## Inputs for Coordination Calculation

A 440 V 60 Hz switchboard feeds a **4-wire distribution board** for small loads such as socket outlets. The switchboard has a fault making capacity of **100kA rms**. After applying diversity factors to the loads the total load current is **90 A**. Moulded case circuit breakers (MCCBs) rated at **16 A** and **32 A** are to be used for the loads.

The installation will use cables having copper conductors and XLPE insulation. The cable from the switchboard to the distribution board is **20 metres in length**.

A typical load cable is **15 metres in length** and will carry a current of **29 A** at a **power factor of 0.85** lagging.

**Ignore the presence of induction motors at the switchboard and find the following:**

- Rating of the incoming circuit breaker.
- Size of the incoming cable.
- Size of the load cable.
- Check that the MCCB coordination is complete.

**The following sequence will be used to calculate the results:**

- Choose the upstream MCCB at the switchboard and its settings
- Choose the incoming feeder cable
- Choose the downstream load MCCB and its settings
- Find the upstream fault source impedance
- Find the cut-off, or let-through, current from the switchboard
- Find the impedance of the incoming cable
- Find the impedance of the load cable
- Find the fault current at the distribution board, point B
- Find the fault current at the beginning of the load cable, point C
- Find the fault current at the end of the load cable, point D
- Check the peak making capacity and peak let-through capacity of the MCCBs chosen above
- Find the highestI-squared-t value for the upstream MCCB
- Calculate a suitable size for the load cable to satisfy the I
^{2}t duty - Calculate the volt-drop in the load cable
- Select the largest conductor size from the above calculations
- Plot the results (coordination curve)

### Let’s dive into solution!

#### 1. Choose the upstream MCCB at the switchboard and its settings

From a manufacturer’s data sheet a **125 A MCCB with an adjustable 100 A thermal release** is chosen. The thermal release is set to **90 A** to match the total load.

#### 2. Choose the incoming feeder cable

From a manufacturer’s data sheet several cables can be compared for the same ambient conditions and laying arrangements. **Their details are:**

- 50 mm
^{2}cable, maximum current 124 A, R = 0.492, X = 0.110 ohms/km. - 70 mm
^{2}cable, maximum current 159 A, R = 0.340, X = 0.106 ohms/km. - 95 mm
^{2}cable, maximum current 193 A, R = 0.247, X = 0.093 ohms/km.

**70 mm**since the rating of the 50 mm

^{2}cable is chosen^{2}cable is just too low.

#### 3. Choose the downstream load MCCB and its settings

From a manufacturer’s data sheet a **32 A MCCB with an adjustable 32 A thermal release** is chosen. The thermal release is set to **29 A** to match its load.

#### 4. Find the upstream fault source impedance

For a prospective symmetrical fault current of **100 kA rms** the **upstream fault source impedance Z _{up}** is:

#### 5. Find the cut-off, or let-through, current from the switchboard

From a manufacturer’s data sheet a **125 A MCCB** has a **let-through current I _{p} of 25 kA peak** for a

**prospective fault current I**.

_{s}of 100 kArms#### 6. Find the impedance of the incoming cable

The **impedance Z _{c1}** of the incoming cable is:

#### 7. Find the impedance of the load cable

The **impedance Z _{c2}** of the incoming cable is:

From a manufacturer’s data sheet several cables can be compared for the same ambient conditions and laying arrangements. Their details are:

- 6 mm
^{2}cable, maximum current 33.8 A, R = 3.91, X = 0.130 ohms/km. - 10 mm
^{2}cable, maximum current 46.7 A, R = 2.31, X = 0.126 ohms/km.

**6 mm**, since its rating is above the

^{2}cable is chosen provisionally**32 A rating of the MCCB that feeds it**.

