{"id":2867,"date":"2024-08-25T19:56:14","date_gmt":"2024-08-25T19:56:14","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2867"},"modified":"2024-08-25T19:56:14","modified_gmt":"2024-08-25T19:56:14","slug":"approximate-expression-for-voltage-regulation","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/approximate-expression-for-voltage-regulation\/","title":{"rendered":"APPROXIMATE EXPRESSION FOR VOLTAGE REGULATION"},"content":{"rendered":"\n

The approximate expression for the no-load secondary voltage is derived in Section 16.1.<\/p>\n\n\n\n

For inductive load<\/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

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

 <\/p>\n\n\n\n

E<\/em>2<\/sub> \u2212 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

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

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

where,<\/p>\n\n\n\n

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

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

 <\/p>\n\n\n\n

% Reg = % resistance drop \u00d7 cos \u0278<\/em>2<\/sub> + % reactance drop \u00d7 sin \u0278<\/em>2<\/sub><\/p>\n\n\n\n

Similarly<\/strong><\/p>\n\n\n\n

(ii) For resistive load: % Reg = % resistance drop<\/p>\n\n\n\n

(iii) For capacitive load<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

% Reg = % resistance drop \u00d7 cos \u0278<\/em>2<\/sub> \u2212 % reactance drop \u00d7 sin \u0278<\/em>2<\/sub><\/p>\n\n\n\n

Example 10.17<\/strong><\/p>\n\n\n\n

A 10 kVA, 2000\/400 V, single-phase transformer has resistance and leakage reactance as follows:<\/p>\n\n\n\n

Primary winding: Resistance = 5.5 \u03a9 Reactance = 12 \u03a9<\/p>\n\n\n\n

Secondary winding: Resistance = 0.2 \u03a9, Reactance = 0.45 \u03a9<\/p>\n\n\n\n

Determine the value of the secondary voltage at full load, 0.8 p.f. lagging, when the primary supply voltage is 2000 V.<\/p>\n\n\n\n

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

Transformer rating = 10 kVA = 10 \u00d7 103 <\/sup>VA<\/p>\n\n\n\n

Primary induced voltage, E<\/em>1 <\/sub>= 2000 V<\/p>\n\n\n\n

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

Primary resistance, R<\/em>1 <\/sub>= 5.5 \u03a9; Primary reactance, X<\/em>1 <\/sub>= 12 \u03a9<\/p>\n\n\n\n

Secondary resistance, R<\/em>2 <\/sub>= 0.2 \u03a9; Secondary reactance, X<\/em>2 <\/sub>= 0.45 \u03a9<\/p>\n\n\n\n

Load p.f.,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

cos \u0278<\/em>2<\/sub> = 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 side,<\/p>\n\n\n\n

\"image\" = K<\/em>2<\/sup>R<\/em>1<\/sub> = (0.2)2<\/sup> \u00d7 5.5 = 0.22\u03a9<\/p>\n\n\n\n

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

R<\/em>es<\/sub> = R<\/em>2<\/sub> + \"image\" = 0.2 + 0.22 = 0.42\u03a9<\/p>\n\n\n\n

Primary reactance referred to secondary side,<\/p>\n\n\n\n

\"image\" = K<\/em>2<\/sup>X<\/em>1<\/sub> = (0.2)2<\/sup> \u00d7 12 = 0.48\u03a9<\/p>\n\n\n\n

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

X<\/em>es<\/sub> = X<\/em>2<\/sub> + \"image\" = 0.45 + 0.48 = 0.93\u03a9<\/p>\n\n\n\n

Load p.f.,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

cos\u0278<\/em>2<\/sub> = 0.8 \u2234 sin\u0278<\/em>2<\/sub> = sincos\u22121<\/sup> 0.8 = 0.6<\/p>\n\n\n\n

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

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

As the primary supply voltage,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>1<\/sub> = E<\/em>1<\/sub> = 2000V<\/p>\n\n\n\n

