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This formula is valid for single phase current ratio
Given primary voltage = 69 kV ; secondary voltage = 12.47 kV
secondary current full load = 1,296.447 A ; Calculate the expected primary current full load
Application of ratio and proportion
| 1296.447 Amp | is 100% of full load rating | 1,037.1576 Amp | is 80% of full load rating |
| 907.51 Amp | is 70% of full load rating | 777.868 Amp | is 60% of full load rating |
| 648.22 Amp | is 50% of full load rating | 518.578 Amp | is 40% of full load rating |
|
Vpri in kV
Vsec in kV
|
: = : |
Isec in Amp
Ipri in Amp
|
Calculator for estimating instantaneous primary single line to ground fault. For example during primary SLG fault the primary voltage instantly drops to about 8 kV. Primary fault current is expected to be about 2,021 Amperes assuming at that instant of time the transformer is carrying full load rating.
| 234.3 Amp | is 100% of full load rating | 187.44 Amp | is 80% of full load rating |
| 164.01 Amp | is 70% of full load rating | 140.58 Amp | is 60% of full load rating |
| 117.15 Amp | is 50% of full load rating | 93.72 Amp | is 40% of full load rating |
|
Vpri in kV
Vsec in kV
|
: = : |
Isec in Amp
Ipri in Amp
|
Calculator for estimating instantaneous secondary single line to ground fault. For example during secondary SLG fault the secondary voltage instantly drops to about 1 kV. Secondary fault current is about 16,167 Amperes
When you get your SEL relay fault current report, you can adjust the voltage to approximate the actual recorded relay fault current data. Use the actual secondary voltage and secondary current measurement from PI historian during fault timestamp.
Application of Current Ratio Formula in Estimating Short Circuit Fault Current
Given the nameplate base rating of a transformer: operational current (ampere) data of 320 MVA , 3 Phase Transformer, 525 / 230 kV as shown in table below.
Estimate the snapshot of temporary transient primary short circuit current, when there is SLG Fault in transmission line primary voltage source due to disturbance using the 10 kV instantaneous voltage drop recorded by PI historian during SLG fault.
| Voltages(kV) | Amperes |
| 525 Primary Voltage | 803.29 Secondary Current |
| 230 Secondary Voltage | 351.92 Primary Current |
| Primary Single Line to Gnd Fault | Instantaneous or snap shot data |
| 50 kV Primary SLG fault | 803.29 Secondary Current |
| 230 Secondary Voltage | 3,695 Short Circuit Primary Current |
In a no loss , ideal transformer , 100% efficient power transfer is usually assumed.
Power transformer efficiency based on efficiency test report is usually from 95% - 98%. So it is reasonable enough to assume a 100% efficiency.
What does 100% power transfer efficiency means? It means the
power generated in primary winding = power generated in secondary winding
Formula recall of power ( P = I V ).
Power Primary winding = Ip Vp
Power Secondary winding = Is Vs
IpVp = Is Vs
By algebraic manipulation.
Vp / Vs = Is / Ip
By applying this knowledge and using PI historian snapshot value, we can estimate the magnitude of short circuit current and estimate the magnetic forces to the primary or secondary winding of a power transformer depending on where the SLG fault happened. Even better if you have Pi historian data, you can create a streaming data of SLG Fault by using PI expression equation.
Formula and Calculator Index
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