Category: electrical Machines

EMF Equation of DC Generator The expression which gives the magnitude of EMF generated in a DC generator is called EMF equation of DC generator. We shall now drive the expression for the EMF induced in a DC generator. Let, $phi $ = flux per pole P = number of poles in the generator Z = no.of armature coductors A = no.of parallel paths N = speed of armature in RPM E = EMF generated Thus, the magnetic flux (in weber) cut by a conductor in one revolution of the armature is given by, $$mathrm{mathit{dphi :=:Ptimes phi }}$$ If N is the number of revolution per minute, then the time (in seconds) taken complete one revolution is, $$mathrm{mathit{dt :=frac{60}{N}}}$$ According to Faraday’s law of electromagnetic induction, the EMF induced per conductor is given by, $$mathrm{mathrm{EMF/conductor}:=:mathit{frac{dphi }{dt}}:=:frac{mathit{Pphi }}{mathrm{left ( {60/mathit{N}} right )}}:=:frac{mathit{Pphi N}}{mathrm{60}}}$$ The total EMF generated in the generator is equal to the EMF per parallel path, which is the product of EMF per conductor and the number of conductors in series per parallel path, i.e., $$mathrm{mathit{E}:=:left ( EMF/Conductor right )times left ( No.:of:conductors/parallel:path right )}$$ $$mathrm{Rightarrow mathit{E}:=:frac{mathit{Pphi N}}{60}times frac{mathit{Z}}{mathit{A}}}$$ $$mathrm{therefore mathit{E}:=:frac{mathit{NPphi N}}{60mathit{A}}:cdot cdot cdot left ( 1 right )}$$ Equation (1) is called the EMF equation of DC generator. For wave winding, $$mathrm{mathrm{Number:of:parllel:paths,}mathit{A}:=:2}$$ $$mathrm{therefore mathit{E}:=:frac{mathit{NPphi Z}}{mathrm{120}}}$$ For lap winding, $$mathrm{mathrm{Number:of:parllel:paths,}mathit{A}:=:mathit{P}}$$ $$mathrm{therefore mathit{E}:=:frac{mathit{Nphi Z}}{mathrm{60}}}$$ For a given DC generator, Z, P and A are constant so that the generated EMF (E) is directly proportional to flux per pole ($phi$) and speed of armature rotation (N). Numerical Example A 6-pole dc generator has 600 armature conductors and a useful flux of 0.06 Wb. What will be the EMF generated, if it is wave connected and lap connected and runs at 1000 RPM? Solution: Given data, No.of poles,P = 6 No.of armature conductors,Z = 600 Flux per pole,$phi$ = 0.06 Wb Speed of armature,N = 1000 RPM For wave-connected generator, $$mathrm{mathit{E}:=:frac{mathit{NPphi Z}}{mathrm{120}}}$$ $$mathrm{Rightarrow mathit{E}:=:frac{1000times6times 0.06times 600}{120}}$$ $$mathrm{therefore mathit{E}:=:1800:V}$$ For lap-connected generator, $$mathrm{mathit{E}:=:frac{mathit{Nphi Z}}{mathrm{60}}}$$ $$mathrm{Rightarrow mathit{E}:=:frac{1000times 0.06times 600}{60}}$$ $$mathrm{therefore mathit{E}:=:600:V}$$ Learning working make money

Transformer on DC In the introductory chapter, we defined an electrical transformer as an AC machine because it works only on alternating current electricity. Therefore, a transformer cannot change (increase or decrease) the value of the DC voltage. In this chapter, we shall know the reason, why a transform does not work on the direct current (DC). Consider an electrical transformer as shown in Figure-1, and it is connected to a battery (or a source of DC voltage) V. When, we apply this DC voltage V to the primary winding of the transformer, it will draw a constant current (DC) and therefore produces a constant magnetic flux flowing through the magnetic core. According to the principle of electromagnetic induction, an EMF can induce in a coil or conductor only when it is subjected to a changing magnetic field, i.e., $$mathrm{mathit{e}:=:mathit{N}frac{mathit{dphi }}{mathit{dt}}}$$ Consequently, the applied DC voltage to the primary winding does not induce EMF in the primary winding or secondary winding. Hence, this discussion proves that a transformer does not work on DC supply. In