Single Phase Induction Motor Module Analysis
Learn about the analysis of a single-phase induction motor module including drawing the equivalent circuit, calculating various parameters, and determining the mechanical power outputs. Explore the worked example and solution provided along with the required equivalent circuit and parameters. Understand calculations for forward and backward fields, power factor, and more in this detailed study of electrical machines.
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Presentation Transcript
EEE 326 - ELECTRICAL MACHINES II SINGLE PHASE INDUCTION MODULE
WORKED EXAMPLE A 23OV, 50Hz, 4 pole , single phase induction motor has the following parameters and is running at a slip of 0.05 ?1= 2.51 ; ? 2= 7.81 ; ??= 150.88 ; ?1= 4.62 ; ? 2= 7.62 ; ??= 30.10 . 1. Draw the EXACT EQUIVALENT CIRCUIT of the Associated Single phase Induction Motor 2. Calculate the following: (i) ??(ii) ??(iii) ?1(iv) ???(v) ?1, Motor circuit (vi) power factor (vii) ??[ forward emf], (viii) ??[ back emf], (ix) ? 2?(x) ? 2? (xi) Airgap power for the forward field, ??? =[? 2?]2? 2 2?Watts
SOLUTION ? 2 (xii) Airgap power for the backward field, ??? =[? 2?]2 2[2 ?]Watts (xiii) Mechanical power output for the forward field, ???? ? = 1 ? ???= [? 2?]2? 2 2. 1 ? Watts ? (xiv) Mechanical power output for the backward field, ???? ? = ? 1 ???= [? 2?]2? 2 2. 1 ? 2 ?Watts
REQUIRED EQUIVALENT CIRCUIT REQUIRED EQUIVALENT CIRCUIT ?? ?? ?? ? ?? ? ?? ?? ? ? ??? ?? ? ?? ?? ? Forward field ? ? ? ?? ? ?? ? ?? ? ? ??? ?? ? ?? ? ???????? ?? ????? ? ? ?
DETERMINATION OF PARAMETERS OF THE EQUIVALENT CIRCUIT DETERMINATION OF PARAMETERS OF THE EQUIVALENT CIRCUIT ?? 2 = 150.88 = 75.44 2 ?? 2 = 30.10 2 = 15.05 ? 2 2?? = 7.81 2[0.05] = 78.10 ; where ? = ??= 0.05 ; ??+ ?? = 2 ? 2 2 = 7.62 2 = 3.81 ? 2 2?? = ? 2 7.81 2[2 ?] = 2[2 0.05] = 2.003
(i)??= [?2 2? + j ?2 2 ]// ?? 2 // j ?? 2 ?? 2?j?? ?? 2+j?? ?? 2 // j ?? ??.??? 2 2 = = 14.48 + j2.872 = 14.76 2 ?2 2? + j ?2 2 = 7.81 + j3.81 [14.48 + j2.872] // [7.81 + j3.81] = [14.48 + j2.872] x [7.81 + j3.81] [14.48 + j2.872] + [7.81 + j3.81] = 12.25 + j2.13 ?? = 12.436 ??.??? (iii) ?1 = ?1 + j ?1 = [2.51+ j 4.62]
2[2 ?] + j ?2 ?2 2 ]// ?? 2 // j ?? (ii)??= [ 2 = [2.003 + j3.81]// [14.48 + j2.876] = [2.003 + j3.81] x [14.48 + j2.876] [2.003 + j3.81] + [14.48 + j2.876] ?? = [2.23 + j2.79] (iii) ?1 = ?1 + j ?1 = [2.51+ j 4.62] (iv) ????? = ?1 + ?? + ?? = {[2.51+ j 4.62] + [12.25 + j2.13] + [2.23 + j2.79]} ????? = [19.22 + j10.28] ?1 230 28.14 (v) ?1 = ????? = 19.22 + j10.28 = 9.30 j4.98 = 10.55A
SOLUTION CONTD (VI) Power factor, ??? = ??? 28.14 = 0.882 ??????? (vii). ??= ?1?? = [9.30 j4.98][12.25 + j2.13] = 124.53 j41.196 = 131V, 18.3? (viii). ??= ?1?? = [9.30 j4.98][2.23 + j2.79] = 34.63 + j14.84 = 37.68V, 23.2? =124.53 j41.196 7.81 + j3.81 ?? = 1.563 j0.604 = 1.6774, 26.3? (ix) ? 2?= ?2 2? + j?2 2 =34.63 + j14.84 2.003 + j3.81 = 6.79 j5.52 = 35A, 39.07? ?? (x) ? 2?= 2[2 ?] + j?2 ?2 2
SOLUTION CONTD (xi) Airgap power for the forward field, ??? = [? 2?]2x [?2 ??? = 219.6 Watts 2?] = [1.677]2x [78.1] ?2 (xii) Airgap power for the forward field, ??? = [? 2?]2x [ ??? = 153.35 Watts 2(2 ?)] = [8.75]2x [2.003] (xiii) Mechanical Power Output for the forward field, ???? .?= 1 ? ??? = 1 0.05 219.6 ?????.?= 208.62 Watts (xiii) Mechanical Power Output for the backward field, ???? .?= ? 1 ??? = 1 0.05 153.35 ?????.?= - 145 Watts
