{"id":2875,"date":"2024-08-25T20:05:10","date_gmt":"2024-08-25T20:05:10","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2875"},"modified":"2024-08-25T20:05:11","modified_gmt":"2024-08-25T20:05:11","slug":"transformer-tests","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/transformer-tests\/","title":{"rendered":"TRANSFORMER TESTS"},"content":{"rendered":"\n

All the transformers are tested before placing them in the field. By performing these tests, we can determine the parameters of a transformer to compute its performance characteristics (like voltage regulation and efficiency.).<\/p>\n\n\n\n

To furnish the required information, open circuit and short circuit tests are conducted conveniently without actually loading the transformer.<\/p>\n\n\n\n

10.23.1  Open-circuit or No-load Test<\/h4>\n\n\n\n

This test is carried out to determine the no-load loss or core loss or iron loss and no-load current I<\/em>0<\/sub> which is helpful in finding the no-load parameters, that is, exciting resistance R<\/em>0<\/sub> and X<\/em>0<\/sub> exciting reactance of the transformer.<\/p>\n\n\n\n

This test is usually carried out on the low-voltage side of the transformer, that is, a watt meter W, a voltmeter V, and an ammeter A are connected in low-voltage winding (say primary). The primary winding is then connected to the normal rated voltage\u00a0V<\/em>1<\/sub>\u00a0and frequency as given on the name plate of the transformer. The secondary side is kept open or connected to a voltmeter V\u2032 as shown in\u00a0Figure 10.31(a).\u00a0Since the secondary (high-voltage winding) is open circuited, the current drawn by the primary is called no-load current\u00a0I<\/em>0<\/sub>\u00a0measured by the ammeter A. The value of no-load current\u00a0I<\/em>0<\/sub>\u00a0is very small usually 2 to 10% of the rated full-load current. Thus, the copper loss in the primary is negligibly small and no copper loss occurs in the secondary as it is open. Therefore, wattmeter reading\u00a0W<\/em>0<\/sub>\u00a0only represents the core or iron losses for all practical purposes. These core losses are constant at all loads. The voltmeter V\u2032 if connected on the secondary side measures the secondary induced voltage\u00a0V<\/em>2<\/sub>.<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Fig. 10.31<\/strong>  (a) Circuit for open circuit test (b) Phasor diagram at no-load<\/p>\n\n\n\n

The ratio of voltmeter readings,\u00a0\"img\"\u00a0gives the transformation ratio of the transformer. The phasor diagram of transformer at no-load is shown in\u00a0Figure 10.31(b).<\/p>\n\n\n\n

Let the wattmeter reading = W<\/em>0<\/sub><\/p>\n\n\n\n

voltmeter reading = V<\/em>1<\/sub><\/p>\n\n\n\n

and ammeter reading = I<\/em>0<\/sub><\/p>\n\n\n\n

Then, iron losses of the transformer P<\/em>i <\/sub>= W<\/em>0<\/sub><\/p>\n\n\n\n

that is, V<\/em>1<\/sub>I<\/em>0<\/sub> cos \u0278<\/em>0<\/sub> = W<\/em>0<\/sub><\/p>\n\n\n\n

No-load power factor,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Working component,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Magnetising component<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

No-load, parameters, that is, Equivalent exciting resistance, \"img\"<\/p>\n\n\n\n

Equivalent exciting reactance, \"img\"<\/p>\n\n\n\n

The Iron losses measured by this test are used to determine transformer efficiency and parameters of exciting circuit of a transformer shown in Figure 10.32<\/a>.<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Fig. 10.32<\/strong>  Transformer at no-load with exciting circuit<\/p>\n\n\n\n

10.23.2  Short Circuit Test<\/h4>\n\n\n\n

This test is carried out to determine the following:<\/p>\n\n\n\n

    \n
  1. Copper losses at full load (or at any desired load). These losses are required for the calculations of efficiency of the transformer.<\/li>\n\n\n\n
  2. Equivalent impedance (Z<\/em>es <\/sub>or Z<\/em>ep<\/sub>), resistance (R<\/em>es<\/sub> or R<\/em>ep<\/sub>) and leakage reactance (X<\/em>es<\/sub> or X<\/em>ep<\/sub>) of the transformer referred to the winding in which the measuring instruments are connected. Knowing equivalent resistance and reactance, the voltage drop in the transformer can be calculated and hence regulation of transformer is determined.<\/li>\n<\/ol>\n\n\n\n

