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    Per Unit Calculation help

    It's been a while since I used per-unit calculation so I am embarrassing to say that I am drawing a blank when I ran into this. I requested the utility for available fault current at the transformer I was give this value.

    250MVA transformer (138kv to 38kV), Z=11% with 100MVA. They told me to use infinite bus for my calculation.

    So, I just need to find the new Z for 250MVA transformer right? Is this formula correct to find new Z?

    Z (new) = Z (old) x ( S (base)/S (new) ) = 11 x ( 100/250 ) = 4.4%

    Is this right? It seem Z is too low for this size transformer. I guess I can go back to my old power system book and verify but I thought asking here would be easier and more accurate. Thanks,

    #2
    close
    z old x s new/s old ~ 27.5%

    you can check by doing sc i calc for each s

    Comment


      #3
      long way
      you may need the actual ohmic value anyways

      vbase 38kv

      100 mva base
      zbase = 38kv^2/100 mva = 14.44 ohm
      zact = 0.11 x 14.44 = 1.5884 ohm per ph or 2.7512 3 ph

      250 mva base
      zbase = 38kv^2/250 mva = 5.7760 ohm
      z pu = 1.5884/5.7760 = 0.2750 pu or 27.5%

      100 mva sc i = 100 mva/(sqrt3 x 38kv 0.11) = 13821 A
      250 mva sc i = 250 mva/(sqrt3 x 38kv x 0.2750) = 13821 A

      38kv / 2.7512 = 13821

      Comment


        #4
        Originally posted by Ingenieur View Post
        long way
        you may need the actual ohmic value anyways

        vbase 38kv

        100 mva base
        zbase = 38kv^2/100 mva = 14.44 ohm
        zact = 0.11 x 14.44 = 1.5884 ohm per ph or 2.7512 3 ph

        250 mva base
        zbase = 38kv^2/250 mva = 5.7760 ohm
        z pu = 1.5884/5.7760 = 0.2750 pu or 27.5%

        100 mva sc i = 100 mva/(sqrt3 x 38kv 0.11) = 13821 A
        250 mva sc i = 250 mva/(sqrt3 x 38kv x 0.2750) = 13821 A

        38kv / 2.7512 = 13821
        Thanks!! I knew I could get the right answer here.

        Comment


          #5
          Originally posted by elec_eng View Post
          It's been a while since I used per-unit calculation so I am embarrassing to say that I am drawing a blank when I ran into this. I requested the utility for available fault current at the transformer I was give this value.

          250MVA transformer (138kv to 38kV), Z=11% with 100MVA. They told me to use infinite bus for my calculation.

          So, I just need to find the new Z for 250MVA transformer right? Is this formula correct to find new Z?

          Z (new) = Z (old) x ( S (base)/S (new) ) = 11 x ( 100/250 ) = 4.4%

          Is this right? It seem Z is too low for this size transformer. I guess I can go back to my old power system book and verify but I thought asking here would be easier and more accurate. Thanks,
          Infinite bus just means that the amount of fault current through the transformer will only be limited by the transformer impedance itself. It appears what the utility gave you is the impedance of the system either on a 100MVA base or they are saying 100MVAsc; can't tell from the post. However, they are telling you to base your design on an infinite bus.

          Thus a 250MVA transformer with impedance of say 8.5% (based on HV KVBIL of 650 and LV KVBIL of 150) will have an MVAsc = 250MVA/0.085 = 2941MVAsc
          @ 38kV that is around 45kAsc.

          However, those numbers may present a problem for 38kV rated equipment. 45KA is out of the range of 38kV gear either vacuum, dead tank or SF6. the limit is 40KA, also the maximum current rating is going to be 3000A but that require fan cooling.

          To limit the current on infinite bus system, the maximum amount of fault current is going to need to be around 38kA (a little margin) thus you need an impedance on the transformer of around 10 to 10.5%. But that doesn't solve the issue of max continuous current which would limit the transformer MVA to around 165MVA or even less depending on manufacturer.
          "Just because you're paranoid, doesn't mean they're not out to get you"

          Comment


            #6
            imo they gave him a 250 mva xfmr that had its pu z tested/measured at 100 mva
            above 10-15% of rating xfmr v, i and hence s are linear, so z is constant

            as most know the method to measure pu z (in this class each one is tested)
            short the sec and apply v source to prim
            increase v until sec i = rated, in this case due to testing equip limitations/safety they only went to 40% of rating
            100 mva or 1520 A, valid since 40%>>10-15%
            the test v applied to achieve this is 0.11 x 138000 = 15180 v
            prim i = 1520/(138/38) = 418.55A
            z = 15180/418.55 = 36.27 ohm 3 ph or 20.94 1 ph

            on a pu base
            zbase = 138kv^2/100 mva = 190.44 ohm (76.1 on a 250 mva basis)
            pu = 20.94/190.44 = 0.1100 or 11%
            or
            vbase = ibase x zbase and vtest = ibase x zact
            combine/sub/rearrange: vtest = vbase/zbase x zact
            vtest/vbase = zact/zbase = pu z = 15180/138000 = 0.11 or 11%

            if they attemped to test at rated 250 mva
            i prim 1045
            i sec 3798
            vtest 2.5 x 15180 = 37950
            test power 100 mva = 11 mva
            test power 250 mva = 68.7 mva
            quite a difference, alot of it heat

            on a 250 mva base
            vtest/vbase = 37950/138000 = 27.5%
            or z = 37950/1045 = 36.32 ohm 3 ph or 20.97 1 ph (same as abv within rounding)
            pu z = 20.97/76.1 = 27.5%


            more than anyone needs to know about pu z lol

            Comment


              #7
              Originally posted by Ingenieur View Post
              imo they gave him a 250 mva xfmr that had its pu z tested/measured at 100 mva
              above 10-15% of rating xfmr v, i and hence s are linear, so z is constant
              Typically, the utility is not going to give you the impedance of a transformer on 100MVA base, mostly because, well they don't need too. They would just supply the nameplate data, but the OP is unclear on this point. They typically will give system data on a 100MVA base, but either provide R and X 3ph and 1ph, or X/R ratio. The fact they said use infinite bus for a 138kV-38KV system is odd from my experience since that is a transmission level voltage and typically those guys have system study info that can be provided.
              "Just because you're paranoid, doesn't mean they're not out to get you"

              Comment


                #8
                Originally posted by kingpb View Post
                Typically, the utility is not going to give you the impedance of a transformer on 100MVA base, mostly because, well they don't need too. They would just supply the nameplate data, but the OP is unclear on this point. They typically will give system data on a 100MVA base, but either provide R and X 3ph and 1ph, or X/R ratio. The fact they said use infinite bus for a 138kV-38KV system is odd from my experience since that is a transmission level voltage and typically those guys have system study info that can be provided.
                the fact they told him to use infinite bus tells me the 11% is the 250 mva xfmr pu z on a 100 mva basis

                if it was sc mva he could calc system impedance (but not resolve it into r + jx without the x/r ratio

                if they are saying 100 mva sc on the primary that makes no sense since mva sc < 250 mva rating

                if 100/0.11 ~ 910 mva primary
                or less than 13 ka sec

                Comment

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