Solve Instrument Transformer Problems
Solve and Answer CT/PT Problems and Questions
Question – 1. For instantaneous protection, find the current transformer ratios in the high and low side of the radial circuit shown in Fig.-1 at the left.
Note: Generally, except for some cases, the current transformer on the wye side of a wye-delta transformer is connected delta while the current transformer on the delta side is connected wye. The configuration is used to compensate for the 30o phase shift introduced by the delta-bank – as explained in Transformer Vector Group Symbol Designation in our Power Transformer page – and blocks the zero sequence current, since the zero sequence component will circulate inside delta connected current transformer preventing relay misoperation in protection circuits.
Ipri = kVA/34.5 x (3)1/2 =5,000/ 34.5 x 1.732 = 83.68Amp
150% x 83.68Amps = 125.51Amps
Use a 200/5 CT ratio
Isec = 5,000/4.16 x (3)1/2 = 694Amps
Use 1000/5 CT ratio.
The protection relay connected to the primary CT protects the 34.5kv cable between the power transformer primary and its circuit breaker.
Note: (The exponent form “1/2” is a square root)
Download the pdf file below for your reference:
Question-2: Determine the secondary terminal voltage of the current transformer that can maintain 20X rated current without exceeding 10% ratio correction factor (rcf). 20X rated current = 5Amp X 20 = 100Amps.
Per ANSI-IEEE C57.13 relaying accuracy classes and burden data:
= 5A X 20 X B-0.5 = 50volts
= 5A X 20 X B-1.0 = 100volts
The CT voltage thus, may be either C50 or C100 for an impedance of 0.5Ω and 1.0Ω respectively per Table 1: IEEE C57.13 relaying accuracy classes and burden data Table 1 below .
The ratio of turns is usually adjusted to compensate for the CT loses. This compensation results to actual secondary output that is higher than the “marked ratio” shown in CT nameplates, which may cause the secondary output to be slightly out of phase with the primary input. This is the phase angle error that is measured in minutes and is positive if the secondary output leads primary input.
Table 1: IEEE C57.13 relaying accuracy classes and burden data
Source: Siemens TechTopics No. 91 Current transformer relaying accuracies – IEEE compared to IEC
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ANSI-IEEE C57.13-1993 standard voltage ratings are 10, 20, 50, 100, 200, 400, and 800 volts with voltage classification “C” or “T”. The designation “C“ is used when the leakage reactance is low and calculated values do not have considerable effect on the CT ratio. For switchgear application, only one C100 can be installed versus two for the C50 rating, since the C100 has a thicker core. The higher CT voltage is costlier; be sure to justify your selection with proper calculations of all the CT’s burden impedance; i.e. the load of current transformer. which is the impedance of the interconnecting cable in its secondary.
C100 CT Class Definition:
“…a C100 CT class means that the CT is capable of passing 20 × 5 A (CT secondary) = 100 A through a standard 1 Ω resistor with a current accuracy less than 10%. In more practical terms, it guarantees a 10% accuracy provided the voltage across the total burden does not exceed 100 V. If the burden is less than 1 Ω, the CT is within 10% accuracy for more than 20 × Isecondary. If the burden is larger then less current will exceeds the accuracy limit.” Ref.: CT Class Definitions GE Power Management No. GET-8418A
Please also see our Instrument Transformer page for selecting current transformer.
Appropriately selecting a current transformer is a tedious process and falls beyond the scope of this article. For your further study download the Siemens and ABB PDF files below:
In order for the CTs to be correctly selected, they must be associated with the type of protection relay to be used whose range of protection in a given network is well defined. Proper organization of the CT/ Relay setup defines the specific position and setting of the selected protections, the CT ratio, their power, accuracy and accuracy limit factor (ALF). Thorough specification of CTs also involves the protection input impedance, wiring impedance and the relay’s operating threshold. Remember that overcurrent protection monitors the magnitude of current, a differential protection compares the magnitude and direction of the incoming and outgoing current, and an earth fault protection considers the sum of the three-phase current.