中壓應用中電流互感器 (踏入50萬瀏覽量第四文)

 本文以外地文章翻譯及修正為香港用語 ：

中壓應用中電流互感器 (CT) 的 IEC 和 NEMA/IEEE 額定值：

量度(MCT)及保護(PCT)用途

首先，讓我們用幾句話來提醒自己一些基礎知識。 這是你必須知道的。 電流互感器 (CT) 旨在產生與初級電流準確成比例的次級電流。 它由一個初級繞組組成，外部母線或電纜穿過它(初級為1匝)，或者它可以有一個初級繞組，引出兩端進行端接。

中壓開關設備電流互感器 (CT) 的 IEC 和 NEMA 額定值：

一個中壓電流互感器最多可以有三個獨立的次級繞組。 整個電流互感器組件用樹脂封裝在絕緣外殼內。 電流互感器用於量度或保護用途。

準確度等級和尺寸取決於個別應用——例如，電力公司計費量度的電流互感器將使用高準確度的電流互感器。

請注意，絕對不能讓使用中的電流互感器使得次級繞組開路，這一點非常重要。 這會產生極高的電壓，因為初級為1匝，如次級是100匝，使會產生100倍的電壓，這樣會對工作進行中的人員構成真正的危險。如在運作中的電流互感器一直開路狀態，使會使它燒毀 - 這也是條例中訂明電流互感器的線路中，不可以加入任何保護器件的原因。

好的，讓我們來看看電流互感器的 IEC 和後來的 NEMA 額定值。 部分評分說明有練習和實例，希望對大家更好的理解有所幫助。

1. IEC 額定值

1.1 額定一次電流：Ipr(A)

電流互感器的初級電流額定值必須大於它所監測的預期最大工作電流。

計費量度的電流互感器的一次電流額定值不應超過最大工作電流的 1.5 倍。 需要選擇保護 CT 的初級電流額定值，以便在故障期間達到保護啟動水平。

Ipr 的標準值為：10、12.5、15、20、25、30、40、50、60、75 A 和這些值的十進制倍數（來源：IEC 60044-1）

1.2 額定二次電流：Isr

CT 的次級電流額定值為 1 A 或 5 A。隨著越來越多的 CT 驅動設備變為數字化，具有 5 A 次級額定值的 CT 變得越來越不常見。 對於較長的次級電纜運行，具有 1 A 次級繞組的 CT 可以最大限度地減小變壓器和次級電纜的尺寸。

Standar1.3變比：Kn

這是次級繞組匝數與初級繞組匝數之比： Kn = Ns/Np = Ipr/Isr

1.4 額定熱短時耐受電流：Ith (kA)

這是 CT 可以承受的最高級別的 rms 初級故障電流，無論是熱的還是動態的，1 秒鐘都不會損壞。在中壓外殼中使用時，Ith 額定值應與整個開關設備的短時耐受額定值相匹配。

1.5 過流係數：Ksi

這是 CT 的短時耐受電流額定值與其初級電流額定值之比：

Ksi = Ith/Ipr

該係數表明製造 CT 的難度。較高的係數意味著物理上較大的 CT，這更難以製造。

如果 Ksi < 100 則易於製造

如果 Ksi 100 ~ 500 則很難製造，有一定的局限性

如果 Ksi > 500，則製造非常困難

1.6 額定一次迴路電壓：Up (kV)

初級電路額定電壓表示 CT 提供的絕緣水平。如果環型 CT 安裝在電纜或套管周圍，則絕緣水平可由電纜或套管提供。

1.7 額定頻率：fr (Hz)

此額定值必須與系統的工作頻率相匹配。標準頻率為 50 Hz 和 60 Hz。小心非常重要，因為 50 Hz CT 可用於 60 Hz 系統，但 60 Hz CT 不能用於 50 Hz 系統。

1.8 額定實際輸出功率（VA）

CT 次級可以提供的最大功率，以保證其準確性和性能。總和 VA（包括電纜、連接器和負載）不得超過 CT 的額定實際輸出功率。標準值為：1、2.5、5、10、15 VA。

電纜負荷可以通過以下方式計​​算： VAcable = k × L/S，其中：

k = 0.44（5 A 次級），= 0.0176（1 A 次級）

L = 電纜的總進/回長度 (m)

