ε rms ε substation HOMEWORK #11 Chapter 29

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1 HOMEWOK # hapter 9 5 f the frequency in the circuit in Figure 9-8 is doubled, the capacitive reactance of the circuit will (a) double, (b) not change, (c) halve, (d) quadruple. Determine the oncept The capacitive reactance of an capacitor varies with the frequency according to ω. Hence, doubling ω will halve. (c) is correct. 8 The impedances of motors, transformers, electromagnets include both resistance inductive reactance. Suppose that phase of the current to a large industrial plant lags the phase of the applied voltage by 5 when the plant is under full operation using.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 60 Hz line voltage at the plant is 40 k. The resistance of the transmission line from the substation to the plant is 5. Ω. The cost per kilowatt-hour to the company that owns the plant is $0.4, the plant pays only for the actual energy used. (a) Estimate the resistance inductive reactance of the plant s total load. (b) Estimate the current in the power lines the voltage at the substation. (c) How much power is lost in transmission? (d) Suppose that the phase that the current lags the phase of the applied voltage is reduced to 8º by adding a bank of capacitors in series with the load. How much money would be saved by the electric utility during one month of operation, assuming the plant operates at full capacity for 6 h each day? (e) What must be the capacitance of this bank of capacitors to achieve this change in phase angle? Picture the Problem We can find the resistance inductive reactance of the plant s total load from the impedance of the load the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it the voltage at the substation by applying Kirchhoff s loop rule to the substation-transmission wires-load circuit. The power lost in transmission can be found from P. We can find the cost savings by finding the difference in the power lost in transmission when the phase angle is reduced to 8. Finally, we can find the capacitance that is required to reduce the phase angle to 8 by first finding the capacitive reactance using the definition of tanδ then applying the definition of capacitive reactance to find. trans 5.Ω trans trans substation f 60 Hz 40 k δ 5

2 (a) elate the resistance inductive reactance of the plant s total load to δ: cosδ L sinδ Express in te of the current in the power lines the voltage at the plant: Express the power delivered to the plant in te of,, δ solve for : P cosδ av Pav () cosδ Substitute to obtain: evaluate : cos P av ( 40 k) δ cos 5 630Ω.3MW evaluate L : ( 630Ω) 0.57 kω L ( 630Ω) 0.7 kω cos 5 57Ω sin 5 66Ω (b) Use equation () to find the current in the power lines: ( 40k) 63A.3MW 63.4A cos 5 Apply Kirchhoff s loop rule to the circuit: sub trans tot 0 Solve for sub : sub ( trans + tot) evaluate sub : sub ( 63.4 A)( 5.Ω+ 630Ω) 40.3 k

3 (c) The power lost in transmission is: P ( 63.4A) ( 5.Ω) trans trans 0.9 kw kw (d) Express the cost savings in te of the difference in energy consumption (P 5 P 8 ) t the per-unit cost u of the energy: Express the power lost in transmission when δ 8 : ( P P ) tu 5 8 P 8 8 trans Find the current in the transmission lines when δ 8 : ( 40 k) Evaluate 8.3MW cos A P : P ( 60.5A) ( 5.Ω) 9.0 kw evaluate : 8 h d $0.4 d month kw h ( 0.9 kw 9.0 kw) 6 30 $ 8 (e) The required capacitance is given by: elate the new phase angle δ to the inductive reactance L, the reactance due to the added capacitance, the resistance of the load : πf L tanδ L tanδ Substituting for yields: πf ( L tanδ) evaluate : π 33µ F - ( 60 s ) 66Ω ( 57Ω) ( tan8 ) An inductor has a reactance of 00 Ω at 80 Hz. (a) What is its inductance? (b) What is its reactance at 60 Hz?

4 Picture the Problem We can use at any frequency. L ωl to find the inductance of the inductor (a) elate the reactance of the inductor to its inductance: L L ωl πfl L πf Solve for evaluate L: L 00Ω 0.99 H π ( 80s ) 0.0H (b) At 60 Hz: ( 60s )( 0.99 H) 0.0kΩ L π 7 A circuit consists of two ideal ac generators a 5-Ω resistor, all connected in series. The potential difference across the terminals of one of the generators is given by (5.0 ) cos(ωt α), the potential difference across the terminals of the other generator is given by (5.0 ) cos(ωt + α), where α π/6. (a) Use Kirchhoff s loop rule a trigonometric identity to find the current in the circuit. (b) Use a phasor diagram to find the current in the circuit. (c) Find the current in the resistor if α π/4 the amplitude of is increased from 5.0 to 7.0. Picture the Problem We can use the trigonometric identity cos θ + cosφ cos θ+ φ cos θ φ ( ) ( ) to find the sum of the phasors then use this sum to express as a function of time. n (b) we ll use a phasor diagram to obtain the same result in (c) we ll use the phasor diagram appropriate to the given voltages to express the current as a function of time. (a) Applying Kirchhoff s loop rule to the circuit yields: + 0 Solve for to obtain: + Use the trigonometric identity cos θ cosφ cos ( θ + φ) cos ( θ φ) to find + : + + ( 5.0) [ cos( ωt α) + cos( ωt+ α)] ( 5) [ cos ( ωt) cos ( α)] π 6 ( 0) cos cosωt ( 8.66) cosωt

5 Substitute for + to obtain: ( 8.66) cosωt 5Ω ( 0.35A) cosωt where 0.35 A ( A) cosωt (b) Express the magnitude of the current in : r The phasor diagram for the voltages is shown to the right. r r Use vector addition to find r : cos30 ( 5.0 ) r r 8.66 r cos30 Substitute for r to obtain: A 5Ω 0.35A cosω ( ) t where 0.35 A

