ch6sol

# ch6sol - Chapter 6 Problems Solutions 6.1 a Synchronous...

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Unformatted text preview: Chapter 6 Problems Solutions 6.1 ( a ) Synchronous speed for 60-Hz excitation is given by [6.12]. 120 / 7200/ s n f p p = = Specific values for s n are tabulated. s n 3600 1800 1200 p 2 4 6 The slip at rated shaft speed should be less than 10 percent. The only 60-Hz synchronous speed possibility that satisfies this criterion is 1800 rpm. ( b ) By [6.13], 1800 1751 0.0272 1800 s m s m s s n n s n ϖ ϖ ϖ--- = = = = ( c ) By [6.18], ( 29 ( 29 0.0272 60 1.633 Hz r f sf = = = ( d ) The output shaft torque is found from P T ϖ = . ( 29 ( 29 5 746 20.34 N m 1751 /30 s s m P T ϖ π = = = ⋅ 6.2 ( a ) By conservation of energy argument, input losses output- = , or ( 29 0.025 0.05 0.025 0.03 in in s P P P- + + + = Solve for in P and evaluate. ( 29 5 746 4287.36 W 1 0.13 0.87 s in P P = = =- Referring to Fig. 6.19, ( 29 ( 29 3 0.025 0.05 0.925 4287.36 g in in P P P =- + = 3 3965.81 W g P = ( b ) Based on [6.46], ( 29 ( 29 ( 29 3 1 1 0.03 3965.81 3846.83 W d g P s P =- =- = ( c ) With both input and output power known, the efficiency is ( 29 5 746 100 100 87% 4287.36 s in P P η = × = × = 93 ( d ) The power factor is determined as the ratio of average power to apparent power. ( 29 ( 29 4287.36 0.897 3 230 12 in in in P PF S = = = 6.3 Referring to Fig. 6.17 and using the results of Prob. 6.2, the total stator ohmic losses are ( 29 1 0.025 0.025 4287.36 107.18 W R in P P = = = Thus, ( 29 1 1 2 2 1 107.18 0.248 3 3 12 R P R I = = = Ω The total core losses can be found using the result of Prob. 6.2. ( 29 0.05 0.05 4287.36 214.37 W c in P P = = = The excitation voltage is needed. Assume 1 V on reference. ( 29 ( 29 1 1 cos cos 0.897 26.23 in PF θ-- = = = ° By KVL, ( 29 ( 29 1 1 1 1 1 230 12 26.23 0.248 1.4 3 E V I R jX j =- + = ∠ ° - ∠ - ° + 1 132.79 10.096 13.75 122.69 13.75 123.46 6.39 E j j =-- =- = ∠ - ° Now, 2 1 3 c c E P R = or ( 29 2 2 1 3 123.46 3E 213.3 214.37 c c R P = = = Ω By KCL, 1 1 2 1 123.46 6.39 123.46 6.39 12 26.23 213.3 37 90 c m E E I I R jX ∠ - ° ∠ - ° ′ =-- = ∠ - ° -- ∠ ° 2 10.62 1.93 10.79 10.3 I j ′ =- = ∠ - ° Using the result of Prob. 6.2, ( 29 2 0.025 0.025 4287.36 107.18 W R in P P ′ = = = 94 Whence, ( 29 ( 29 2 2 2 2 2 107.18 0.307 3 3 10.79 R P R I ′ ′ = = = Ω ′ The value of input reactive power is given by ( 29 ( 29 ( 29 sin 3 230 12 sin 26.23 2112.78 VARs in in Q S θ = = ° = The input reactive power must also equal the sum of the reactive power supplied to each inductive element. ( 29 2 2 2 1 1 2 2 1 3 3 3 / in m Q I X I X E X ′ ′ = + + or ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 1 1 1 2 2 2 2 2112.78 3 12 1.4 3 123.46 /37 3 3 / 3 3 10.79 in m Q I X E X X I---- ′ = = ′ 2 0.779 X ′ = Ω 6.4 The file im_data.m must contain the correct values of equivalent circuit parameters. If the first and third through sixth lines of code following the comment statement “% Empirical adjustment of R2pr & X2pr for fr variation” are active (no leading %), then R2pr and X2pr are adjusted for increase in slip; otherwise, R2pr and X2pr are...
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## This document was uploaded on 01/20/2012.

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ch6sol - Chapter 6 Problems Solutions 6.1 a Synchronous...

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