S = 銅纜截面積（mm2）

計量儀表負擔：

計量儀表（數字）= 1 VA（大約）

計量儀表（電磁或感應）= 3 VA（大約）

傳感器（自供電）= 3 VA（大約）

保護儀器負擔：

保護儀表（數字）= 1 VA（大約）

保護儀表（電磁過電流）= 3-10 VA（大約）

1.8.1 練習

練習 #1 – 使用 2.5 mm2 銅電纜將帶有 1 A 次級的 CT 連接到 10 m 外的電磁電流表。

計算 CT 所需的最小 VA 額定值。

VAcable = k × L/S = 0.0176 × 20/2.5 = 0.14 VA

VA 安培計 = 3 VA

VAtotal = 0.14 + 3 = 3.14 VA

總負擔為 3.14 VA。使用 5 VA CT。

練習 #2 – 使用 1.5 mm2 銅纜將具有 5 A 次級電流的 CT 連接到 2 m 外的數字保護繼電器。

計算 CT 所需的最小 VA 額定值。

VAcable = k × L/S = 0.44 × 4/1.5 = 1.17 VA

VA 安培計 = 1 VA

VAtotal = 1.17 + 1 = 2.17 VA

總負擔為 2.17 VA。使用 2.5 VA CT。

1.9 計量類

計量等級表示 CT 二次電流在額定一次電流的 5% 至 125% 時的準確度。高於此水平，CT 開始飽和，次級電流被削波以保護連接的計量儀器的輸入。

一般計量 CT 將使用計量等級 CL 0.5 – 1.0

收入計量 CT 將使用計量等級 CL 0.2 – 0.5

在哪裡：

飽和

線性工作範圍，精度等級公差

1.10 防護等級 CT

保護等級 CT 在高過載水平下提供初級到次級電流的線性變換。這一特性使其適合與過電流保護繼電器一起使用。

繼電器跳閘設置通常為最大負載電流的 10~15 倍，該水平應落在 CT 二次電流曲線的線性部分。如果在達到繼電器跳閘水平之前 CT 飽和，則故障將無法檢測到，從而導致設備損壞和人員嚴重危險。

最常用的保護等級是 5PX，其中 X 是準確度限制因子 (ALF) 或額定初級電流的倍增因子。次級電流在額定初級電流下準確度為 +/-1%，在 X 倍額定初級電流下準確度為 +/-5%。

典型的保護等級 CT 額定值為 5P10、5P15、5P20。

在哪裡：

飽和

線性工作範圍，精度等級公差

理想保護設定跳閘區 50%~100% ALF

1.10.1 示例

200/1 A CT 的防護等級為 5P15。次級電流保證線性高達額定初級電流的 15 倍。次級電流在 200 A 初級電流時為 1 A (+/-1%)，在 3000 A 初級電流時為 15 A (+/-5%)。

為保證運行，任何過流脫扣設置應在 7.5 ~ 15 A 二次電流之間。

1.11 電流互感器的選擇

選擇 CT 的主要考慮因素是初級和次級電流比、實際輸出功率 (VA) 和精度等級。 二次選擇考慮因素是額定一次電壓、頻率和熱短時耐受電流。

1.11.1 初級和次級電流比

額定初級電流：Ipr (A)

源 額定初級電流 Ipr (A)

來自變壓器的 Ipr ≥ 1.0-1.25 標稱源電流

變壓器饋線 Ipr ≥變壓器額定一次電流的 1.0-1.25

饋線至電機 Ipr ≥ 電機滿載電流的 1.0-1.5

饋線至電容器組 Ipr ≥ 1.3-1.5 的標稱電容器電流

額定二次電流：Isr (A)

使用 1 A 和 5 A 進行本地安裝

使用 1 A 進行遠程安裝

1.11.2 實際輸出功率（VA）

CT 的實際輸出額定值必須是高於 CT 次級預期總負載的下一個最高標稱尺寸。 總負擔是輸出電纜、連接器和儀器的總和。

1.11.3 類類型

使用計量等級 CT 進行計量和指示。 更高等級的 CT 在初級和次級電流之間提供更高的準確度。

對基於電流的保護繼電器輸入使用 5PX 保護等級 CT。 必須選擇 ALF，使繼電器跳閘點位於次級電流曲線的線性部分，介於 ALF 的 50% 和 100% 之間。