The **impedance Z _{c2}** of the load cable is:

#### 8. Find the fault current at the distribution board, point B

From a manufacturer’s data sheet **the contact impedance** data for low voltage MCCBs are:

MCCB (Rating in Amps) | Resistance (in Ohms) | Reactanse (in Ohms at 60Hz) |

16 | 0.01 | neglect |

20 | 0.008 | neglect |

25 | 0.0065 | neglect |

32 | 0.005 | 0.000009 |

50 | 0.0027 | 0.000016 |

63 | 0.002 | 0.000025 |

80 | 0.0014 | 0.000042 |

100 | 0.0011 | 0.00007 |

125 | 0.0008 | 0.0001 |

160 | 0.00055 | 0.00015 |

200 | 0.0004 | 0.0002 |

250 | 0.00029 | 0.00027 |

320 | 0.0002 | 0.0004 |

Hence **the upstream MCCB impedance Z _{m1}** is

**0.0008 + j 0.0001 ohms**. Therefore the fault impedance Z

_{fb}is:

The **fault making current I _{fb}** is:

Where **V _{p}** is the line-to-neutral voltage. Locate

**the point R**for

**26,195 A**on the prospective curve in Figure 1.

#### 9. Find the fault current at the beginning of the load cable, point C

Hence the downstream MCCB impedance** Z _{m2} is 0.005+j0.000009 ohms**. Add this to

**Z**to give the fault impedance

_{fb}**Z**as:

_{fc}The fault making current **I _{fc}** is:

**the point S**for

**17,443 A**on the prospective curve in Figure 1.

#### 10. Find the fault current at the end of the load cable, point D

**Add Z _{c2} to Z_{fc}** to give the fault impedance

**Z**as:

_{fd}The **fault making current I _{fd}** is:

**the point U for 3473 A**on the prospective curve in Figure 1.

#### 11. Check the peak making capacity and peak let-through capacity of the MCCBs chosen above

The following manufacturer’s data are typical for **125 A** and **32 A MCCBs**:

MCCB Rating | Making capacity | Let-through capacity kA _{peak} (cut-off) | |

kA_{rms} | kA_{peak} | ||

32 A | 95 | 209 *** | 6.0 |

125 A | 132 | 290 *** | 25.0 |

******* Approximate values of the doubling factor taken to be **2.2**

Hence the peak making capacity of the **32 A MCCB** is well in excess of the let-through peak current of the **125 A MCCB**.

#### 12. Find the highest I^{2}t value for the upstream MCCB

Locate two points **P** and **Q** on the curve of the upstream MCCB as follows,

Point | Current in p.u. | Current in Amps | Time in seconds | I^{2}t |

P | 14 | 406 | 6 | 989016.0 |

Q | 602 | 17,450 | 0.0016 | 487204.0 |

**Hence I ^{2}t at P exceeds that at Q.**

#### 13. Calculate a suitable size for the load cable to satisfy the I^{2}t duty

For XLPE cables the **‘k factor’** for the I^{2}t is** 143**. The cross-sectional area A is:

**10 mm**.

^{2}#### 14. Calculate the volt-drop in the load cable

The usual limit to volt-drop in three-phase cables feeding static loads is **2.5% at full load**.

Where, **I _{flc} = 29 A**,

**L = 15 m**and

**φ = 54.5495 degrees**. For a

**6 mm**cable the volt-drop is found to be:

^{2}**2.5%**.

#### 15. Select the largest conductor size from the above calculations

Comparing the conductor sizes found in **13.** and **14.** gives the **larger as 10 mm ^{2}**, and this size should be used. Revise the calculation of the fault current

**I**.

_{fd}**The impedance Z**of the load cable is:

_{c2}Add **Z _{c2}** to

**Z**to give

_{fc}**the fault impedance Z**as:

_{fd}The **fault making current I _{fd}** is:

#### 16. Plot the results

The results are plotted in Figure 1.

**Refrence //** Switchgear and Motor Control Centres – Handbook of Electrical Engineering: For Practitioners in the Oil, Gas and Petrochemical Industry by Alan L. Sheldrake (Download here)

Hello, just a small question, in the introduction the main cable size is given as 20m so why was it used as 25m to calculate the impedance in number 6 ?