# Thread: November 2010 Paper 2 Exam Question 1 and Question 2 (ANSWERS)

1. ## Installation Rules Paper 2 (November 2010 Exam Question 1 and Question 2 -ANSWERS)

PAPER 2 QUESTION 1 OF NOVEMBER 2010
THREE PHASE AC CIRCUITS: PROSPECTIVE SHORT CIRCUIT CURRENT

1. Calculate the PSCC for a three phase fault if the following information is given:

*Transformer capacity 200 kVA at 400/250 volt
*Transformer protection 200 amp per phase
*70 mm2 4 core copper cable + ECC of 100 meters
*Z% of the transformer is 5% ohm

2. Calculate the touch voltage at the load if the ECC is the same size as one of
the phase conductors

3. What is the total loop impedance for the circuit?

1. 2.26 kA

2. 12.748 V

3. 0.1037418 Ohm

Hint For Question 3:
Z(Total) = Z(Transformer) + Z(Phase Conductor) + Z(PEN or PE Conductor)

PAPER 2 QUESTION 2 OF NOVEMBER 2010 VOLTAGE DROP CALCULATION

A 1 kW single phase electrical motor is used as part of a lift system to hoist and deliver a 4 tons of bricks per hour at a building site.
If the supply voltage is 230 V at the basement of the lift with the dumping floor level 65 m higher, calculate the following:

2.1 The percentage volt drop just before the load is dumped (7 Marks)**

NOTE: Take a motor power factor of 0,9 with a hoist speed of 100 r/min through the reduction gearbox when fully loaded. Use a 3 core PVC armoured copper cable of 4 mm2 for the entire installation

2.2 The resistance of the ECC (1 Mark)**

2.3 The touch voltage at the motor if a 5 amp circuit breaker is used (2 Marks)**

2.1 1.50%

2.2 0.3575 Ohm

2.3 3.575V

I have not provided you with the full solution for this section but I have provided answers, try them out and if you have any problems email me and I will provide you with the full solution (Provided that you email me your attempt).

You may visit the installation Rules Study Mate Website http://www.installationrulesstudymate.co.za or send email to info@installationrulesstudymate.co.za

To find out how you can get the full solution to the above questions and many more.

Your Personal Tutor: Installation Rules Paper 1 and Paper 2
Leonard Fikani Ndlovu
Cell: 072 634 8547  Reply With Quote

2. i would be interested in seeing how this works...like a refresher course for us old ballies...i havent got time to review this product now but i will take the time to check it out when i have some time.

are the regulations upto date with the latest amendments?

do you have have a similar thing for master electricians?

could i use this product in a study group?

just from what i have read it seems to cover the whole sans 1042-1  Reply With Quote

3. If you mail me a trial version I'll put it through its paces and write a review for you :-)  Reply With Quote

4. Please tell me if I am heading in the correct direction...I have done this almost 30 years ago.. and I am just trying for no.1 at his stage...

zTransformer = V²/ (P x 1000) x 5/100 = (400)² / (200 x 1000) x (5/100)
= 0.0309 Ohms (i think)

Get R and X from Table E1 on page 307 for 70mm² cable
zECC = (L x √ (R2 + X2)) /1000 = (100 x √ (0.31)2 + (0.074)2)/1000
= 0.03187 Ohms

zConductor (Phase) = zECC (PE or PEN) = 0.3187 Ohms
zTotal =zTransformer + zConductor (Phase) + zECC (PE or PEN)
zTotal = 0.04 + 0.03187 + 0.03187 = 0.10374 Ohm
PSCC = V/√3 x ZTotal = 400/1.732 x 0.31874 =

2.26 kA  Reply With Quote

5. 123] Well done! You are 100% correct.  Reply With Quote

6. Hi AndyD, that will be great but I do not have the trial version at the moment. I am thinking of your offer. Thank you  Reply With Quote

7. Thanks murdock, I will post the full solution soon.  Reply With Quote

8. there are programs you can purchase online which cover all aspects of the code without having to do any calculations yourself...you feed it info it does all the calculations for you then gives you an answer...price around R7000 for the full package.  Reply With Quote

9. ## INSTALLATION RULES NOVEMBER 2010 PAPER 2 QUESTION 1 (ANSWER)