1.11.4 練習

為以下變壓器輸入和饋線電路選擇合適的 CT。

在哪裡：

1. 變壓器來電總掣：

中壓/中壓變壓器 (TXR1)：5 MVA，36/11 kV，10% Z

瞬時過電流脫扣設置 = 15 × In 用於斷開 CT1-2 的數字保護繼電器 (OC1)

電磁電流表（A）由 CT1-1 驅動

2.變壓器後總掣：

中壓/低壓變壓器 (TXR2)：2 MVA，11/0.4 kV，5% Z

瞬時過電流脫扣設置 = 10 × In 用於斷開 CT2 的數字保護繼電器 (OC2)

練習 1 – 測量變壓器輸入電路的 CT1-1：

第 1 步 – 計算變壓器 TXR1 標稱次級電流：In (A)

In = S/(√3 × U) = 5000/(√3 × 11) = 262 A

TXR1 的次級電流為 262 A

第 2 步 – 計算最大值。 CT1 安裝時的預期短路電流：Isc (A)

忽略任何電源線或母線阻抗：

Isc = In × 100/Z = 262 × 100/10 = 2620 A

CT1 的最大預期短路電流為 2620 A

第 3 步 – 選擇計量 CT1-1 額定值：

一次額定電流：Ipr = (1.0-1.25) × In = (1.0-1.25) × 262 A

使用 300 A 的額定值

二次額定電流：Isr

使用 1 A 的額定值

短時耐受等級：Ith ≥ Isc

使用 10 kA 的額定值

一次迴路電壓：Up≥U

使用 12 kV 的額定值

實際輸出功率：對於電磁式儀表，通常 > 3 VA

使用 5 VA（這允許 2 VA 用於電纜負擔等）

精度等級

使用 Class 1.0（一般計量的通用類）

練習 2 – 變壓器輸入電路的保護 CT1-2：

第 1 步 – 選擇計量 CT 和保護 CT 共有的額定值

一次/二次額定電流：使用300/1 A

短時耐受額定值 [Ith]：使用 10 kA 額定值

初級電路電壓 [Up]：使用 12 kV 額定值

第 2 步 – 選擇實際輸出功率

實際輸出功率：對於數字式保護繼電器，通常 > 1VA

使用 2.5 VA（這允許 1.5 VA 用於電纜負擔等）

第 3 步 – 計算防護等級 5PX

1.11.4 練習

為以下變壓器輸入和饋線電路選擇合適的 CT。

初級額定電流 Ipr = (1.0 – 1.25) × In = (1.0 – 1.25) × 105

使用 150 A 的額定值

二次額定電流 Isr

使用 1 A 的額定值

短時耐受等級，Ith ≥ Isc

使用 10 kA 的額定值

初級電路電壓Up≥U

使用 12 kV 的額定值

實際輸出功率：對於數字型保護繼電器，通常 > 1 VA。

使用 2.5 VA（這允許 1.5 VA 用於電纜負擔等）

第 4 步 – 計算防護等級 5PX

保護繼電器 OC2 的瞬時跳閘電流水平設置為 10 × In

ITRIP = 10 × 105 = 1050 A（初級電流）

注意：在大多數數字保護繼電器中，跳閘電流水平是根據次級電流設置的。在這種情況下

ISEC = 3900/300 × 1 = 13.1 A

CT 次級的瞬時跳閘電流水平為 7 A

跳閘電流水平應介於精度限制因子 (ALF) 的 100% 至 50% 之間。使用 10 (5P10) 的 ALF，1050 A 的跳閘電流水平落在 100% 到 50% ALF 的範圍內，因此 5P10 保護等級 CT 是合適的。

