MVA Method Short Circuit Calculation

MVA Method Short Circuit Calculationhttp://s.pangonilo.com/index.php/tutorials Assumptions for MVA Method Short Circuit Calculations For Short Circuit calculations y y y y y 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 @ 100MVA KVASC = (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 = 0.113 KVASC = (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 @ 55 OC, 66.5 / 74.5 MVA @ 65OC, OA/FA, Z = 9% @ 50MVA KVASC = 50 x 1000 / 0.09 = 555,555 kVA Reactor: KVASC = (1000 x KV2 ) / Z (ohms) Example: Reactor 11 kV, 0.125 ohms KVASC = 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 = (0.0258 2 + 0.0272 )0.5 x 150 / 1000 = 0.0056 ohms KVASC = 0.402 x 1000 / 0.0056 = 28,571 kVA The total KVAs in series (KVAtotal) is the reciprocal sum or inverse sum of all series KVAs. The total KVAs in parallel (KVAtotal) is the arithmetic sum of all parallel KVAs in parallel. KVAs in parallel. .Combining KVAs KVAs in series. 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. 33/11KV. 9% Z . 250 MVAsc Transformer 1: 10 MVA. we shall be presenting a short circuit study of a power system. Figure 1 Utility: 33KV. MVA Method Short Circuit Calculation In this example. say if you have two 500KVA transformers in parallel. 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. what is the total KVA? It will be 1000KVA and not 250KVA. The "Equivalent MVA" are: Transformers and Motors Generators Cables and Reactors . the sources of short circuit current are 1.6KV/400V.6KV.113 Transformer 2: 5 MVA. 7% Z Motor 1: 5MVA (Lumped). 6. 17% Z 400V Bus Motor 4: 300 KVA (Lumped). 11/6.8 MVA (Lumped). Utility 2. X"d = 0. 17% Z Motor 5: 596 KVA (Lumped). 25% Z In the event of a short circuit. 17% Z 6. Generators 3.6KV Bus Transformer 3: 2 MVA.11KV Bus Generator: 3MVA. Motors Static loads such as heaters and lighting do not contribute to short circuit. 6% Z Motor 3: 6. 41 MVA 6.43 MVA Motor 1: MVAsc = 5 / 0.113 = 35.596 / 0.07 = 71.11 MVA 11KV Bus Generator: MVAsc = 4 / 0.8 / 0.In Figure 1.09 = 111.33 MVA Motor 3: MVAsc = 6. I have calculated the Equivalent MVAs of each equipment.38 MVA .17 = 29.25 = 2. writing it below the ratings.3 / 0.6KV Bus Transformer 3: MVAsc = 2 / 0.17 = 40 MVA 400V Bus Motor 4: MVAsc = 0.06 = 33.4 MVA Transnformer 2: MVAsc = 5 / 0.76 MVA Motor 5: MVAsc = 0.17 = 1. Figure 2 Utility: MVAsc = 250MVA Transformer 1: MVAsc = 10 / 0. 49 MVA MVAsc @ 400V = 1/ (1 / 87.14 MVA At 400V Motors Motor 3: MVAsc = 24.76 + 2.41 = 141. At Transformer 1: MVAsc @ 33KV = 250 MVA MVAsc @ 11KV = 1/ (1 / 250 + 1 /111.76 + 2.33) = 24.14 x 2.11) = 76.6KV = 1/ (1 / 141.49 + 40 = 87.76 / ( 1.4 + 29.43) = 47.38 ) = 13.6KV = 47.Figure 3 Upstream Contribution Starting from the utility. combine MVAs writing each one above the arrows.14 x 1.38 / ( 1.49 + 1 / 33.68 MVA MVAsc @ 6.87 + 35.49 MVA At Transformer 3: MVAsc @ 6.38 ) = 10.88 MVA .68 + 1 / 71.26 MVA Motor 4: MVAsc = 24.87 MVA At Transformer 2: MVAsc @ 11KV = 76. 14 + 1 / 33.68 + 27.6KV = 3. Ifault @ 11KV = 168.92 + 1 /111.41 + 35. negative sequence and zero sequence impedances need to be calculated.732 x 11) = 8.68 + 1 / 71. at the 400V Bus.43) = 27.11 MVA At Transformer 1: MVAsc @ 11KV = 27.79 MVA From Generator : MVAsc = 35. Example: 11 KV Bus: From Transformer 1: MVAsc = 76.11 + 29. the negative sequence and zero sequence impedance are equal to the positive sequence impedance.14 MVA MVAsc @ 6. on Transformer 3 contributes to the zero sequence current. how about single phase faults? For single phase faults. 