fact, connecting a DC supply to the primary winding of a transformer could be dangerous. The equivalent primary winding circuit of a transformer connected to the DC voltage is shown in Figure-2. In this case, there is no self-induced EMF in the primary winding to oppose the applied voltage V (according to Lenz’s law), and the current in the primary winding is given by, $$mathrm{mathit{I_{mathrm{1}}}:=:frac{mathit{V}}{mathit{R_{mathrm{1}}}}}$$ Where, $mathit{R_{mathrm{1}}}$ is the resistance of the primary winding. Due to very small value of R1, the current $mathit{I_{mathrm{1}}}$ through the primary winding will be very large. This large current will cause the overheating and burning of the transformer or fuses will blow. Therefore, we must not connect the primary winding of a transformer to the DC supply because it may damage the transformer or may cause an electrical accident. Learning working make money

Construction of DC Machines An electromechanical device which can convert direct current (dc) electricity into mechanical energy or mechanical energy into direct current (dc) electricity is known as a DC machine. If the DC machine converts DC electrical energy into mechanical energy, it is known as DC motor. If the machine converts mechanical energy into DC electrical energy, then it is known as a DC generator. Both DC motor and DC generator have the similar construction. A typical DC machine consists of the following major parts − Yoke or Frame Armature Field System Commutator Brushes Bearings The schematic diagram of a DC machine is shown below − Let us now discuss each of these components in greater detail. Yoke or Frame The yoke is the outer frame of the DC machine. It is made up of such materials that have high permeability and high mechanical strength. In practice, the yoke of DC machine is made up of cast steel. The yoke or frame of the DC machine serves the following main purposes − It protects the internal machine parts like armature, windings, field poles, etc. against mechanical damages. The yoke houses the magnetic field system. It provides a low reluctance path to the working magnetic flux. It supports the rotor or armature through bearings. Armature In DC machines (motor or generator), armature is a system of conductors or coils that can rotate freely on the supporting bearings. The working torque and EMF are developed in coils of the armature. The armature consists of two main parts namely, armature core and armature winding. The armature core is a solid cylindrical structure, made up of high permeability thin silicon steel laminations. On the outer periphery of the core slots are cut to carry the armature winding. The armature winding is made up of copper wires. The armature winding of DC machine is generally former wound. Depending upon the end connections of the armature conductors, the armature winding may be of two types namely lap winding and wave winding. The type of winding decides the voltage and current rating of the machine. In case of the lap winding, the number of parallel paths (A) for current to flow are equal to the number of poles (P) in the machine. On the other hand, for wave winding, the number of parallel paths (A) are equal to 2. Field System Field system is the part of a DC machine which produces the working magnetic flux in the machine. It is basically a system of electromagnets which is excited by a DC supply. In case of DC machine, the field system is a stationary part of the machine and it is bolted to the yoke or frame of the machine. There are three main parts of a field system in dc machines namely pole core, pole shoes, and field coils. The pole core is made up of thin steel laminations. One end of the pole core is bolted to the frame and other end has pole shoe. The pole core carries the field