{"id":2702,"date":"2024-08-24T20:58:02","date_gmt":"2024-08-24T20:58:02","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2702"},"modified":"2024-08-24T20:58:03","modified_gmt":"2024-08-24T20:58:03","slug":"cumulative-and-differential-compound-wound-generators","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/24\/cumulative-and-differential-compound-wound-generators\/","title":{"rendered":"Cumulative and Differential Compound-wound Generators"},"content":{"rendered":"\n

In compound-wound DC generators, the field is produced by the shunt as well as series winding. Generally, the shunt field is stronger than the series field. When the series field assist the shunt field, the generator is called as cumulative compound-wound generator (see\u00a0Fig. 11.21(a)). However, when the series field opposes the shunt field, the generator is known as differential compound-wound generator (see\u00a0Fig. 11.21(b)).<\/p>\n\n\n\n

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

Fig. 11.21<\/strong>  (a) Cumulatively compound
(b) Differentially compound<\/p>\n\n\n\n

Example 11.11<\/strong><\/p>\n\n\n\n

A 200 V, 8-pole, lap-connected DC shunt generator supplied 60, 40 W, 200 V lamps. It has armature and field circuit resistances of 0.2 \u03a9 and 200 \u03a9, respectively. Calculate the generated emf, armature current, and current in each armature conductor.<\/p>\n\n\n\n

(U.P.T.U. 2004<\/strong>–05)<\/strong><\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.22.<\/p>\n\n\n\n

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

Fig. 11.22<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Here, V<\/em> = 200 V; P<\/em> = 8; A<\/em> = P<\/em> = 8 (lap winding);<\/p>\n\n\n\n

R<\/em>a<\/sub> = 0.2 \u03a9; R<\/em>sh<\/sub> = 200 \u03a9<\/p>\n\n\n\n

Load = 60 \u00d7 40 = 2,400 W<\/p>\n\n\n\n

Load current, \"image\"<\/p>\n\n\n\n

Shunt field current, \"image\"<\/p>\n\n\n\n

Armature current, I<\/em>a<\/sub> = I<\/em>L<\/sub> + I<\/em>sh<\/sub><\/p>\n\n\n\n

 <\/p>\n\n\n\n

= 12 + 1 = 13 A<\/p>\n\n\n\n

Generated emf, E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub> = 200 + 13 \u00d7 0.2<\/p>\n\n\n\n

 <\/p>\n\n\n\n

= 202.6 V<\/p>\n\n\n\n

Current in each armature conductor, \"image\"<\/p>\n\n\n\n

Example 11.12<\/strong><\/p>\n\n\n\n

A 20 kW, 200 V shunt generator has an armature resistance of 0.05 \u03a9 and a shunt field resistance of 200 \u03a9. Calculate the power developed in the armature when it delivers rated output.<\/p>\n\n\n\n

(U.P.T.U. 2006<\/strong>–07)<\/strong><\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.23.<\/p>\n\n\n\n

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

Fig. 11.23<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

For shunt generator: I<\/em>a<\/sub> = I<\/em>L<\/sub> + I<\/em>sh<\/sub><\/p>\n\n\n\n

Power delivered to load, P<\/em>L<\/sub> = VI<\/em>L<\/sub><\/p>\n\n\n\n

Power developed in armature, P<\/em>g<\/sub> = E<\/em>g<\/sub>I<\/em>a<\/sub><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

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

I<\/em>a<\/sub> = 100 + 1 = 101 A<\/p>\n\n\n\n

E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub> = 200 + 101 \u00d7 0.05 = 205.05 V<\/p>\n\n\n\n

P<\/em>g<\/sub> = E<\/em>g<\/sub> I<\/em>a<\/sub> = 205.05 \u00d7 101 = 20,710.05 W = 20.71 kW<\/p>\n\n\n\n

Example 11.13<\/strong><\/p>\n\n\n\n

A 4-pole DC generator with wave-connected armature has 41 slots, and 12 conductors\/slot. Armature resistance R<\/em>a<\/sub> = 0.5 \u03a9; shunt resistance is R<\/em>sh<\/sub> = 200 \u03a9; flux per pole = 125 mWb; and speed N<\/em> = 1,000 rpm. Calculate voltage across 10 \u03a9 load resistance across the armature terminal.<\/p>\n\n\n\n

(U.P.T.U. 2007<\/strong>–08)<\/strong><\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.24.<\/p>\n\n\n\n

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

Fig. 11.24<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Here, P<\/em> = 4; R<\/em>a<\/sub> = 0.5 \u03a9; R<\/em>sh<\/sub> = 200 \u03a9; \u0278<\/em> = 125 mWb = 125 \u00d7 10\u2212<\/sup>3<\/sup> Wb; N<\/em> = 1,000 rpm; R<\/em>L<\/sub> = 10 \u03a9<\/p>\n\n\n\n