100%(ALF) = 1.0 × 10 × 150 = 1500 A

50%(ALF) = 0.5 × 10 × 150 = 750 A

我們可能會注意到 750 ≤ 1050 ≤ 1500 A

使用防護等級 5P10

2. NEMA/IEEE 評級

這些額定值通常用於在北美裝置中製造或使用的電流互感器。除了規定的初級與次級標稱電流比外，該器件還具有格式中的整體準確度等級。

AC-CR-BU

在哪裡：

AC = 準確度等級

CR = 等級

BU = 最大負載（歐姆）

2.1 準確度等級

指定次級電流相對於初級額定電流的準確度。只有在不超過最大負荷的情況下才能保證這種準確性。

準確度等級 100% 初級電流下的容差

1.2 ±1.2 %

0.6 ±0.6 %

0.5 ±0.5 %

0.3 ±0.3 %

1.11.4 練習

為以下變壓器輸入和饋線電路選擇合適的 CT。

2.2 等級評定

指定設備的預期應用。

B = 用於計量應用

H = 用於保護應用。 CT 次級精度保證為額定初級額定電流的 5 至 20 倍

2.3 負擔

允許連接到電流互感器次級的最大負載，以保證精度等級。最大負擔包括二次電纜/電線、連接器和負載。

對於 5 A 次級，下表將負載（以歐姆為單位）轉換為 VA。

Ω 0.04 0.06 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36 0.40 0.48 0.56 0.64 0.72 0.80

弗吉尼亞州 1 1.5 2 3 4 5 6 7 8 9 10 12 14 16 18 20

塔貝拉

2.4 例子

0.5-B-0.1

此示例表明電流互感器的精度為 ±0.5%，最大允許二次負載為 0.1 Ω（或 5 A 二次 CT 上為 2.5 VA）。這是一種計量級額定電流互感器。

1.2-H-0.2

此示例表明電流互感器的精度為 ±1.2%，最大允許次級負載為 0.2 Ω（或 5 A 次級 CT 上的 5 VA）。這是一種防護等級額定電流互感器。

資料來源：

Aucom 的中壓應用指南

James C. Burke 的變電站工程

變電站中電流互感器和電線尺寸的選擇——Sethuraman Ganesan； ABB 公司

原文

IEC and NEMA/IEEE ratings of current transformers (CTs) in medium voltage applications

Metering and protection purpose

First, let’s remind ourselves of the basics in a few sentences. That is something you must know. A current transformer (CT) is designed to produce a secondary current which is accurately proportional to the primary current. It consists of a single primary winding, which an external busbar or cable runs through, or it can have a single primary bar, brought out to two ends for termination.

IEC and NEMA ratings of current transformers (CTs) for a medium voltage switchgear (photo credit: Energie Technik Becker GmbH)

A medium voltage current transformer can have up to three independent secondary winding sets. The entire current transformer assembly is encapsulated in resin, inside an insulated casing. Current transformers are used for metering or protection purposes.

The accuracy class and size depends on the individual application – for example, revenue metering would use high accuracy metering CTs.

Just to note, it’s very important to never leave the secondary winding of a CT open circuit. This creates extremely high voltages which pose a real danger to personnel.

Ok, let’s get on the IEC and later NEMA ratings of a current transformer. Some rating explanations have exercises and real examples, which I hope it will help for better understanding.

1. IEC Ratings

1.1 Rated primary current: I_pr (A)

The primary current rating of a CT must be greater than the expected maximum operating current it is monitoring.

Metering CT’s primary current rating should not exceed 1.5 times the maximum operating current. Protection CT’s primary current rating needs to be chosen so that the protection pick-up level is attained during a fault.

Standard values for I_pr are: 10, 12.5, 15, 20, 25, 30, 40, 50, 60, 75 A, and decimal multiples of these values (source: IEC 60044-1)

1.2 Rated secondary current: Isr

The secondary current rating of a CT is either 1 A or 5 A. CTs with a 5 A secondary rating are becoming less common as more CT driven equipment becomes digital. For long secondary cable runs, CTs with 1 A secondary windings can minimize the transformer and secondary cable size.

1.3 Transformer ratio: K_n

This is the ratio of secondary to primary winding turns: K_n = N_s/N_p = I_pr/I_sr

1.4 Rated thermal short-time withstand current: I_th (kA)

This is the highest level of rms primary fault current which the CT can endure, both thermally and dynamically, for 1 second without damage. When used in a medium voltage enclosure, the I_th rating should match the short-time withstand rating of the entire switchgear.

1.5 Overcurrent coefficient: K_si

This is the ratio of a CT’s short-time withstand current rating to its primary current rating:

K_si = I_th/I_pr

This coefficient indicates how difficult it would be to manufacture a CT. A higher coefficient means a physically larger CT, which is more difficult to manufacture.