NOVEMBER 2010 INSTALLATION RULES PAPER 2 QUESTION 1

THREE PHASE AC CIRCUITS: PROSPECTIVE SHORT CIRCUIT CURRENT

1. Calculate the PSCC for a three phase fault if the following information is given:
*Transformer capacity 200 kVA at 400 volts
*Transformer protection 200 amp per phase
*70 mm2 4 core copper cable + ECC of 100 meters
*Z% of the transformer is 5% ohm (8 Marks)

2. Calculate the touch voltage at the load if the ECC is the same size as one of
the phase conductors (1 Mark)

3. What is the total loop impedance for the circuit? (1 Mark)

1. We are required to calculate the PSCC for a three phase fault:

Given:
Transformer capacity 200 KVA at 400 volts
Transformer protection 200 amp per phase
70mm² 4 core copper cable + ECC of 100 meters.
Z % of the transformer 5% ( ohm)
In section 8.5.4 on Page 271 of the Regulation book we have:

Z(Transformer) = (V x V)/(P x 1000) x (Z%/100)

Where,

Z(Transformer)  is the source transformer impedance (in ohms)

P  is the power of the transformer (in kVA)

Z%  is the rated short-circuit impedance voltage of the transformer, expressed as a percentage

Z(Transformer) = (V x V)/(P x 1000) x (Z%/100)

= (400V x 400V)/(200kVA x 1000) x (5/100)

= 0.04 Ohms

Next let us obtain the values of R and X from Table E1 on page 307 for a 70mm2 cable and apply the formula on page 272.

Z(ECC)= (L x √ (R x R + X x X)) /1000

= (100 x √ (0.31 x 0.31) + (0.074 x 0.074))/1000

= 0.0318709 Ohms

Z(Conductor) = Z(ECC) [PE or PEN] = 0.0318709 Ohms

Z(Total) =Z(Transformer) + Z(Conductor) + Z(ECC)

Z(Total) = 0.04Ohms + 0.0318709 Ohms + 0.0318709Ohms

= 0.1037418 Ohms

PSCC = V/((√3) x Z(Total))

= 400/((1.7320508) x (0.1037418))

= 2.2261055 kA

2. We are required to calculate the touch voltage at the load if the ECC is the same size as one of the phase conductors

We have R of a 70mm² cable to be 0.31 Ohms/Km from page 307 and we are told that the R(ECC) is the same as one of the phase conductor which is 0.31 Ohms/Km

Touch Voltage (T.V.) = (2 x I(CB) x R(ECC) *CB - Circuit Breaker

= (2 x 200A x 0.31Ω/Km x 100m)/1000

= 12.4 V

3. We are required to calculate the total loop impedance for the circuit

Z(Total) =Z(Transformer) + Z(Conductor) + Z(ECC)

Z(Total) = 0.04 Ohms + 0.03187 Ohms + 0.03187 Ohms = 0.1037418 Ohms

http://www.installationrulesstudymate.co.za
fikani2010@hotmail.com
Wishing you all the best:  Reply With Quote

10. ## INSTALLATION RULES NOVEMBER 2010 PAPER 2 QUESTION 2 (ANSWER)

INSTALLATION RULES PAPER 2 EXAM QUESTION 2 (NOVEMBER 2010)

VOLTAGE DROP CALCULATION

A 1 kW single phase electrical motor is used as part of a lift system to hoist and deliver 4 tons of bricks per hour at a building site.

If the supply voltage is 230 V at the basement of the lift with the dumping floor level 65 m higher, calculate the following:

2.1 The percentage volt drop just before the load is dumped

NOTE: Take a motor power factor of 0,9 with a hoist speed of 100 r/min through the reduction gearbox when fully loaded. Use a 3 core PVC armoured copper cable of 4 mm2 for the entire installation (7 Marks)

2.2 The resistance of the ECC (1 Mark)

2.3 The touch voltage at the motor if a 5 amp circuit breaker is used (2 Marks)

2.1 We are required to calculate the percentage voltage drop just before the load is dumped.

Given:

The Supply Voltage is 230V (voltage measured between phase and neutral)

