{"id":2828,"date":"2024-08-25T19:12:47","date_gmt":"2024-08-25T19:12:47","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2828"},"modified":"2024-08-25T19:12:47","modified_gmt":"2024-08-25T19:12:47","slug":"two-wattmeter-method-balanced-load","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/two-wattmeter-method-balanced-load\/","title":{"rendered":"\u00a0\u00a0TWO-WATTMETER METHOD (BALANCED LOAD)"},"content":{"rendered":"\n

The two-wattmeter method can be explained somewhat more clearly by considering a balanced load. In this case, we shall prove that power measured by the two wattmeters (i.e.,sum of two wattmeter readings) is equal to \"image\"V<\/em>L<\/sub>I<\/em>L<\/sub> cos\u0278<\/em>, which is the actual power consumed in a three- phase balanced load.<\/p>\n\n\n\n

The connection diagram for a three-phase balanced load (connected in star) is shown in\u00a0Figure 8.37(a). Considering load to be an inductive one, the phasor diagram is shown in\u00a0Figure 8.37(b). The three-phase voltages\u00a0V<\/em>RN<\/sub>,\u00a0V<\/em>YN<\/sub>,and\u00a0V<\/em>BN<\/sub>displaced by an angle of 120\u00b0 electrical are shown in phasor diagram. The phase currents lag behind their respective phase voltages by an angle\u00a0\u0278<\/em>.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 8.37<\/strong>  (a) Connection of two wattmeters for power measurement in balanced 3-phase load (b) Phase diagram<\/p>\n\n\n\n

Current through current coil (c.c.) of W<\/em>1<\/sub> = I<\/em>R<\/sub><\/p>\n\n\n\n

Potential difference across potential coil (p.c.) of \"image\"<\/p>\n\n\n\n

To obtain V<\/em>RB<\/sub>, reverse the phasor V<\/em>BN<\/sub>and add it vectorially to phasor V<\/em>RN<\/sub>,<\/em> as shown in Figure 8.37(b).<\/a><\/p>\n\n\n\n

The phase difference between V<\/em>RB<\/sub>and I<\/em>R<\/sub>is (30\u00b0 \u2212 \u0278<\/em>).<\/p>\n\n\n\n

Power measured by wattmeter<\/p>\n\n\n\n

W<\/em>1 <\/sub>= V<\/em>RB<\/sub>I<\/em>R<\/sub> cos(30 \u2212 \u0278<\/em>)<\/p>\n\n\n\n

Current through c.c. of W<\/em>2<\/sub> = I<\/em>Y<\/sub><\/p>\n\n\n\n

Potential difference across p.c. of \"image\"<\/p>\n\n\n\n

To obtain\u00a0V<\/em>YB<\/sub>, reverse the phasor\u00a0V<\/em>BN<\/sub>and add it vectorially to phasor\u00a0V<\/em>YN<\/sub>as shown in\u00a0Figure 8.37(b).<\/p>\n\n\n\n

The phase difference between V<\/em>YB<\/sub>and I<\/em>Y<\/sub>is (30\u00b0 + \u0278<\/em>).<\/p>\n\n\n\n

Power measured by wattmeter, W<\/em>2<\/sub> = V<\/em>YB<\/sub>= V<\/em>BR<\/sub>= V<\/em>L<\/sub><\/p>\n\n\n\n

Wattmeter reading, W<\/em>1<\/sub> = V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 \u2212 \u0278<\/em>)<\/p>\n\n\n\n

Wattmeter reading, W<\/em>2<\/sub> = V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 + \u0278<\/em>)<\/p>\n\n\n\n

Sum of the two wattmeter readings,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

W<\/em>1<\/sub> + W<\/em>2<\/sub> = V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 \u2212 \u0278<\/em>) + V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 + \u0278<\/em>)<\/p>\n\n\n\n

<\/a>= V<\/em>L<\/sub> I<\/em>L<\/sub>[cos (30\u00b0 \u2212 \u0278<\/em>) + cos (30\u00b0 + \u0278<\/em>)]<\/p>\n\n\n\n

