Role of Electrical technology in our daily Life
The electrical technology doing a very important role in our day to day life. We need service to electrical appliances at every time like when we are sleeping We need fan/AC/Heating equipments for pleasant relaxation.for cooking microwave oven induction heater etc. Electrical technology are also using many fields of touching our life like Railway system, traction system, healthcare systems, industrial & entertainments. many switchgears are using to generate control and supply of electricity.
Power plant is the place where Electricity is generated
Generated Electric power is stap up with the help of transformer.Transformer is a static electrical machine which can increase or decrease voltage on same frequency. step up transformer is used for transmission of Electric power on high voltage & step down transformer is used for distribution of electric power on low voltage. It is also used for correction of voltage in distribution.
Transformer
1 Power in Single Phase Circuits Let v = Vm cos(ωt) = √2V cos(ωt) is the voltage applied to a R-L circuit and i = Im cos(ωt−θ) = √2I cos(ωt−θ) is the current flowing in it (V and I are the rms values of the voltage and current respectively). The power at any instant of time is p = vi = VmIm cos(ωt)cos(ωt−θ) p = VmIm 2 cosθ + VmIm 2 cos(2ωt−θ) p = VmIm 2 cosθ + VmIm 2 cos(2ωt)cosθ + VmIm 2 sinθsin(2ωt) A careful study of the above equation reveals the following characteristics: • The average value of power = P = Vm Im 2 cosθ = V I cosθ Watts. In AC circuits, when we talk of power rating, we are usually referring to the average power. For example, a “100 W” bulb implies that average power output of the bulb is 100 W. • The frequency of instantaneous power is twice the frequency of the voltage and current. Note that instantaneous power may be negative for a portion of each cycle, even if the load is passive. In a passive network, negative power at a given instant implies that energy stored in inductors or capacitors is being given back to the supply at that instant. The fact that the instantaneous power varies with time explains why some single phase mo- tor driven appliances such as refrigerators, which are fed from single phase supply experience vibration, and hence require resilient mountings to prevent excessive vibration of the equipment. • If the load is purely resistive, the voltage and current are in phase, which means that θ = 0. The above equation reduces to p = P + P cos(2ωt). In this case instantaneous real power can never be negative, i.e., power cannot be given back to the supply from a purely resistive network. • If the circuit is purely reactive ( either capacitive or inductive, θ = ±90o), the expression for instantaneous power is p = VmIm 2 sin(2ωt). The average power is zero. Therefore in a purely reactive circuit there is no average power dissipation. Energy is drawn from the supply for a part of a cycle is stored, and given back in another part of a cycle. However, since there is a current flow, the supply system must be designed to source such currents. Therefore it is useful to define a quantity called “reactive power” as follows:
Q = VmIm 2 sinθ = VI sinθ • Dimensionally P and Q are same. However, in order to distinguish between average (also called “real power”) and reactive power, we use the term ‘VAr’ for reactive power. Q is positive for inductive load and negative for capacitive load. In other words inductors demand or absorb reactive power while capacitors furnish or deliver the reactive power. For purely resistive circuits, Q=0.
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If v and i are represented by phasors V and I, then it is easily seen that S = P + jQ = V I∗ cosθ = P √P2+Q2
S is called the complex power and cosθ is called the power factor. Power factor is said to be leading if θ is negative (capacitive circuits), and lagging for if θ is positive (inductive circuits). Dimensionally, complex power is the same as real and reactive power. However, in order to distinguish complex power from real and reactive power, we use the term ‘volt amps’ or ‘VA’. The magnitude of complex power (|S|= S =|V I∗|=√P2 + Q2 = VI), is referred to as apparent power. S indicates the capacity of the source required to supply the load. The apparent power specifications of source and load equipment can be more important than the average power specifications. This is because any electrical equipment requires appropriate design of conductor size, amount of dielectric (insulation) and magnetic material to handle both current and voltage magnitudes. Thus S has a direct bearing on size and cost of any electrical equipment. Since average power represents the useful output of the load equipment (essentially energy converting devices like motors, heaters, lighting), operating such equipment close to unity power factor is desirable (i.e, P should be as close to S as possible, or Q should be minimized). The vector diagram for a lagging load is shown in Fig.1 (a). The current vector lags the voltage vector by and angle θ. In Fig. 1(b) the current phasor is resolved along x and y axes and in Fig.1(c) these components are multiplied by V. The resulting triangle is known as power triangle. The sides of the triangle are P, Q and S, where,
P = VI cosθ Watt Q = VI sinθ VAr S = VI VA
1.1 Measurement of Power in single phase circuit Power consumed by single phase load can be measured using a wattmeter. This meter comprises of the following: • Two identical current coils. Depending upon the magnitude of load current these coils are con- nected either in series or in parallel ( if the current rating of one coil is less than the load current, these coils are connected in parallel, otherwise they are connected in series). The two ends of these coils are marked as ‘M’ which stands for Mains, and ‘L’ for load. The current coils are connected in series with the load. • A voltage coil with different voltage tapings. Depending upon the voltage applied to the load suitable taping is used. The two ends of this coil are marked as ‘C’ which stands for common and ‘V’ stands for voltage. This coil is connected across the load. The terminals C and M are connected together.