3 Core Copper Cable is 4mm2 (R = 5.5 Ω/Km and X = 0.093 Ω/Km from Table E1 page 307

***For the simplicity of calculations and the accuracy of results and for the fact that the value of X is negligible when compared to that of R. Let us ignore the influence of X in the installation and use Z (Impedance) = R)***

The Length of the Cable is 65m

True Power of a Single phase motor is 1kW

Motor Power Factor (cos ϕ) is 0.9

Fv is the multiplication factor determined from Table E3 on page 311which is 2

Let Vd be the Voltage Drop to be calculated. We can use about SIX different methods to calculate the Voltage Drop of the installation:

The only unknown from all SIX methods is the load current (I). This tells us that to be able to answer this question we need to determine the current drawn by the motor.

The current drawn by the motor can be calculated using the following formulae:

For a Single phase supply the Power (P) is given by:

P = V x I x cos ϕ

it follows that,

I = P/(V x cos ϕ)

= 1000W/(230V x 0.9)

= 1000W/207V

= 4.8309178A

Please Note: For a Three phase supply the Power (P) is given by:

P = √3 x V x I x cos ϕ

Now that we have the value of (I) we are ready to calculate the voltage drop. We are going to use all SIX methods to authenticate our answer.

METHOD 1
Use this formula from page 308

Vd = (Fv x I x R x L)/1000 [Volts]

= (2 x 4.8309178A x 5.5 Ω x 65m)/1000

= 3.4541061V

The percent voltage drop is (3.454106V/230V) x 100% =1.50178% which is roughly 1.5%

METHOD 2
Use this formula:

Vd = I x R [Volts]

= (4.8309178A x 5.5 Ω x 65m)/1000

= 1.727053V

Note: We need to consider the voltage drop on Neutral as well which is equal to that of the phase.

Use the formula:

Vd (Neutral) = I x R [Volts]

= (4.8309178A x 5.5 Ω x 65m)/1000

= 1.727053V

Vd (Total) = 1.727053V + 1.727053V

= 3.454106V

The percent voltage drop is (3.454106V/230V) x 100% =1.50178% which is roughly 1.5%

METHOD 3

Let us turn to page 105 using Table 6.3(b) determine the value of mV/A/m of a 4mm2 3 Core Copper Cable which is 11 (Single phase circuit)

Use this formula:

Vd = ((mV/A/m) x A x m)/1000 [Volts]

= (11 x 4.8309178A x 65m)/1000

= 3.4541061V

The percent voltage drop is (3.4541061V/230V) x 100% =1.50178% which is roughly 1.5%

METHOD 4

Use this formula:

Vd = (I x ρL)/A [Volts]

= (4.8309178A x 0.0223 x 65m)/4mm2

= 1.7506027V

Note: We need to consider the voltage drop on Neutral as well which is equal to that of the phase.

Use this formula:

Vd (Neutral) = (I x ρL)/A [Volts]

= (4.8309178A x 0.0223 x 65m)/ 4mm2

= 1.7506027V

Vd (Total) = 1.7506027V + 1.7506027V

= 3.5012054V

The percent voltage drop is (3.5012054V/230V) x 100% =1.52226% which is roughly 1.5%

METHOD 5

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 [Volts] [it is on page 308]

The phase angle θ of the load is determined by power factor = cos θ. The resistance R and the reactance X may be obtained from the values obtained in Table E1 page 307.

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

= (2 x 4.8309178A x (5.5Ω x 0.9 + 0.093Ω x 0.9) x 65m)/1000

= (9.6618356A x (4.95 Ω + 0.0837 Ω) x 65m)/1000

= (9.6618356A x (5.0337 Ω) x 65m)/1000

= 3.1612608V

The percent voltage drop is (3.1612608V/230V) x 100% =1.37446% which is roughly 1.4%

METHOD 6
Use this formula:

Voltage drop = (1.72 x I x R) / √3 [Volts]

= (1.72 x 4.8309178A x 5.5 Ω x 65m)/ 1000 x 1.7320508

= (2970.5313)/1732.0508 V

= 1.715037V

Note: We need to consider the voltage drop on Neutral as well which and is essentially the same as that of the phase:
Use the formula:

Vd (Neutral) = (1.72 x I x R) / √3 [Volts]

= (1.72 x 4.8309178A x 5.5 Ω x 65m)/ 1000 x 1.7320508

= (2970.5313)/1732.0508 V = 1.715037V

Vd (Total) = 1.715037V + 1.715037V = 3.430074V

The percent voltage drop is (3.430074V/230V) x 100% =1.49133% which is roughly 1.5%

Conclusion: Most of our methods gave the percentage voltage drop of about 1.5% except when we considered the value of X the voltage drop was about 1.4%.

