{"id":2871,"date":"2024-08-25T20:00:00","date_gmt":"2024-08-25T20:00:00","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2871"},"modified":"2024-08-25T20:00:01","modified_gmt":"2024-08-25T20:00:01","slug":"efficiency-of-a-transformer","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/efficiency-of-a-transformer\/","title":{"rendered":"\u00a0\u00a0EFFICIENCY OF A TRANSFORMER"},"content":{"rendered":"\n

The efficiency of a transformer is defined as the ratio of output to the input power, the two being measured in same units (either in watts or in kW).<\/p>\n\n\n\n

Transformer efficiency, \"img\"<\/p>\n\n\n\n

or<\/p>\n\n\n\n

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

Where V<\/em>2<\/sub> = Secondary terminal voltage<\/p>\n\n\n\n

I<\/em>2<\/sub> = Full-load secondary current<\/p>\n\n\n\n

cos \u0278<\/em>2<\/sub> = p.f. of the load<\/p>\n\n\n\n

P<\/em>i<\/sub> = lron losses = Hysteresis losses + eddy current losses        (constant losses)<\/p>\n\n\n\n

P<\/em>c<\/sub> = Full-load copper losses = I<\/em>2<\/sup>2<\/sub>R<\/em>es<\/sub>        (variable losses)<\/p>\n\n\n\n

If x <\/em>is the fraction of the full load, the efficiency of the transformer at this fraction is given by the relation<\/p>\n\n\n\n

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

The copper losses vary as the square of the fraction of the load.<\/p>\n\n\n\n

10.21  CONDITION FOR MAXIMUM EFFICIENCY<\/h5>\n\n\n\n

The efficiency of a transformer at a given load and p.f. is expressed by the relation<\/p>\n\n\n\n

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

The terminal voltage V<\/em>2<\/sub> is approximately constant. Thus, for a given p.f., efficiency depends upon the load current I<\/em>2<\/sub>. In expression (i), the numerator is constant and the efficiency will be maximum, if denominator is minimum. Thus, the maximum condition is obtained by differentiating the quantity in the denominator with respect to the variables I<\/em>2<\/sub> and equating that to zero, that is,<\/p>\n\n\n\n

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

or<\/p>\n\n\n\n

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

that is,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

Copper losses = iron losses<\/p>\n\n\n\n

Thus, the efficiency of a transformer will be maximum when copper (or variable) losses are equal to iron (or constant) losses.<\/p>\n\n\n\n

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

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

From equation (ii), the value of output current I<\/em>2<\/sub> at which the efficiency of the transformer will be maximum is given by<\/p>\n\n\n\n

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

<\/a>If x <\/em>is the fraction of full-load kVA at which the efficiency of the transformer is maximum.<\/p>\n\n\n\n

Then, copper losses = x<\/em>2<\/sup>P<\/em>c<\/sub> (where P<\/em>c<\/sub> is the full-load Cu losses)<\/p>\n\n\n\n

Iron losses = P<\/em>i<\/sub><\/p>\n\n\n\n

For maximum efficiency, \"img\"<\/p>\n\n\n\n

\u2234 Output kVA corresponding to maximum efficiency<\/p>\n\n\n\n

= x <\/em>\u00d7 full-load kVA = full-load kVA \u00d7 \"img\"<\/p>\n\n\n\n

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

Example 10.24<\/strong><\/p>\n\n\n\n

The primary and secondary windings of a 500 kVA transformer have resistance of 0.42 \u03a9 and 0.0011 \u03a9, respectively. The primary and secondary voltages are 600 V and 400 V, respectively. The iron loss is 2.9 kW. Calculate the efficiency at half full load at a power factor of 0.8 lagging.<\/em><\/p>\n\n\n\n

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

Transformer rating, = 500 kVA<\/p>\n\n\n\n

Primary resistance, R<\/em>1<\/sub> = 0.42 \u03a9<\/p>\n\n\n\n

Secondary resistance, R<\/em>2<\/sub> = 0.0011 \u03a9<\/p>\n\n\n\n

Primary voltage, E<\/em>1<\/sub> = 6600 V<\/p>\n\n\n\n

Secondary voltage, E<\/em>2<\/sub> = 400 V<\/p>\n\n\n\n

Iron losses, P<\/em>i<\/sub> = 2.9 kW<\/p>\n\n\n\n

Fraction of the load, x <\/em>= \"img\" = 0.5<\/p>\n\n\n\n

Load p.f., cos \u0278<\/em> = 0.8 lagging<\/p>\n\n\n\n

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

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

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

Total resistance referred to secondary, R<\/em>es<\/sub> = R<\/em>2<\/sub> +\"image\" = 0.0011 + 0.00154 = 0.00264 \u03a9<\/p>\n\n\n\n

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

Copper losses at full load, P<\/em>c<\/sub> = \"image\"R<\/em>es<\/sub> = (1250)2<\/sup> \u00d7 0.00264<\/p>\n\n\n\n