The reading of this meter is proportional to
Vapp IFl cos6 (IFl & V app) where, Vapp is the voltage applied to voltage coil, & IFl is current flowing through the current coil.
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2 Three Phase System
The generation, transmission, distribution and utilization of large blocks of electric power are accom- plished by means of three-phase circuits. A balanced three phase system consists of three sinusoidal voltages of identical amplitudes and frequency but are out of phase with each other by 120o. In dis- cussing three-phase circuits, standard practice is to refer the three phase as A, B, and C ( or R,Y, B). Furthermore, the A-phase ( or R-phase) is almost always used as the reference phase. Because the phase voltages are out of phase by 120o, two possible phase relationships can exist between the A phase ( or R phase) voltage and the B and C phase voltages ( Y and B). One possibility is for the B phase to lag the A phase by 120o ( phase C will be lagging B phase by 120o, or lead A phase by 120o). In this case the phase relationship is known as ABC (or RYB) phase sequence. The only other possibility is for the C phase to lag A phase ( or B phase to lead A phase by 120o). This phase relationship is known as ACB sequence. In phasor notation, two possible sets of balanced phase voltages are: V a = V6 0; V b = V6 −120o; V c = V6 −240o; and V a = V6 0; V c = V6 −120o; V b = V6 −240o = V6 +120o. An important characteristic of balanced three phase voltages is that the sum of the voltages is zero (V a + V b + V c = 0). Three phase voltage sources can be either Y ( star) or delta connected. These connections are shown in Fig.2 (a) and (b) respectively. Let V an, V bn and V cn are the phase voltages ( the voltage between line and neutral). Now there are 4 wires - three for three lines and fourth one for the neutral. All the single phase loads are connected between any one of the lines and neutral. The relationship between the line-line voltage (V ab, V bc and V ca) and phase voltage can be derived as follows: Referring to Fig.2(a), the line-line voltages for ABC sequence are: V ab = V an −V bn = V6 0−V6 −120o =√3V6 30o Similarly, V bc =√3V6 −90o V ca =√3V6 −210o In three phase system the load is said to be balanced if the magnitude and phase of the load im- pedance connected to each phase is the same. For the Fig.2 (b), the phase current ( the current in each phase of the load) is
IA = V an Zan = V6 0 Z6 θ = I6 −θ IB = V bn Zbn = V6 −120 Z6 θ = I6 −(120 + θ) IC = V cn Zcn = V6 −240 Z6 θ = I6 −(240 + θ) Line current is the current in each phase of the line. In star connected system, phase currents & line currents are identical. The complete vector diagram for a three phase system feeding a star connected load is shown in fig.3(a). For delta connected system shown in Fig. 2(b), line to line voltage is the same as phase voltage. However, the relationship between the phase currents and line currents can be determined as follows: The phase currents are given by: Iab = V ab Zab = V6 0 Z6 θ = I6 −θ Ibc = V bc Zbc = V6 −120 Z6 θ = I6 −(120 + θ) Ica = V ca Zca = V6 −240 Z6 θ = I6 −(240 + θ) 3
We can write the line currents in terms of the phase currents by direct application of KCL: IA = Iab −Ica = I6 −θ−I6 −(240 + θ) =√3I6 −(30 + θ)I B = Ibc −Iab =√3I6 −(150 + θ) & IC = Ica −Ibc =√3I6 −(270 + θ) The complete vector diagram for a three phase system feeding a delta connected load is shown in fig.3(b).