winding. The pole shoe is a projected part of the pole core and has a large area of cross-section. Pole shoes help in spreading the magnetic flux uniformly in the air gap, and offers low reluctance path to the magnetic flux. Also, it supports the field winding. The field coil or winding is made up of copper wire. The field winding is former wound and inserted around the pole core. When field windings are excited by DC supply, they become electromagnets and produce magnetic flux in the machine. Commutator The commutator is one of the important parts of the DC machine. It is basically mechanical rectifier. It is a cylindrical shaped device and is made up of copper. The outer periphery of the commutator has V-shaped slots to carry commutator segments. Where, the commutator segments are copper bars inserted in the slots. These segments are insulated from each other by mica. The commutator is mounted on the shaft of the DC machine on one side of the armature. The armature conductors are connected to the commutator segments with the help of copper lugs. The commutator performs the following two major functions − In a DC generator, it collects the current from the armature conductor. In a DC motor, it supplies the current to the armature conductors. It converts the alternating current of the armature into unidirectional current in the external circuit with the help of brushes, and vice-versa. Brushes Brushes are used to make an electrical connection with the rotating commutator. These collect (or supply) current from (or to) the moving commutator. Brushes are usually made up of carbon. They are housed in brush holders and are in contact with the commutator surface with the help of spring pressure. Bearings Bearings are used in the DC machine to reduce the frictional losses. Thus, the main function of bearings in the DC machine is to support the machine shaft with minimum friction. In DC machines, ball bearings or roller bearings are commonly used. Learning working make money

Efficiency of Transformer Transformer Efficiency The ratio of the output power to the input power in a transformer is known as efficiency of transformer. The transformer efficiency is represented by Greek letter Eta ($eta $). $$mathrm{mathrm{Efficiency,}eta :=:frac{Output:Power}{Input:Power}}$$ From this definition, it appears that we can determined the efficiency of a transformer by directly loading the transformer and measuring the input power and output power. Although, this method of efficiency determination has the following disadvantages − In practice, the efficiency of a transformer is very high, and a very small error (let say 1%) in input and output wattmeters may give ridiculous results. Consequently, this method may give efficiency more than 100%. In this method, the transformer is loaded, hence a considerable amount of power is wasted. Therefore, this method becomes uneconomical for large transformers. It is very difficult to find a load which is capable of absorbing all of the output power. This method does not provide any information about losses in the transformer. Thus, due to these limitations, the direct-loading method is rarely used to determine the efficiency of a transformer. In practice, we use open-circuit and short-circuit tests to find out the transformer efficiency. For a practical transformer, the input power is given by, $$mathrm{mathrm{Input:power}:=:mathrm{Output:power:+:Losses}}$$ Therefore, the transformer efficiency can also be calculated using the following expression − $$mathrm{eta :=:frac{Output:power}{Output:power:+:Losses}}$$ $$mathrm{Rightarrow eta :=:frac{VAtimes Power:Factor}{left ( VAtimes Power:Factor right ):+:Losses}}$$ Where, $$mathrm{mathrm{Output:power}:=:VAtimes Power:factor}$$ And, losses can be determined by transformer tests. Efficiency