68 MVA At Transformer 2: MVAsc @ 6.41 = 168.87 + 91. At Transformer 3: MVAsc @ 400V = 1.79 / (1.38 + 29. MVAsc @ 33KV = 1/ (1 / 91.33) = 3.11) = 50. Z1 = Z2 = Z0 or MVA1 = MVA2 = MVA0 At the 400V Bus . I combined MVAs writing each one below the arrows.79 MVA From Transformer 2: MVAsc = 141. If = 3 (I1 + I2 + I0) Examining the circuit in Figure 1.3 MVA To determine the Faults Current at any bus on the power system.86 kA All we have done above are three phase faults.38 = 4. add the MVA values above and below the arrows.92 MVA Note: Two downstream plus the generator contribution.6KV = 1/ (1 / 4.4 = 91. the motor contribution to short circuit is the sum of the MVAs of the lumped motors Motor 3 and Motor 4.79MVA From Motor 1: MVAsc = 139.76 + 2.4 + 133. For transformers.68 + 40 = 43.92 = 168. you may ask.68 MVA MVAsc @ 11KV = 1/ (1 / 43. In this bus. positive sequence. The sum should be the same on any branch.39 = 168.11 = 168.Downstream Contribution Starting from the bottom (400V Bus).79 MVA This is a check that we have done the correct calculation. 732 x 6.49 / (1.26 kA Conclusion: This example illustrates that using the MVA Method of Short Circuit Calculation.33 ) MVAsc =3 x 9.49 MVA If = 83.83 = 83.79 / (1. We will calculate the Load Flow and Voltage dips for starting a 150MVA motor at the 400V Bus when all loads are running. we will be discussing about Load Flow and Voltage Dips during motor starting.79 MVA If = 29. combining KVAs and Short Circuit Calculations for three (3) phase and single phase faults.1 / MVAsc =1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0) 1 / MVAsc = 1/3 (1 / 28.28 + 1 / 33. In this part of the tutorial.4) = 43 kA At 6.17 + 1 / 91. except that the 2 x 150 MVa motors at the 400V Bus are now separate. we have discussed the importance of Short Circuit Study.6KV Bus 1 / MVAsc = 1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0) 1 / MVAsc = 1/3 (1 / 91. MVA Method Load Flow Calculation In previous tutorials for the MVA method. .43) MVAsc = 3 x 27.93 = 29.28 + 1 / 28. it will be very easy to calculate the fault current at any node within a power system. We shall be using the same single line diagram as in the previous tutorial.732 x 0.17 + 1 / 71.6) = 7. 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 Transformer: .75 Starting Motor : MVAS = 6 x Motor HP / 1000 = 6 x Motor KW / 0. 87 + 35.7 % or 306. 3. In this Load Flow Calculation.97 / (2. In Load Flow. the upstream contribution is equal to the downstream. Available MVAs are equal to the ratio of the MVAsc of each branch. Here are the steps in calculating the values in above Figure 2: I. we assume that a 150 MVA motor is being started at the 400V Bus after all loads has been started.49 + 5.Reactor & Cable: Static Loads: MVASC = Actual MVA The differences between a Short Circuit Calculation and Load Flow Calculation are: 1.27 MVA Downstream: MVAt = 7. Starting Motor MVAs are equal to 6 times rated Equivalent MVAs.4 = 112. It means that operational procedures should be in placed to have efficient plant operations. 33KV Bus: MVAsc = 250MVA 11KV Bus 10MVA Transformer: . At 11KV Bus: Upstream: MVAt = 76. Downstream Values: Note: These are the values above the arrow. Motor MVAs are equal to the rated Equivalent MVAs. 2.49 MVA Voltage dip at the 400V Bus during motor starting Vd = 2. 