    This test is usually carried out on the high-voltage side of the transformer, that is, a wattmeter W, voltmeter V,\u00a0<\/em>and an ammeter A are connected in high-voltage* winding (say secondary). The other winding (primary) is then short circuited by a thick strip or by connecting an ammeter A\u2032 across the terminals as shown in\u00a0Figure 10.33. A low voltage at normal frequency is applied to the high-voltage winding with the help of on autotransformer so that full-load current flows in\u00a0both the windings, measured by ammeters A and A\u2032. Low voltage is essential, failing which an excessive current will flow in both the windings that may damage them.<\/p>\n\n\n\n

    Since a low voltage (usually 5 to 10% of normal rated voltage) is applied to the transformer winding, therefore, the flux set-up in the core is very small about \"img\" th to \"img\" th of normal flux.<\/p>\n\n\n\n

    The iron losses are negligibly small due to low value of flux as these losses are approximately proportional to the square of the flux. Hence, wattmeter reading\u00a0W<\/em>c\u00a0<\/sub>only represents the copper losses in the transformer windings for all practical purposes. The applied voltage\u00a0V<\/em>2sc\u00a0<\/sub>is measured by the voltmeter V which circulates the current\u00a0I<\/em>2sc\u00a0<\/sub>(usually full-load current) in the impedance\u00a0Z<\/em>es<\/sub>\u00a0of the transformer to the side in which instruments are connected as shown in\u00a0Figure 10.33.<\/p>\n\n\n\n

    Let the wattmeter reading = W<\/em>c<\/sub><\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Fig. 10.33<\/strong>  Circuit for short circuit test<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Fig. 10.34<\/strong>  Phasor diagram at short circuit<\/p>\n\n\n\n

    voltmeter reading = V<\/em>2sc<\/sub><\/p>\n\n\n\n

    and ammeter reading = I<\/em>2sc<\/sub><\/p>\n\n\n\n

    Then, full-load copper losses of the transformer,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Equivalent resistance referred to secondary, \"img\"<\/p>\n\n\n\n

    From phasor diagram as shown in\u00a0Figure 10.34<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    I<\/em>2sc<\/sub> Z<\/em>es<\/sub> = V<\/em>2sc<\/sub><\/p>\n\n\n\n

    \u2234Equivalent impedance referred to secondary, Z<\/em>es <\/sub>= V<\/em>2sc<\/sub>\/I<\/em>2sc<\/sub><\/p>\n\n\n\n

    Equivalent reactance referred to secondary, \"img\"<\/p>\n\n\n\n

    After calculating R<\/em>es <\/sub>and X<\/em>es<\/sub>, the voltage regulation of the transformer can be determined at any load and power factor.<\/p>\n\n\n\n

    Example 10.35<\/strong><\/p>\n\n\n\n

    Open-circuit and short-circuit tests were conducted on a 50 kVA, 6360\/240 V, 50 Hz, single-phase transformer in order to find its efficiency. The observations during these tests are as follows:<\/p>\n\n\n\n

    O.C. test: Voltage across primary winding = 6360 V; primary current = 1.0 A, and power input = 2 kW.<\/p>\n\n\n\n

    <\/a>S.C. test: Voltage across primary = 180 V; current in secondary winding = 175 A, and power input = 2 kW.<\/p>\n\n\n\n

    Calculate the efficiency of the transformer, when supplying full load at p.f. of 0.8 lagging.<\/p>\n\n\n\n

    (A.U.)<\/strong><\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    O.C. test:<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    S.C.test:<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    I<\/em>2sc<\/sub> = 175 A; W<\/em>c<\/sub> = 2000 W<\/p>\n\n\n\n

    \u2234 Cu loss at full load, \"img\"<\/p>\n\n\n\n

    \u2234<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Example 10.36<\/strong><\/p>\n\n\n\n

    A 15 kVA, 440\/230 V, 50 Hz, single-phase transformer gave the following test results:<\/p>\n\n\n\n

    Open Circuit (L.V. side)<\/td>250 V, 1.8 A, 95 W.<\/td><\/tr>
    Short circuit test (H.V. side)<\/td>80 V, 12.0 A, 380 W.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Compute the parameters of the equivalent circuit referred to L.V. side.<\/p>\n\n\n\n

    (P.T.U. May 2005)<\/strong><\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    Transformer rating = 15 kVA; E<\/em>1<\/sub> = 440 V; E<\/em>2<\/sub> = 230 V; f <\/em>= 50 Hz<\/p>\n\n\n\n

