The only way is through electromagnetic interference from the 13.8kV. If your 13.8 kV cables are shielded with the shield grounded at one end only, it shouldn't be an issue to electronics. If shielded and both ends are grounded, potential sheath currents through the ground can influence the...
Good point. It would be impossible to pull cables through conduit with all of those bends. The ceiling runs could be Teck cable. The ones on the wall look like EMT.
I am using the ABB AMVAC-27 in an application. The tripping is done via an under voltage coil in the breaker fed via 120VAC (UPS Powered). The tripping is done by an 86 LOR relay that opens the 120VAC to the UV coil.
To Open or Close the breaker, there are 120VAC Close and Open Coils in the...
For utilities, there is 'A' and 'B' fully redundant protection. Each relay has the same protection functions and 'B' relay backs-up 'A'. There are also two tripping circuits for each breaker and separate CT's for each of 'A' and 'B'. They also use different manufacturers in case there is a...
Try a software package called Elecdes CAD Design software. You can generate schematics and automatically generate wiring diagrams from the schematics.
If it's just SLD's, try Easy-Power or SKM. All component data is stored and can be displayed as you wish on the drawings. These programs are...
Looks like a lot of work for the contractor to install all of the EMT bends. It's probably so there won't be shadows from the lights. Usually, the EMT is run straight and the lights hung a little below.
If the system is ungrounded, fault current will be based on capacitive coupling between ground and the ungrounded neutral. The capacitance will determine the amplitude of L-G fault current. A low capacitance value will result in higher capacitive reactance Xc = 1/jwc. Therefore it will result in...
A way to visualize this is to consider that the torque angle of a generator, which determines the real and reactive power generated in the stator, is determined by the advance in degrees of the field poles relative to the stator poles. If you increase power input with generated voltage constant...
To increase Vs or terminal voltage at the generator, you must increase the generated voltage (ie: Field) however with a constant power input from the prime mover, this increase in excitation voltage will decrease the torque angle and therefore the real power output of the generator will remain...
If you refer to IEEE 1584, there is an equation that calculates arcing current in air (Ia) using the variables of voltage, gap between conductors and 3 phase bolted fault current (Ibf). It is part of the calculations for incident energy Ie measured in cals/cm squared.
From the arcing current...