• If K_si < 100 it’s easy to manufacture
• If K_si 100 ~ 500 it’s difficult to manufacture, with certain limitations
• If K_si > 500 it’s  extremely difficult to manufacture

1.6 Rated primary circuit voltage: U_p (kV)

The primary circuit voltage rating indicates the level on insulation provided by the CT. If a ring type CT is installed around a cable or bushing, the insulation level can be provided by the cable or bushing.

1.7 Rated frequency: f_r (Hz)

This rating must match the system’s operating frequency. Standard frequencies are 50 Hz and 60 Hz. It’s very important to be cautios, because a 50 Hz CT can be used on a 60 Hz system, but a 60 Hz CT cannot be used on a 50 Hz system.

1.8 Rated real output power (VA)

The maximum power a CT secondary can deliver, to guarantee its accuracy and performance. The total sum VA (including cable, connectors and load) must not exceed the rated real output power of the CT. Standard values are: 1, 2.5, 5, 10, 15 VA.

Cable burden can be calculated the following way: VA_cable = k × L/S, where:

• k = 0.44 for 5 A secondary, = 0.0176 for 1 A secondary
• L = total feed/return length of cable (m)
• S = cross sectional area of copper cable (mm²)

Metering instrument burden:

• Metering instrument (digital) = 1 VA (approx.)
• Metering instrument (electromagnetic or induction) = 3 VA (approx.)
• Transducer (self powered) = 3 VA (approx.)

Protection instrument burden:

• Protection instrument (digital) = 1 VA (approx.)
• Protection instrument (electromagnetic overcurrent) = 3-10 VA (approx.)

1.8.1 Exercises

Exercise #1 – A CT with a 1 A secondary is connected to an electromagnetic ammeter located 10 m away, using 2.5 mm² copper cable.

Calculate the minimum required VA rating of the CT.

• VA_cable = k × L/S = 0.0176 × 20/2.5 = 0.14 VA
• VA_ammeter = 3 VA
• VA_total = 0.14 + 3 = 3.14 VA

The total burden is 3.14 VA. Use a 5 VA CT. 

Exercise #2 – A CT with a 5 A secondary is connected to a digital protection relay located 2 m away, using 1.5 mm² copper cable.

Calculate the minimum required VA rating of the CT.

• VA_cable = k × L/S = 0.44 × 4/1.5 = 1.17 VA
• VA_ammeter = 1 VA
• VA_total = 1.17 + 1 = 2.17 VA

The total burden is 2.17 VA. Use a 2.5 VA CT. 

1.9 Metering class

A metering class indicates the accuracy of the CT secondary current at 5 to 125% of rated primary current. Above this level, the CT starts to saturate and the secondary current is clipped to protect the inputs of a connected metering instrument.

• General metering CT would use a metering class CL 0.5 – 1.0
• Revenue metering CT would use a metering class CL 0.2 – 0.5

Where:

1. Saturation
2. Linear operating range, at accuracy class tolerance

1.10 Protection class CT

A protection class CT provides a linear transformation of the primary to secondary current at high overload levels. This characteristic makes them suitable for use with overcurrent protection relays.

A relay trip setting is normally 10~15 times the maximum load current and this level should fall on the linear part of the CT secondary current curve. If a CT saturates before the relay trip level is reached, the fault will remain undetected, leading to equipment damage and serious danger to personnel.

The most commonly used protection class is a 5PX, where X is the accuracy limit factor (ALF) or multiplication factor of the rated primary current. The secondary current is +/-1% accurate at rated primary current and +/-5% accurate at X times rated primary current.

Typical protection class CT ratings are 5P10, 5P15, 5P20.

Where:

1. Saturation
2. Linear operating range, at accuracy class tolerance
3. Ideal protection setting trip zone 50%~100% ALF

1.10.1 Example

A 200/1 A CT has a protection class rating of 5P15. The secondary current is guaranteed to be linear up to 15 times the rated primary current. The secondary current will be 1 A (+/-1%) at 200 A primary current and 15 A (+/-5%) at 3000 A primary current.

For guaranteed operation, any overcurrent trip setting should be between 7.5 ~ 15 A secondary current.

1.11 Selection of current transformers

The main considerations for selecting a CT are the primary and secondary current ratio, real output power rating (VA) and accuracy class. Secondary selection considerations are rated primary voltage, frequency and thermal short-time withstand current.