2.1 Power in three phase circuits: In a balanced three phase system, the magnitude of each line-to-neutral voltage (also line-line) is the same, as is the magnitude of each phase current. Therefore, the expression for total power in terms of phase current and phase voltage is given by: P = 3 V I cosθ Watt, θ = 6 (I&V) In-terms of line voltage & line current the above expression becomes: P =√3 VL IL cosθ Watt, θ = 6 (I & V) In the above expression V and I are the rms values of the phase voltage and phase current respectively. We can also calculate reactive power (Q) and complex power (S). For balanced load, the expressions for Q and S are given by: Q =√3 VL IL sinθ VAr, θ = 6 ( I & V), S =√3 VL IL VA All the above equations are valid for both the connections (Y or Delta).
2.2 Measurement of Power in Three phase circuit Real power in balanced (or unbalanced ) 3 phase circuits can be measured by using two wattmeters connected in any two lines of a three-phase three wire system. In Fig.4(a) the current coils are connected in lines A and C, with the voltage coil reference connections in line B. The current flowing through the current coil of the wattmeter connected in line A is IA and that in line C is IC. Similarly, the voltage applied to the voltage coils is V ab and V cb (and not V bc) respectively. Their readings will be: WA = Vab IA cos6 ( IA & V ab) and WB = Vcb IC cos6 ( IC & V cb) From Fig.3 (a or b) angle between Vab and IA is (30+θ) and between Vcb and Ic is (30−θ). Therefore the wattmeter readings will be: W1 =√3 V IA cos(30 + θ) W2 =√3 V IC cos(30−θ) W1 + W2 = 3 V I cosθ = √3 VL IL cosθ Watt
Note that if θ is greater than 60o, W1 will be negative. In actual power measurement it is possible to encounter such a situation when one of the wattmeters has negative read- ing. The point to remember is that the total power in three phase circuit is obtained by adding the two wattmeter readings, taking the signs of the readings into account. It may be worthwhile to identify the following cases: • If θ is zero (load is purely resistive), W1 = W2 and both are positive. • For θ = 60o lag, W1 = 0. Similarly for θ = 60o lead, W2 = 0. Note that you will be using a Power analyzer to measure power. It is also possible to measure other quantities such as voltage, current, Q, S, power factor etc. using this equipment.
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3 Power factor Improvement:
Electrical service to industrial customers is given as a three-phase supply as opposed to the single phase power supplied to residential and small commercial customers. Most electric loads are reactive in na- ture and have lagging power factor (less than unity). Transmission lines, transformers and generators of the electric power utility have to carry the lagging reactive power of the load. Transformers, the distribution systems and the generators are all rated in kVA or MVA. An improvement in the power factor leads to a reduction in KVA for the same real (average) power supplied. This leads to a release in some of the generation and transmission capability so that it can be used to serve more customers. Generally, power is distributed through transmission lines ( in the city of Mumbai, power is distributed through underground cables) and the voltage at one end of these lines is maintained constant. These transmission lines have finite resistance and inductance. From Fig.1 it can be inferred that as the power factor falls, for a fixed amount of power, the magnitude of current flowing through these lines increases. As a result the voltage at the load end drops. Also, loss in the transmission line increases thereby decreasing the overall efficiency. In order to discourage the bulk consumers drawing a large amount of reactive power from the source, the utilities charge higher tariff if the power factor falls below a specified value. Hence, to improve the power factor, capacitors in three-phase banks are connected to the system, such that the combination of the plant load and the capacitor banks presents a load to the serving utility which is nearer to unity power factor. This is explained with the help of Fig.5. In the Fig.5(a), a R-L load is being supplied by the source and a capacitor is connected across the load. The vector diagram for this combination is shown in Fig.5(b). It is known that the capacitor draws a leading current (or supplies lagging current) from the source. If the magnitude of the current drawn by the capacitor is equal to the quadrature component of the load current (Iy = IL sinθ), the source will then supply only the active component of current (Ix). In that case there is a significant reduction in the current flowing through the trans- mission line ( without the capacitor the current flowing in the transmission line is ‘IL’ and with the capacitor it is IL cosθ). This results in the decrease in voltage drop and power loss in the transmission line.
