How to Find The Suitable Size of Cable & Wire for Electrical
Wiring Installation? With Examples
Remember that it's
far very crucial to choose proper wire size while sizing
a wire for electrical installations.
An inappropriate size of wire for heavy loads current can
also create chaos which results in failure of the
electrical system, hazardous fire and serious injuries.
Voltage Drop in Cables
Whenever the current flows through the conductor, there will
be a voltage drops in that conductor. Normally, voltage drop may be neglected
for small length of conductors but in case of a lower diameter and long length
conductors we should consider a significant voltage drops for proper wiring
installation and future load management.
According to Institute of Electrical and Electronics
Engineers (IEEE) rule B-23, at any point between a power supply terminal and
installation, voltage drop should not increase above 2.5% of provided
(supply) voltage
Example:
Let us assume , the supply voltage is 230V AC,
then the value of permissible voltage drop should be;
- Permissible
Voltage Drop = 230 x (2.5/100) = 5.75V
Similarly, if the supply voltage is 110V AC, the Permissible
Voltage Drop should be not more than 2.75V ( 110V x 2.5%).
In electrical wiring circuits, for sub circuits the value of
voltage drop should be half of that permissible voltage drop.
Normally, the voltage drop is expressed in Ampere per meter
(A / m)
Tables & Charts for Proper Cable & Wire Sizes
Below are the important tables which you should follow for
determining the proper size of cable for Electrical Wiring Installation
How to calculate Voltage Drop in a Cable?
Step no. 1: Calculate the maximum allowable voltage drop.
Step no. 2: Calculate
the load current
Step no. 3: After
finding the load current select a proper cable from table 1
Step no. 4: from
table 1 find the voltage drop of the cable and multiply with the length of
cable
Step no. 5: Now
multiply this calculated value of volt drop by load factor where;
Load factor = Load Current to be taken by Cable/ Rated
Current of Cable given in the table.
This is
the value of Volt drop in the cables when load current flows through it.
Step no. 6: If
the calculated value of voltage drop is less than the value calculated in step
(1) (Maximum allowable voltage drop), than the size of selected cable is proper
Step no. 7: If
the calculated value of voltage drop is greater than the value calculated in
step (1) (Maximum allowable voltage drop), than calculate voltage drop for the
next (greater in size) cable and so on until the calculated value of voltage
drop became less than the maximum allowable voltage drop calculated in step
(1).
Example :
For Electrical wiring in a building, Total load is 5kW and
the length of the cable from Main panel to sub circuit is 40 feet. Supply
voltages is 230V and temperature is 40°C.
Find the suitable size of cable which is going through conduits .
Solution:-
Ø Total Load = 5kW
Ø Let us assume at max 20% overload occurs i.e.
1.2*5kW=6kW or 6000W
Now for 6000W load current will
flow i.e. 6000/230= 26.08A
ØNow we have to select the size of cable from
table 1 for 26.08A current which is 7/0.036 (28 Amperes). It means we can use 7/0.036 cable
according to table 1.
Ø Now check the selected (7/0.036) cable with
temperature factor in Table 3, so the temperature factor is 0.94 (in table 3) at 40°C
and current carrying capacity of
(7/0.036) is 28A, therefore, current carrying capacity of this cable at 40°C
(104°F) would be
Current rating for 40°C = 28 x 0.94 = 26.32 Amp.
Maximum current carrying capacity is 26.32 A and
actual is 26.08A.Hence this size of cable (7/0.036) is also suitable.
Ø
Now find the voltage drop for 100 feet for this
(7/0.036) cable is 7V, But in our case, the length of cable is 40 feet. Therefore,
the voltage drop for 40 feet cable would be;
Actual Voltage drop for 40 feet = (7 x 40/100) x (26.08/28)
= 2.608V
And Allowable voltage drop = (2.5 x 220)/100 = 5.5V
Here The Actual Voltage Drop (2.608V) is less than that of
maximum allowable voltage drop of 5.5V. Therefore, the most
suitable cable size is (7/0.036) for that given load.