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.0 = 12. Static Loads are included in the calculations.97 + 0.8 V The above calculation indicates that when all loads are already running.9) = 76. Upstream Val es: At 400V B s: At 6.6KV B s: M : M : At 11KV B s: M : M : 4M : .4M M l t t : 4M t l l t t l t i l tM t t i t M t l t t M t t II. Phase to Ground Fault   The calculation will be the same for the other buses. The faults currents for three (3) phase unbalance faults are as follows: 1. Let us now calculate the bus voltages: @400V B s: MVA Method Unbalance Fault Calculation In the M A Method Short Circuit Calculation tutorial we have discussed how to calculate the three (3) phase and phase to ground fault currents. (N t I will l           t ¢¡   Let us now go back to the downstream calculation where we stopped.(N t c lc l ti n f y t d ) I will l tt y t ) ¦ ¥ ¡¤£¡ t i ©     §  ¨    £§ £ .At 33KV B s: 10M Transformer: At t is point we have completed all the val es in the upstream M A contri ution for load flow. Three (3) Phase Fault 2. 3. Phase to phase Fault 2. Phase to Phase Fault 4. Phase to phase to Ground Fault Figure 1 In the above example. Phase to Phase to Ground Fault Using the same example as in the MVA Method Short Circuit Calculation tutorial. we will calculate the 1. we calculated the Single Phase to Ground Faults At the 400V Bus . 28 + 1 / 33.732 x 6.732 x 0.17 ) /2 MVAsc =78.49 / (1.4) = 43 kA At 6.732 ( MVAsc1) / 2 MVAsc = 1.732 ( 91.43) MVAsc = 3 x 27.79 / (1.93 = 29.5 / (1.6) = 7.4 kA At 6.91 kA .732 x 6.33 ) MVAsc =3 x 9.6) = 6.MVAsc =3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0) MVAsc = 3 (1 / 28.96 MVA If = 78.6KV Bus MVAsc = 1.4) = 35.28 ) /2 MVAsc =24.732 ( MVAsc1) / 2 MVAsc = 1.17 + 1 / 71.732 ( 28. we will be able to calculate the following unbalanced faults.26 kA Knowing the values of MVA1.17 + 1 / 91.732 x 0.28 + 1 / 28.49 MVA Iff = 83.79 MVA If = 29.83 = 83.96 / (1. Phase to phase Faults At the 400V Bus Using MVAs instead of KVAs MVAsc = 1.5 MVA If = 24.6KV Bus MVAsc = 3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0) MVAsc = 3 (1 / 91. MVA2 and MVA0. 28 + 28.4) = 45.43) MVAsc =77 MVA If = 77 / (1.6 / (1.43 / (91.17 + 71.28 + 33.28 x 33.732 x 0.4 kA At 6.6KV Bus MVAsc = 3 x 91.732 x 6.17 + 91.33 ) MVAsc = 31.46 MVA If = 31.At the 400V Bus Using MVAs instead of KVAs MVAsc = 3 x 28.73 kA .17 x 71.33 / (28.6) = 6. 1 28.108 0.111 0.108 0.61 10.00 This article is the second part on the discussion about voltage drop calculation.0800 0.499 0.069 0.64 5.103 0. Selecting the correct size of cable could lower initial cost.185 0.140 0.134 0.0992 15.3 30.0808 0.6 9. lower operating cost.0933 0. usually ohms per kilometer( /km) for IEC or miles( /mi) for North American Standards.6 9.0850 0.163 0.5 2.108 0.499 0.15 8.108 0.0983 0.0825 0.0800 0.99 3.087 0.3 60.0808 0.118 0.214 0.0858 0.97 2.43 11.674 0. .99 3.94 10. better voltage regulation notwithstanding the safety factor.271 0.0867 0.121 0.175 0.105 0.069 0.5 36.148 0.118 0.Voltage Drop Calculation This article is the first of a series in the selection of electrical cables.64 5. Cable selection is an important part of electrical installation design.271 0.106 0.175 0.936 0.140 0.2 51.936 0.1 10. We shall be presenting an IEC standard equipment in our sample calculation.173 0.87 11. 600/1000 V.92 7.153 0.344 0.105 0. 1 to 6 x single.0808 0.10 10.48 0.105 0.0892 0.48 0.112 0. 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.087 0.7 23.38 8.35 1.0795 46.97 2.5 4 6 10 16 25 35 50 70 95 120 150 185 240 300 400 15. The following table illustrate the typical parameters for industrial cables Table 1.674 0.344 0.214 0.81 13. 