How to select cable size for various loads with calculation with real life examples?

Voltage drop in Cables

Voltage drop is a important factor to consider while selecting cable size. Every cable have internal resistance and this resistance is directly proportional to it's length and inversely proportional to it's diameter[R = ρ (L/a)]. It can be neglect in small diameter and small length wire but it must have to consider in long cable or big diameter cable. As per standard voltage drop between power supply terminal and installation not be increase above 2.5% of supply voltage. i.e. If supply voltage is 220 V then voltage drop not be increase then the 5.5 V. Voltage drop also occur in between installation point(distribution box) & sub-circuit or final sub-circuits and this drop should be half of that allowable voltage drop. i.e. 2.75 V of 5.5 V. Their are two method to describe voltage drop, The SI(System International and Metric System) and FPS(Foot Pound System). In SI system voltage drop describe in A/m and in FPS system it describe as 100 feet.

Tables  for Suitable Cable & Wire Sizes

For proper size of cable or wire the selection tables are given below:

Current carrying capacity (In Amp)	No. of wire and thickness of each wire	Area (In mm2)	Current carrying capacity (In Amp)	No. of wire and thickness of each wire	Area (In mm2)
11	1/1.13	1	11	1/0.044	0.0015
13	1/1.38	1.5	13	3/0.029	0.002
18	1/1.78	2.5	16	3/0.036	0.003
24	7/0.85	4	21	7/0.029	0.0045
31	7/1.04	6	28	7/0.036	0.007
42	7/1.35	10	34	7/0.044	0.01
56	7/1.70	16	43	7/0.052	0.0145
73	7/2.14	25	56	7/0.064	0.0225
90	19/1.53	35	66	19/0.044	0.03
145	19/1.78	50	77	19/0.052	0.04
185	19/2.14	70	105	19/0.064	0.06
230	19/2.52	95	180	19/0.83	0.1

Table 1 - Current rating of Cu cables at 30°C or 86F

Current carrying capacity (In Amp)	No. of wire and thickness of each wire	Area (In mm2)	Current carrying capacity (In Amp)	No. of wire and thickness of each wire	Area (In mm2)
3	16/0.20	0.5	3	14/0.0076	0.0006
6	24/0.20	0.75	6	23/0.0076	0.001
10	32/0.20	1	13	40/0.0076	0.00017
13	40/0.20	1.25	18	70/0.0076	0.003
15	48/0.20	1.5	24	110/0.0076	0.004
20	80/0.20	2.5	31	162/0.0076	0.007

Table 2 - Current rating of flexible cords Cu cables at 30°C or 86F

Temp Factor	1.02	1	0.97	0.94	0.91	0.88	0.77	0.63
Temp F	77	86	95	104	113	122	1131	140
Temp °C	25	30	35	40	45	50	55	60
Temp K	298.15	303.15	308.15	313.15	318.15	323.15	328.15	333.15

Table 3 - Temperature factor

Inside Trunking & Conduit				Conductor
Single phase one cable AC & DC		Three phase three or four core cable		No. & Diameter of wires (In Inch)	Cross Section Area (In Inch2)
Current Rating (A)	Volt Drop/100 Feet (V)	Current Rating (A)	Volt Drop/100 Feet (V)
11	14	9	9.8	1/0.044	0.0015
13	12	11	9.1	3/0.029	0.002
16	11	14	7.7	3/0.036	0.003
21	8.4	18	6.4	7/0.029	0.0045
28	7	23	5.3	7/0.036	0.007
34	5.7	28	4.1	7/0.044	0.01
43	5.5	36	4	7/0.052	0.0145
56	4.8	48	3.5	7/0.064	0.0225
66	4.3	56	3.2	19/0.044	0.03
77	3.6	65	2.7	19/0.052	0.04
105	3.4	88	2.5	19/0.064	0.06

Table 4 - Cable size, Current ratings with Voltage drop (British System)

Inside Trunking & Conduit				Conductor
Single phase one cable AC & DC		Three phase three or four core cable		No. & Diameter of wires (In mm)	Cross Section Area (In mm2)
Current Rating (A)	Volt Drop/Amp-meter(mVolt)	Current Rating (A)	Volt Drop/Amp-meter(mVolt)
11	41	9	35	1/1.13	1
13	28	12	24	1/1.38	1.5
18	17	16	15	1/1.78	2.5
24	11	22	9.1	7/0.85	4
31	7	27	6	7/1.04	6
40	4.1	37	3.6	7/1.35	10
53	2.6	47	2.2	7/1.70	16
60	1.7	53	1.5	7/2.14	25
74	1.2	65	1	19/1.53	35

Table 5 - Cable size, Current ratings with Voltage drop (SI/Decimal/Metric System)

Steps to find out voltage drop in a cable

Step 1-Find the maximum allowable voltage drop.

