ELL100: INTRODUCTION TO ELECTRICAL ENGG. Course Instructors: J.-B. Seo, S. Srirangarajan, S.-D. Roy, and S. Janardhanan Department of Electrical Engineering, IITD
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Generation, Transmission, & Distribution 1. Power Plant generates 3-phase AC power 2. Transmission Substation steps the voltage up to 69,000 - 765,000 V 3. High Voltage Transmission Lines moves power over long distances on the transmission grid 6. Distribution Power Lines Moves power over distribution grid to points-of-use steps the voltage down to 4160 -36,000V 5. Distribution Substation 4. Subtransmission Steps the voltage down to 22,000 – 69,000 V
Distribution Power Lines Small Substation Distribution Power Lines Industrial Uses 1200 A 2400 – 15,000 Volts 6. Distribution Power Lines Commercial Uses 480Y/277 Volts Moves power over distribution grid to points-of-use Residential Uses 120/240 Volts
Distribution Power Lines Small Substation Distribution Power Lines Industrial Uses 1200 A 2400 – 15,000 Volts 6. Distribution Power Lines Commercial Uses 480Y/277 Volts Moves power over distribution grid to points-of-use Residential Uses 120/240 Volts
Distribution Power Lines Small Substation Distribution Power Lines Industrial Uses 1200 A 2400 – 15,000 Volts 6. Distribution Power Lines Commercial Uses 480Y/277 Volts Moves power over distribution grid to points-of-use Residential Uses 120/240 Volts
Power Dissipated in an Electricity Distribution System AC Electricity Why high voltage? Power Dissipated in an Electricity Distribution System 150 miles 120 Watt Light bulb Power Plant on Colorado River 12 Volt Connection Box Estimate resistance of power lines: - say 0.001 Ohms per meter, times 200 km = 0.001 W/m 2105 m = 20 Ohms The current required by a single bulb using P = VI so I = P/V = 120 Watts/12 Volts = 10 Amps (!) Power in transmission line: P = I2R = 102 20 = 2,000 Watts!! “Efficiency” is e = 120 Watts/4120 Watts = 0.3%!!! Lecture 7
Power Dissipated in an Electricity Distribution System AC Electricity Why high voltage? Power Dissipated in an Electricity Distribution System 150 miles 120 Watt Light bulb Power Plant on Colorado River 12 Volt Connection Box Estimate resistance of power lines: - say 0.001 Ohms per meter, times 200 km = 0.001 W/m 2105 m = 20 Ohms The current required by a single bulb using P = VI so I = P/V = 120 Watts/12 Volts = 10 Amps (!) Power in transmission line: P = I2R = 102 20 = 2,000 Watts!! “Efficiency” is e = 120 Watts/4120 Watts = 0.3%!!! Lecture 7
AC Electricity Why high voltage? The thing that kills us most is the high current through the (fixed resistance) transmission lines Need less current to reduce power: I2R But our appliance needs a certain amount of power P = VI so less current demands higher voltage Solution is high voltage transmission Repeating the previous calculation with 12,000 Volts delivered to the house draws only I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving P = I2R = (0.01)220 = 2010-4 Watts, so P = 0.002 Watts of power dissipated in transmission line Efficiency in this case is e = 120 Watts/120.004 = 99.996% Lecture 7
AC Electricity Why high voltage? The thing that kills us most is the high current through the (fixed resistance) transmission lines Need less current to reduce power: I2R But our appliance needs a certain amount of power P = VI so less current demands higher voltage Solution is high voltage transmission Repeating the previous calculation with 12,000 Volts delivered to the house draws only I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving P = I2R = (0.01)220 = 2010-4 Watts, so P = 0.002 Watts of power dissipated in transmission line Efficiency in this case is e = 120 Watts/120.004 = 99.996% Lecture 7
three-phase “live” wires AC Electricity Why high voltage? three-phase “live” wires to house 500,000 230,000 138,000 69,000 7–13,000 long-distance neighborhood But having high voltage in each household is a recipe for disaster sparks every time you plug something in/risk of fire/not cat-friendly Need a way to step-up/step-down voltage at will can’t do this with DC, so go to AC Lecture 7
Instantaneous and average power Instantaneous power absorbed by the element: where
Instantaneous and average power Average power absorbed by the element: and
Instantaneous and average power Average power absorbed by the element: and
Instantaneous and average power Inductive load: current lags 90 degree behind voltage
Average power for a general impedance load
Example Find the power absorbed by each element and The average power at the voltage source
Example Find the power absorbed by each element and The average power at the voltage source
Example Find the power absorbed by each element and The average power at the voltage source
Example Find the power absorbed by each element The voltage across the resistor The average power at the resistor The voltage across the current source The average power at the current source
Example Find the power absorbed by each element The voltage across the resistor The average power at the resistor The voltage across the current source The average power at the current source
Example Find the power absorbed by each element The voltage across the capacitor The average power at the inductor The voltage across the inductor The average power at the inductor
Example Find the power absorbed by each element The voltage across the capacitor The average power at the inductor The voltage across the inductor The average power at the inductor
Maximum power transfer What impedance will absorb the maximum amount of power? circuit
Maximum power transfer
Maximum power transfer When the load impedance is a complex conjugate of the Zth, the maximum power is transferred to the load
Maximum power transfer What impedance will absorb the maximum amount of power? circuit
Example – 1
Example – 1
Example – 1
Example – 1
Example – 1
Example – 1
Example – 2 Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.