The electrical technology doing a very important role in our day to day life. We need service to electrical appliances at every time like when we are sleeping We need fan/AC/Heating equipments for pleasant relaxation.for cooking microwave oven induction heater etc. Electrical technology are also using many fields of touching our life like Railway system, traction system, healthcare systems, industrial & entertainments. many switchgears are using to generate control and supply of electricity.
Power plant is the place where Electricity is generated
Generated Electric power is stap up with the help of transformer.Transformer is a static electrical machine which can increase or decrease voltage on same frequency. step up transformer is used for transmission of Electric power on high voltage & step down transformer is used for distribution of electric power on low voltage. It is also used for correction of voltage in distribution.
Transformer
1 Power in Single Phase Circuits Let v = Vm cos(ωt) = √2V cos(ωt) is the voltage applied to a R-L circuit and i = Im cos(ωt−θ) = √2I cos(ωt−θ) is the current flowing in it (V and I are the rms values of the voltage and current respectively). The power at any instant of time is p = vi = VmIm cos(ωt)cos(ωt−θ) p = VmIm 2 cosθ + VmIm 2 cos(2ωt−θ) p = VmIm 2 cosθ + VmIm 2 cos(2ωt)cosθ + VmIm 2 sinθsin(2ωt) A careful study of the above equation reveals the following characteristics: • The average value of power = P = Vm Im 2 cosθ = V I cosθ Watts. In AC circuits, when we talk of power rating, we are usually referring to the average power. For example, a “100 W” bulb implies that average power output of the bulb is 100 W. • The frequency of instantaneous power is twice the frequency of the voltage and current. Note that instantaneous power may be negative for a portion of each cycle, even if the load is passive. In a passive network, negative power at a given instant implies that energy stored in inductors or capacitors is being given back to the supply at that instant. The fact that the instantaneous power varies with time explains why some single phase mo- tor driven appliances such as refrigerators, which are fed from single phase supply experience vibration, and hence require resilient mountings to prevent excessive vibration of the equipment. • If the load is purely resistive, the voltage and current are in phase, which means that θ = 0. The above equation reduces to p = P + P cos(2ωt). In this case instantaneous real power can never be negative, i.e., power cannot be given back to the supply from a purely resistive network. • If the circuit is purely reactive ( either capacitive or inductive, θ = ±90o), the expression for instantaneous power is p = VmIm 2 sin(2ωt). The average power is zero. Therefore in a purely reactive circuit there is no average power dissipation. Energy is drawn from the supply for a part of a cycle is stored, and given back in another part of a cycle. However, since there is a current flow, the supply system must be designed to source such currents. Therefore it is useful to define a quantity called “reactive power” as follows:
Q = VmIm 2 sinθ = VI sinθ • Dimensionally P and Q are same. However, in order to distinguish between average (also called “real power”) and reactive power, we use the term ‘VAr’ for reactive power. Q is positive for inductive load and negative for capacitive load. In other words inductors demand or absorb reactive power while capacitors furnish or deliver the reactive power. For purely resistive circuits, Q=0.
1
If v and i are represented by phasors V and I, then it is easily seen that S = P + jQ = V I∗ cosθ = P √P2+Q2
S is called the complex power and cosθ is called the power factor. Power factor is said to be leading if θ is negative (capacitive circuits), and lagging for if θ is positive (inductive circuits). Dimensionally, complex power is the same as real and reactive power. However, in order to distinguish complex power from real and reactive power, we use the term ‘volt amps’ or ‘VA’. The magnitude of complex power (|S|= S =|V I∗|=√P2 + Q2 = VI), is referred to as apparent power. S indicates the capacity of the source required to supply the load. The apparent power specifications of source and load equipment can be more important than the average power specifications. This is because any electrical equipment requires appropriate design of conductor size, amount of dielectric (insulation) and magnetic material to handle both current and voltage magnitudes. Thus S has a direct bearing on size and cost of any electrical equipment. Since average power represents the useful output of the load equipment (essentially energy converting devices like motors, heaters, lighting), operating such equipment close to unity power factor is desirable (i.e, P should be as close to S as possible, or Q should be minimized). The vector diagram for a lagging load is shown in Fig.1 (a). The current vector lags the voltage vector by and angle θ. In Fig. 1(b) the current phasor is resolved along x and y axes and in Fig.1(c) these components are multiplied by V. The resulting triangle is known as power triangle. The sides of the triangle are P, Q and S, where,
P = VI cosθ Watt Q = VI sinθ VAr S = VI VA
1.1 Measurement of Power in single phase circuit Power consumed by single phase load can be measured using a wattmeter. This meter comprises of the following: • Two identical current coils. Depending upon the magnitude of load current these coils are con- nected either in series or in parallel ( if the current rating of one coil is less than the load current, these coils are connected in parallel, otherwise they are connected in series). The two ends of these coils are marked as ‘M’ which stands for Mains, and ‘L’ for load. The current coils are connected in series with the load. • A voltage coil with different voltage tapings. Depending upon the voltage applied to the load suitable taping is used. The two ends of this coil are marked as ‘C’ which stands for common and ‘V’ stands for voltage. This coil is connected across the load. The terminals C and M are connected together.