= 4125 W = 4.125 kW<\/p>\n\n\n\n

Efficiency of transformer at any fraction (x) of the load,<\/p>\n\n\n\n

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

<\/a>Example 10.25<\/strong><\/p>\n\n\n\n

In a 25 kVA, 2000\/200 V power transformer, the iron and full-load copper losses are 350 W and 400 W, respectively. Calculate the efficiency at unity power factor at (i) full load and (ii) half load.<\/em><\/p>\n\n\n\n

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

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

Where<\/p>\n\n\n\n

 <\/p>\n\n\n\n

cos \u0278<\/em> = 1; P<\/em>i<\/sub> = 350 W; P<\/em>c <\/sub>= 400 W<\/p>\n\n\n\n

    \n
  1. At full load x <\/em>= 1\u2234\"img\"<\/li>\n\n\n\n
  2. At half-load; x <\/em>= 0.5\u2234\"img\"<\/li>\n<\/ol>\n\n\n\n

    Example 10.26<\/strong><\/p>\n\n\n\n

    A 220\/400 V, 10 kVA, 50 Hz, single-phase transformer has copper loss of 120 W at full load. If it has an efficiency of 98% at full load, unity power factor, determine the iron losses. What would be the efficiency of the transformer at half full load at 0.8 p.f. lagging?<\/p>\n\n\n\n

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

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

    When<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    x <\/em>= 1\/2 and cos \u0278<\/em> = 0.8<\/p>\n\n\n\n

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

    Example 10.27<\/strong><\/p>\n\n\n\n

    The efficiency of a 400 kVA, single-phase transformer is 98.77% when delivering full load at 0.8 power factor and 99.13% at half load and unity power factor. Calculate (i) the iron loss (ii) the full-load copper loss.<\/p>\n\n\n\n

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

    Efficiency of a transformer at any fraction x <\/em>of the load<\/p>\n\n\n\n

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

    <\/a>Case I: <\/strong>x <\/em>= 1; cos \u0278<\/em> = 0.8; \u03b7<\/em>x <\/sub>= 98.77%<\/p>\n\n\n\n

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

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

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

     <\/p>\n\n\n\n

    P<\/em>i<\/sub> + P<\/em>c <\/sub>= 3.985 kW        (10.5)<\/p>\n\n\n\n

    Case II: <\/strong>x <\/em>= 0.5; cos \u0278<\/em> = 1; \u03b7<\/em>x <\/sub>= 99.13<\/p>\n\n\n\n

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

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

    or<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    P<\/em>i<\/sub> + 0.25 P<\/em>c<\/sub> = 1.755 kW        (10.6)<\/p>\n\n\n\n

    Subtracting\u00a0eq. (10.6)\u00a0from\u00a0(10.5), we get<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    0.75 P<\/em>c<\/sub> = 2.23 kW or P<\/em>c <\/sub>= 2.973kW<\/p>\n\n\n\n

    and<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    P<\/em>i<\/sub> = 3.985 \u2212 2.973 = 1.012kW<\/p>\n\n\n\n

    Example 10.28<\/strong><\/p>\n\n\n\n

    A 440\/110 V transformer has an effective primary resistance of 0.3 \u03a9 and a secondary resistance of 0.02 \u03a9. If iron loss on normal input voltage is 150 W, calculate the secondary current at which maximum efficiency will occur. What is the value of this maximum efficiency for unity power factor load?<\/p>\n\n\n\n

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

    Primary resistance, R<\/em>1<\/sub> = 0.3 \u03a9<\/p>\n\n\n\n

    Secondary resistance, R<\/em>2<\/sub> = 0.02 \u03a9<\/p>\n\n\n\n

    Iron losses, P<\/em>i <\/sub>= 150 W<\/p>\n\n\n\n

    Load power factor, cos \u0278<\/em> = 1<\/p>\n\n\n\n

    Primary induced voltage, E<\/em>1<\/sub> = 440 V<\/p>\n\n\n\n

    Secondary induced voltage, E<\/em>2<\/sub> = 110 V<\/p>\n\n\n\n

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

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

    Primary resistance referred to secondary,<\/p>\n\n\n\n

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

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

    R<\/em>es<\/sub> = R<\/em>2<\/sub> + \"image\" = 0.02 + 0.01875 = 0.03875 \u03a9<\/p>\n\n\n\n

    We know the condition for max, efficiency is<\/p>\n\n\n\n

    Copper losses = Iron losses, that is, \"image\"R<\/em>es<\/sub> = P<\/em>i<\/sub><\/p>\n\n\n\n

    Secondary current at which the efficiency is maximum,<\/p>\n\n\n\n

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

    <\/a>The maximum efficiency, \"img\"<\/p>\n\n\n\n

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

    Example 10.29<\/strong><\/p>\n\n\n\n

    The efficiency of a 1000 kVA, 110\/220 V, 50 Hz. Single-phase transformer is 98.5% at half load and 0.8 power factor leading and 98.9% at full-load unity power factor. Determine (i) iron loss and (ii) full-load copper loss.<\/em><\/p>\n\n\n\n