from Transformer Tests When we perform transformer tests, the following results are obtained − From open-circuit test − $$mathrm{mathrm{Full:load:iron:loss}:=:mathit{P_{i}}}$$ From short-circuit test − $$mathrm{mathrm{Full:load:copper:loss}:=:mathit{P_{c}}}$$ Therefore, the total losses at full load in a transformer are $$mathrm{mathrm{Total:FL:losses}:=:mathit{P_{i}+:P_{c}}}$$ Now, we are able to determine the full-load efficiency of the transformer at any power factor without actual loading the transformer. $$mathrm{mathit{n_{FL}}:=:frac{(VA)_{mathit{FL}}times Power:factor}{[(VA)_{mathit{FL}}times Power:factor]+:mathit{P_{i}}+mathit{P_{c}}}}$$ Also, the transformer efficiency at any load equal to x × full load. Where, x is the fraction of loading. In this case, the total losses corresponding to the given load are, $$mathrm{(Total:losses)_{x}:=:mathit{P_{i}+:x^{mathrm{2}}mathit{P_{c}}}}$$ It is because, the iron loss ($mathit{P_{i}}$) is the constant loss and hence remains the same at all loads, while the copper loss is proportional to the square of the load current. $$mathrm{thereforeeta _{x}:=: frac{mathit{x}times (VA)_{mathit{FL}}times Power:factor}{[mathit{x}times (VA)_{mathit{FL}}times Power:factor]+:mathit{P_{i}}+:x^{mathrm{2}}mathit{P_{c}}}}$$ Condition for Maximum Efficiency For a given transformer, we have, $$mathrm{mathrm{Output:power}:=:mathit{V_{mathrm{2}}I_{mathrm{2}}cosphi _{mathrm{2}}}}$$ Let the transformer referred to secondary side, then Ro2 is the total resistance of the transformer. The total copper loss is given by, $$mathrm{mathit{P_{c}}:=:mathit{I_{mathrm{2}}^{mathrm{2}}mathit{R_{omathrm{2}}}}}$$ Therefore, the transformer efficiency is given by, $$mathrm{eta :=:frac{mathit{V_{mathrm{2}}}I_{mathrm{2}}cosphi _{mathrm{2}}}{mathit{V_{mathrm{2}}I_{mathrm{2}}cosphi _{mathrm{2}}}+mathit{P_{i}}+mathit{I_{mathrm{2}}^{mathrm{2}}}R_{omathrm{2}}}}$$ On rearranging the expression, we get, $$mathrm{eta :=:frac{mathit{V_{mathrm{2}}}cosphi _{mathrm{2}}}{mathit{V_{mathrm{2}}cosphi _{mathrm{2}}}+left ( mathit{frac{P_{i}}{I_{mathrm{2}}}} right )+mathit{I_{mathrm{2}}}R_{omathrm{2}}}:=:mathit{frac{V_{mathrm{2}}cosphi _{mathrm{2}}}{D}}:cdot cdot cdot (1)}$$ In practice, the secondary voltage V2 is approximately constant. Hence, for a load of given power factor, the transformer efficiency depends upon the load current (I2). From the equation (1), we can see that the numerator is constant and for the efficiency to be maximum, the denominator (D) should be minimum, i.e. $$mathrm{mathit{frac{d(D)}{dI_{mathrm{2}}}}:=:0}$$ $$mathrm{Rightarrowmathit{frac{d}{dI_{mathrm{2}}}}left [ mathit{V_{mathrm{2}}cosphi _{mathrm{2}}}+left ( mathit{frac{P_{i}}{I_{mathrm{2}}}}right )+mathit{I_{mathrm{2}} R_{0mathrm{2}}} right ]:=:0}$$ $$mathrm{Rightarrow 0-left ( mathit{frac{P_{i}}{I_{mathrm{2}}}} right )+mathit{R_{omathrm{2}}}:=:0}$$ $$mathrm{Rightarrow mathit{P_{i}}:=:mathit{I_{mathrm{2}}^{mathrm{2}}R_{omathrm{2}}}}$$ $$mathrm{Rightarrow mathrm{Iron:loss}:=:Copper:loss}$$ Therefore, the transformer efficiency for a given power factor will be maximum when the constant iron loss is equal to the variable copper loss. The maximum efficiency at any load is given by, $$mathrm{mathit{eta _{max}}:=:frac{mathit{xtimes (VA)_{mathit{FL}}times mathrm{Power:factor}}}{[mathit{xtimes (VA)_{mathit{FL}}}times Power:fctor]+:2mathit{P_{i}}}}$$ Also, the load current (I2) corresponding to the maximum efficiency of transformer is, $$mathrm{mathit{I_{mathrm{2}}}:=:sqrt{frac{mathit{P_{i}}}{R_{o2}}}}$$ Numerical Example In a 100 kVA transformer, the iron loss is 450 W and full-load copper loss is 900 W. Find the transformer efficiency at full load and the maximum efficiency of the transformer, where the load power factor is 0.8 lagging. Solution Given data, Full load VA = 100 kVA = 