Example – 2 Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.
Example – 2 Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.
Effective or RMS value Effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current — Effective value of a periodic signal is its root mean square (rms) value. — For a sinusoidal , its rms value is
Effective or RMS value The average power in terms of rms value: The average power absorbed by a resistor R: Determine the rms value of the current waveform shown below If the current is passed through a 2-Ohm resistor, find the average power absorbed by the resistor
Effective or RMS value The average power in terms of rms value: The average power absorbed by a resistor R: Determine the rms value of the current waveform shown below If the current is passed through a 2-Ohm resistor, find the average power absorbed by the resistor
Apparent power and power factor (volt-amperes, VA) Power factor (pf) Inductive element, lagging power factor (the current lags the voltage) Capacitive element, leading power factor (the current leads the voltage) Recall the impedance power factor is the angle of the load impedance
Example Determine the power factor of the entire circuit below. Find the average power delivered by the source. Total impedance is Power factor: The rms value of the current: The average power supplied by the source:
Example Determine the power factor of the entire circuit below. Find the average power delivered by the source. Total impedance is Power factor: The rms value of the current: The average power supplied by the source:
Example Determine the power factor of the entire circuit below. Find the average power delivered by the source. Total impedance is Power factor: The rms value of the current: The average power supplied by the source:
Complex power The complex power absorbed by the ac load and with : the average power in watt delivered to a load (only useful power) : a measure of energy exchange between the source and the reactive par of the load (volt-ampere reactive: VAR)
Complex power The complex power absorbed by the ac load and with : the average power in watt delivered to a load (only useful power) : a measure of energy exchange between the source and the reactive par of the load (volt-ampere reactive: VAR)
Complex power (p.211) The complex power absorbed by the ac load Based on the reference voltage signal, Leading power factor (or load): Current leads voltage (capacitive load) Lagging power factor (or load): Current lags voltage (Inductive load) Apparent power Power triangle
Example – 1 A load Z draws 12 kVA at a power factor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance (kW) (kVA)
Example – 2 A coil is to be represented by a linear model consisting of inductance L in series with resistance R. When a 60-Hz current of 2 A (rms) is supplied to the coil, the voltmeter across the coil reads 26 V (rms). A wattmeter indicates 20 W delivered to the coil. Determine L and R. and Reactive power
Example – 3 (p.212) An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor. From power factor we have (kVA)
Example – 3 (p.212) An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor. From power factor we have (kVA)
Example – 3 (p.212) An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor. The power factor gives
Example – 3 (p.212) An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor. (kVA)
AC power conservation Whether the loads are connected in series or in parallel (or in general), the total power supplied by the source equals the total power delivered to the load. Thus, in general, for a source connected to N loads, The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.
Example Find the real and reactive power absorbed by the source, the line and the load i) ii)
Example Find the real and reactive power absorbed by the source, the line and the load iii) iv)
Example Find the real and reactive power absorbed by the source, the line and the load v)
Power correction factor (Original)
Power correction factor Example: When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.
Example – 1 When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.
Example – 2 (p.213) An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Power factor of the previous example is to be corrected to 0.9 lagging. Specify the necessary auxiliary equipment.
Example – 2 (p.213) Power factor of the previous example is to be corrected to 0.9 lagging. Specify the necessary auxiliary equipment. The necessary correction is