ALTERNATIVE WAY OF CALCULATING LOAD CURRENT OF THE MOTOR

We can use the following formulae to calculate the full load current (I) drawn by the motor:

Please Note: Formula (a)* is given for your information, just in case next time they decide to use a Three phase motor, but because today we are using a Single phase motor therefore we are going to employ formula (b)* to help us with this Question.

For a Three phase motor: I = P x 1000 / (√3 x V x η x cos ϕ) .(a)*

For a Single phase motor: I = P x 1000 / (V x η x cos ϕ) .(b)*

Where,

I  is the current drawn by the motor in Amps

P  is the True Power of the motor in kW

V  is the voltage between phases if Three phase motor is used. For a Single phase motor V is the voltage between phase and neutral (in Volts).

cos ϕ  is the power factor = True Power in kW/Apparent Power in kVA

Therefore,

Apparent Power in kVA = True Power in kW/Power Factor

= 1kW/0.9

= 1.1kVA

So, if we were given the value of Apparent Power in kVA we would be able to determine

True Power in kW = Apparent Power in kVA x cos ϕ (in kW)

Let us assume that the motor efficiency η (which is explained below) is 1(since when the motor is running (close to or in) its full speed  that is when fully loaded, its efficiency approaches 100%. The Single phase motor have a Centrifugal switch which is wired in series with the start winding and the Capacitor, and the Run winding which is in parallel to the Cap, the Switch and the Start so that when the motor is running at about 75% or above of its full speed the Centrifugal switch opens so as to leave the motor running only with the Run winding.)

Take Note: The True Power in kW of a motor indicates its rated equivalent mechanical power output whereas; the Apparent Power in kVA supplied to the motor is the input Power and is a function of the output (True Power), the motor efficiency and the Power Factor. What this tells us is that we can say:

Apparent Power in kVA = True Power in kW/Motor efficiency x cos ϕ

It follows that,

η- The motor efficiency = output Power in kW/input Power in kW

In other words,

η- The motor efficiency = True Power in kW/Apparent Power in kVA x cos ϕ (in kW)

Since we have already calculated the Apparent Power above we can conclude that:
the motor efficiency = 1kW/(1.1kVA x 0.9) = 1.01010 which is = 1

Now using formula (b)* to calculate (I) in amps:
For a Single phase motor: I = P x 1000 / (V x η x cos ϕ) (b)*
P = 1kW
V = 230V
η = 1
cos ϕ =0.9

We have,
I = P x 1000/(V x η x cos ϕ) (b)*

I = 1kW x 1000/(230V x 1 x 0.9)

= 1000kW/207V

= 4.8309178A
To calculate the percentage voltage drop, you can apply any of the above SIX methods. I hope you followed through.

2.2 We are required to calculate the resistance of the ECC

Given:
L = 65m 4mm2 of a copper (Cu) cable

From Table E1 page 307 R = 5.5 Ω/Km for a 4mm2 Cu

NB: We must express resistance in Ohms. We proceed as follows:

R (in ohms) = Value of R in Ω/Km x L (in meters) where [1Km = 1000m]

So we can say,

R = (Value of R in Ω x L in meters)/1000 [Ohms]

= (Value of R in Ω x L in meters)/1000 [Ohms]

= (5.5 Ω x 65m)/1000 [Ohms]

= 0.3575Ohms

2.3 We are required to calculate the touch voltage at the motor if a 5 amp circuit breaker is used.

What do we have?:

I(CB)  Circuit breaker size = 5 amp ***

R(ECC)  The resistance of the earth continuity conductor which is equal to the other conductors as we are using a 3 Core cable on a Single phase circuit = 0.3575 Ω

Touch Voltage (T.V.) = 2 x I(CB) x R(ECC)

= 2 x 5A x 0.3575 Ω

=10A x 0.3575 Ω

= 3.575V

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