    Open circuit test (L.V. side); V<\/em>2 <\/sub>= 250 V; I<\/em>0<\/sub> = 1.8 A; W<\/em>0<\/sub> = 95 W<\/p>\n\n\n\n

    Short-circuit test (H.V. side); V<\/em>1(sc)<\/sub> = 80 V; I<\/em>1(sc)<\/sub> = 12 A; W<\/em>c <\/sub>= 380 W<\/p>\n\n\n\n

    From open-circuit test performed on L.V. side<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n

    Exciting resistance,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Exciting reactance,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    From short-circuit test performed on H.V. side<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n

    <\/a>Transformation ratio,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Transformer resistance and reactance referred to L.V. (secondary) side<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    R<\/em>es <\/sub>= R<\/em>ep<\/sub> \u00d7 K<\/em>2<\/sup> = 2.639 \u00d7 (0.5227)2<\/sup> = 0.7211 \u03a9<\/p>\n\n\n\n

    X<\/em>es <\/sub>= X<\/em>ep<\/sub> \u00d7 K<\/em>2<\/sup> = 6.122 \u00d7 (0.5227)2<\/sup> = 1.673 \u03a9<\/p>\n\n\n\n

    Example 10.37<\/strong><\/p>\n\n\n\n

    The following test data are obtained on a 5 kVA, 220\/440 V single-phase transformer; O.C. test \u2212220 V, 2 A, 100 W on L.V. side; S.C. test \u2212 40 V. 11.4 A, 200 W on H.V. side. Determine the percentage efficiency and regulation at full load 0.9 p.f. lag.<\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    From O.C. test, Iron losses, P<\/em>i<\/sub> = 100 W<\/p>\n\n\n\n

    From S.C. test, Copper losses, W<\/em>c<\/sub> = 200 W (at the load at which test is performed)<\/p>\n\n\n\n

    Full-load current on H.V. side, \"img\"<\/p>\n\n\n\n

    that is, S.C test is performed at full load since I<\/em>2sc <\/sub>= I<\/em>2<\/sub><\/p>\n\n\n\n

    Full-load copper loss, P<\/em>c <\/sub>= W<\/em>c <\/sub>= 200 W<\/p>\n\n\n\n

    Efficiency,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n

    From S.C. test:<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n

    Here cos \u0278<\/em>2<\/sub> = 0.9; sin \u0278<\/em>2<\/sub> = sin cos\u22121<\/sup> 0.9 = 0.4359<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    E<\/em>2<\/sub> = V<\/em>2<\/sub> + I<\/em>2 <\/sub>R<\/em>es<\/sub> cos \u0278<\/em>2<\/sub> + I<\/em>2<\/sub>X<\/em>es<\/sub>sin \u0278<\/em>2<\/sub><\/p>\n\n\n\n

    = 440 + 11.4 \u00d7 1.539 \u00d7 0.9 + 11.4 \u00d7 3.153 \u00d7 0.4359 = 471.46 V<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Example 10.38<\/strong><\/p>\n\n\n\n

    The O.C. and S.C. tests on a 5 kVA, 230\/160 V, 50 Hz, transformer gave the following data.<\/p>\n\n\n\n

    O.C. test (H.V. side) \u2212 230 V, 0.6 A, 80 watt<\/p>\n\n\n\n

    <\/a>S.C. test (L.V. side) \u2212 6 V, 15 A, 20 watt<\/p>\n\n\n\n

    Calculate the efficiency of transformer on full load at 0.8 p.f. lagging.<\/p>\n\n\n\n

    (P.U. June, 1996)<\/strong><\/p>\n\n\n\n

    Solution:<\/em><\/p>\n\n\n\n

    From open circuit test, iron losses, P<\/em>i <\/sub>= 80 W<\/p>\n\n\n\n

    As short-circuit test is performed on L.V. side, I<\/em>2sc <\/sub>= 15 A<\/p>\n\n\n\n

    Full-load secondary current,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n

    Copper losses measured at S.C. test, W<\/em>c<\/sub> = 20 W<\/p>\n\n\n\n

    Full-load copper losses, \"img\"<\/p>\n\n\n\n

    Efficiency of transformer at full load 0.8 p.f. lagging<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
    \"img\"\/<\/figure>\n\n\n\n

