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Shop Practice with Electrical Code: Voltage Drop and Short Circuit Calculation Residential Single Phase System Jayson Bryan E. Mutuc, REE, RME BSEE-PUP MSEE, Major in Power Systems -MIT (candidate)
Voltage Drop 2.10.2.1 (a) (1) FPN No. 4: To provide reasonable efficiency of operation of electrical equipment, branch-circuit conductors should be sized to prevent a voltage drop not to exceed three percent. In addition, the maximum total voltage drop on both feeders and branch circuits should not exceed five percent.
Voltage Drop Calculation Voltage Drop Calculation: IR %VD = x100% Voltage Supply (Vs )
where: I – line current, amperes R – line ac resistance, ohms
Purpose of Short Circuit Calculation The purpose of short circuit calculation is to know the rating the interrupting capacity of the overcurrent protective device In circuit breaker there are 3 important parameters when it comes to residential design, ampere trip, ampere frame and interrupting rating Ampere trip – the current rating the circuit breaker set to trip Ampere frame – insulation housing for maximum long time current available in physical Interrupting Rating – the highest current at rated voltage that a device is intended to interrupt under standard test conditions.
Analogies to Short Circuit Condition Normal Current Operation
Short Circuit Condition with Inadequate Interrupting Rating
Analogies to Short Circuit Condition Short Circuit with Adequate Interrupting Rating
Calculation of Short Circuit Currents Basic Short Circuit Calculation Procedure Step 1: Determine transformer full-load amperes from either: a) Name plate b) Table c) Formula Step 2: Find the transformer multiplier Note: most % Z of xformer is multiplied by .9
Formula 𝟏 ∅ 𝐭𝐫𝐚𝐧𝐬𝐟. 𝐈𝐅𝐋𝐀 =
𝐊𝐕𝐀 𝐱 𝟏𝟎𝟎 𝐄𝐋−𝐋
EL-L = Line to Line Voltage
𝟏𝟎𝟎 𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐢𝐞𝐫 = 𝐓𝐫𝐚𝐧𝐬𝐟. % 𝐙
Calculation of Short Circuit Currents Step 3: Determine transformer let-through short-circuit current Step 4: Calculate “f” factor Note: IL-N = 1.5 x IL-L at transformer terminals
Formula 𝐈𝐋−𝐋 = 𝐭𝐫𝐚𝐧𝐬.𝐅𝐋𝐀 𝐱 𝐦𝐮𝐥𝐭𝐢𝐩𝐥𝐢𝐞𝐫 1Φ line-to-line
𝐟=
𝟐 𝐱 𝐋 𝐱 𝐈𝐋−𝐋 𝐂 𝐱 𝐧 𝐱 𝐄𝐋−𝐋
(L-L) faults 1 Φ line-to-neutral
𝐟=
𝟐 𝐱 𝐋 𝐱 𝐈𝐋−𝑵 𝐂 𝐱 𝐧 𝐱 𝐄𝐋−𝑵
(L-N) faults L – length (feet) of conduit to the fault. C – conduit constant (table B)
Subscript L-L means Line to Line
n – number of conductors per phase
Subscript L-N means Line to Neutral
I – available short circuit current in amperes at beginning of the circuit.
Calculation of Short Circuit Currents Step 5: Calculate “M” or take Step 6: Compute the available short circuit current (RMS symmetrical)
Formula
𝟏 𝐌= 𝟏+𝐟
𝐈𝐒𝐂𝐀 = 𝐈(𝐋−𝐋 𝐨𝐫 𝐋−𝐍) 𝐱 𝐌
Short Circuit Sample Computation All faults are line to line single phase fault
Fault #1 Step 1 IFLA =
KVA x 100 EL−L
=
50 x 1000 240
= 208.33 A
Step 2 Multiplier =
100 Transf.% Z
=
100 .9(2)
= 55.55
Step 3 IL−L = trans.FLA x multiplier = 208.33 x 55.55 = 11572.92 A Single phase short circuit current at Transformer Secondary Step 4 2 x L x IL−L 2 𝑥20 𝑥 11572.92 f= = C x n x EL−L 22185 𝑥 2 𝑥 240
= 0.04347125185
Short Circuit Sample Computation Step 5
1 1 M= = = 0.958 1+f 1 + 0.043
All faults are line to line single phase fault
Step 6 ISCA = I(L−L or L−N) x M = 11572.92 x 0.958 = 11090.7863 A Fault #2 (Use ISCA at Fault #1 to calculate) Step 4 2 x L x IL−L 2 𝑥20 𝑥 11090.7863 f= = C x n x EL−L 5907 𝑥 2 𝑥 240 = 0.156 Step 5 M=
1 1 = = 0.864 1+f 1 + 0.156
Step 6 ISCA = I(L−L or L−N) x M = 11090.7863 x 0.864 = 9590.257 A
Use Circuit Breaker with 10 KAIC rating
Table A. Short – Currents Available from Various Size Transformers
