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HOMEWORK 03 - SOLUTION
- Problem 2-2. [30 points] SOLUTION a. The impedances of the excitation branch are already referred to H.V. side. We only need to transfer the secondary series impedances. Turns ratio, a = 8000480 = 16. 7 R′ s = a^2 Rs = (16.7)^2 (0.05) = 13.89Ω X s′ = a^2 Xs = (16.7)^2 (0.06) = 16.73Ω Req = Rp + R′ s = 32 + 13.89 = 45.89Ω Xeq = Xp + X s′ = 45 + 16.73 = 61.73Ω
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
- 7 = 2.^50 ∠^ −^36.^8
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%
- Problem 2-3. [40 points] SOLUTION a. Using open-circuit test values, we can compute the impedances of the excitation branch. θ = cos−^1 ( (^) VocPocIoc ) = cos−^1 ( (^23030) × 0. 45 ) = 73. 15 o^ lagging. Yoc = (^) VIococ = 0.^45 ∠ 230 −^73.^15 = 5. 67 × 10 −^4 − j 1. 875 × 10 −^3 Ω−^1 Zoc = (^) Y^1 oc = 1763 + j 533 .09Ω → Rc = 1763Ω, XM = 533.09Ω Using short-circuit test values, we can compute the series impedances: θ = cos−^1 ( (^) VscPscIsc ) = cos−^1 ( (^19).^421 ×.^38. 7 ) = 75. 3 o^ lagging. ZSE = (^) Isc∠V−sc 75. 3 o = 2. 19 ∠ 75 .3 = 0.557 + j 2 .118Ω → Req = 0.557Ω, Xeq = 2.118Ω
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 =^
- 72 440 = 32W η = (^) 800+32+10^800. 49 × 100% = 94.9%