1.11.1 Primary and secondary current ratio

Rated primary current: I_pr (A)

Source	Rated primary current Ipr (A)
Incomer from transformer	Ipr ≥ 1.0-1.25 of nominal source current
Feeder to transformer	Ipr ≥ 1.0-1.25 of transformer’s rated primary current
Feeder to motor	Ipr ≥ 1.0-1.5 of motor full load current
Feeder to capacitor bank	Ipr ≥ 1.3-1.5 of nominal capacitor current

Rated secondary current: I_sr (A)

• Use 1 A and 5 A for local installation 
• Use 1 A for remote installation 

1.11.2 Real output power (VA)

The real output rating of the CT must be the next highest nominal size above the expected total burden on the CT secondary. Total burden is the sum of output cable, connectors and instruments.

1.11.3 Class type

Use a metering class CT for metering and indication. A higher class CT gives greater accuracy between the primary and secondary currents.

Use a 5PX protection class CT for current based protection relay inputs. The ALF must be selected so that the relay trip point lies on the linear part of the secondary current curve, between 50% and 100% of the ALF.

1.11.4 Exercise

Select appropriate CTs for the following transformer incomer and feeder circuits.

Where:

1. Transformer Incomer:

MV/MV transformer (TXR1): 5 MVA, 36/11 kV, 10% Z

Instantaneous overcurrent trip setting = 15 × In for digital protection relay (OC1) driven off CT1-2

Electromagnetic ammeter (A) is driven off CT1-1

2.Transformer Feeder:

MV/LV transformer (TXR2): 2 MVA, 11/0.4 kV, 5% Z

Instantaneous overcurrent trip setting = 10 × In for digital protection relay (OC2) driven off CT2

Exercise 1 – Metering CT1-1 for transformer incomer circuit:

Step 1 – Calculate transformer TXR1 nominal secondary current: In (A)

In = S/(√3 × U) = 5000/(√3 × 11) = 262 A

The secondary current for TXR1 is 262 A 

Step 2 – Calculated max. expected short circuit current at CT1 installation: Isc (A)

Ignoring any power cable or busbar impedances:

Isc = In × 100/Z = 262 × 100/10 = 2620 A

The maximum expected short circuit current at CT1 is 2620 A 

Step 3 – Select metering CT1-1 ratings:

Primary rated current: Ipr = (1.0-1.25) × In = (1.0-1.25) × 262 A

Use a rating of 300 A 

Secondary rated current: Isr

Use a rating of 1 A 

Short-time withstand rating: Ith ≥ Isc

Use a rating of 10 kA 

Primary circuit voltage: Up ≥ U

Use a rating of 12 kV 

Real output power: Typically > 3 VA for electromagnetic type meter

Use 5 VA (this allows 2 VA for cable burden, etc.) 

Accuracy Class

Use Class 1.0 (common class for general metering) 

Exercise 2 –  Protection CT1-2 for transformer incomer circuit:

Step 1 – Select ratings common to both the metering and protection CTs

Primary/secondary rated current: Use 300/1 A 

Short-time withstand rating [Ith]: Use 10 kA rating 

Primary circuit voltage [Up]: Use 12 kV rating 

Step 2 – Select real output power

Real output power: typically > 1VA for digital type protection relay

Use 2.5 VA (this allows 1.5 VA for cable burden, etc.) 

Step 3 – Calculate protection class 5PX

The instantaneous trip current level of protection relay OC1 is set to 15 × In.

ITRIP = 15 × 262 = 3930 A (primary current) 

Note: In most digital protection relays, the trip current levels are set with respect to the secondary current. In this case

ISEC = 3900/300 × 1 = 13.1 A

The instantaneous trip current level for the CT secondary is 13.1 A 

The trip current level should fall between 100 to 50% of the accuracy limit factor (ALF). Using an ALF of 10 (5P10), the trip current level of 3930 A falls outside the range 100% to 50% ALF, so a 5P10 protection class CT is not suitable.

100%(ALF) = 1.0 × 10 × 300 = 3000 A 

50%(ALF) = 0.5 × 10 × 300 = 1500 A 

We may notice that 1500 ≤ 3930 ≥ 3000 A. Using an ALF of 15 (5P15), the trip current level of 3930 A falls within the range 100% to 50% ALF so a 5P15 protection class CT is suitable.

