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the solutions are as follows
Divide the building load into two halves – supplly each halves with individual generator of 20% extra KVA and utiility service provider with ATS
Use Bus risers for very large buildings UPTO 20Floor and above.
Laurent said:
@all other visitors: please share also your experience here with others, it will help to highlight the risks related to electrical short circuits
Shorts cause many issues, One that I was privy to occurred in a large factory. During the installation of cables onto a cable tray one of the large supply cables suddenly pulled loose from the tray, and jumped approx 5 feet to the side causing a worker to receive a crushing blow to the side of his body, The force was sufficient to break the collarbone and cause additional injuries to the shoulder and arm regions. on investigation the cause turned out to be twofold.
1. A high fault current due to an insulation failure at the end of the cable run (not close to where anyone was working) created a large magnetic field arround the cable causing the cable to act like a solinoid, and move to the side, with sufficient force to cause injuries to the workers in the area.
2. The cable had not been secured sufficiently on original installation, and over many years the clips securing it had degraded and left the cable insecure.
Testing verified that electrical short circuit protection operated correctly, and that the supply had in fact been isolated inside the traditional 0.4 seconds required by our legislation. Subsequent testing and remedial action, also identified most of the tray earthing had failed over the years, and the tray was in very poor overall condition. It was then replaced, the earths renewed and the cables secured.
People often overlook the fact that high currents can cause mechanical forces withing cables, and all cables in the proximity of other cables (especially in Multi-Phase installations) must be physically secured to prevent the mechanical forces from causing sudden movement of the cables in the tray.
Pl check whether the same problem exist in your neighbour also. In that case you have to contact your distributor of supply otherwise install automatic voltage regulator at your end.
duclampv said:
could you explain to me the conditons of 2 transformers operate parallel?
Parallel Operation of Transformers
The essential conditions for successful parallel operation of transformers are:
I. Transformation or turn-ratios and voltage ratings should be the same.
II. Polarities of the transformers should be the same.
III. Percent impedances of the transformers should be the same.
IV. Ratios of resistance to reactance should be the same.
V. Phase displacement between primary and secondary windings of the transformers should be the same.
VI. Phase sequences or vector group of the transformers should be the same.
ally said:
Dear All,
I need your help on the following matter
I always did electrical installation design in industries, commercial and residential buildings.
The problems comes when i want to choose the distribution voltage.
What i did is to select the nearby distribution system either 33kV or 11kV and stepped it down to 0.415kV.
I NEED A FORMULA ON HOW TO CALCULATE THE REQUIRED DISTRIBUTION SYSTEM TO BE USED AND THE
CRITERIALS USED TO SELECT IT.
Anyone with clear details need your help. ( i will be happy if Egineer GHALI will participate on this)
Regards.
To begin with 33 & 11kv are high voltages,
you can use these voltages in secondary transmission and distribution networks in a power system depending on your supply utility voltage level(330/132/66/33kv for zambia's primary,secondarytertiary transmission networks and distribution networks).
33kv is handled via overhead lines and is economic for such while 11kv depends on application and costs of its transformation to other voltage levels.
Why and where to use any voltage level depends on the voltage level requirements of your drives,have worked with a 2.5MW Compressor at 11kv voltage.
The main worry is the cost of insulation on your equipment which makes the cost high all together .
for 0.4kv, this is used for consumer purposes mainly and few industrial machines use the system voltage maybe single phase motors and for lighting purposes.
The other worry is the switchgear to use for a particular voltage level e.g you need a VCB or ABCB or SF6 for 33kv which is expensive and VCB or OCB for 11kv with VCB and ABCB being expensive to buy and replace with additional maintenance costs for ABCB.
anant said:
i want the ans for this question
anant said:
i want the ans for this question
you need to make sure you have a complete signal cable from plc to sensor or actuator or what you are controlling.
ensure you use 0.7 0r 0.5 or 1 sqr mm signal cable and avoid making connections on one cable.
as same problem with me when we installed the plc to the system,
we used the small battery as the back up power of the plc,
the battery nominal volt is depending the plc voltage,
or you can used Capasitor with big uF,
hope this will solved the problem
My name Istvan fro Hungary and I met with this problem in my practice. If the protection are designed well . look aroud the installation. The output of the MGE ups (tree phase ) mostly has a neutral in the output , and there is an earthing . One case a costumer led the tree phase and the earthing from the UPS to his devices , and the star connetted output transformer's star point were not conneted. There arised a virtual star point , the earthing wire , and the different phase voltage set by the ratio of the the loads by phases . The line voltage were constant, only the phase voltage were changing between 100 and 500 V . The result was same what You described.
The main function of power supply is to convert AC to DC; as shown in fig 1, the first stage is to make full wave rectifier to the AC signals by using bridge rectifier, filter the rectified wave by using filtering capacitor and finally select the appropriate voltage regulator to generate pure DC signal.
Current rating for bridge rectifier should be suitable with load current, also; the designer should consider the voltage drop across each diode, which is normally equal to 0.7v. Note that only two diodes are required when usingcenter-tapped transformer.
The following formula is used to calculate the capacitance value for the filtering capacitor:
. . . (1)C: Capacitor value.
Vp: Peak voltage. (“Bridge output max voltage”)
f: Frequency of the AC supply.
i: Load current.Note that the above equation for 10% ripple voltage.
Tip:
It is possible to build your power supply without adding a regulator, when you want to power up an electrical motor, a small ripple in voltage will not affect the performance, unlike the electronic circuits and devices!Designing of +12v 5A Power Supply:
To calculate the voltage required and the transformer secondary winding we first determine the input voltage for the regulator, which is 15v, plus a 10% of this value for ripple. For a regular transformer we have to consider a bridge rectifier, as a result; we will add 1.4v. So the secondary winding should be 15+1.5+1.4=18.9 lets say 18v @ 5 Amps. Now we will calculate the capacity of the filtering capacitor. By using equation number 1 and assuming that f=60Hz, we will get C=5 x 5 / ((18-1.4) x f) =25,100µF
you have to make the bulding two seprate parts every part take power from one geneator,
that means you connected 11 storey with one generator and ather 12 storey with another
. . . (1)