Introduction to Electric Systems Course Code: EE 111 Course Code: EE 111 Department: Electrical Engineering Department: Electrical Engineering Instructor Name: B G Fernandes Instructor Name: B.G. Fernandes E ‐ mail id: bgf @ee iitb ac in E ‐ mail id: bgf @ee.iitb.ac.in EE 111: Introduction to Electric Systems Lecture 9 Tue Aug B.G.Fernandes 1/10 11 2009
Sub ‐ Topics: • Concept of Reactive Power • Power factor improvement EE 111: Introduction to Electric Systems Lecture 9 Tue Aug B.G.Fernandes 2/10 11 2009
Review • Instantaneous power in 1 ‐ Ø circuit pulsates at 2F Instantaneous power in 1 Ø circuit pulsates at 2F • Single phase motors require special resilient mountings • Complex power S = VI * = P + jQ • Complex power, S = VI = P + jQ P active power = VI cos θ W = I 2 R θ = ∠ I V = I 2 X Q Q VI sin θ VAr VI sin θ VAr I X S = VI * VA • In DC if P & V are known I can be determined • In DC, if P & V are known, I can be determined • However in AC, P, V & cos Ѳ or V & S should be known to determine I to determine I EE 111: Introduction to Electric Systems Lecture 9 Tue Aug B.G.Fernandes 3/10 11 2009
For a given P • I drawn by load as cos θ (P.F.) ⇒ Drop in the line ⇒ I 2 R , I 2 X in the line and therefore ‘S’ of source ⇒ I 2 R loss in source also increases ⇒ I R loss in source also increases • V L < V s for lagging & unity P.F. • V L ≤ V s or V L >V S for leading P.F. EE 111: Introduction to Electric Systems Lecture 9 Tue Aug B.G.Fernandes 4/10 11 2009
= ∠ > I I |V | |V | Source P.F.= cos s L s L V S S If P.F. is leading If the P.F. is unity > |V | |V | ≥ ≥ ≤ ≤ |V | |V | |V | |V | |V | |V | |V | |V | or or s s L L s L L s L L Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 5/10 11 2009
Concept of Reactive Power 1 kVA, 200V, 50 Hz Generator k ⇒ Rated current = 5A ⇒ Assume load = 1 kVAr, 200V ⇒ Current drawn by the load = 5A ⇒ Current drawn by the load = 5A = rated current of generator ⇒ Power consumed by the load =0 P d b th l d 0 Assume generator is ideal (losses = 0) ∴ Input power = o/p power + loss = 0 ⇒ No input is required. p q Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 6/10 11 2009
⇒ If losses are taken into account, I/P power = losses I/P power = losses ⇒ Source has the capability to supply P ⇒ If additional load (P) is connected ⇒ If additional load (P) is connected ≠ ⇒ = + ≥ 2 2 For any I a 0 I I 5 5 s a ⇒ Fuse will operate ⇒ Fuse will operate ⇒ Else I 2 R losses , RMS value l ⇒ Temperature rise ⇒ Cooling requirement ⇒ Cooling requirement Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 7/10 11 2009
⇒ Though ‘L’ does not consume any P, source capacity can not be utilized to cater other loads not be utilized to cater other loads ⇒ Energy meter reading α power α VI cos θ α I cos θ Load is drawing I L ⇒ Energy Meter reading α I L cos θ ⇒ ead g α L cos θ e gy e e ∴ Tariff α I L cos θ Who pays for I L sin θ ? Who pays for I L sin θ ? ⇒ Fuel (power input) to supply Q may not be required Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 8/10 11 2009
⇒ Utilities may not be able to cater other loads ⇒ Returns are low ⇒ Returns are low ⇒ Utilization and returns are maximum at unity p.f ⇒ Load requires reactive power ⇒ Load requires reactive power ⇒ Generate reactive power locally ⇒ Capacitor draws leading I ⇒ If I C = I L sin θ , I s = I L cos θ ⇒ If I = I sin θ I = I cos θ ⇒ Overhead line loss & drop ⇒ Voltage profile at load end also improves Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 9/10 11 2009
For a small consumer, power tariff depends only on ‘P’ ⇒ No direct benefit ⇒ Cross ‐ section of cable may reduce if ‘Q’is generated locally (one time investment) ⇒ For Bulk consumer : ⇒ For Bulk consumer : Power tariff depends on ‘P’ & ‘Q’ ⇒ Expected to maintain P.F. within permissible limit Lecture 9 Tue Aug EE 111: Introduction to Electric Systems B.G.Fernandes 10/10 11 2009
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