100 × 1000 VA Iron loss,Pi = 450 W Copper loss,Pc = 900 W cos$mathit{phi _{mathrm{2}}}$ = 0.8 Transformer efficiency at full-load − $$mathrm{mathrm{Total:losses}:=:450:+:900:=:1350:W}$$ $$mathrm{mathit{eta _{mathit{FL}}}:=:frac{(VA)_{mathit{FL}}times Power:factor}{[(VA)_{mathit{FL}}times Power:factor]+:Total:losses}}$$ $$mathrm{Rightarrow mathit{eta _{mathit{FL}}}:=:frac{100times 1000times 0.8}{(100times 1000times 0.8)+1350}:=:frac{80000}{81350}:=:0.9834}$$ $$mathrm{therefore eta _{mathit{FL}}:=:0.9834times 100%:=:98.34%}$$ Maximum efficiency of the transformer − For maximum efficiency, $$mathrm{mathrm{Iron:loss}:=:Copper:Loss}$$ $$mathrm{therefore eta _{mathit{max}}:=:frac{(VA)_{mathit{FL}}times Power:factor}{[(VA)_{mathit{FL}}times Power:factor]+2mathit{P_{i}}}}$$ $$mathrm{Rightarrow eta _{mathit{max}}:=:frac{100times 1000times 0.8}{(100times 1000times 0.8)+(2times 450)}:=:0.9888}$$ $$mathrm{therefore eta _{mathit{max}}:=:0.9888times 100%:=:98.88%}$$ Learning working make money

Turns Ratio and Voltage Transformation Ratio As discussed in the previous chapter, the EMF of equation of a transformer is given by, $$mathrm{mathit{E}:=:4.44:mathit{fphi _{m}:N}}$$ For primary winding, $$mathrm{mathit{E_{mathrm{1}}}:=:4.44:mathit{fphi _{m}:N_{mathrm{1}}}:cdot cdot cdot (1)}$$ For secondary winding, $$mathrm{mathit{E_{mathrm{2}}}:=:4.44:mathit{fphi _{m}:N_{mathrm{2}}}:cdot cdot cdot (2)}$$ Turns Ratio of Transformer From equations (1) and (2), we have, $$mathrm{frac{mathit{E_{mathrm{1}}}}{mathit{E_{mathrm{2}}}}:=:frac{mathit{N_{mathrm{1}}}}{mathit{N_{mathrm{2}}}}:=mathrm{a}::cdot cdot cdot (3)}$$ The constant “a” is known as the turns ratio of the transformer. It may be defined as under, The ratio of number of turns in the primary winding the number of turns in the secondary winding of a transformer is known as turns ratio. Voltage Transformation Ratio of Transformer The ratio of output voltage to the input voltage of transformer is known as voltage transformer ratio, i.e., $$mathrm{mathrm{Transformation: Ratio}:=:frac{Output :Voltage}{Input :Voltage}}$$ Thus, if V1 is the input voltage and V2 is the output voltage of a transformer, then its transformation ratio is given by, $$mathrm{mathrm{Transformation: Ratio}:=:frac{mathit{V_{mathrm{2}}}}{mathit{V_{mathrm{1}}}}:cdot cdot cdot (4)}$$ For an ideal transformer, V1 = E1 and V2 = E2, then $$mathrm{mathrm{Transformation: Ratio}:=:frac{mathit{V_{mathrm{2}}}}{mathit{V_{mathrm{1}}}}:=:frac{mathit{E_{mathrm{2}}}}{mathit{E_{mathrm{1}}}}:=::frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}}:=:frac{1}{a}cdot cdot cdot (5)}$$ However, in a practical transformer, there is a small difference between V1 and E1, and V2 and E2, due to winding resistances. Although, this difference is very small so for analysis purposes, we take V1 = E1 and V2 = E2. Numerical Example (1) A transformer with 1000 primary turns and 400 secondary turns is supplied from a 220 V AC supply. Calculate the secondary voltage and the volts per turn. Solution Given data, $$mathrm{mathit{N_{mathrm{1}}}:=:1000:mathrm{and}:mathit{N_{mathrm{2}}}:=:400}$$ $$mathrm{mathit{V_{mathrm{1}}}:=:220:V}$$ The turns ratio of transformer is, $$mathrm{frac{mathit{V_{mathrm{1}}}}{mathit{V_{mathrm{2}}}}:=:frac{mathit{N_{mathrm{1}}}}{mathit{N_{mathrm{2}}}}}$$ $$mathrm{Rightarrow mathit{V_{mathrm{2}}}:=:mathit{V_{mathrm{1}}}times frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}}:=:220times frac{400}{1000}}$$ $$mathrm{thereforemathit{V_{mathrm{2}}}:=:88:mathrm{Volts}}$$ The volts per turn is given by, $$mathrm{mathrm{For: primary: winding}:=:frac{mathit{V_{mathrm{1}}}}{mathit{N_{mathrm{1}}}}:=:frac{200}{1000}:=:0.22:mathrm{Volts}}$$ $$mathrm{mathrm{For: Secondary: winding}:=:frac{mathit{V_{mathrm{2}}}}{mathit{N_{mathrm{2}}}}:=:frac{88}{400}:=:0.22:mathrm{Volts}}$$ Hence, from this example, it is clear that the volts per turn for a transformer remain the same on both primary and secondary windings. Numerical Example (2) A transformer with an output voltage of 2200 V is supplied at 220 V. If the secondary winding has 2000 turns, then calculate the number of turns in primary winding. Solution Given data, $$mathrm{mathit{V_{mathrm{1}}}:=:200:mathit{V}:mathrm{and}:mathit{V_{mathrm{2}}}:=:2200:mathit{V}}$$ $$mathrm{mathit{N_{mathrm{2}}}:=:2000:mathrm{turns}}$$ The turns ratio of transformer is, $$mathrm{frac{mathit{V_{mathrm{1}}}}{mathit{V_{mathrm{2}}}}:=:frac{mathit{N_{mathrm{1}}}}{mathit{N_{mathrm{2}}}}}$$ $$mathrm{Rightarrow {mathit{N_{mathrm{1}}}}:=:mathit{N_{mathrm{2}}}:times :frac{mathit{V_{mathrm{1}}}}{mathit{V_{mathrm{2}}}}:=:mathrm{2000}:times :frac{220}{2200}:=:mathrm{200:turns}}$$ Learning working make money

Singly-Excited and Doubly Excited Systems Excitation means providing electrical input to an electromechanical energy conversion device such as electric motors. The excitation produces working magnetic field in the electrical machine. Some electrical machines require single electrical input whereas some others require two electrical inputs. Therefore, depending on the number of electrical inputs to electromechanical energy conversion systems, they can be classified into two types − Singly-Excited System Doubly-Excited System Singly Excited System As its name implies, a singly-excited system is one which consists of only one electrically energized coil to produce working magnetic field in the machine or any other electromechanical energy conversion device. Hence, the singly-excited system requires only one electrical input. A singly excited system consists of coil wound around a magnetic core and is connected to a voltage source so that it produces a magnetic field. Due to this magnetic field, the rotor (or moving part) which is made up of ferromagnetic material experiences a torque which move it towards a region where the magnetic field is stronger, i.e., the torque exerted on the rotor tries to position it such that it shows minimum reluctance in the path of magnetic flux. The reluctance depends upon the rotor angle. This torque is known as reluctance torque or saliency torque because it is caused due to saliency of the rotor. Analysis of Singly Excited System We made following assumption to analyze the singly-excited system − For any rotor position, the relationship between flux linkage ($psi $) and current ($mathit{i}$) is linear. The coil has negligible leakage flux, which means all the magnetic flux flows through the main magnetic path. Hysteresis loss and eddy-current loss are neglected. All the electric fields are neglected and the magnetic field is predominating. Consider the singly-excited system as shown in Figure-1. If R is the resistance of the coil circuit, the by applying KVL, we can write the voltage equation as, $$mathrm{mathit{v:=:iR:+:frac{mathit{dpsi }}{mathit{dt}}}cdot cdot cdot (1)}$$ On multiplying equation (1) by current $mathit{i}$, we have, $$mathrm{mathit{vi:=:i^{mathrm{2}}R:+mathit{i}:frac{mathit{dpsi }}{mathit{dt}}}cdot cdot cdot (2)}$$ We are assuming initial conditions of the system zero and integrating the equation (2) on both side with respect to time, we obtain, $$mathrm{int_{0}^{mathit{T}}mathit{vi:dt}:=:int_{0}^{mathit{T}}left ( i^{mathrm{2}}mathit{R}:+mathit{i}:frac{mathit{dpsi }}{mathit{dt}} right )mathit{dt}}$$ $$mathrm{Rightarrowint_{0}^{mathit{T}}mathit{vi:dt}:=:int_{0}^{mathit{T}}mathit{i^{mathrm{2}}R:dt}:+:int_{0}^{psi }mathit{i:dpsi }cdot cdot cdot (3)}$$ Equation-3 gives the total electrical energy input the singly-excited system and it is equal to two parts namely, First part is the electrical loss ($mathit{W_{el}}$). Second part is the