Number of slots = 41<\/p>\n\n\n\n

Number of conductors\/slot = 12<\/p>\n\n\n\n

Armature conductors, Z<\/em> = 41 \u00d7 12 = 492<\/p>\n\n\n\n

Number of parallel paths, A<\/em> = 2 (wave winding)<\/p>\n\n\n\n

Generated emf, \"image\"<\/p>\n\n\n\n

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

=2,050V<\/p>\n\n\n\n

Load current, \"image\"<\/p>\n\n\n\n

Shunt field current, \"image\"<\/p>\n\n\n\n

Armature current, \"image\"<\/p>\n\n\n\n

<\/a>Now,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub><\/p>\n\n\n\n

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

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

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

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

Example 11.14<\/strong><\/p>\n\n\n\n

A 4-pole DC shunt generator with wave-wound armature has 40 slots each having 12 conductors. Armature resistance is 1 \u03a9 and shunt field resistance is 200 \u03a9. The flux per pole is 25 mWb. If a load of 50 \u03a9 is connected across the armature terminals, calculate the voltage across the load when the generator is driven at 1,000 rpm. What will be the load voltage if the generator is lap wound?<\/p>\n\n\n\n

(U.P.T.U. 2006<\/strong>–07)<\/strong><\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.25.<\/p>\n\n\n\n

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

Fig. 11.25<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Here, P<\/em> = 4, R<\/em>a<\/sub> = 1 \u03a9; R<\/em>sh<\/sub> = 200 \u03a9; \u0278<\/em> = 25 mWb = 25 \u00d7 10\u2212<\/sup>3<\/sup> Wb; N<\/em> = 1,000 rpm<\/p>\n\n\n\n

Number of slots = 40; number of conductors\/slot = 12<\/p>\n\n\n\n

Total armature conductors,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

Z<\/em> = 40 \u00d7 12 = 480<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

R<\/em>L<\/sub> = 50 \u03a9<\/p>\n\n\n\n

Number of parallel path, A<\/em> = 2 (for wave winding)<\/p>\n\n\n\n

Generated emf,<\/p>\n\n\n\n

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

Load current,<\/p>\n\n\n\n

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

Shunt field current,<\/p>\n\n\n\n

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

Armature current,<\/p>\n\n\n\n

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

Now,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub><\/p>\n\n\n\n

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

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

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

If the generator is lap wound, then A<\/em> = P<\/em> = 4<\/p>\n\n\n\n

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

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

 <\/p>\n\n\n\n

E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub><\/p>\n\n\n\n

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

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

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

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

Example 11.15<\/strong><\/p>\n\n\n\n

A 4-pole DC shunt generator with a wave-wound armature having 390 conductors has to supply a load of 500 lamps each of 100 W at 250 V. Allowing 10 V for the voltage drop in the connecting leads between the generator and the load and brush drop of 2 V. Calculate the speed at which the generator should be driven. The flux per pole is 30 mWb and the value of R<\/em>a<\/sub> = 0.05 \u03a9 and R<\/em>sh<\/sub> = 65 \u03a9.<\/p>\n\n\n\n

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

The conventional circuit diagram of the DC shunt generator is shown in\u00a0Figure 11.26.<\/p>\n\n\n\n

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

Fig. 11.26<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Total load = 500 \u00d7 100 W<\/p>\n\n\n\n

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

Voltage drop in leads, V<\/em>L<\/sub> = 10 V<\/p>\n\n\n\n

Voltage across shunt field winding,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>sh<\/sub> = V<\/em> + V<\/em>L<\/sub> = 250 + 10 = 260 V<\/p>\n\n\n\n

I<\/em>sh<\/sub> = V<\/em>sh<\/sub>\/R<\/em>sh<\/sub> = 260\/65 = 4 A<\/p>\n\n\n\n

I<\/em>a<\/sub> = I<\/em>L<\/sub> + I<\/em>sh<\/sub> = 200 + 4 = 204 A<\/p>\n\n\n\n

Armature drop = I<\/em>a<\/sub> R<\/em>a<\/sub> = 204 \u00d7 0.05 = 10.2 V<\/p>\n\n\n\n

Total brush drop, 2v<\/em>b<\/sub> = 2 V<\/p>\n\n\n\n

Generated emf, E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub> + V<\/em>L<\/sub> + 2<\/p>\n\n\n\n

= 250 + 10.2 + 10 + 2 = 272.2 V<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