3 or 4 cores Single cores in trefoil Resistance at 90oC Reactance (ohm/km) at 50Hz 3 cores Resistance at 90oC (ohm/km) Reactance at 50Hz Approximate armouring resistance GSWB at 60oC 3â 4 cores (ohm/km) Nominal conductor area (mm2) 1. Cable parameters are provided in values per unit length.35 1. Cu/EPR/CSP/GSWB run on open trays or enclosed in air.101 0.125 0. 6764 x ( 0.009696 x 0.0 x 0.006738 ) x 100% %Vd = 6.1386 volts/phase DF = 1080.9682 Let us assume the sending voltage remains constant at 400 volts.0 x 0.009696 x 0.0808 x 120 / 1000 = 0.072 ohm/km respectively at 90OC and 50 Hz. In actual practice.02568 x 0. .009696 ohms/phase For the 110kW motor the starting current is: I = 6 × 180. If the sending end line-to-line voltage is 400 volts. Find the percentage volt-drop on starting the motor. therefore sinØ = 0. The starting power factor is 0.0 x 0.9682 ) x 100% %Vd = 4.74 volts/phase Typical minimum starting voltage for motors is 80% x rated voltage.0. the minimum starting voltage is raised to 85% to provide a factor for safety. Since the calculated volt-drop in our example is less than 20%. the motor will accelerate to full speed easily. we shall be comparing several cable sizes.02568 x 0.25 + 0.15% Therefore.9682 = 10.0 x 0.2500.9336 volts/phase BE = 1080. the specific resistance r and reactance x for the cable are 0.6805 volts/phase EF = 1080.2500 = 6.009696 x 0.0615 ) / 1. In practice. In our example in Voltage Calculation.25 lagging.00642 + 0. AB = 1080.02568 o hms/phase X = xl = 0.197 and 0.0 A The power factor is: cosØ = 0.9682 = 26. since the actual motor parameters are not known during the design stage. Solution For the 120 mm2 cable the series impedance is: R = rl = 0. Vr = 400 x ( 1 .0 = 1080.732 Vr = 216. selecting the cable that provides optimum design consideration. we have only considered a single cable and checked only the voltage drop at starting.0 x ( 0.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.6179 volts/phase %Vd = ( 1.02568 x 0.214 x 120 / 1000 = 0.2500 = 2.732 / 400 ) x 1080. we have considered 5 cables sizes.0867 0. however. there are more that need to be considered such as operating temperature and short-circuit withstand.7% 9.0257 0.0825 0.0103 0. why we selected a motor in our example.9% 4.4% 8. The 50 mm2 cable satisfy all conditions and cost of the project.0104 0. You might be asking. Cable Selection.0599 0.499 0. 90 mm2 and 120 mm2 cables satisfy all our design conditions.493 Motor Starting I starting = 6 * 180 = 1080 A PFstarting = cos Østarting = 0. The reason.0413 0.0809 0.2% 11.344 0. selecting any of these cables will make the installation more expensive than required.2% 2.674 0. thus there will be multiple considerations when selecting the cable for a motor circuit. A motor circuit load varies from the starting to the normal running condition.0097 5.271 0. Cable Selection Size r50 x50 R X %Vdrunning %Vdstarting 35 50 70 95 120 0. The 70 mm2.25 sin Østarting = 0.In this example. 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 volts Motor Running: Irunning = 180 A PFrunning = cos Ørunning = 0.214 0.0808 0.87 sin Ø running = 0.4% In Table 2.0102 0.0099 0.0850 0. we shall be using several cables from 35 mm2 up to 120 mm2. Likewise. 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. not only the starting condition but the running condition of the motor as well. .968 Using the formula Table 2. In our next article. we shall be discussing about cable operating temperatures.3% 7.0858 0.6% 2. In Table 2 Cable Selection.0325 0. Cable selection considerations does not stop here.1% 14. we shall be considering.5% 3. a motor is a dynamic load.