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The course gives an introduction to various principles and applications pertaining to electrical machines. The course covers magnetic circuits, single phase transformer and equivalent circuit, auto transformer, basic concepts of electromechanical energyc onversion, DC and AC machines modeling and, steady state analysis. It includes: Impedances, Excitation, Branch, Transfer, Secondary, Series, Ratio, Transformer, Rated, Power
Typology: Exercises
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b. For transforming the equivalent circuit to a per unit system, we need to establish a base value. One widely accepted base value is the rated value of that transformer. For the given transformer the rated impedance value can be found from the rated power and voltage values as shown below. Ip,rated = (^) VSp,ratedrated = 800020 k = 2. 5 A
Zbase = V Ip,ratedp,rated = (^80002). 5 = 3200Ω
Zpu = Z Zactualbase
→ Req,pu = 453200.^98 = 0.014Ω → Xeq,pu = 613200.^73 = 0.019Ω → RC,pu = 2503200 k = 78.13Ω → RM,pu = 320030 k = 9.38Ω
c. VL = 480V , θ = cos−^1 (0.8) = 36. 8 o^ Current is lagging voltage. Current in the secondary winding is: IS = (^) VSL = 20480 k = 41. 67 ∠ − 36. 8 oA
Secondary current referred to primary is: I S′ = I aS = 41.^67 ∠−^36.^8
o
oA
Primary voltage VP can be found by applying KVL on the loop in the circuit drawn in part a. VP = aVS + (Req + jXeq )I S′ VP = 16. 7 × 480 + (45.9 + j 61 .73)(2. 50 ∠ − 36. 8 o) VP = 8200. 5 ∠ 0. 38 oV Voltage Regulation is given by: V R = Vnl V−f lV f l× 100% = 82008000 −^8000 = 2.5%
d. Efficiency η = (^) PoutP+outPlost × 100% 1
2
Pout = S cos θ = 12k cos(0.8) = 16kW PCu = I′ s^2 Req = (2.5)^2 × 45 .98 = 286. 8 W PCore = V (^) P^2 RC =^
8200 250 k = 268.^9 W → η = (^16) k+286^16 .k8+268. 9 × 100% = 96.6%
Since the test data were taken from primary side (high voltage side), we need to transfer the impedances to secondary side so as to get the equivalent circuit referred to the low voltage side.
The turns ratio, n = 230115 = 2 To transform the impedances to secondary side, Z′^ = (^) aZ 2 → R′ eq = R aeq 2 = 0.139Ω → X eq′ = Xeq a^2 = 0.529Ω → R′ C = R aC 2 = 440Ω → X M′ = X aM 2 = 133.3Ω
b. 0.8 PF lagging. Vp a = (Req^ +^ Xeq^ )Is^ +^ Vs Is = (^) VSs = 1000115 = 8. 69 ∠ − 36. 9 oA Vp a = (0.139 +^ j^0 .529)^ ×^ (8.^69 ∠^ −^36 .9) + 115 Vp a = 118.^7 ∠^1.^419 V V R = Vnl V−f lV f l× 100% = 118115 −^115 × 100% = 2.6%
c. η = (^) PoutP+outPlost Pout = Scosθ = 1000 × 0 .8 = 800W Plost = PCu + PCore PCu = I^2 s Req = (8.69)^2 × 0 .139 = 10. 49 W PCore =
Vp a
2 Rc =^