= V<\/em>L<\/sub> I<\/em>L<\/sub>[cos 30\u00b0 cos \u0278<\/em>+ sin 30\u00b0 sin \u0278<\/em>+ cos 30\u00b0 cos \u0278<\/em>\u2212 sin 30\u00b0 sin \u0278<\/em>]<\/p>\n\n\n\n

= V<\/em>L<\/sub> I<\/em>L<\/sub>(2 cos 30\u00b0 cos\u0278<\/em> ) \"image\"<\/p>\n\n\n\n

= Total power absorbed by a three-phase balanced load (P<\/em>)<\/p>\n\n\n\n

P <\/em>= W<\/em>1<\/sub> + W<\/em>2<\/sub><\/p>\n\n\n\n

Thus, the sum of the readings of the two wattmeters is equal to the power absorbed in a three-phase balanced load.<\/p>\n\n\n\n

8.16.1  Determination of Power Factor from Wattmeter Readings<\/h4>\n\n\n\n

We have seen that<\/p>\n\n\n\n

W<\/em>1<\/sub> + W<\/em>2 <\/sub>= \"image\"V<\/em>L<\/sub>I<\/em>L<\/sub>cos\u0278<\/em>        (8.1)<\/p>\n\n\n\n

W<\/em>1<\/sub> \u2212 W<\/em>2 <\/sub>= V<\/em>L<\/sub>I<\/em>L<\/sub>(cos(30\u00b0 \u2212 \u0278<\/em>) \u2212 cos(30\u00b0 + \u0278<\/em>))<\/p>\n\n\n\n

=2V<\/em>L<\/sub>I<\/em>L<\/sub>sin30sin\u0278<\/em>= V<\/em>L<\/sub>I<\/em>L<\/sub>sin\u0278<\/em>        (8.2)<\/p>\n\n\n\n

Dividing\u00a0Equation (8.2)\u00a0by (8.1), we get<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

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

\"image\"\/<\/figure>\n\n\n\n

Power factor<\/p>\n\n\n\n

cos\u0278<\/em> = cos tan\u22121<\/sup>\u0278<\/em> = \"image\"<\/p>\n\n\n\n

8.16.2  Determination of Reactive Power from Two Wattmeter Readings<\/h4>\n\n\n\n

Multiplying\u00a0Equation (8.2)\u00a0by\u00a0\"image\", we get<\/p>\n\n\n\n

\"image\"(W<\/em>1<\/sub> \u2212 W<\/em>2<\/sub>) = \"image\" V<\/em>L<\/sub>I<\/em>L<\/sub>sin\u0278 = P<\/em>r<\/em><\/sub><\/p>\n\n\n\n

Reactive power,<\/p>\n\n\n\n

P<\/em>r<\/sub> = \"image\"(W<\/em>1<\/sub> \u2212 W<\/em>2<\/sub>)<\/p>\n\n\n\n

8.17  EFFECT OF POWER FACTOR ON THE TWO WATTMETER READINGS<\/h5>\n\n\n\n

For lagging power factor, the wattmeter readings are<\/p>\n\n\n\n

 <\/p>\n\n\n\n

W<\/em>1<\/sub> = V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 \u2212 \u0278<\/em>) and W<\/em>2<\/sub> = V<\/em>L<\/sub> I<\/em>L<\/sub>cos (30\u00b0 + \u0278<\/em>)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

It is clear that the readings of the two wattmeters do not only depend upon load, but they also depend upon the phase angle \u0278<\/em>or the power factor of the load.<\/p>\n","protected":false},"excerpt":{"rendered":"

The two-wattmeter method can be explained somewhat more clearly by considering a balanced load. In this case, we shall prove that power measured by the two wattmeters (i.e.,sum of two wattmeter readings) is equal to VLIL cos\u0278, which is the actual power consumed in a three- phase balanced load. The connection diagram for a three-phase balanced load […]<\/p>\n","protected":false},"author":1,"featured_media":2812,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[411],"tags":[],"class_list":["post-2828","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-three-phase-ac-circuits"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/ac-voltage-source.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2828","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=2828"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2828\/revisions"}],"predecessor-version":[{"id":2829,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2828\/revisions\/2829"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2812"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}