in two different plants, both are
without OLTC but i am facing problem of voltage fluctuation
at LV side from only one transformer.
i have to change its taps manually twice in a day.

Another transformer delivering voltage is ok.

Please guide me about the problem

Answer

Voltage fluctuation it means voltage drop. It could be a
lot of factors producing big voltage drop.
The voltage drop in transformer -transformer size does not
fit the load and then at load peak the voltage drops under
the permissible limit and in less-or no load-the voltage
raises up to the secondary EMF [no-load voltage]- mainly if
there are many motor starts per hours.
If the supply-medium voltage-is variable-medium voltage
cable too small.
If the feeder from transformer to switchgear is undersized
the main voltage drop could be here.
A complete analysis could find the cause.

Question { 4974 }

we use Dyn11 trafo for distribution purpose, if it`s possible
for replacing Dyn1. if it`s change means what will happen?

Answer

If this transformer works in parallel with other
transformer of the same connection type it is not possible
–the angle between supplied voltage will be 300 degrees [or
60 degrees] and will provoke a short-circuit equal to
Vstator/2/Ztrf .
If the transformer works alone it will be ok [no any
difference]

Question { 5246 }

Why in KVA diesel engine? Why not KW

Answer

Sheik answer is almost correct with some remarks:
Diesel Motor rating is in Horse Power [HP] since this is a
motor. The DG-Diesel Generator -the generator rating- is in
kVA [the actual [active] power depends on p.f. [cosfi] and it
is measured in kW indeed. This power is variable depending on
required –and produced power but the maximum will be for p.f.
=1.Then the kW=kVA .The rated power is the maximum output
power which in indefinite time will not damage the generator.

Question { Vizag Steel, 15789 }

two transformers of the same type, using the same grade of
iron and conducting material, are designed to work at same
flux and current densities, but the linear dimension of one
are two times of the other in all respects. The ration of
kVA of the two transformer closely equal to................?
options: 16, 8, 4, 2

Answer

According to Rudolf Richter “Electric Machines” vol III
Transformers the rated apparent power will be n^2 times this
of the smaller dimensions [So]:
Sn=n^2*So that means for twice in dimensions the rated could
be 4 times the rated of the smaller.

Question { 4921 }

Hydro P.P. run @ 600 rpm,8pole, Nucler P.P @ 1500rpm 4 pole ,
ThermlP.P @ 3000 rpm 2 pole.Why this diffrence??? when we can
generate electricity @ 600 rpm then Why we had incresed speed
in NPP & TPP???

Answer

The synchronous a.c. generator rpm[rotations per minute]
depends on the mechanical drive rated rpm –Diesel Motor
could be high speed –for 60 Hz systems up to 1800 [I did
not see 3600] and the lower limit may be 100-150 rpm.Low
speed D.G. consume per kwh is lower so for 40 years will be
a huge advantage.
The dimensions [and weight] of the motor-and of the
generator too-decrease as the speed increase [and the price
also].The turbo-generator is 2 poles generator usually as
this is the minimum equipment and installation cost.

Question { 4921 }

Hydro P.P. run @ 600 rpm,8pole, Nucler P.P @ 1500rpm 4 pole ,
ThermlP.P @ 3000 rpm 2 pole.Why this diffrence??? when we can
generate electricity @ 600 rpm then Why we had incresed speed
in NPP & TPP???

Answer

In my opinion, the fuel consumption depends-mainly- on the
load and the losses. Most of losses are copper losses and
iron core losses.
These losses does not depends on rotor velocity directly-
but since the rotor speed will increase ventilation and so
increase the cooling one may increase the current for the
same conductor cross section and so the losses could be
more but as a percentage -referring to the load-will be
less. Only bearing friction and ventilation will be more.
However as the total efficiency is higher for higher
velocity these losses are not significant. In no way the
consumption will be more-usually less-for turbine drive
generator. For Diesel-Generator will be more, a little, but
this still does not justified the equipment and
installation elevated cost.Diesel Motor low-speed can use
cheap oil-slow-burning oil.

Question { 3126 }

in short circuit. test of a
transformer 5 to 10%
ofvrated voltage is used why??? if
we use rated
voltafe then what will be?

Answer

The 5-10% of rated voltage represents the short-circuit
voltage that means if you connect all the secondary
terminals of the transformer together-it is a “short-
circuit”. If you supply the primary winding terminals this
short-circuit voltage you’ll get the rated current-primary
and secondary.
If the supply voltage is the rated voltage you’ll get the
short-circuit current [10-20 times the rated].

Question { 6179 }

what is the ohmic value of transformers

Answer

I think it is the active [real] part of the transformer short-
circuit impedance.
Let’s say uk% it is the short-circuit voltage-4-25% percent of
rated voltage supplied at primary terminals when secondary
terminal are short-circuited and the current will be the rated
current. Then the short-circuit impedance Zk=R+jX
=uk%/100*Vrated/sqrt(3)/Irated –for 3 phases transformer. If
we’ll know kxr= X/R [6-40 ] then X=Zk/sqrt(1+1/kxr^2)
and R=X/kxr.

Question { Engro, 6196 }

10Hp, 3-phase, 415v, 50Hz induction motor.
The above motor is to run on Star-Delta connection, so how
do i make following hardware selection
1) MCB, 2) Contactor, 3) Wire Gage (Depending on P-P current
or Line Current), 4) OLR,
Please tell in detail how to do the above, so as to get the
ans. of it even if the Hp rating changes.

Answer

10 hp=7.5 kW. The manufacturer has to indicate the starting
current.
Let's say the motor is designed for star/delta starting
then according to IEC 60034-12 Tab2 for 6.3 kw up to 25 kw
S1=12 then maximum permissible starting current will be
Istart=12*7.5*1000/1.73/415=125.2 A.
The short-circuit protection has to be set at
1.25*125.2=156.5 A.
The circuit breaker rating has to be 150 A. The motor
overload has to be set at 1.25*Irated and also will trip
the breaker if the current of 2*Irated will last more than
the permissible stall time [let’s say 4 sec].
The supply cable allowable current [see IEC 60364-5-52]
shall be 1.25*Irated.
Usually a 7.5 kW motor at 415 V rated current will be 13-16
A then the cable permissible minimum should be 20 A.
According to the above mentioned standard 2.5 mm^2 copper
conductor PVC insulated will be ok.
You have to check the voltage drop also: no more than 3%
rated and 10% at start. If DV=1.73*I*[R*cos(fi)+X*sin
(fi)].R,X =total resistance-at 70 dgr.C- and reactance
[ohm].
cos(fi)=0.85 [approx.] rated and 0.3 at start. Sin(fi)=
(1-cos(fi)^2)^0.5.
In this case if the cable length will be 80 m 4 mm^2 copper
conductor has to be employed.

Question { Enercon, 5880 }

what do you mean electrical earth pit?

Answer

Usually a vertical grounding electrode –made of copper,
copperweld steel, stainless steel or else-and length of 3-20 m
is driven in earth and the upper end will be exposed in a pit
covered by a lid- about 0.3-0.6 m deep and 0.3 m diameter.
This pit facilitates electrode connection with horizontal
grounding conductor and as a check point of electrode state-
grounding [earthing] resistance measurements and other.