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

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

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

Example 11.16<\/strong><\/p>\n\n\n\n

A 4-pole DC shunt generator with a shunt field resistance of 100 \u03a9 and an armature resistance of 1 \u03a9 has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. If a load resistance of 10 \u03a9 is connected across the armature terminals and the generator is driven at 1,000 rpm, calculate the power absorbed by the load.<\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.27.<\/p>\n\n\n\n

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

Fig. 11.27<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Generated emf, \"image\"<\/p>\n\n\n\n

<\/a>Line current, \"image\" (where V<\/em> is terminal voltage)<\/p>\n\n\n\n

Shunt field current, \"image\"<\/p>\n\n\n\n

Armature current, \"image\"<\/p>\n\n\n\n

Using the relation, E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub>; 252 = V<\/em> + 0.11 V<\/em> \u00d7 1.0<\/p>\n\n\n\n

Terminal voltage, V<\/em> = 227 V<\/p>\n\n\n\n

Load current,<\/p>\n\n\n\n

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

Power absorbed by the load, P<\/em> = VI<\/em>L<\/sub> = 227 \u00d7 22.7 = 5.153 kW<\/p>\n\n\n\n

Example 11.17<\/strong><\/p>\n\n\n\n

A short-shunt cumulative compound DC generator supplies 7.5 kW at 230 V. The shunt field, series field, and armature resistance are 100, 0.3 and 0.4 \u03a9, respectively. Calculate the induced emf and the load resistance.<\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.28.<\/p>\n\n\n\n

From\u00a0Figure 11.29,<\/p>\n\n\n\n

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

Fig. 11.28<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

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

I<\/em>a<\/sub> = I<\/em>L<\/sub> + I<\/em>sh<\/sub> = 32.61 + 2.39 = 35 A<\/p>\n\n\n\n

Induced emf, E<\/em>g<\/sub> = V<\/em> + I<\/em>L<\/sub> R<\/em>se<\/sub> + I<\/em>a<\/sub> R<\/em>a<\/sub><\/p>\n\n\n\n

 <\/p>\n\n\n\n

= 230 + 32.61 \u00d7 0.3 + 35 \u00d7 0.4 = 253.78 V<\/p>\n\n\n\n

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

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

Example 11.18<\/strong><\/p>\n\n\n\n

A 20 kW compound generator works on full-load with a terminal voltage of 230 V. The armature, series, and shunt field resistance are 0.1, 0.05, and 115 \u03a9, respectively. Calculate the generated emf when the generator is connected as shunt.<\/p>\n\n\n\n

(Pb. Univ. Dec. 1994)<\/strong><\/p>\n\n\n\n

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

The conventional circuit is shown in\u00a0Figure 11.29.<\/p>\n\n\n\n

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

Fig. 11.29<\/strong>  Conventional diagram as per data<\/p>\n\n\n\n

Load = 20 kW = 20 \u00d7 103 <\/sup>W<\/p>\n\n\n\n

V<\/em> = 230 V; R<\/em>a<\/sub> = 0.1 \u03a9; R<\/em>se<\/sub> = 0.05 \u03a9<\/p>\n\n\n\n

 <\/p>\n\n\n\n

<\/a>R<\/em>sh<\/sub> = 115 \u03a9<\/p>\n\n\n\n

Line current,<\/p>\n\n\n\n

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

Shunt field current,<\/p>\n\n\n\n

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

Armature current, I<\/em>a<\/sub> = I<\/em>L<\/sub> + I<\/em>sh<\/sub> = 86.96 + 2 = 88.96 A<\/p>\n\n\n\n

Generated emf, E<\/em>g<\/sub> = V<\/em> + I<\/em>a<\/sub> R<\/em>a<\/sub> + I<\/em>a<\/sub> R<\/em>se<\/sub><\/p>\n\n\n\n

= 230 + 88.96 \u00d7 0.1 + 88.96 \u00d7 0.05<\/p>\n\n\n\n

= 243.3 V<\/p>\n","protected":false},"excerpt":{"rendered":"

In compound-wound DC generators, the field is produced by the shunt as well as series winding. Generally, the shunt field is stronger than the series field. When the series field assist the shunt field, the generator is called as cumulative compound-wound generator (see\u00a0Fig. 11.21(a)). However, when the series field opposes the shunt field, the generator […]<\/p>\n","protected":false},"author":1,"featured_media":2700,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[414],"tags":[],"class_list":["post-2702","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-dc-machines-generators-and-motors"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/direct-current-generator-components-fig1.webp","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2702","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=2702"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2702\/revisions"}],"predecessor-version":[{"id":2703,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2702\/revisions\/2703"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2700"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2702"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2702"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}