<\/a>Secondary induced voltage,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

E<\/em>2<\/sub> = KE<\/em>1<\/sub> = 0.2 \u00d7 2000 = 400V<\/p>\n\n\n\n

Using the expression;<\/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> \u2212 I<\/em>2<\/sub>X<\/em>es<\/sub> sin\u0278<\/em>2<\/sub><\/p>\n\n\n\n

Secondary terminal voltage,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>2<\/sub> = E<\/em>2<\/sub> \u2212 I<\/em>2<\/sub>R<\/em>es<\/sub> cos\u0278<\/em>2<\/sub> \u2212 I<\/em>2<\/sub>X<\/em>es<\/sub> sin\u0278<\/em>2<\/sub><\/p>\n\n\n\n

= 400 \u2212 25 \u00d7 0.42 \u00d7 0.8 \u2212 25 \u00d7 0.93 \u00d7 0.6 = 400 \u2212 8.4 \u2212 13.95 = 377.65 A<\/p>\n\n\n\n

Example 10.18<\/strong><\/p>\n\n\n\n

The ratio of turns of a single-phase transformer is 8, the resistance of the primary and the secondary windings are 0.85 \u03a9 and 0.012 \u03a9, respectively, and the leakage reactance of these windings are 4.8 \u03a9 and 0.07 \u03a9, respectively. Determine the voltage to be applied to the primary to obtain a current of 150 A in the secondary when the secondary terminal are short circuited. Ignore the magnetizing current.<\/em><\/p>\n\n\n\n

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

Ratio of turns,<\/p>\n\n\n\n

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

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

 <\/p>\n\n\n\n

R<\/em>1<\/sub> = 0.85;<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

X<\/em>1<\/sub> = 4.8\u03a9<\/p>\n\n\n\n

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

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

Secondary resistance<\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>2<\/sub> = 0.012\u03a9<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

X<\/em>2<\/sub> = 0.07\u03a9<\/p>\n\n\n\n

Secondary resistance referred to primary, \"img\"<\/p>\n\n\n\n

Equivalent resistance referred to primary, R<\/em>ep<\/sub> = R<\/em>1<\/sub> + \"image\" = 0.85 + 0.768 = 1.618\u03a9<\/p>\n\n\n\n

Secondary reactance referred to primary, \"img\"<\/p>\n\n\n\n

Equivalent reactance referred to primary, X<\/em>ep<\/sub> = X<\/em>1<\/sub> + \"image\" = 4.8 + 4.48 = 9.28\u03a9<\/p>\n\n\n\n

Equivalent impedance referred to primary, \"img\"<\/p>\n\n\n\n

= 9.42 \u03a9<\/p>\n\n\n\n

Short circuit current referred to primary, I<\/em>1(SC)<\/sub> = KI<\/em>2(SC)<\/sub> = \"img\" \u00d7 150 = 18.75 A<\/p>\n\n\n\n

Voltage applied to the primary under short circuit condition,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>1(SC)<\/sub> = I<\/em>1(SC)<\/sub> \u00d7 Z<\/em>ep<\/sub> =18.75 \u00d7 9.42 = 176.625 V<\/p>\n","protected":false},"excerpt":{"rendered":"

The approximate expression for the no-load secondary voltage is derived in Section 16.1. For inductive load   E2 = V2 + I2Res cos\u02782 + I2Xes sin\u02782 or   E2 \u2212 V2 = I2Res cos\u02782 + I2Xes sin\u02782 or where, \u2234   % Reg = % resistance drop \u00d7 cos \u02782 + % reactance drop \u00d7 sin \u02782 Similarly (ii) For resistive load: % Reg = % resistance drop (iii) For capacitive load \u2234   […]<\/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-2867","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\/2867","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=2867"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2867\/revisions"}],"predecessor-version":[{"id":2868,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2867\/revisions\/2868"}],"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=2867"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2867"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2867"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}