Volt Drop

This is a 2391 question

DURING THE COURSE OF AN INSPECTION ONE OF THE SINGLE PHASE RADIALS supplying an item of fixed equipment is to be checked for volt drop.

45A full load, 80m length, 10mm singles,70 degree thermo singles, steel conduit, mohms/m at 20 degrees=1.83(p166 OSG)

Apparently this is a closed exam. How can this be calculated without reference to the tables??

Originally Posted by buttonmoon

This is a 2391 question

DURING THE COURSE OF AN INSPECTION ONE OF THE SINGLE PHASE RADIALS supplying an item of fixed equipment is to be checked for volt drop.

45A full load, 80m length, 10mm singles,70 degree thermo singles, steel conduit, mohms/m at 20 degrees=1.83(p166 OSG)

Apparently this is a closed exam. How can this be calculated without reference to the tables??

You wouldn't need reference to the tables as all info has been provided.

To get VD the calculation would be (mΩ/m x 1.2 x 2) x Ib x L ÷ 1000:

(1.83 x 1.2 x 2) x 45 x 80 ÷ 1000 = 15.8V (6.9%)

You multiply the conductor resistance by 1.2 to correct it for 70°C, then multiply by 2 to include the neutral.

Now multiply by 45A and then by 80m and divide by 1000.

This is how the values in the VD tables are worked out (mΩ/m x 1.2 x 2 = mV/A/m). Remember that and you won't go far wrong.



Calculation from the VD tables is mV/A/m x Ib x L ÷ 1000:

4.4 x 45 x 80 ÷ 1000 = 15.8V (6.9%)

This is no good as it's over 11.5V (5%)