SAMPLE EXAMPLE:

Name FOUR different methods of how the voltage drop of a cable can be determined and give examples for each method.

ANSWER:

EXAMPLE: Suppose a balanced three phase load of 100A is to be supplied using a 35mm2 copper cable over a distance of 100m. Calculate the Voltage Drop (USING FOUR DIFFERENT METHODS).

Let Vd stand for Voltage Drop to be calculated.

We are given: I = 100A and L = 100m (35mm2 of a copper (Cu) cable)

From Table E1 page 307 R = 0.63 Ω/Km for a 35mm2 Cu. NB: We must express resistance in Ohms, to do this use:

R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)

METHOD 1

Use this formula taken from page 308 of the Regulation Book:

Vd = (Fv x I x R x L) /1000

Where,

Vd is the voltage drop, in volts

Fv is the multiplication factor determined from Table E3 on page 311 of the Regulation Book and Fv = 1 for a balanced three-phase load

I is the current in amperes (100A)

R is the resistance in ohms per kilometer (Ω/Km) from Table E1 and it is 0.63 Ω/Km for a 35mm2 copper cable

L is the length in meters (100m)

Vd = (Fv x I x R x L) /1000

So substituting values we have

Vd = (1 x 100A x 0.63 Ω/Km x 100m)/1000 = 6.30 V

METHOD 2

Use: Vd = I x R

Where,

Vd is the voltage drop, in volts

I is the current in amperes

The value of R in ohms/Km from Table E1 is 0.63 Ω/Km for a 35mm2 copper cable

Now determine R in ohms using:

R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)

R = (0.63 Ω/1000m) x 100m = 0.063 Ω

Vd = I x R

So substituting values we have

Vd = I x R = 100 A x 0.063 Ω = 6.30 V

METHOD 3

Use this formula Vd = (mV/A/m) x A x m [Volts]

Where,

mV is the unit of voltage in milli-volts. NB 1V = 1000mV

A is the unit of current in amps [100A]

m is the unit of length of the cable in meters [100m]

(mV/A/m) = 1.10 for a 35mm2 copper (Cu) determined from Table 6.3(b) on page 105 or Table 6.4(b) on page 107 of the Regulation Book â€“ The difference between the two tables is that Table 6.3(b) makes provision for a 1mm2 whereas Table 6.4(b) does not. In the exam you will most probably be given Table 6.3(b) [Please familiarize yourself with the use of these Tables and also Table 6.4(a) â€“ Current Carrying Capacity]. NB: From the Tables 6.3(b) or 6.4(b) take the value of z (the impedance) corresponding to the given cable size and not the value of r (resistance) or x (the reactance).

Vd = (mV/A/m) x A x m [Volts]

So substituting values we have

Vd = (mV/A/m) x A x m = 1.10mV/A/m x 100A x 100m = 11000mV = 11V

For phase to neutral Vd = 11V/√3 = 6.35V *[NB: See section 6.2.7 page 120]

METHOD 4

Use Vd = (I x pL)/A

Where,

A is area of cable in this case 35mm2

p = 0.0223, the resistivity of copper conductors

Vd = (I x pL)/A

So substituting values we have

Vd = (I x pL)/A = (100A x 0.0223 x100m) / 35 mm2 = 6.37 V

ADDITIONAL METHODS

In the case of a load with impedance, the voltage drop can be calculated from the following

Vd = (Fv x I x (R cosθ + X cosθ) x L)/1000

The phase angle θ of the load is determined by power factor = cos θ

The reactance X may be obtained from cable manufactures

OR

Voltage drop = (1.72 x I x R) / √3

OR Alternatively the value of the Voltage Drop may be derived from a monogram

END OF SAMPLE EXAMPLE:

APRIL 2010 EXAM QUESTION

QUESTION 1: SANS 10142 PART 1 OF 2008: ANNEXURE E: VOLT DROP (COMPULSORY)

1.1 With reference to FIGURE 1, on the DIAGRAM SHEET (attached), calculate the following:

1.1.1 The total volt drop of the supply (4)

1.1.2 The total resistance of the supply cable (1)

1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method. (3)

1.2 By making use of TABLE E1 (attached) only, calculate the actual length of a 10 mm2 copper ECC conductor if the total resistance of the conductor is 0,22 ohm. (2)

[10]

ANSWER TO QUESTION 1 (PRIL 2010 PAPER II:

1.1. With reference to FIGURE 1, on the DIAGRAM SHEET (attached), calculate the following:

1.1.1. The total volt drop of the supply (4 Marks)

Assumption: The installation uses copper cables

NB: We know that the voltage drop shall not exceed 11.5V [5% of 230V] according to the code.

