Case study Fault MVA (megavolt-ampere) calculation: checking tool


Figure 1 shows an electrical system deriving a 415V three-phase 50Hz a.c. supply from a 350MVA (fault MVA) 66kV grid supply. Usuffi x


denotes the impedance voltage of the transformer and sub-transient reactance of the


standby generator in percentage. The voltage values shown are line voltage and the ratings shown against the transformers and standby generators are the full load rating of the equipment. The steps for applying the Fault MVA calculation method for the system shown in Figure 1 are as follows: Step 1: Fault MVA at location FA = 350MVA


Step 2: Fault MVA of transformer T1 = Step 3: Fault MVA of 0.1Ω 11kV distribution cable = Step 4: Fault MVA of transformer T2 = Step 5: Fault MVA of 0.01Ω 415V LV incoming cable =


MVATX Uk–Tx


MVATX Uk–Tx


Step 6: Fault MVA of generator G = Step 7: The single line diagram is then converted into an equivalent Fault MVA single line diagram (see Figure 2) Step 8: At location FA


MVAG Uk–G


= 0.7 = 4.667MVA 0.15


, the fault MVA = 350MVA, fault kA =


Step 9: At location FB the fault MVA = (∑ MVA )-1 = 1


fault kA =


MVAFault 3 x VLL


= 175.945 x 106


Step 10: When the mains is healthy, at location FC


at location FC fault kA = 3 x 11 x 103 = 9.235kA


the fault MVA = ( 1 + 1 + 1 = 7.81 x 106


MVAFault 3 x VLL


Step 11: When the mains is down and the supply at location FC


fault MVA = 4.667MVA – the result is from Step 6. Hence, at location FC


, the generator fault kA =


175.945 15.556 17.223 )-1 = 7.81MVA 3 x 0.415 x 103 = 10.867kA


is derived only from the standby generator, the


MVAFault 3 x VLL


= 4.667 x 106 3 x 0.415 x 103 = 6.493kA


Step 12: Since when the mains supply is restored, the mains and the standby generator will be running in parallel for a short period before the standby generator is taken offl ine, the distribution equipment should be designed to cope with the sum of the maximum fault currents supplied by both energy sources. Therefore, maximum designed fault kA for the switchboard and circuit breakers at location Fc should have a fault capacity not less than (10.867kA + 6.493kA)=17.36kA. Maximum fault MVA is therefore 12.48MVA.


Step 13: When the mains supply is restored, because of the short-time paralleling arrangement, fault contributions from the standby generating set need to be added to location FB


.


The fault MVA due to standby generator = therefore, at location FB


fault kA =


MVAFault 3 x VLL


( 1 + 1 + 1 = 2.971 x 106


4.667 15.556 17.223 )-1= 2.971MVA 3 x 11 x 103 = 0.156kA


Step 14: Therefore location FB should have a fault capacity not less than (9.235kA + 0.156kA)=9.391kA due to fault current contributions from the standby generator. Maximum fault MVA is therefore 178.92MVA.


Step 15: At location FA the fault current contributions from the standby generator is attenuated to 25.8A. The fault withstand capacities should not be less than 3.088kA and 353MVA.


The design check results are summarised thus: Fault location


FA Line Voltage


3-phase fault kA 3-phase fault MVA


66kV


3.088kA 353MVA


FB 11kV 9.391kA 178.92MVA FC 415V 17.36kA 12.48MVA Fault MVA Calculation results for the electrical system shown in Figure 1 58 CIBSE Journal September 2010 www.cibsejournal.com


MVAFault 3 x VLL


= 350 x 106 ( 1 + 1 + 1 3 x 66 x 103 = 3.062kA 350 500 1210)-1= 175.945MVA = 25 = 500MVA


= 0.7 = 15.556MVA 0.045


0.05 (VLL)2 ZCable


(VLL)2 ZCable


= 4152 0.01


= 17.223MVA


= 110002 0.1


= 1210MVA


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