MVA Method Short Circuit Calculationhttp://s.pangonilo.com/index.php/tutorials

Assumptions for MVA Method Short Circuit Calculations

For Short Circuit calculations

The Power Utility, generators and all motors are sources of electrical energy or short circuit currents.

Transformers, reactors and cables limit short circuit currents. Capacitors and static loads such as heaters and lighting do not contribute to short circuit

current. If actual motor impedance are not known

o All motors 37kW or less are lumped and assigned an impedance Z = 25%.o All motors abover 37kW are lumped and assigned an impedance Z = 17%.

For motors, 1 HP = 0.75 kW = 1 KVA

Short Circuit KVA of Circuit Elements

Utility: KVASC = Utility FAULT DUTY (KVA)Example:Fault Duty = 0.04 pu @ 100MVAKVASC = (100/0.04) x 1000 = 2,500,000 kVA

Generator: KVASC = (100 x KVAG) / %Z = KVAG / X"d

Example:Generator 50 MVA, 11 000 V, X"d = 0.113KVASC = (50 x 1000) / 0.113 = 442,478 kVA

Motor: KVASC = (100 x KVAM) / %Z = KVAM / X"d

Example:Motor 1500 HP, 4000V, FLA = 193, X"d = 0.167.KVASC = 1500 / 0.167 = 9000 kVA

Transformer: KVASC = (100 x KVAT) / %Z = KVAT / Zpu

Example:Transformer 132kV / 11kV, 3PH, 50/56MVA @ 55OC, 66.5 / 74.5 MVA @ 65OC, OA/FA, Z = 9% @ 50MVAKVASC = 50 x 1000 / 0.09 = 555,555 kVA

Reactor: KVASC = (1000 x KV2) / Z (ohms)Example:Reactor 11 kV, 0.125 ohmsKVASC = 112 x 1000 / 0.125 = 1,523,520 kVA

Cable: KVASC = (1000 x KV2) / Z (ohms)Example:Cable : 3/C - 185 mm2, 400V, 150 m, R = 0.0258 / km, X = 0.027 / km

Z = (0.02582 + 0.0272)0.5 x 150 / 1000 = 0.0056 ohmsKVASC = 0.402 x 1000 / 0.0056 = 28,571 kVA

Combining KVAsKVAs in series. The total KVAs in series (KVAtotal) is the reciprocal sum or inverse sum of all series KVAs.

KVAs in parallel. The total KVAs in parallel (KVAtotal) is the arithmetic sum of all parallel KVAs in parallel.

KVA total = (KVA1 x KVA3) / ( KVA1+KVA3) + ( KVA2 x KVAn) / KVA2+KVAn)

To give an example to clarify that KVA in parallel are are added, say if you have two 500KVA transformers in parallel, what is the total KVA? It will be 1000KVA and not 250KVA.

MVA Method Short Circuit Calculation

In this example, we shall be presenting a short circuit study of a power system. Motors are are already lumped with ratings 37kW and below assigned an impedance value of 25% while larger motors are 17%. A 4MVA generator is also included into the system to augment the utility.

Figure 1

Utility: 33KV, 250 MVAsc

Transformer 1: 10 MVA, 33/11KV, 9% Z

11KV BusGenerator: 3MVA, X"d = 0.113Transformer 2: 5 MVA, 11/6.6KV, 7% ZMotor 1: 5MVA (Lumped), 17% Z

6.6KV BusTransformer 3: 2 MVA, 6.6KV/400V, 6% ZMotor 3: 6.8 MVA (Lumped), 17% Z

400V BusMotor 4: 300 KVA (Lumped), 17% ZMotor 5: 596 KVA (Lumped), 25% Z

In the event of a short circuit, the sources of short circuit current are1. Utility

2. Generators3. Motors

Static loads such as heaters and lighting do not contribute to short circuit.

The "Equivalent MVA" are:

Transformers and Motors

Generators

Cables and Reactors

In Figure 1, I have calculated the Equivalent MVAs of each equipment, writing it below the ratings.