useful electrical energy which is the sum of field energy ($mathit{W_{f}}$) and output mechanical energy ($mathit{W_{m}}$). Therefore, symbolically we may express the Equation-3 as, $$mathrm{mathit{W_{in}}:=:mathit{W_{el}}:=:left (mathit{W_{f}} :+:mathit{W_{m}} right )}cdot cdot cdot (4)$$ The energy stored in the magnetic field of a singly-excited system is given by, $$mathrm{mathit{W_{f}}:=: int_{0}^{psi }mathit{i:dpsi }:=:int_{0}^{psi }frac{psi }{mathit{L}}mathit{dpsi }:=:frac{psi ^{mathrm{2}}}{2mathit{L}}cdot cdot cdot (5)}$$ For a rotor movement, where the rotor angle is $mathit{theta _{m}}$, the electromagnetic torque developed in the singly-excited system is given by, $$mathrm{mathit{tau _{e}}:=:frac{mathit{i^{mathrm{2}}}}{mathrm{2}}frac{mathit{partial L}}{mathit{partial theta _{m}}}cdot cdot cdot (6)}$$ The most common examples of singly-excited system are induction motors, PMMC instruments, etc. Doubly Excited System An electromagnetic system is one which has two independent coils to produce magnetic field is known as doubly-excited system. Therefore, a doubly-excited system requires two separate electrical inputs. Analysis of Doubly Excited System A doubly-excited system consists of two main parts namely a stator and a rotor as shown in Figure-2. Here, the stator is wound with a coil having a resistance R1 and the rotor is wound with a coil of resistance R2. Therefore, there are two separated windings which are excited by two independent voltage sources. In order to analyze the double-excited system, the following assumption are made − For any rotor position the relationship between flux-linkage ($psi$) and current is linear. Hysteresis and eddy current losses are neglected. The coils have negligible leakage flux. The electric fields are neglected and the magnetic fields are predominating. The magnetic flux linkages to two windings are given by, $$mathrm{psi _{mathrm{1}}:=:mathit{L_{mathrm{1}}i_{mathrm{1}}}:+:mathit{Mi_{mathrm{2}}}}cdot cdot cdot (7)$$ $$mathrm{psi _{mathrm{2}}:=:mathit{L_{mathrm{2}}i_{mathrm{2}}}:+:mathit{Mi_{mathrm{2}}}}cdot cdot cdot (8)$$ Where, M is the mutual inductance between two windings. By applying KVL, we can write the equations of instantaneous voltage for two coils as, $$mathrm{mathit{v}_{mathrm{1}}:=:mathit{i_{mathrm{1}}R_{mathrm{1}}}:+:frac{mathit{dpsi _{mathrm{1}}}}{mathit{dt}}}cdot cdot cdot (9)$$ $$mathrm{mathit{v}_{mathrm{2}}:=:mathit{i_{mathrm{2}}R_{mathrm{2}}}:+:frac{mathit{dpsi _{mathrm{2}}}}{mathit{dt}}}cdot cdot cdot (10)$$ In case of doubly-excited system, the energy stored in the magnetic field is given by, $$mathrm{mathit{W_{f}}:=:frac{1}{2}mathit{L_{mathrm{1}}i_{mathrm{1}}^{mathrm{2}}}:+:frac{1}{2}mathit{L_{mathrm{2}}i_{mathrm{2}}^{mathrm{2}}}:+:mathit{Mi_{mathrm{1}}i_{mathrm{2}}}cdot cdot cdot (11)}$$ And, the electromagnetic torque developed in the doubly excited system is given by, $$mathrm{mathit{tau _{e}}:=:frac{mathit{i_{mathrm{1}}^{mathrm{2}}}}{mathrm{2}}frac{mathit{dL_{mathrm{1}}}}{mathit{dtheta _{m}}}:+:frac{mathit{i_{mathrm{2}}^{mathrm{2}}}}{mathrm{2}}frac{mathit{dL_{mathrm{2}}}}{mathit{dtheta _{m}}}:+:mathit{i_{mathrm{1}}i_{mathrm{2}}}frac{mathit{dM}}{mathit{dtheta _{m}}}cdot cdot cdot (12)}$$ In Equation-12, the first two terms are the reluctance torque and the last term gives the co-alignment torque due to interaction of two fields. The practical examples of doubly-excited system are synchronous machines, tachometer, separately-excited DC machines, etc. Learning working make money