The reading of this meter is proportional to
Vapp IFl cos6 (IFl & V app) where, Vapp is the voltage applied to voltage coil, & IFl is current flowing through the current coil.
2
2 Three Phase System
The generation, transmission, distribution and utilization of large blocks of electric power are accom- plished by means of three-phase circuits. A balanced three phase system consists of three sinusoidal voltages of identical amplitudes and frequency but are out of phase with each other by 120o. In dis- cussing three-phase circuits, standard practice is to refer the three phase as A, B, and C ( or R,Y, B). Furthermore, the A-phase ( or R-phase) is almost always used as the reference phase. Because the phase voltages are out of phase by 120o, two possible phase relationships can exist between the A phase ( or R phase) voltage and the B and C phase voltages ( Y and B). One possibility is for the B phase to lag the A phase by 120o ( phase C will be lagging B phase by 120o, or lead A phase by 120o). In this case the phase relationship is known as ABC (or RYB) phase sequence. The only other possibility is for the C phase to lag A phase ( or B phase to lead A phase by 120o). This phase relationship is known as ACB sequence. In phasor notation, two possible sets of balanced phase voltages are: V a = V6 0; V b = V6 −120o; V c = V6 −240o; and V a = V6 0; V c = V6 −120o; V b = V6 −240o = V6 +120o. An important characteristic of balanced three phase voltages is that the sum of the voltages is zero (V a + V b + V c = 0). Three phase voltage sources can be either Y ( star) or delta connected. These connections are shown in Fig.2 (a) and (b) respectively. Let V an, V bn and V cn are the phase voltages ( the voltage between line and neutral). Now there are 4 wires - three for three lines and fourth one for the neutral. All the single phase loads are connected between any one of the lines and neutral. The relationship between the line-line voltage (V ab, V bc and V ca) and phase voltage can be derived as follows: Referring to Fig.2(a), the line-line voltages for ABC sequence are: V ab = V an −V bn = V6 0−V6 −120o =√3V6 30o Similarly, V bc =√3V6 −90o V ca =√3V6 −210o In three phase system the load is said to be balanced if the magnitude and phase of the load im- pedance connected to each phase is the same. For the Fig.2 (b), the phase current ( the current in each phase of the load) is
IA = V an Zan = V6 0 Z6 θ = I6 −θ IB = V bn Zbn = V6 −120 Z6 θ = I6 −(120 + θ) IC = V cn Zcn = V6 −240 Z6 θ = I6 −(240 + θ) Line current is the current in each phase of the line. In star connected system, phase currents & line currents are identical. The complete vector diagram for a three phase system feeding a star connected load is shown in fig.3(a). For delta connected system shown in Fig. 2(b), line to line voltage is the same as phase voltage. However, the relationship between the phase currents and line currents can be determined as follows: The phase currents are given by: Iab = V ab Zab = V6 0 Z6 θ = I6 −θ Ibc = V bc Zbc = V6 −120 Z6 θ = I6 −(120 + θ) Ica = V ca Zca = V6 −240 Z6 θ = I6 −(240 + θ) 3
We can write the line currents in terms of the phase currents by direct application of KCL: IA = Iab −Ica = I6 −θ−I6 −(240 + θ) =√3I6 −(30 + θ)I B = Ibc −Iab =√3I6 −(150 + θ) & IC = Ica −Ibc =√3I6 −(270 + θ) The complete vector diagram for a three phase system feeding a delta connected load is shown in fig.3(b).
