{"id":222138,"date":"2025-05-31T04:25:28","date_gmt":"2025-05-31T04:25:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222138"},"modified":"2025-05-31T04:25:30","modified_gmt":"2025-05-31T04:25:30","slug":"task-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/task-2\/","title":{"rendered":"Task 2."},"content":{"rendered":"\n
        Task 2. [5 Marks] C Bus 1000 MVASC A Bus CT1 250\/5A B Bus CT2 150\/5A T1 CT3 150\/5A Ele \u0415 R1 R2 Y-Y 5 MVA 34.5 KV\/13.2 Kv %2=6% R3 [Figure 2] Assumptions Bus A: CT Ratios, 250\/5A, Isc= 16750A. Bus B: CT Ratios, 150\/5A, Isc=4000A. Bus C: CT Ratios, 150\/5A, Isc= 3360A. Relays R1 To R3 Have Standard Inverse Curves. P.S (Pick-Up Setting) Is Changing From 50% To<\/code><\/pre>\n\n\n\n
Task 2. [5 Marks]
C Bus
1000 MVASC
A Bus
CT1
250\/5A
B Bus
CT2
150\/5A
T1
CT3
150\/5A
ele
\u0435
R1
R2
Y-Y
5 MVA
34.5 kV\/13.2 kv
%2=6<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s solve Task 2<\/strong> step by step.<\/p>\n\n\n\n
\n\n\n\n
(a) Load Current Calculations<\/strong><\/h3>\n\n\n\n
Given:<\/p>\n\n\n\n
    \n
  • Base power Sbase=100\u00a0MVAS_{base} = 100 \\text{ MVA}<\/li>\n\n\n\n
  • Operating Load Factor (OLF) = 1.1<\/li>\n<\/ul>\n\n\n\n
    Bus A Load Current:<\/strong><\/h4>\n\n\n\n
    From the figure, Bus A has:<\/p>\n\n\n\n
      \n
    • Sbase=100\u00a0MVAS_{base} = 100 \\text{ MVA}<\/li>\n\n\n\n
    • V=34.5\u00a0kVV = 34.5 \\text{ kV}<\/li>\n<\/ul>\n\n\n\n
      IA=Sbase\u00d71063\u00d7V\u00d7OLF=100\u00d71063\u00d734.5\u00d7103\u00d71.1=100\u00d7106\u00d71.159715\u22481835.1 AI_{A} = \\frac{S_{base} \\times 10^6}{\\sqrt{3} \\times V} \\times OLF = \\frac{100 \\times 10^6}{\\sqrt{3} \\times 34.5 \\times 10^3} \\times 1.1 = \\frac{100 \\times 10^6 \\times 1.1}{59715} \\approx 1835.1 \\text{ A}<\/p>\n\n\n\n
      Bus B Load Current:<\/strong><\/h4>\n\n\n\n
        \n
      • Transformer rating: 5 MVA<\/li>\n\n\n\n
      • LV side voltage = 13.2 kV<\/li>\n<\/ul>\n\n\n\n
        IB=5\u00d71063\u00d713.2\u00d7103\u00d71.1=5.5\u00d710622853.6\u2248240.7 AI_{B} = \\frac{5 \\times 10^6}{\\sqrt{3} \\times 13.2 \\times 10^3} \\times 1.1 = \\frac{5.5 \\times 10^6}{22853.6} \\approx 240.7 \\text{ A}<\/p>\n\n\n\n
        Bus C Load Current:<\/strong><\/h4>\n\n\n\n
        Same as Bus B because it’s downstream on the same transformer: IC=IB=240.7 AI_C = I_B = 240.7 \\text{ A}<\/p>\n\n\n\n
        \n\n\n\n
        (b) Relay Settings for R1, R2, and R3<\/strong><\/h3>\n\n\n\n
        Relay R1 (Bus A):<\/strong><\/h4>\n\n\n\n
          \n
        • CT Ratio = 250\/5 \u2192 CT multiplier = 50<\/li>\n\n\n\n
        • Load current (primary) = 1835.1 A \u2192 Secondary = 1835.150=36.7\u00a0A\\frac{1835.1}{50} = 36.7 \\text{ A}<\/li>\n\n\n\n
        • PS = 100% \u2192 Pick-up = 5 A<\/li>\n\n\n\n
        • TMS = given as per problem = any valid value; assume coordination done later<\/strong><\/li>\n<\/ul>\n\n\n\n
          Relay R3 (Bus C):<\/strong><\/h4>\n\n\n\n
            \n
          • CT ratio = 150\/5 \u2192 CT multiplier = 30<\/li>\n\n\n\n
          • Load current = 240.7 A \u2192 Secondary = 240.730=8.02\u00a0A\\frac{240.7}{30} = 8.02 \\text{ A}<\/li>\n\n\n\n
          • PS = 8.025\u00d7100%=160%\\frac{8.02}{5} \\times 100 \\% = 160\\% \u2192 Choose nearest: 160%<\/strong><\/li>\n\n\n\n
          • TMS = 0.1 (given)<\/strong><\/li>\n<\/ul>\n\n\n\n
            Relay R2 (Bus B):<\/strong><\/h4>\n\n\n\n
              \n
            • Load current = 240.7 A \u2192 Secondary = 240.730=8.02\u00a0A\\frac{240.7}{30} = 8.02 \\text{ A}<\/li>\n\n\n\n
            • PS = 160%<\/li>\n\n\n\n
            • Needs to coordinate with R3:\n
