Forum Replies Created
-
Posts
-
“The issue is the elements are apparently tripping alarms on the ship because they are showing resistance to ground.”
Probably the circuit breaker (if they were using one instead of fuse) exceeded its maximum current carrying capacity because the rating of that “tubular heating elements” exceeded the rating of the circuit breaker. They could have solve the problem without replacing that “element” unless that “element” was already defective. I could tell you how but I don’t know the specs of that “tubular heating element.”
Hello,
The process is :
1º Calculate de current in the circuit
2º Get a cable manufactures Current rating tables for the type of cable that you will use
3º Choose a cable with a current rating bigger than the circuit current
Iz (Imax from the cable) > Is (circuit)
you must be aware that the value in the tables are usually for 20ºC or 30ºC, so youi may have to correct the value, be also aware
that te value taken form the table is for a single cable alone, so if you have more cables together in the same conduit you must
correct the value.
4º Choose a protecion for the circuit
In > Is and In*1,45 < Iz*1,45
5º Verify that the length of the cable is not a problem for the Voltage level in the end of the circuit.
This is only some guide lines…..
Hi ,
what u calculated its currect but how taken cable size
which method u used for calculating the cable size
Excellent topic. I jus want to add something that I guess is important, maybe you already talked about. It is very important to have a good grounding system , otherwise the SPD is not going to do its work properly, and for those star systems it is important to make a reference at the entrance of the system (Neutral to earht conection) in order to minimaze the transients.
akhilesh said:
if new method u know then plz send msg me.plz…………..dears………
Power factor is that of the load & not of the alternator.Therfore there is no question to improve the pf of the altertnator. Off course the power factor at which the alternator is feeding the load can be improved by reducing the RMVA loading of the alternator.
Hi,
P(kW) = HP x 0.746
I = P *10^3 / (1,73*400*0,8)
P=5,6 kW => I=10, A ; wire 2,5 mm2, Circuit breaker In=10A
P=7,5 kW => I=13,5 A ; wire 2,5 mm2, Circuit breaker In=16A
P=11 kW => I=19,9 A ; wire 4 mm2, Circuit breaker In=20A
P=15 kW => I=27 A ; wire 6 mm2, Circuit breaker In=32A
P=18,6 kW => I=33,6; ; wire 6 mm2, Circuit breaker In=40A
P=57,7 kW => I=104,2 A ; wire 25 mm2, Circuit breaker In=125AAn other big advantage is that one doesn’t need Cord End terminals to end a flexible (stranded) cable.
Flexible cable in a screw clamp will losen after a while because the thin wires will reorganise themselves. Especially under influence of vibrations. In Belgium it’s even not aloud to use flexible cable in a screw clamp without Cord End terminals.
Flexible cables in cage clamps give no problem since the spring permanently tightens the cable.
Cage clamps are also faster, but they are more expensive.
I have only used WAGO cage clamps (up to 95mm²) and have had no negative experience with them.
that’s right dude! you’ve just posted a better explanation.. :)
akhilesh said:
if new method u know then plz send msg me.plz…………..dears………
to decrease excitation voltage of the alternater
Hello,
Step down, Step Up…….
It is a transformer, it has a primary and a secundary, two levels of tension and same power (if excluded the losses in the iron core and in the wiring), so if you have the same power and diferent tension levels, off course the wires are thicker in the side of the lower tension.
It is a reversible machine, you must respect the tension and current rating, it doesn´t mather if you use it as a step up or a step down transformer.
You get a transformer off let’s say 15kV/0,4kV 630kVA and conect a load of 630 kVA (many loads used in 0,4kV) using the transformer as step down, than you get in the secudary (0,4kV) 909 A .
Later on you decide to use it as a step up transformer, then you conect 0,4kV to the former secundary, now is the primary (primary, secundary it is not the level but were you conect the power input and power output), you also conect a load 630kVA in the secundary (15kV) side, off course it can not be the same load used in the step down operation, same power but diferent level of tension. Then you get in the secundary (15kV) 24,2A.
Same power transfer from primary to secundary (630kVA) working as step up or step down, same efficiency, same insulation problems, same losses.
Hello!
You can draw load from the 33kV side but take note the current is much lesser and thinner wire compared to the secondary side.
Since it is a step down transformer the load side is the secondary, the 11 kV. Thus, it has thicker conductor in it (higher current rating) intended to handle a load.
In addition, you could not enjoy the full capacity of the transformer if you use the primary as the load side.
Anyway, the answer for your question is “YES”. :)
> http://www.biricreations.blogspot.com
SEREYBOTH said:
Hello,
I am cambodia.I am not well in writing English,working at Siem Reap diesel power plant(10MW).Now my country has many lightnings during raining that kills people, damages eletrical overhead transmission line(22kv) and low voltage electric devices.
could you help me, how protection,how to get a good skill in surge protection ! ! !
Dear Cambodia,
I,m dinesh kumar represting an XXXXXXX, we here provide Power Quality solutions to such a type of poor line i.e. surge solution.
kindly send your problem in detail with Electrical Pannel size.
We will surely provide u solution within time frame.
rgds
Dinesh
XXXXXXXXXXX
(This comment is edited by moderator team)
(Please note that any publicity without added value is deleted. Thank You!)
For Parallel Circuit, the general formula is:
1/Rt = 1/R1 + 1/R2 + 1/R3 + ….+ 1/Rn where “n” is the number of resistor connected in parallel. This means that the reciprocal of the total resistance is equal to the sum of the reciprocals of the given resistances. For two resistors, 1/Rt = 1/R1 + 1/R2. Simplifying this equation, Rt = (R1*R2)/(R1 + R2). For three resistors in parallel, 1/Rt = 1/R1 + 1/R2 + 1/R3. Also equal to Rt = R1*R2*R3/(R2*R3 + R1*R3 + R1*R2). And so on…
