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  1. mard57

    Can being in the same room as 13.6KV damage electronics?

    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...
  2. mard57

    33/11 kv substation

    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.
  3. mard57

    ABB AMVAC in California and how do you do the control design

    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...
  4. mard57

    Protection System

    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...
  5. mard57

    Operational Single Line Diagrams for Distribution Utilities

    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...
  6. mard57

    33/11 kv substation

    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.
  7. mard57

    Arc impedance

    Not sure. You'll have to do some research yourself.
  8. mard57

    Arc impedance

    European Standard EN 50110: Operation of electrical installations The IEEE 1584 calculations should be the same in the European Standard.
  9. mard57

    Danger of electrocution from ungrounded system

    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...
  10. mard57

    Power transfer question

    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...
  11. mard57

    Power transfer question

    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...
  12. mard57

    Arc impedance

    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...

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