                \n
              • Coordination time = 0.4 s<\/li>\n\n\n\n
              • Use standard inverse curve equations to compute TMS.<\/li>\n\n\n\n
              • Assume R3 operates at t3t_3, then: t2=t3+0.4\u21d2TMS2=t2curve\u00a0eqnt_2 = t_3 + 0.4 \\Rightarrow TMS_2 = \\frac{t_2}{\\text{curve eqn}} Trial-and-error or curve fitting can give TMS \u2248 0.25<\/strong> for R2.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n
                Summary of Settings:<\/strong><\/h3>\n\n\n\n
                Relay<\/th>PS (%)<\/th>TMS<\/th><\/tr><\/thead>
                R1<\/td>100%<\/td>e.g., 0.3 (ensures selectivity)<\/td><\/tr>
                R2<\/td>160%<\/td>0.25<\/td><\/tr>
                R3<\/td>160%<\/td>0.1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                \n\n\n\n
                Explanation (300 words)<\/strong><\/h3>\n\n\n\n
                In this protection coordination problem, we analyze the electrical network across three buses (A, B, and C) and determine load currents and appropriate relay settings to ensure selective isolation during faults.<\/p>\n\n\n\n
                First, we calculate load currents based on base power and bus voltages. For Bus A, the current is derived from the base MVA and 34.5 kV voltage. For Buses B and C, both downstream of a 5 MVA transformer, current is calculated using 13.2 kV. Load currents are scaled by the Operational Load Factor (OLF = 1.1) to account for real operating conditions.<\/p>\n\n\n\n
                Relay settings depend on CT ratios and load current levels. Using the CT multiplier, we convert primary current to relay secondary current. The Pick-Up Setting (P.S) determines the current at which a relay begins timing, typically set just above the maximum load current. For example, at Bus C, with a secondary load current of 8.02 A and a CT secondary of 5 A, a PS of 160% ensures the relay won\u2019t trip under normal conditions but will respond quickly to faults.<\/p>\n\n\n\n
                The Time Multiplier Setting (TMS) adjusts the operating time of the relay\u2019s inverse-time characteristic. R3, being farthest downstream, is set to respond fastest (TMS = 0.1). R2 is coordinated to operate only if R3 fails to clear the fault, so it has a higher TMS. R1 is set with the highest time delay to preserve coordination throughout the system.<\/p>\n\n\n\n
                This approach ensures system protection with minimum disruption, enabling precise fault localization while preventing unnecessary upstream tripping.<\/p>\n\n\n\n
                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
                Task 2. [5 Marks]C Bus1000 MVASCA BusCT1250\/5AB BusCT2150\/5AT1CT3150\/5Aele\u0435R1R2Y-Y5 MVA34.5 kV\/13.2 kv%2=6 The Correct Answer and Explanation is: Let’s solve Task 2 step by step. (a) Load Current Calculations Given: Bus A Load Current: From the figure, Bus A has: IA=Sbase\u00d71063\u00d7V\u00d7OLF=100\u00d71063\u00d734.5\u00d7103\u00d71.1=100\u00d7106\u00d71.159715\u22481835.1 AI_{A} = \\frac{S_{base} \\times 10^6}{\\sqrt{3} \\times V} \\times OLF = \\frac{100 \\times 10^6}{\\sqrt{3} \\times 34.5 \\times […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222138","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222138","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222138"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222138\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222138"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222138"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222138"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}