Step 2-Find load current and according to load current, select a proper cable from table 1(which current rating nearest to the load current).

Step 3-Find the voltage drop in meter or 100feet according it's rated current from table 1.

Step 4-Calculate the voltage drop for the actual length of wiring circuit according to it's rated current by using these formulas.

Voltage drop (In per meter) = (Actual length of circuit * voltage drop for 1m) /100

Voltage drop (In 100feet) = (Actual length of circuit * volt drop for 100ft) /100

Step 5-Now, find out voltage drop by using load factor

Load factor = Load current of cable/ Rated current of cable

Find out the rated current of cable from table and this value of load factor is voltage drop in the cables when the current is passing through it. If this value of voltage drop is less then the maximum allowable voltage drop which calculated in step 1, then the selected size of cable is proper but if it is greater then the step 1 max. allowable voltage drop then calculate the voltage drop greater size until it value becomes less then the step 1 voltage drop.

Examples - Find out proper cable size for given load

To determine cable size keep in mind these rules of voltage drop.

• For a given load, there should be 20% extra scope of current for additional, emergency or future needs (except the known value of current).
• From Energy meter to DB, voltage drop should be 1.25% and 2.5% for final sub-circuit should not exceed 2.5% of Supply voltage.
• Use temperature factor for considering change in temperature from table 3 and consider load factor while finding cable size.
• Selecting cable size consider the wiring system (Temperature is low in open wiring system then the conduit wiring)
• Consider diversity factor in wiring installation

Solved examples for cable size selection

Example 1 - For Electrical wiring installation in a building, Total load is 4.5kW and total length of cable from energy meter to sub circuit distribution board is 35 feet. Supply voltages are 220V and temperature is 40°C (104°F). Find the most suitable size of cable from energy meter to sub circuit if wiring is installed in conduits. 

Solution-(British System)

Total Load = 4.5 kW = 4.5 x 1000 W = 4500 W and 20% additional load = 4500 x (20/100) = 900 W

∴Total Load = 4500 W + 900 W = 5400 W

Total Current = I = P/V = 5400 W /220 V =24.5 A

For the load current the size of cable is 7/0.036(28 A) from table. Now, check the temperature factor from table 3, it is 0.94 at 40°C and current carrying capacity of cable size 7/0.036 is 28 A. So,

Current rating for 40°C = 28 x 0.94 = 26.32 A
It  is less then the current carrying capacity of selected cable which is 28 A, therefore it is also suitable with respect to temperature.

Now find the voltage drop for 100 feet for this (7/0.036) cable from Table 4 which is 7 V, But in our case, the length of cable is 35 feet.  Therefore, the voltage drop for 35 feet cable would be;

Actual Voltage drop for 35 feet = (7 x 35/100) x (24.5/28) = 2.1 V

And Allowable voltage drop = (2.5 x 220)/100 = 5.5 V

Here, The actual voltage drop (2.1 V) is less than that of maximum allowable voltage drop of 5.5 V. Therefore, the cable size (7/0.036) is suitable for that given load.

Example 2- What type and size of cable suits for given situation

Load = 5.8 kW

Volts = 230 V

Length of Circuit = 35 meter

Temperature = 35°C (95°F)

Solution:-( SI / Metric / Decimal System )

Load = 5.8 kW = 5800 W

Voltage = 230 V

Current = I = P/V = 5800 / 230 = 25.2 A

20% additional load current = (20/100) x 5.2 A = 5 A

Total Load Current = 25.2 A + 5 A = 30.2 A

From the table 1, size of cable is 7/1.04 for 31 A and temperature factor is 0.97 for 35°C from table 3.

Current rating for 35°C = 31 x 0.97 = 30 A

It  is less then the current carrying capacity of selected cable which is 31 A, therefore it is also suitable with respect to temperature.

