C.S.P.3: Lesson #8: Calculation of fault currents

Question 1

why short-circuit calculations are necessary for electrical systems?

Answer 1

They are needed to determine the standard ampere rating of the OCPD.

Question 2

critical point where short-circuit calculations should be completed in a system?

Answer 2

Industrial control panels

Panelboards, motor control centers, transfer switches

Service entrance equipment

Question 3

reasons why short-circuit studies involve calculating a bolted 3-phase fault condition?

Answer 3

Three phases bolted together create a near-zero impedance connection.

A “worst-case” (highest current) condition that results in maximum 3-phase thermal and mechanical stress in the system is established.

A “worst-case” condition is typically what is needed to ensure proper equipment ratings, such as OCPD interrupting rating and equipment short-circuit current ratings (SCCR)

Question 4

major source(s) of fault current contribution?

Answer 4

On-site generators

Utility source

Question 5

true or false?

When a short-circuit condition occurs, running motors may contribute four to six times their normal full load into the fault, adding to the magnitude of the available fault current.

Answer 5

true

Question 6

The first step to determine the fault current at any point in the system is to draw a(n) ? showing all the sources of the available fault current.

Answer 6

one-line diagram

Question 7

true or false?

Short-circuit calculations are performed without overcurrent protective devices in the system. Calculations are done as though these devices are replaced with copper bars to determine the available fault current.

Answer 7

true

The point is that when doing a short-circuit calculation, the OCPDs are not considered in the calculations. Only after the calculations are determined are the OCPDs considered. For instance, does each OCPD have an interrupting rating equal to or greater than the available fault current, or do the OCPDs protect the circuit devices or equipment for the available fault current?

Question 8

Various methods have been developed to calculate the available fault current. All are based on ? .

Answer 8

Ohm’s Law

Question 9

When calculating the multiplier for transformer available fault current, the transformer impedance (%Z) is multiplied by ? to determine the worst-case condition to achieve the highest available fault current.

Answer 9

0.9

The marked impedance values on a transformer may vary ± 10% from the actual values determined by UL product standard. To establish a worst-case condition requires multiplying % Z by 0.9.

Question 10

study math

Answer 10

study mAath