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ch7sol

# ch7sol - Chapter 7 Problems Solutions 7.1 For any...

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Chapter 7 Problems Solutions 7.1 For any particular value of f I , 3 / L s L V X I = where L V is read from the OCC and L I is read from the SCC. The OCC and SCC data from Fig. 7.16 are used in the following MATLAB program Xs1.m to calculate and plot s X vs. f I . % % Xs1.m - Uses OCC/SCC data from Fig 7.16 to solve % Problem 7.1. clear; VL=[0 2.85 5.7 8.55 10.9 12.9 14.7 16.2 17.5 ... 18.6 19.3 19.8 20.2]*1000; If=[0 100 200 300 400 500 600 700 800 900 1000 ... 1100 1200]; IL=21500/1200*If; m=length(VL); for i=2:m; Xs(i)=VL(i)/sqrt(3)/IL(i); end plot(0,0,If(2:m),Xs(2:m)); grid xlabel('Field current, A'); ylabel('Xs, ohms'); 118

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7.2 Since insufficient information is available to determine r E , use r an E V 2245 . Enter the OCC curve of Fig. 7.16 with 13.8 kV L V = to find 490 A fu I = and 550 A fs I = . By [7.26], 490 0.891 550 fu s fs I k I = = = From [7.27], ( 29 ( 29 0.15 0.85 0.15 0.85 0.891 0.898 0.815 s su s su X X k X = + = + × = This value for s X is slightly larger than the value of Example 7.3 due to the fact that the saturation factor s k is only applied to X φ , the portion of s X that is affected by satura- tion. This problem has emphasized that the method used to determine s X in Example 7.3 is an approximate method. Although the error was about 2.4 percent in this case, the error becomes greater as the level of saturation increases. 7.3 The value of per phase stator winding resistance at 150ºC is found as 2 1 234.5 234.5 sac s sdc sdc R T R R R T + = × × + ( 29 234.5 150 0.03 1.04 0.0471 234.5 20 + = × = + With the stator resistance nonzero in value, 3 13,800 3 0.796 10,000 / / OC s SC V Z I = = = Thus, ( 29 ( 29 2 2 2 2 0.796 0.0471 0.7946 s s s X Z R = - = - = ( 29 0.15 0.15 0.7946 0.1192 s X X = = = l 0.7946 0.1192 0.6753 s X X X φ = - = - = l Since s s R X = , the values of X l and X φ differ little from those of Example 7.3. 7.4 First, determine the rated stator current and its associated phase angle. ( 29 6 250 10 10,459.2 A 3 3 13,800 LR L S I V × = = = ( 29 ( 29 1 1 cos cos 0.8 36.87 PF θ - - = = = ° Assume an V on the reference; then by KVL, ( 29 13,800 0 10,459.2 36.87 0.815 90 3 f an a s E V jI X = + = ∠ ° + ∠ - ° ° 119
cratio.m is used to evaluate the second term of f E ; then csum.m is applied to add the two terms of f E . 14,720.4 27.8 V f E = ° The screen session for computation of f E is shown below. 7.5 Use X φ and X l from Example 7.3 and an V , a I , f E of Example 7.4. Apply KVL to Fig. 7.13 a with 0 s R = . ( 29 13,800 0 10,459 36.87 0.135 90 3 r an a E V jI X = + = ∠ ° + ∠ - ° ° l 120

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8886.93 7.30 r E = ° 27.19 7.30 19.89 r f r E E δ = ∠ - ∠ = - = ° By [7.31], ( 29 ( 29 ( 29 ( 29 ( 29 3 3 14,573.6 8886.93 3 sin sin 19.89 120 0.661 f r d r s E E T X φ δ ϖ π = = ° 5 3 5.305 10 N m d T = × The answer is identical to the result of Example 7.5. 7.6 The per phase equivalent circuit and the associated phasor diagram for analysis are shown below. ( a ) The rated voltage, current, and PF angle are found. 208 120 V 3 an V = 2245 ( 29 5000 13.88 A 3 3 208 T a L S I V = = = ( 29 1 cos 0.866 30 θ - = = ° Assume an V on the reference and apply KVL to find ( 29 120 0 13.88 30 8 90 200.13 28.72 V f an a s E V jI X = + = ∠ ° + ∠ - ° ° = ° Thus, 200.13 V f E = ( b ) The torque angle is given by 28.72 f an E V δ = ∠ - ∠ = ° 121
( c ) Since 0 s R = , [7.33] is applicable. For this 1800 rpm, 60-Hz machine, 188.49 rad /s s ϖ = .

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