    Example 10.39<\/strong><\/p>\n\n\n\n

    The following results were obtained on a 50 kVA transformer:<\/p>\n\n\n\n

      \n
    1. Open-circuit tests: Primary voltage 3300 V, secondary voltage 415 V, power 430 W<\/li>\n\n\n\n
    2. Short-circuit test: Primary voltage 124 V, primary current 15.3 A, primary power 525 W secondary current full-load value.<\/li>\n<\/ol>\n\n\n\n

      Calculate:<\/p>\n\n\n\n

        \n
      1. The efficiency at full load and at half load for 0.7 power factor.<\/li>\n\n\n\n
      2. The voltage regulation for power factor 0.7: (i) lagging (ii) leading<\/li>\n\n\n\n
      3. The secondary terminal voltages corresponding to (a) and (b)<\/li>\n<\/ol>\n\n\n\n

        (P.T.U. Dec. 2007)<\/strong><\/p>\n\n\n\n

        Solution:<\/em><\/p>\n\n\n\n

        Rating of transformer = 50 kVA; Power factor = 0.7<\/p>\n\n\n\n

        Open-circuit test (primary): V<\/em>1<\/sub> = 3300 V; V<\/em>2<\/sub> = 415 V; W<\/em>0<\/sub> = 430 W<\/p>\n\n\n\n

        Short-circuit test (primary): V<\/em>1(SC)<\/sub> = 124 V; I<\/em>1(SC)<\/sub> = 15.3 A; W<\/em>c <\/sub>= 525 W<\/p>\n\n\n\n

        Short-circuit test is performed at full-load secondary current,<\/p>\n\n\n\n

        \u2234 Full-load copper losses,<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        P<\/em>c<\/sub> = W<\/em>c<\/sub> = 525 W<\/p>\n\n\n\n

        Iron losses,<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        P<\/em>i<\/sub> = W<\/em>0<\/sub> = 430 W<\/p>\n\n\n\n

        <\/a>When p.f., cos \u0278<\/em> = 0.7<\/p>\n\n\n\n

          \n
        1. Full \u2212 load efficiency, \"img\" \"img\"Efficiency at half load, \"img\"\"img\"<\/li>\n\n\n\n
        2. Transformer impedance referred to primary, \"img\"Transformer resistance referred to primary, \"img\"Transformer reactance referred to primary, \"img\"\"img\"At 3300 V, primary full-load current, \"img\"For p.f., cos \u0278<\/em> = 0.7 lag; sin \u0278<\/em> = sin cos\u22121<\/sup>0.7 = 0.714E<\/em>1<\/sub> = V<\/em>1 <\/sub>\u2212 I<\/em>1 <\/sub>R<\/em>ep<\/sub> cos \u0278<\/em> \u2212 I<\/em>1<\/sub>X<\/em>cp <\/sub>sin \u0278<\/em>= 3300 \u2212 15.15 \u00d7 2.243 \u00d7 0.7 \u2212 15.15 \u00d7 7.783 \u00d7 0.714= 3300 \u2212 23.787 \u2212 84.189 = 3192 V\"img\"For p.f., cos \u0278<\/em> = 0.7 leading; sin \u0278<\/em> = sin cos\u22121<\/sup>0.7 = 0.714\"image\" = V<\/em>1<\/sub>\u2212 I<\/em>1 <\/sub>R<\/em>ep<\/sub>cos \u0278<\/em> + I<\/em>1<\/sub> X<\/em>cp <\/sub>sin \u0278<\/em>= 3300 \u2212 15.15 \u00d7 2.243 \u00d7 0.7 + 15.15 \u00d7 7.783 \u00d7 0.714= 3300 \u2212 23.787 + 84.189 = 3360 V\"img\"<\/li>\n\n\n\n
        3. <\/a>Secondary terminal voltage at 0.7 p.f. lagging: \"img\"Secondary terminal voltage at 0.7 p.f. leading\"img\"<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

          All the transformers are tested before placing them in the field. By performing these tests, we can determine the parameters of a transformer to compute its performance characteristics (like voltage regulation and efficiency.). To furnish the required information, open circuit and short circuit tests are conducted conveniently without actually loading the transformer. 10.23.1  Open-circuit or No-load […]<\/p>\n","protected":false},"author":1,"featured_media":2841,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[413],"tags":[],"class_list":["post-2875","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-single-phase-transformers"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/power-transformer.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2875","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=2875"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2875\/revisions"}],"predecessor-version":[{"id":2876,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2875\/revisions\/2876"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2841"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2875"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2875"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2875"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}