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The Impedances of the Excitation Part 1-Electro Mechanical Systems-Assignment, Exercises of Electromechanical Systems and Devices

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

2011/2012

Uploaded on 07/31/2012

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HOMEWORK 03 - SOLUTION
1. 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=8000
480 = 16.7
R0
s=a2Rs= (16.7)2(0.05) = 13.89Ω
X0
s=a2Xs= (16.7)2(0.06) = 16.73Ω
Req =Rp+R0
s= 32 + 13.89 = 45.89Ω
Xeq =Xp+X0
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 =Srated
Vp,rated =20k
8000 = 2.5A
Zbase =Vp,rated
Ip,rated =8000
2.5= 3200Ω
Zpu =Zactual
Zbase
Req,pu =45.98
3200 = 0.014Ω
Xeq,pu =61.73
3200 = 0.019Ω
RC,pu =250k
3200 = 78.13Ω
RM,pu =30k
3200 = 9.38Ω
c. VL= 480V,θ= cos1(0.8) = 36.8oCurrent is lagging voltage.
Current in the secondary winding is: IS=S
VL=20k
480 = 41.6736.8oA
Secondary current referred to primary is: I0
S=IS
a=41.6736.8o
16.7= 2.5036.8oA
Primary voltage VPcan be found by applying KVL on the loop in the circuit drawn in part a.
VP=aVS+ (Req +jXeq )I0
S
VP= 16.7×480 + (45.9 + j61.73)(2.5036.8o)
VP= 8200.50.38oV
Voltage Regulation is given by: V R =Vnl Vfl
Vfl
×100% = 82008000
8000 = 2.5%
d. Efficiency η=Pout
Pout+Plost
×100%
1
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HOMEWORK 03 - SOLUTION

  1. 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

  1. 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%

  1. 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 =^

  1. 72 440 = 32W η = (^) 800+32+10^800. 49 × 100% = 94.9%