6 (c) The phasor diagram is shown to the right. Note that the phase angle between r r is now 90. r 90 o α δ α r Use the Pythagorean theorem to find r : r r r r ( 5.0 ) + ( 7.0 ) Express as a function of t: r cos( ω t+δ) where δ α α 45 ( ) 7.0 tan rad evaluate : 8.60 cos 5Ω ( ωt+ 0.65rad) ( 0.34A) cos( ωt+ 0.7 rad) 36 A two conductor transmission line simultaneously carries a superposition of two voltage signals, so the potential difference between the two conductors is given by +, where (0.0 ) cos(ω t) (0.0 ) cos(ω t), where ω 00 rad/s ω rad/s. A.00 H inductor a.00 kω shunt resistor are inserted into the transmission line as shown in Figure 9-3. (Assume that the output is connected to a load that draws only an insignificant amount of current.) (a) What is the voltage ( out ) at the output of the transmission line? (b) What is the ratio of the low-frequency amplitude to the high-frequency amplitude at the output? Picture the Problem We can express the two output voltage signals as the product of the current from each source.00 kω. We can find the

7 currents due to each source using the given voltage signals the definition of the impedance for each of them. (a) Express the voltage signals, out observed at the output side of the transmission line in te of the potential difference across the resistor:, out Evaluate : ( 0.0) cos00t - (.00 kω) + ( 00s )(.00H) 00 [ ] 4 ( 0.0) cos0 t 4 (.00 kω) + ( 0 s )(.00H) [ ] ( 9.95mA) cos t 0 4 ( 0.995mA) cos t Substitute for to obtain:, out (.00 kω)( 9.95 ma) ( 9.95) cos00t where ω 00 rad/s ω rad/s..00 kω ma, out ( )( ) 4 ( 0.995) cos0 t cos00t cos0 4 t (b) Express the ratio of,out to,out :, out, out : 64 A series L circuit that has an inductance of 0 mh, a capacitance of.0 µf, a resistance of 5.0 Ω is driven by an ideal ac voltage source that has a emf of 00. Find (a) the resonant frequency (b) the root-mean-square current at resonance. When the frequency is 8000 rad/s, find (c) the capacitive inductive reactances, (d) the impedance, (e) the root-mean-square current, (f) the phase angle. Picture the Problem We can use ω0 L to find the resonant frequency of the circuit, to find the current at resonance, the definitions of L to find these reactances at ω 8000 rad/s, the definitions of to find the impedance current at ω 8000 rad/s, the definition of the phase angle to find δ.

8 (a) Express the resonant frequency ω 0 in te of L : ω 0 L ω 0 evaluate ω 0 : ( 0mH)(.0µ F) rad/s (b) elate the current at resonance to the impedance of the circuit at resonance: 4A max 00 ( 5.0Ω) (c) Express evaluate L at ω 8000 rad/s: ω ( 8000s )(.0µ F) 6.50 Ω 63Ω L ω L ( 8000s )( 0 mh) 80Ω (d) Express the impedance in te of the reactances, substitute the results from (c), evaluate : ( L ) + ( 5.0Ω) + ( 80Ω 6.5Ω) 8.Ω 8Ω (e) elate the current at ω 8000 rad/s to the impedance of the circuit at this frequency: 3.9A max 00 ( 8.Ω) (f) δ is given by: evaluate δ: δ tan L 80Ω 6.5Ω δ tan 5.0Ω 74

9 75 A resistor a capacitor are connected in parallel across an ac generator (Figure 9-43) that has an emf given by cos ωt. (a) Show that the current in the resistor is given by ( /) cos ωt. (b) Show that the current in the capacitor branch is given by ( / ) cos(ωt + 90º). (c) Show that the total current is given by cos(ωt + δ), where tan δ / /. Picture the Problem Because the resistor the capacitor are connected in parallel, the voltage drops across them are equal. Also, the total current is the sum of the current through the resistor the current through the capacitor. Because these two currents are not in phase, we use phasors to calculate their sum. The amplitudes of the applied voltage the currents are equal to the magnitude of the phasors. That is r, r, r,, r,. (a) The ac source applies a voltage given by cosωt. Thus, the voltage drop across both the load resistor the capacitor is: The current in the resistor is in phase with the applied voltage: cosωt, cosωt Because, : t cosω (b) The current in the capacitor leads the applied voltage by 90 : ( t+ ) cosω 90, Because, : cos( ωt+ 90 ) (c) The net current is the sum of the currents through the parallel branches: +

10 Draw the phasor diagram for the circuit. The projections of the phasors onto the horizontal axis are the instantaneous values. The current in the resistor is in phase with the applied voltage, the current in the capacitor leads the applied voltage by 90. The net current phasor is the sum of the r r r branch current phasors( + ). r δ ωt r r r max The current through the parallel combination is equal to, where is the impedance of the combination: From the phasor diagram we have: cos t where ( ω δ ),, +, + + where + Solving for yields: where + From the phasor diagram: cos( ω t+δ) where tanδ 80 A transformer has 400 turns in the primary 8 turns in the secondary. (a) s this a step-up or a step-down transformer? (b) f the primary is connected to a 0 voltage source, what is the open-circuit voltage across the secondary? (c) f the primary current is 0.00 A, what is the secondary current, assuming negligible magnetization current no power loss? Picture the Problem Let the subscript denote the primary the subscript the secondary. We can decide whether the transformer is a step-up or step-down

11 transformer by examining the ratio of the number of turns in the secondary to the number of te in the primary. We can relate the open-circuit voltage in the secondary to the primary voltage the turns ratio. (a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer. (b) elate the open-circuit voltage, in the secondary to the voltage, in the primary: evaluate :, (c) Because there are no power losses: evaluate, : N,, N ( 0 ).40,,,,,,,,, 0.40 ( 0.00 A) 5.00 A,