Unformatted text preview: ECE 333 Homework 2 Solutions :: Spring 2010
2.2 A 120V, 60 Hz source supplies current to a 1 µF capacitor, a 7.036 H inductor and a 120 ! resistor, all wired in parallel (illustrated in Figure P 2.2). a. Find the rms value of current through the capacitor, the inductor and the resistor. Note : I ! rms ; i ! time domain V source ! 120 ! 0o Capacitor: 1 1 o ZC! ! % 6 ! 2652.8 &% 90 ' j " C j 2 # 60$1$10 o V source 120 & 0 o IC! ! ! 0.045& 90 A ( I C ) rms *! 0.0452 A o ZC 2652.8&%90 Inductor: Z L ! j " L ! j 2 # 60$7.036 ! 2651 & 90o ' V 120 & 0o I L ! source ! ! 0.045 &% 90 o A ( I L ) rms * ! 0.0452 A o ZL 2651 & 90 Resistor: Z R! 120 ' V 120 & 0 o I R! source ! !1 A ( I R ) rms * ! 1 A ZR 120 b. Write the current through each component as a function of time i ! I ! 2 cos )" t + ,* Capacitor: # i C ! I C ! 2 cos ) 2 # f t + , * ( i C ) t *! 0.0452 ! 2 cos ) 2 # 60 t + * 2 Inductor: # i L ! I L ! 2 cos ) 2 # f t +, * ( i L ) t *! 0.0452 ! 2 cos ) 2 # 60 t % * 2 Resistor: i R ! I R ! 2 cos ) 2 # f t + , * ( i R ) t *! 1 ! 2 cos ) 2 # 60 t * c. What is the total current delivered by the 120 V source (both RMS value and as a function of time)? # # i T ! i C + i L+ i R! 0.0452 ! 2 cos 2 # 60 t + + cos 2 # 60 t % + ! 2 cos ) 2 # 60 t * 2 2 i T ! ! 2 cos ! 2 ! 60 t ! A o o o o I T ! 0.0452 & 90 + 0.0452 &% 90 + 1 & 0 !1 & 0 A ( I T ) rms *! 1 A !! "! "! 2.4 Find the RMS value of voltage for the sawtooth waveform shown in Figure P2.4 (recall from calculus that T 1 the average value of a periodic function is given by  ) t *! ! f ) t * dt . f T0 From the waveform given in the figure, the function and period can be found by inspection. f ) t *! 2t ; 0 . t . 1 s T ! 1 sec Given the function and the period the RMS can be found using the following process: 1. Square the function: 2 2 f ) t *! 4 t 2. Find the mean of the squared function: T 1 2 ) t *! 1 / f 2 ) t * dt ! 1 / 4t 2 dt ! 4 t 3 from 0 to 1 ! 4 f T0 10 3 3 3. Take the square root of the previous result: ! f2 ) t *! 4 01.155 V RMS 3 In one step: ! 1 V RMS ! / f 2 ) t * dt ! 1 / 4 t 2 dt ( V RMS ! 4 V T0 10 3 ! T ! 1 ! 2.7 Suppose a utility charges its large industrial customers $0.04/kWh for energy plus $7/month per peak kVA (demand charge). Peak kVA means the highest level drawn by the load during the month. If a customer uses an average of 750 kVA during a 720hr month, with 1000 kVA peak, what would be their monthly bill if their power factor (pf) is 0.8? How much money could be saved each month if their real power is the same but their power factor is corrected to 1.0? Total cost per month before the power factor correction: Average real power used: P avg ! S avg! pf ! 750 kVA!0.8 ! 600 kW Usage charge for a 720hr month: hours cost 720 h $ 0.04 $ 17,280 C U ! P1 1 ( 600 kW 1 1 ! month kWh month kWh month Demand charge given a peak load of 1000 kVA: cost $7 $ 7000 C D ! S peak1 !1000 kVA peak1 ! month1S peak month1peak kVA month Total cost per month (sum of usage charge and demand charge): $ 24,280 C T ! C U + C D! $ 17,280+ $ 7000! month Cost savings when power factor is corrected to 1.0: Average real power remains constant: P ! 600 kW $ 17,280 Similarly, usage charge for a 720hr month remains constant: C U ! month Before power factor correction, the peak real power was 1000 kVA!0.8! 800 kW . Since the power factor adjustment does not affect real power, this value will remain constant and can be used to find the new peak apparent power (S): S peak ) new*! P peak! pf ! 800 kW!1.0 !800 kVA Demand charge after power factor correction: $7 $ 5600 C D ) new *! 800 kVA peak1 ! month1peak kVA month Total cost per month after power factor correction: $ 22,880 C T ) new*!C U + C D ) new*! $ 17,280+ $ 5600! month Cost savings per month after power factor correction: $ 1400 2 C ! C T %C T ) new * ! $ 24,280% 22,880 ! month 2.9 A motor with power factor 0.6 draws 4200 watts of real power at 240 volts from a 10 kVA transformer. Suppose a second motor is needed and it too draws 4200 watts. With both motors on line, the transformer will be overloaded unless a power factor correction can be made. a. What power factor would be needed to be able to continue to use the same 10 kVA transformer? 2 2 2 2 2 2 2 S ! P + Q ( Q ! ! S % P ( Q ! ! 10000 % 8400 !Q ! 5425 VAR %1 Q % 1 5425 pf ! cos ) tan ) ** ( cos ) tan ) **! 0.84 lag P 8400 P 8400 ! 0.84 lag Alternate solution: pf ! ! S 10000 b. How many kVAR would need to be added to provide the needed power factor? From (a), the required reactive power can be found as below: S 2! P 2 !Q 2 ! Q required !! S 2 ! P 2 ! Q !! 10000 2!8400 2!Q required !5425 VAR max The reactive power consumed when the power factor is 0.6 can be found by first finding the apparent power (S) for the two motors: P 8400 S! ! ! 14,000 kVA pf 0.6 The reactive power can now be found as below: %1 %1 Q ! sin ) ,*! Q ! S1sin ) cos ) pf ** ( 14000 sin ) cos )0.6 **! 11,200 VAR The amount of kVARs needed to correct the power factor to 0.84: Q add ! Q %Q required ! 11200% 5425! 5.774 kVAR c. How much capacitance would be needed to provide the kVAR correction? From equation (2.46) [assume frequency to be 60 Hz]: VARs 5774 C! 2( 2 ! 266 3 F "V 2 # 601240 2.10 A transformer rated at 1000 kVA is operating near capacity as it supplies a load that draws 900kVA with a power factor of 0.7. a. How many kW of real power is being delivered to the load? P ! S! pf !900 kVA!0.7 ! P !630 kW b. How much additional load (in kW of real power) can be added before the transformer reaches its full rated kVA (assume the power factor remains 0.70)? Remaining apparent power: S remain ! S rated % S used ! 1000 % 900! 100 kVA Using the same relation as in (a): P add ! S remain1 pf ( P ! 70 kW c. How much additional power (above the amount in a) can the load draw from this transformer without exceeding its 1000 kVA rating if the power factor is corrected to 1.0? When the power factor is 1, all of the power drawn from the transformer will be real so 1000 kW can be drawn without exceeding the ratings. S max ! S rated11.0! 1000 kVA11.0! 1000 kVA Therefore, 370 kW can be added: P ) pf !1.0* ! P ) pf !0.6* !1000 ! 630 ! 370 kW ...
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 Spring '09
 Overby
 Electric power, AC power, Power factor, $0.04, 1 sec, 70 kW

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