Figure 2

Utility: MVAsc = 250MVATransformer 1: MVAsc = 10 / 0.09 = 111.11 MVA

11KV BusGenerator: MVAsc = 4 / 0.113 = 35.4 MVATransnformer 2: MVAsc = 5 / 0.07 = 71.43 MVAMotor 1: MVAsc = 5 / 0.17 = 29.41 MVA

6.6KV BusTransformer 3: MVAsc = 2 / 0.06 = 33.33 MVAMotor 3: MVAsc = 6.8 / 0.17 = 40 MVA

400V BusMotor 4: MVAsc = 0.3 / 0.17 = 1.76 MVAMotor 5: MVAsc = 0.596 / 0.25 = 2.38 MVA

Figure 3

Upstream ContributionStarting from the utility, combine MVAs writing each one above the arrows.

At Transformer 1:MVAsc @ 33KV = 250 MVAMVAsc @ 11KV = 1/ (1 / 250 + 1 /111.11) = 76.87 MVA

At Transformer 2:MVAsc @ 11KV = 76.87 + 35.4 + 29.41 = 141.68 MVAMVAsc @ 6.6KV = 1/ (1 / 141.68 + 1 / 71.43) = 47.49 MVA

At Transformer 3:MVAsc @ 6.6KV = 47.49 + 40 = 87.49 MVAMVAsc @ 400V = 1/ (1 / 87.49 + 1 / 33.33) = 24.14 MVA

At 400V MotorsMotor 3: MVAsc = 24.14 x 1.76 / ( 1.76 + 2.38 ) = 10.26 MVAMotor 4: MVAsc = 24.14 x 2.38 / ( 1.76 + 2.38 ) = 13.88 MVA

Downstream Contribution

Starting from the bottom (400V Bus), I combined MVAs writing each one below the arrows. In this bus, the motor contribution to short circuit is the sum of the MVAs of the lumped motors Motor 3 and Motor 4.

At Transformer 3:MVAsc @ 400V = 1.76 + 2.38 = 4.14 MVAMVAsc @ 6.6KV = 1/ (1 / 4.14 + 1 / 33.33) = 3.68 MVA

At Transformer 2:MVAsc @ 6.6KV = 3.68 + 40 = 43.68 MVAMVAsc @ 11KV = 1/ (1 / 43.68 + 1 / 71.43) = 27.11 MVA

At Transformer 1:MVAsc @ 11KV = 27.11 + 29.41 + 35.4 = 91.92 MVANote: Two downstream plus the generator contribution.MVAsc @ 33KV = 1/ (1 / 91.92 + 1 /111.11) = 50.3 MVA

To determine the Faults Current at any bus on the power system, add the MVA values above and below the arrows. The sum should be the same on any branch.Example:11 KV Bus:From Transformer 1: MVAsc = 76.87 + 91.92 = 168.79 MVAFrom Generator : MVAsc = 35.4 + 133.39 = 168.79 MVAFrom Transformer 2: MVAsc = 141.68 + 27.11 = 168.79MVAFrom Motor 1: MVAsc = 139.38 + 29.41 = 168.79 MVA

This is a check that we have done the correct calculation.Ifault @ 11KV = 168.79 / (1.732 x 11) = 8.86 kA

All we have done above are three phase faults, you may ask, how about single phase faults?

For single phase faults, positive sequence, negative sequence and zero sequence impedances need to be calculated.

If = 3 (I1 + I2 + I0)

Examining the circuit in Figure 1, at the 400V Bus, on Transformer 3 contributes to the zero sequence current.

For transformers, the negative sequence and zero sequence impedance are equal to the positive sequence impedance.