Now find the voltage drop for per ampere meter for this (7/1.04) cable from Table 5 which is 7 mV, But in our case, the length of cable is 35 meter.  Therefore, the voltage drop for 35 meter cable would be;

Actual Voltage drop for 35 meter = mV x I x L = (7/1000) x 30×35 = 7.6 V

Allowable voltage drop = (2.5 x 230)/100 = 5.75 V

Here, the actual voltage drop is greater than then the maximum allowable voltage drop of 5.75 V. Therefore, the cable size (7/1.04) is not suitable for that given load. Now, find out voltage drop again by selecting higher cable size. Let's select 7/1.35 and calculate again.

From table 5, cable rating for selected cable size is 40 A and voltage drop is 4.1 mV.  So, actual voltage drop for 35 meter is

Actual Voltage drop for 35 meter =(4.1/1000) x 40×35 = 7.35 V = 5.74 V

Now, The actual voltage drop is less than that of maximum allowable voltage drop. Therefore, the cable size (7/1.04) is suitable for that given load.


Example 3- Following Loads are connected in a building:

For Sub-Circuit 1

2 lamps each o 1000 W and

4 fans each of 80 W

2 TV each of 120 W

For Sub-Circuit 2

6 Lamps each of 80 W and

5 sockets each of 100 W

4 lamps each of 800 W

If supply voltages are 230 V then calculate circuit current and Cable size for each Sub-Circuit?

Solution:

Total load of Sub-Circuit 1 = (2 x 1000) + (4 x 80) + (2×120) = 2000 W + 320 W + 240 W = 2560 W

Current for Sub-Circuit 1 = I = P/V = 2560/230 = 11.1 A

Total load of Sub-Circuit 2 = (6 x 80) + (5 x 100) + (4 x 800) = 480 W + 500 W + 3200 W= 4180 W

Current for Sub-Circuit 2 = I = P/V = 4180/230 = 18.1 A

From the table, cable size for sub circuit 1 is 3/.029” (13 A) or 1/1.38 mm (13 A) and for sub-circuit 2 is 7/.029” (21 A) or 7/0.85 mm (24 A).

Total Current drawn by both Sub-Circuits = 11.1 A + 18.1 A = 29.27 A

So, cable size for main-circuit is 7/.044” (34 A).

Example 4-A 10 H.P (7.46 kW) three phase squirrel cage induction motor of continuous rating using Star-Delta starting is connected through 400 V supply by three single core PVC cables run in conduit from 250 feet (76.2 m) away from multi-way distribution fuse board. Its full load current is 19 A. Average summer temperature in Electrical installation wiring is 35°C (95°F). Calculate the size of the cable for motor?

Solution:

Motor load = 10 HP = 10 x 746 = 7460 W (∵1 HP = 746 W)

Supply Voltage = 400 V (3-Phase)

Length of cable = 250 feet (76.2 m)

Motor full load Current = 19 A

Temperature factor for 35°C (95°F) = 0.97 (From Table 3)

Size of cable for 19 A full load current is 7/0.36” (For 23 A) in the table. Here, 3-phase system and voltage drop is 5.3 V fro 100 feet. According to table 4 cable size 7/0.036 can be use. From the table 3, the temperature factor is 0.97 at 35°C and current carrying capacity of selected wire is 23 A. So,

Current rating for 40°C (104°F) = 23 x 0.97 = 22.31 A

It  is less then the current carrying capacity of selected cable which is 23 A, therefore it is also suitable with respect to temperature.

Load factor = 19/23 = 0.826

From the table 4, voltage drop is 5.3 V for selected cable at 100 feet and here, length of cable is 250 feet. therefore,

Actual voltage drop (250 feet) = (5.3 x 250/100) x 0.826 = 10.94 V

And maximum Allowable voltage drop = (2.5/100) x 400 V= 10 V

Here, the actual voltage drop is greater than then the maximum allowable voltage drop of 10 V. Therefore, the cable size (7/0.036) is not suitable for that given load. Now, find out voltage drop again by selecting higher cable size. Let's select 7/0.044 cable size and calculate again.

Actual voltage drop (250 feet) = (4.1/100) x 250 x 0.826 = 8.46V

And Maximum allowable voltage drop = (2.5/100) x 400V= 10V

Now, The actual voltage drop is less than that of maximum allowable voltage drop. Therefore, the cable size (7/0.044 is suitable for that given load.