Given:

Point of control (DB): Supply voltage is 230V

The resistive load of 50A is 65m away from the point of control [I=50A, L=65m]

From the Point of control to the cable JOINT we have 40m of 16 mm2 copper cable [R = 1.4 Ω/Km, given this from Table E1]

From the cable JOINT to the load we have 25m of 10 mm2 copper cable [R = 2.2 Ω/Km, given this from Table E1]

Use this formula from page 308 Vd = (Fv x I x R x L) /1000

Where,

Vd is the voltage drop, in volts

Fv is the multiplication factor determined from Table E3 on page 311 we have Fv = 2 for a single-phase load

I is the current in amperes

R is the resistance in ohms per kilometer (Ω/Km) from Table E1

L is the length in meters [40m and 25m lengths of different cables]

Vd = (Fv x I x R x L) /1000

So substituting values we have:

Vd = (2 x 50A x 1.4 x 40m)/1000 + (2 x 50A x 2.2 x 25m)/1000 = 5.6V + 5.5V = 11.10 V

1.1.2 The total resistance of the supply cable (1 Mark)

R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)

Resistance of a 16 mm2 = 1.4 Ω/Km x 40m = 1.4 Ω/1000m x 40m = 0.056 Ω

Resistance of a 10 mm2 = 2.2 Ω/Km x 25m = 2.2 Ω/1000m x 25m = 0.055 Ω

Total Resistance of the supply cable = Resistance of a 16 mm2 + Resistance of a 10 mm2=0.056 Ω + 0.055 Ω = 0.111 Ω

1.1.3 Test the answer in QUESTION 1.1.1 above by making use of any other method. (3 Marks)

Alternative method 1:

Use: Vd = I x R

Where,

Vd is the voltage drop, in volts

I is the current in amperes [which is 50A]

The value of R in ohms/Km from Table E1 is 1.4 Ω/Km for a 16mm2 and 2.2 Ω/Km for a 10mm2 copper cable

We have already determined R in ohms from the previous question [0.111 Ω]

Vd = I x R

So substituting values we have

Vd = I x R = 50A x 0.111 Ω = 5.55 V

NB: Remember that we have to account for the voltage drop on neutral as well, which is 5.55V in this case.

Therefore, Total Voltage Drop = 5.55V + 5.55V = 11.10V

Alternative method 2:

Use Vd = (I x pL)/A

Where,

A is area of cable [16mm2 and 10mm2]

p = 0.0223, the resistivity of copper conductors

Vd = (I x pL)/A

So substituting values we have

Vd on a 16mm2 cable = (I x pL)/A = (50A x 0.0223 x40m) / 16mm2 = 2.7875V

Vd on a 10mm2 cable = (I x pL)/A = (50A x 0.0223 x25m) / 10mm2 = 2.7875V

Vd of the supply = Vd on a 16mm2 + Vd on a 10mm2 = 2.7875V + 2.7875V = 5.575V

Again remember that we have to account for the voltage drop on neutral as well, which is 5.575V in this case.

Therefore, Total Voltage Drop = 5.575V + 5.575V = 11.15V

Alternative method 3:

*** Use this formula Vd = (mV/A/m) x A x m [Volts]

Determine the mV/A/m either from Table 6.3(b) or 6.4(b)

mV/A/m for a 16mm2 is 2.8

mV/A/m for a 10mm2 is 4.4

Vd = (mV/A/m) x A x m [Volts]

So substituting values we have:

Vd on a 16mm2 cable = mV/A/m x A x m = 2.8 mV/A/m x 50A x 40m = 5600 mV = 5.60V

Vd on a 10mm2 cable = mV/A/m x A x m = 4.4 mV/A/m x 50A x 25m = 5500 mV = 5.50V

Vd of the supply = Vd on a 16mm2 + Vd on a 10mm2 = 5.60V + 5.50V = 11.10V

*** Please note that neither Table 6.3(b) nor Table 6.4(b) was given in the exam, so we would not be able to use this method to substantiate our answer, but it is given here to prove that our answer is indeed right.

1.2. By making use of TABLE E1 (attached) only, calculate the actual length of a 10 mm2 copper ECC conductor if the total resistance of the conductor is 0,22 ohm. (2 Marks)

Recall that resistance in Ohms can also be obtained by using the following method:

R = Value of R in Ω/Km (for a given cable size from page 307 of the Regulation Book) x L (in meters). We also have to convert Km to meters: 1Km = 1000m. Therefore, R in Ohms = Value of R in Ω/1000m x L (in meters)

We are given 0.22 ohms and we have to calculate the actual length of a 10 mm2 ECC, we can apply the above formula:

We are given R = 0.22 ohm

From Table E1, R in Ω/Km is 2.2 for a 10 mm2

The only unknown now is the length (L) of the cable

R = Value of R in Ω/1000m x L (in meters)

So substituting values we have

0.22 ohm = 2.2 ohm/1000m x L

0.22 ohm = 0.00022 ohm/m x L

L = 0.22 ohm / (0.00022 ohm/m)

L = 100m

Therefore, the actual length of a 10 mm2 ECC is 100m for a given resistance value

Refer to figure 4 and calculate the following:

1. Calculate the minimum cable size for the loads.

2. Touch voltage at L3.

3. PSCC at L1

NOTE: Use a three core cable and ignore suppliers transformer and mains supply characteristics.

FIGURE 4

CIRCUIT DIAGRAM NOT ACCORDING TO SCALE

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