Z1 = Z2 = Z0 orMVA1 = MVA2 = MVA0

At the 400V Bus

1 / MVAsc =1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)1 / MVAsc = 1/3 (1 / 28.28 + 1 / 28.28 + 1 / 33.33 )MVAsc =3 x  9.93 = 29.79  MVA

If = 29.79 / (1.732 x 0.4) = 43 kA

At 6.6KV Bus

1 / MVAsc = 1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)1 / MVAsc = 1/3 (1 / 91.17 + 1 / 91.17 + 1 / 71.43)MVAsc = 3 x 27.83  = 83.49 MVA

If = 83.49 / (1.732 x 6.6) = 7.26 kA

Conclusion:This example illustrates that using the MVA Method of Short Circuit Calculation, it will be very easy to calculate the fault current at any node within a power system.

MVA Method Load Flow Calculation

In previous tutorials for the MVA method, we have discussed the importance of Short Circuit Study, combining KVAs and Short Circuit Calculations for three (3) phase and single phase faults. In this part of the tutorial, we will be discussing about Load Flow and Voltage Dips during motor starting.

We shall be using the same single line diagram as in the previous tutorial, except that the 2 x 150 MVa motors at the 400V Bus are now separate. We will calculate the Load Flow and Voltage dips for starting a 150MVA motor at the 400V Bus when all loads are running.

In Load Flow Calculations, the "Equivalent MVA" of equipment are:

Utility: MVAsc = Utility Fault Duty ( MVAsc )

Generator : MVAsc = 100 x MVAG / %Z = MVAG / X"d

Running Motor: MVAM = Motor HP / 1000 = Motor KW / 0.75

Starting Motor : MVAS = 6 x Motor HP / 1000 = 6 x Motor KW / 0.75

Transformer:

Reactor & Cable:

Static Loads: MVASC = Actual MVA

The differences between a Short Circuit Calculation and Load Flow Calculation are:1. Motor MVAs are equal to the rated Equivalent MVAs.2. Starting Motor MVAs are equal to 6 times rated Equivalent MVAs.3. Static Loads are included in the calculations.

In this Load Flow Calculation, we assume that a 150 MVA motor is being started at the 400V Bus after all loads has been started.

In Load Flow, the upstream contribution is equal to the downstream. Available MVAs are equal to the ratio of the MVAsc of each branch.

At 11KV Bus:Upstream: MVAt = 76.87 + 35.4 = 112.27 MVADownstream: MVAt = 7.49 + 5.0 = 12.49 MVA

Voltage dip at the 400V Bus during motor starting

Vd = 2.97 / (2.97 + 0.9) = 76.7 % or 306.8 V

The above calculation indicates that when all loads are already running, starting a 150MVA motor at the 400V Bus is not possible as the voltage dip is lower that the 85% which is the minimum starting voltage for a motor. It means that operational procedures should be in placed to have efficient plant operations.

Here are the steps in calculating the values in above Figure 2:

I. Downstream Values:Note: These are the values above the arrow.33KV Bus:

MVAsc = 250MVA

11KV Bus

10MVA Transformer:

4MVA Generator:

MVAsc = 35.4MVA

From here, we are not able to go any further since we need values from downstream to be able to calculate the equivalent MVA between the 5MVa transformer and the 5MVa motor.

II. Upstream Values:

At 400V Bus:

At 6.6KV Bus:

2MVA Transformer:

5MVA Transformer:

At 11KV Bus:

5MVA Transformer:

10MVA Transformer:

4MVA Transformer:

At 33KV Bus:

10MVA Transformer:

At this point, we have completed all the values in the upstream MVA contribution for load flow.

Let us now go back to the downstream calculation where we stopped. (Note: I will leave this calculation for you to do).

Let us now calculate the bus voltages:@400V Bus:

The calculation will be the same for the other buses. (Note: I will leave that to you too).

MVA Method Unbalance Fault CalculationIn the MVA Method Short Circuit Calculation tutorial, we have discussed how to calculate the three (3) phase and phase to ground fault currents.