2.1 Power in three phase circuits: In a balanced three phase system, the magnitude of each line-to-neutral voltage (also line-line) is the same, as is the magnitude of each phase current. Therefore, the expression for total power in terms of phase current and phase voltage is given by: P = 3 V I cosθ Watt, θ = 6 (I&V) In-terms of line voltage & line current the above expression becomes: P =√3 VL IL cosθ Watt, θ = 6 (I & V) In the above expression V and I are the rms values of the phase voltage and phase current respectively. We can also calculate reactive power (Q) and complex power (S). For balanced load, the expressions for Q and S are given by: Q =√3 VL IL sinθ VAr, θ = 6 ( I & V), S =√3 VL IL VA All the above equations are valid for both the connections (Y or Delta).
2.2 Measurement of Power in Three phase circuit Real power in balanced (or unbalanced ) 3 phase circuits can be measured by using two wattmeters connected in any two lines of a three-phase three wire system. In Fig.4(a) the current coils are connected in lines A and C, with the voltage coil reference connections in line B. The current flowing through the current coil of the wattmeter connected in line A is IA and that in line C is IC. Similarly, the voltage applied to the voltage coils is V ab and V cb (and not V bc) respectively. Their readings will be: WA = Vab IA cos6 ( IA & V ab) and WB = Vcb IC cos6 ( IC & V cb) From Fig.3 (a or b) angle between Vab and IA is (30+θ) and between Vcb and Ic is (30−θ). Therefore the wattmeter readings will be: W1 =√3 V IA cos(30 + θ) W2 =√3 V IC cos(30−θ) W1 + W2 = 3 V I cosθ = √3 VL IL cosθ Watt
Note that if θ is greater than 60o, W1 will be negative. In actual power measurement it is possible to encounter such a situation when one of the wattmeters has negative read- ing. The point to remember is that the total power in three phase circuit is obtained by adding the two wattmeter readings, taking the signs of the readings into account. It may be worthwhile to identify the following cases: • If θ is zero (load is purely resistive), W1 = W2 and both are positive. • For θ = 60o lag, W1 = 0. Similarly for θ = 60o lead, W2 = 0. Note that you will be using a Power analyzer to measure power. It is also possible to measure other quantities such as voltage, current, Q, S, power factor etc. using this equipment.
4
3 Power factor Improvement:
Electrical service to industrial customers is given as a three-phase supply as opposed to the single phase power supplied to residential and small commercial customers. Most electric loads are reactive in na- ture and have lagging power factor (less than unity). Transmission lines, transformers and generators of the electric power utility have to carry the lagging reactive power of the load. Transformers, the distribution systems and the generators are all rated in kVA or MVA. An improvement in the power factor leads to a reduction in KVA for the same real (average) power supplied. This leads to a release in some of the generation and transmission capability so that it can be used to serve more customers. Generally, power is distributed through transmission lines ( in the city of Mumbai, power is distributed through underground cables) and the voltage at one end of these lines is maintained constant. These transmission lines have finite resistance and inductance. From Fig.1 it can be inferred that as the power factor falls, for a fixed amount of power, the magnitude of current flowing through these lines increases. As a result the voltage at the load end drops. Also, loss in the transmission line increases thereby decreasing the overall efficiency. In order to discourage the bulk consumers drawing a large amount of reactive power from the source, the utilities charge higher tariff if the power factor falls below a specified value. Hence, to improve the power factor, capacitors in three-phase banks are connected to the system, such that the combination of the plant load and the capacitor banks presents a load to the serving utility which is nearer to unity power factor. This is explained with the help of Fig.5. In the Fig.5(a), a R-L load is being supplied by the source and a capacitor is connected across the load. The vector diagram for this combination is shown in Fig.5(b). It is known that the capacitor draws a leading current (or supplies lagging current) from the source. If the magnitude of the current drawn by the capacitor is equal to the quadrature component of the load current (Iy = IL sinθ), the source will then supply only the active component of current (Ix). In that case there is a significant reduction in the current flowing through the trans- mission line ( without the capacitor the current flowing in the transmission line is ‘IL’ and with the capacitor it is IL cosθ). This results in the decrease in voltage drop and power loss in the transmission line.
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