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

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

    Here, Rating of transformer = 1000 kVA<\/p>\n\n\n\n

    We know,<\/p>\n\n\n\n

    \"img\"\/<\/figure>\n\n\n\n
      \n
    1. Where % \u03b7<\/em>0.5 <\/sub>= 98.5; x <\/em>= 0.5; p.f. = 0.8 leading\u2234\"img\"or\"img\"<\/li>\n\n\n\n
    2. When %\u00a0\u03b7<\/em>fl\u00a0<\/sub>= 98.8;\u00a0x\u00a0<\/em>= 1; p.f. = 1\u00a0\"img\"or\"img\"Subtracting\u00a0eq. (10.7)\u00a0from (10.8), we get\u00a00.75\u00a0P<\/em>c<\/sub>\u00a0= 6000 or\u00a0P<\/em>c<\/sub>\u00a0= 8000WFrom\u00a0eq. (10.8), we get\u00a0P<\/em>i<\/sub>\u00a0= 12100 \u2212 8000 = 4100W<\/li>\n<\/ol>\n\n\n\n

      Example 10.30<\/strong><\/p>\n\n\n\n

      A 50 kVA transformer on full load has a copper loss of 600 and iron loss of 500 W.<\/em> Calculate the maximum efficiency and the load at which it occurs.<\/em><\/p>\n\n\n\n

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

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

      Efficiency will be maximum when Cu. loss = lron loss = 500 W<\/p>\n\n\n\n

      Fraction at which the efficiency is maximum, \"img\"<\/p>\n\n\n\n

      <\/a>Load at which the efficiency is maximum, that is,<\/p>\n\n\n\n

      Output<\/p>\n\n\n\n

       <\/p>\n\n\n\n

      = x <\/em>\u00d7 kVA = 0.9128 \u00d7 50 = 45.64 kVA<\/p>\n\n\n\n

      = 45.64 \u00d7 1 = 45.64 kW (since cos \u0278<\/em> = 1)<\/p>\n\n\n\n

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

      Example 10.31<\/strong><\/p>\n\n\n\n

      In a 25 kVA, 1100\/400 V, single-phase transformer, the iron and copper losses at full load are 350 W and 400 W,<\/em> respectively. Calculate the efficiency on unity power at half load.<\/p>\n\n\n\n

      Determine the load at which maximum efficiency occurs.<\/em><\/p>\n\n\n\n

      (May, 1984)<\/strong><\/p>\n\n\n\n

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

      Transformer rating = 25 kVA<\/p>\n\n\n\n

      Iron losses, P<\/em>i<\/sub> = 350 W<\/p>\n\n\n\n

      Full-load copper losses, P<\/em>c <\/sub>= 400 W<\/p>\n\n\n\n

      Load power factor cos \u0278<\/em> = 1<\/p>\n\n\n\n

      Fraction of the load x <\/em>= \"img\" = 0.5<\/p>\n\n\n\n

      Efficiency of transformer at any fraction of the load,<\/p>\n\n\n\n

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

      Output kVA corresponding to maximum efficiency<\/p>\n\n\n\n

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

      Output power or load on maximum efficiency.<\/p>\n\n\n\n

       <\/p>\n\n\n\n

      = output kVA for max efficiency \u00d7 p.f.<\/p>\n\n\n\n

      = 23.385 \u00d7 1 = 23.385kW<\/p>\n\n\n\n

      Example 10.32<\/strong><\/p>\n\n\n\n

      A 100 kVA, 2 winding transformer has an iron loss of 1 kW and a cu loss on a normal output current of 1.5 kW. Calculate the kVA loading at which the efficiency is maximum and its efficiency at this loading: (i) at unit p.f. and (ii) at 8 p.f. lagging.<\/em><\/p>\n\n\n\n

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

      Here, rated capacity = 100 kVA; Iron loss, P<\/em>i<\/sub> = 1 kW<\/p>\n\n\n\n

      Full-load Cu loss, P<\/em>c<\/sub> = 1.5 kW<\/p>\n\n\n\n

      Output kVA corresponding to maximum efficiency<\/p>\n\n\n\n

      \"img\"\/<\/figure>\n\n\n\n
        \n
      1. <\/a>At unity p.f. \"img\"<\/li>\n\n\n\n
      2. At 0.8 p.f. lagging \"img\"<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

        The efficiency of a transformer is defined as the ratio of output to the input power, the two being measured in same units (either in watts or in kW). Transformer efficiency,  or Where V2 = Secondary terminal voltage I2 = Full-load secondary current cos \u02782 = p.f. of the load Pi = lron losses = Hysteresis losses + eddy current losses        (constant […]<\/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-2871","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\/2871","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=2871"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2871\/revisions"}],"predecessor-version":[{"id":2872,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2871\/revisions\/2872"}],"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=2871"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2871"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}