Answer Posted / prasad
Dear Friend,
Why do we do the calculation?
The proper sizing of an electrical (load bearing) cable is important to ensure that the cable can:
Operate continuously under full load without being damaged
Withstand the worst short circuits currents flowing through the cable
Provide the load with a suitable voltage (and avoid excessive voltage drops)
(optional) Ensure operation of protective devices during an earth fault
When to do the calculation?
This calculation can be done individually for each power cable that needs to be sized, or alternatively, it can be used to produce cable sizing waterfall charts for groups of cables with similar characteristics (e.g. cables installed on ladder feeding induction motors).
General Methodology
All cable sizing methods more or less follow the same basic six step process:
1) Gathering data about the cable, its installation conditions, the load that it will carry, etc
2) Determine the minimum cable size based on continuous current carrying capacity
3) Determine the minimum cable size based on voltage drop considerations
4) Determine the minimum cable size based on short circuit temperature rise
5) Determine the minimum cable size based on earth fault loop impedance
6) Select the cable based on the highest of the sizes calculated in step 2, 3, 4 and 5
In this example, we will size a cable for a 415V, 37kW three-phase motor from the MCC to the field.
Step 1: Data Gathering
The following data was collected for the cable to be sized:
Cable type: Cu/PVC/GSWB/PVC, 3C+E, 0.6/1kV
Operating temperature: 75C
Cable installation: above ground on cable ladder bunched together with 3 other cables on a single layer and at 30C ambient temperature
Cable run: 90m (including tails)
Motor load: 37kW, 415V three phase, full load current = 61A, power factor = 0.85
Protection: aM fuse of rating = 80A, max prospective fault I2t = 90 A2s , 5s melt time = 550A
Step 2: Cable Selection Based on Current Rating
Suppose the ambient temperature derating is 0.89 and the grouping derating for 3 bunched cables on a single layer is 0.82. The overall derating factor is 0.89 0.82 = 0.7298. Given that a 16 mm2 and 25 mm2 have base current ratings of 80A and 101A respectively (based on Reference Method E), which cable should be selected based on current rating considerations?
The installed current ratings for 16 mm2 and 25 mm2 is 0.7298 80A = 58.38A and 0.7298 101A = 73.71A respectively. Given that the full load current of the motor is 61A, then the installed current rating of the 16 mm2 cable is lower than the full load current and is not suitable for continuous use with the motor. The 25 mm2 cable on the other hand has an installed current rating that exceeds the motor full load current, and is therefore the cable that should be selected.
Step 3: Voltage Drop
Suppose a 25 mm2 cable is selected. If the maximum permissible voltage drop is 5%, is the cable suitable for a run length of 90m?
A 25 mm2 cable has an ac resistance of 0.884 Ω/km and an ac reactance of 0.0895 Ω/km. The voltage drop across the cable is:
A voltage drop of 7.593V is equivalent to , which is lower than the maximum permissible voltage dorp of 5%. Therefore the cable is suitable for the motor based on voltage drop considerations.
Step 4: Short Circuit Temperature Rise
The cable is operating normally at 75C and has a prospective fault capacity (I2t) of 90,000 A2s. What is the minimum size of the cable based on short circuit temperature rise?
PVC has a limiting temperature of 160C. Using the IEC formula, the short circuit temperature rise constant is 111.329. The minimum cable size due to short circuit temperature rise is therefore:
In this example, we also use the fuse for earth fault protection and it needs to trip within 5s, which is at the upper end of the adiabatic period where the short circuit temperature rise equation is still valid. Therefore, it's a good idea to also check that the cable can withstand the short circuit temperature rise for for a 5s fault. The 80A motor fuse has a 5s melting current of 550A. The short circuit temperature rise is thus:
Therefore, our 25 mm2 cable is still suitable for this application.
Step 5: Earth Fault Loop Impedance
Suppose there is no special earth fault protection for the motor and a bolted single phase to earth fault occurs at the motor terminals. Suppose that the earth conductor for our 25 mm2 cable is 10 mm2. If the maximum disconnection time is 5s, is our 90m long cable suitable based on earth fault loop impedance?
The 80A motor fuse has a 5s melting current of 550A. The ac resistances of the active and earth conductors are 0.884 Ω/km and 2.33 Ω/km) respectively. The reactances of the active and earth conductors are 0.0895 Ω/km and 0.0967 Ω/km) respectively.
The maximum length of the cable allowed is calculated as:
The cable run is 90m and the maximum length allowed is 108m, therefore our cable is suitable based on earth fault loop impedance. In fact, our 25 mm2 cable has passed all the tests and is the size that should be selected.
Waterfall Charts
Sometimes it is convenient to group together similar types of cables (for example, 415V PVC motor cables installed on cable ladder) so that instead of having to go through the laborious exercise of sizing each cable separately, one can select a cable from a pre-calculated chart.
These charts are often called "waterfall charts" and typically show a list of load ratings and the maximum of length of cable permissible for each cable size. Where a particular cable size fails to meet the requirements for current carrying capacity or short circuit temperature rise, it is blacked out on the chart (i.e. meaning that you can't choose it).
Preparing a waterfall chart is common practice when having to size many like cables and substantially cuts down the time required for cable selection.
A professional, fully customisable Excel spreadsheet template for cable sizing waterfall charts can be purchased from Tradebit.
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