The faults currents for three (3) phase unbalance faults are as follows:

1. Three (3) Phase Fault

2. Phase to Ground Fault

3. Phase to Phase Fault

4. Phase to Phase to Ground Fault

Using the same example as in the MVA Method Short Circuit Calculation tutorial, we will calculate the

1. Phase to phase Fault2. Phase to phase to Ground Fault

Figure 1

In the above example, we calculated the Single Phase to Ground Faults

At the 400V Bus

MVAsc =3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)

MVAsc = 3 (1 / 28.28 + 1 / 28.28 + 1 / 33.33 )

MVAsc =3 x 9.93 = 29.79 MVA

If = 29.79 / (1.732 x 0.4) = 43 kA

At 6.6KV Bus

MVAsc = 3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)

MVAsc = 3 (1 / 91.17 + 1 / 91.17 + 1 / 71.43)

MVAsc = 3 x 27.83 = 83.49 MVA

Iff = 83.49 / (1.732 x 6.6) = 7.26 kA

Knowing the values of MVA1, MVA2 and MVA0, we will be able to calculate the following unbalanced faults.

Phase to phase Faults

At the 400V Bus

Using MVAs instead of KVAs

MVAsc = 1.732 ( MVAsc1) / 2

MVAsc = 1.732  ( 28.28 ) /2

MVAsc =24.5 MVA

If = 24.5 / (1.732 x 0.4) = 35.4 kA

At 6.6KV Bus

MVAsc = 1.732 ( MVAsc1) / 2

MVAsc = 1.732  ( 91.17 ) /2

MVAsc =78.96 MVA

If = 78.96 / (1.732 x 6.6) = 6.91 kA

At the 400V Bus

Using MVAs instead of KVAs

MVAsc = 3  x 28.28 x 33.33 / (28.28 + 28.28 + 33.33 )

MVAsc = 31.46 MVA

If = 31.6 / (1.732 x 0.4) = 45.4 kA

At 6.6KV Bus

MVAsc = 3 x 91.17 x 71.43 / (91.17 + 91.17 + 71.43)

MVAsc =77 MVA

If = 77 / (1.732 x 6.6) = 6.73 kA

Voltage Drop Calculation

This article is the first of a series in the selection of electrical cables.

Cable selection is an important part of electrical installation design. Selecting the correct size of cable could lower initial cost, lower operating cost, better voltage regulation notwithstanding the safety factor.

Cable parameters are provided in values per unit length, usually ohms per kilometer(Ω/km) for IEC or miles(Ω/mi) for North American Standards.

The following table illustrate the typical parameters for industrial cables

Table 1. 600/1000 V. Cu/EPR/CSP/GSWB run on open trays or enclosed in air. 1 to 6 x single, 3 or 4 cores

Nominalconductorarea (mm2)

Single coresin trefoil 3 cores

Approximatearmouringresistance

Resistanceat 90oC

(ohm/km)Reactance

at 50Hz

Resistanceat 90oC

(ohm/km)Reactance

at 50Hz

GSWB at 60oC3â4 cores��(ohm/km)

1.5 15.6 0.185 15.6 0.118 46.22.5 9.64 0.173 9.64 0.111 51.34 5.99 0.163 5.99 0.108 60.36 3.97 0.153 3.97 0.105 30.510 2.35 0.148 2.35 0.0983 36.716 1.48 0.134 1.48 0.0933 23.125 0.936 0.125 0.936 0.0892 28.135 0.674 0.121 0.674 0.0867 10.4350 0.499 0.118 0.499 0.0858 11.8170 0.344 0.112 0.344 0.0850 13.6195 0.271 0.108 0.271 0.0825 10.87120 0.214 0.106 0.214 0.0808 11.92150 0.175 0.105 0.175 0.0808 7.38185 0.140 0.105 0.140 0.0808 8.15240 0.108 0.103 0.108 0.0800 8.94300 0.087 0.101 0.087 0.0800 10.10400 0.069 0.0992 0.069 0.0795 10.00

This article is the second part on the discussion about voltage drop calculation. We shall be providing real world example for you to be able to appreciate it better as it may have already been a part of what you have already done or currently doing in your design.

We shall be presenting an IEC standard equipment in our sample calculation.

