ProblemsAndSolutions_Ch-7

- fr 7-6 fe A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed of 3440 r/min Calculate the

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116 ( ) ( ) re 7-6. A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed of 3440 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation (4-57)]? S OLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at no- load conditions is % 56 . 0 % 100 3600 3580 3600 % 100 sync nl sync nl = × = × = n n n s ( ) ( ) ,nl 0.0056 60 Hz 0.33 Hz fs f == = The slip and electrical frequency at full load conditions is % 44 . 4 % 100 3600 3440 3600 % 100 sync nl sync fl = × = × = n n n s ( ) ( ) ,fl 0.0444 60 Hz 2.67 Hz f = The speed regulation is % 1 . 4 % 100 3440 3440 3580 % 100 SR fl fl nl = × = × = n n n

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126 () M M X X j R jX R jX Z + + + = 1 1 1 1 TH ( ) [] M M M M X X j R X X j R X X j R jX R jX Z + + + + + = 1 1 1 1 1 1 1 1 TH [ ] [ ] 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 1 TH M M M M M M M X X R X X X X X R j X R X X R X X R Z + + + + + + + = 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 TH TH TH M M M M M M X X R X X X X X R j X X R X R jX R Z + + + + + + + = + = The Thevenin resistance is 2 1 2 1 2 1 TH M M X X R X R R + + = . If 1 R X M >> , then 2 1 2 1 2 1 M M X X X X R + + + , so 2 1 1 TH ¸ ¸ ¹ · ¨ ¨ © § + M M X X X R R The Thevenin reactance is 2 1 2 1 2 1 2 1 2 1 TH M M M M X X R X X X X X R X + + + + = . If 1 R X M >> and 1 X X M >> then M M M X X X R X X 2 1 2 1 2 1 + >> and 2 1 2 2 1 R X X X M M >> + , so 1 2 2 1 TH X X X X X M M = 7-13. Figure P7-1 shows a simple circuit consisting of a voltage source, two resistors, and two reactances in series with each other. If the resistor R L is allowed to vary but all the other components are constant, at what value of R L will the maximum possible power be supplied to it? Prove your answer. ( Hint: Derive an expression for load power in terms of V , R S , X S , R L and X L and take the partial derivative of that expression with respect to R L .) Use this result to derive the expression for the pullout torque [Equation (7-52)]. S OLUTION The current flowing in this circuit is given by the equation L L S S L jX R jX R + + + = V I
127 () ( ) 2 2 L S L S L X X R R V I + + + = The power supplied to the load is ( ) 2 2 2 2 L S L S L L L X X R R R V R I P + + + = = ( ) ( ) 22 2 2 SL S L L L S L RR X X VV R P R X X ªº ª º ++ + + ¬ ¼ ¬¼ = + To find the point of maximum power supplied to the load, set L R P / = 0 and solve for L R . ( ) 20 S L L X X R + += ( ) 2 S L L R RX X R R R + = + 2 2 2 SS L LS L S L L R RR R X X R + + =+ 2 2 2 L R X R + = 2 LL R XX R = Therefore, for maximum power transfer, the load resistor should be 2 2 L S S L X X R R + + = 7-14. A 440-V 50-Hz six-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R 1 = 0.082 R 2 = 0.070 X M = 7.2 X

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This note was uploaded on 11/10/2011 for the course EEL 3216 taught by Professor Brooks during the Fall '08 term at FSU.

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- fr 7-6 fe A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed of 3440 r/min Calculate the

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