100%(ALF) = 1.0 × 15 × 300 = 4500 A

50%(ALF) = 0.5 × 15 × 300 = 2250 A

We may notice that 2250 ≤ 3930 ≤ 4500 A. Use protection class 5P15 

Exercise 3 –  Protection CT2 for transformer feeder circuit:

Step 1 – Calculate transformer TXR2 nominal primary current: In (A)

In = S/(√3 × U) = 2000/(√3 × 11) = 105 A

The primary current for TXR2 is 105 A 

Step 2 – Calculated maximum expected short circuit current at CT2 installation: Isc (A)

Ignoring any power cable or busbar impedances

Isc = In × 100/Z = 105 × 100/5 = 2100 A

The maximum expected short circuit current at CT2 is 2100 A 

Step 3 – Select protection CT2 ratings

Primary rated current Ipr = (1.0 – 1.25) × In = (1.0 – 1.25) × 105

Use a rating of 150 A 

Secondary rated current Isr

Use a rating of 1 A 

Short-time withstand rating, Ith ≥ Isc

Use a rating of 10 kA 

Primary circuit voltage Up ≥ U

Use a ratings of 12 kV 

Real output power: Typically > 1 VA for digital type protection relay.

Use 2.5 VA (this allows 1.5 VA for cable burden, etc.) 

Step 4 – Calculate protection class 5PX

The instantaneous trip current level of protection relay OC2 is set to 10 × In

ITRIP = 10 × 105 = 1050 A (primary current) 

Note: In most digital protection relays, the trip current levels are set with respect to the secondary current. In this case

ISEC = 3900/300 × 1 = 13.1 A

The instantaneous trip current level for the CT secondary is 7 A 

The trip current level should fall between 100 to 50% of the accuracy limit factor (ALF). Using an ALF of 10 (5P10), the trip current level of 1050 A falls within the range of 100% to 50% ALF so a 5P10 protection class CT is suitable.

100%(ALF) = 1.0 × 10 × 150 = 1500 A 

50%(ALF) = 0.5 × 10 × 150 = 750 A 

We may notice that 750 ≤ 1050 ≤ 1500 A 

Use protection class 5P10 

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2. NEMA/IEEE Ratings

These ratings are typically used for current transformers manufactured or used in North American installations. As well as a stated primary to secondary nominal current ratio, the device also carries an overall accuracy rating in the format.

AC-CR-BU

Where:

AC = accuracy class

CR = class rating

BU = maximum burden (ohms)

2.1 Accuracy class

Designates the accuracy of the secondary current with respect to the primary rated current. This accuracy is only guaranteed provided the maximum burden is not exceeded.

Accuracy class Tolerance at 100% primary current

1.2 ±1.2 %

0.6 ±0.6 %

0.5 ±0.5 %

0.3 ±0.3 %

2.2 Class rating

Designates the intended application of the device.

B = for metering applications

H = for protection applications. The CT secondary accuracy is guaranteed at 5 to 20 times the nominal primary rated current

2.3 Burden

The maximum load allowed to be connected to the current transformer secondary, to guarantee the accuracy class. The maximum burden includes secondary cable/wire, connectors and the load.

The following table converts burden in ohms to VA, for a 5 A secondary.

Ω 0.04 0.06 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36 0.40 0.48 0.56 0.64 0.72 0.80

VA 1 1.5 2 3 4 5 6 7 8 9 10 12 14 16 18 20

tabela

2.4 Examples

0.5-B-0.1

This example indicates a current transformer with an accuracy of ±0.5%, and a maximum allowable secondary burden of 0.1 Ω (or 2.5 VA on a 5 A secondary CT). This is a metering class rated current transformer.

1.2-H-0.2

This example indicates a current transformer with an accuracy of ±1.2%, and a maximum allowable secondary burden of 0.2 Ω (or 5 VA on a 5 A secondary CT). This is a protection class rated current transformer.

Sources:

Medium Voltage Application Guide by Aucom

Electric Power Substations Engineering By James C. Burke

Selection of current transformers and wire sizing in substations – Sethuraman Ganesan; ABB Inc.

https://electrical-engineering-portal.com/iec-nema-ieee-ratings-current-transformers-cts#rated-primary-current

https://electrical-engineering-portal.com/iec-nema-ieee-ratings-current-transformers-cts

https://www.sciencedirect.com/topics/engineering/magnetization-curve