A 120 mm2 3core XLPE insulated cable 120m in length supplies a 110 kW induction motor that has a starting current of 6 times the full-load current of 180 amps. The starting power factor is 0.25 lagging.

If the sending end line-to-line voltage is 400 volts, the specific resistance r and reactance x for the cable are 0.197 and 0.072 ohm/km respectively at 90OC and 50 Hz.

Find the percentage volt-drop on starting the motor.

Solution

For the 120 mm2 cablethe series impedance is:R = rl = 0.214 x 120 / 1000 = 0.02568 ohms/phaseX = xl = 0.0808 x 120 / 1000 = 0.009696 ohms/phase

For the 110kW motorthe starting current is:I = 6 × 180.0 = 1080.0 AThe power factor is:cosØ = 0.2500, therefore sinØ = 0.9682

Let us assume the sending voltage remains constant at 400 volts,

AB = 1080.0 x 0.02568 x 0.2500 = 6.9336 volts/phaseBE = 1080.0 x 0.02568 x 0.9682 = 26.6805 volts/phaseEF = 1080.0 x 0.009696 x 0.9682 = 10.1386 volts/phaseDF = 1080.0 x 0.009696 x 0.2500 = 2.6179 volts/phase

%Vd = ( 1.732 / 400 ) x 1080.0 x ( 0.02568 x 0.25 + 0.009696 x 0.9682 ) x 100%%Vd = 4.6764 x ( 0.00642 + 0.006738 ) x 100%%Vd = 6.15%

Therefore,Vr = 400 x ( 1 - 0.0615 ) / 1.732Vr = 216.74 volts/phase

Typical minimum starting voltage for motors is 80% x rated voltage. Since the calculated volt-drop in our example is less than 20%, the motor will accelerate to full speed easily.

In practice, since the actual motor parameters are not known during the design stage, the minimum starting voltage is raised to 85% to provide a factor for safety.

In our example in Voltage Calculation, we have only considered a single cable and checked only the voltage drop at starting. In actual practice, we shall be comparing several cable sizes, selecting the cable that provides optimum design consideration.

In this example, we shall be using several cables from 35 mm2 up to 120 mm2. Likewise, we shall be considering, not only the starting condition but the running condition of the motor as well. You might be asking, why we selected a motor in our example. The reason, a motor is a dynamic load. A motor circuit load varies from the starting to the normal running condition, thus there will be multiple considerations when selecting the cable for a motor circuit.

In Table 2 Cable Selection, we have considered 5 cables sizes, all cables except one satisfy our design consideration which are the following:Voltage drop during starting < 15%Voltage drop during running < 5%

Again we will be using values from our previous example which are the following:Voltagesending end = 400 voltsMotor Running:Irunning = 180 APFrunning = cos Ørunning = 0.87sin Ørunning = 0.493

Motor Starting Istarting = 6 * 180 = 1080 APFstarting = cos Østarting = 0.25sin Østarting = 0.968

Using the formula

Table 2. Cable SelectionSize r50 x50 R X %Vdrunning %Vdstarting

35 0.674 0.0867 0.0809 0.0104 5.9% 14.2%50 0.499 0.0858 0.0599 0.0103 4.5% 11.7%70 0.344 0.0850 0.0413 0.0102 3.2% 9.4%95 0.271 0.0825 0.0325 0.0099 2.6% 8.3%120 0.214 0.0808 0.0257 0.0097 2.1% 7.4%

In Table 2. Cable Selection, the 35 mm2 cable passes the motor starting voltage drop but fails the motor running voltage drop which makes this cable to be eliminated from our selection list.

The 70 mm2, 90 mm2 and 120 mm2 cables satisfy all our design conditions, however, selecting any of these cables will make the installation more expensive than required. The 50 mm2 cable satisfy all conditions and cost of the project.

Cable selection considerations does not stop here, there are more that need to be considered such as operating temperature and short-circuit withstand. In our next article, we shall be discussing about cable operating temperatures.