Losses in a Transformer The following power losses may occur in a practical transformer − Iron Loss or Core Loss Copper Loss or I2R Loss Stray Loss Dielectric Loss In a transformer, these power losses appear in the form of heat and cause two major problems − Increases the temperature of the transformer. Reduces the efficiency of the transformer. Iron Loss or Core Loss Iron loss occurs in the magnetic core of the transformer due to flow of alternating magnetic flux through it. For this reason, the iron loss is also called core loss. We generally use the symbol ($mathit{P_{i}}$) to represent the iron loss. The iron loss consists of hysteresis loss ($mathit{P_{h}}$) and eddy current loss ($mathit{P_{e}}$). Thus, the iron loss is given by the sum of the hysteresis loss and eddy current loss, i.e. $$mathrm{mathrm{Iron:loss,}mathit{P_{i}}:=:mathrm{Hysteresies:loss(mathit{P_{h}})}:+:mathrm{Eddy:current:loss(mathit{P_{e}})}}$$ The hysteresis loss and eddy current loss (or iron loss) are determined by performing the open-circuit test on the transformer. The empirical formulae for the hysteresis loss and eddy current loss are given by, $$mathrm{mathit{P_{h}}:=:mathit{k_{h}f:B_{m}^{x}}:cdot cdot cdot (1)}$$ $$mathrm{mathit{P_{e}}:=:mathit{ke:B_{m}^{mathrm{2}}:f^{mathrm{2}}t^{mathrm{2}}}:cdot cdot cdot (2)}$$ Where, The exponent of Bm, i.e. “x” is called the Steinmetz’s constant. Depending on the properties of the core material, its value is ranging from 1.5 to 2.5. kh is a proportionality constant whose value depends upon the volume and quality of the material of core. ke is a proportionality constant which depend on the volume and resistivity of material of the core. f is the frequency of the alternating flux in the core. Bm is the maximum flux density in the core. t is the thickness of each core lamination. Therefore, the total iron loss or core loss can also be written as, $$mathrm{mathit{P_{i}}:=:mathit{k_{h}f:B_{m}^{x}}:+:mathit{ke:B_{m}^{mathrm{2}}:f^{mathrm{2}}t^{mathrm{2}}}:cdot cdot cdot (3)}$$ Since the input voltage to the transformer is approximately equal to the induced voltage in the primary winding, i.e. $$mathrm{mathit{V_{mathrm{1}}}:=:mathit{E_{mathrm{1}}}:=:4.44:mathit{fphi _{m}N_{mathrm{1}}}}$$ $$mathrm{Rightarrow mathit{V_{mathrm{1}}}:=:4.44:mathit{f:B_{m}AN_{mathrm{1}}}}$$ Where, A is the cross-sectional area of the transformer core, N1 is the number of turns in the primary winding and f is the supply frequency. $$mathrm{therefore mathit{B_{m}}:=:frac{mathit{V_{mathrm{1}}}}{4.44mathit{fAN_{mathrm{1}}}}:cdot cdot cdot (4)}$$ Hence, from equations (1) & (4), we get, $$mathrm{mathit{P_{h}}:=:mathit{k_{h}f}left ( frac{mathit{V_{mathrm{1}}}}{4.44mathit{fAN_{mathrm{1}}}} right )^{x}}$$ $$mathrm{Rightarrow mathit{P_{h}}:=:mathit{k_{h}f}left ( frac{mathrm{1}}{4.44mathit{AN_{mathrm{1}}}} right )^{x}cdot left ( frac{mathit{V_{mathrm{1}}}}{mathit{f}} right )^{x}}$$ $$mathrm{Rightarrow mathit{P_{h}}:=:mathit{k_{h}}left ( frac{mathrm{1}}{4.44mathit{AN_{mathrm{1}}}} right )^{x}cdot mathit{V_{mathrm{1}}^{x}}:mathit{f^{(mathrm{1}-x)}}:cdot cdot cdot (5)}$$ Thus, Equation (5) shows that the hysteresis loss depends upon both input voltage and supply frequency. Again, from equations (2) & (4), we get, $$mathrm{mathit{P_{e}}:=:mathit{k_{e}f^{mathrm{2}}t^{mathrm{2}}}left ( frac{mathit{V_{mathrm{1}}}}{4.44mathit{fAN_{mathrm{1}}}} right )^{mathrm{2}}}$$ $$mathrm{Rightarrow mathit{P_{e}}:=:mathit{k_{e}left ( frac{mathit{V_{mathrm{1}}}}{mathrm{4.44}mathit{AN_{mathrm{1}}}} right )^{mathrm{2}}mathit{t^{mathrm{2}}}:cdot cdot cdot mathrm{(6)}}}$$ Hence, from equation (6), we can conclude that the eddy current loss in the transformer is proportional to the square of the input voltage and is independent of the supply frequency. Therefore, the total core loss can also be written as, $$mathrm{mathit{P_{i}}:=:mathit{k_{h}left ( frac{mathrm{1}}{mathrm{4.44}mathit{AN_{mathrm{1}}}} right )^{mathrm{2}}cdot mathit{V_{mathrm{1}}^{mathit{x}}f^{(mathrm{1-x})}}:+:mathit{k_{e}}left ( frac{V_{mathrm{1}}}{mathrm{4.44}mathit{AN_{mathrm{1}}}} right )^{mathrm{2}}mathit{t^{mathrm{2}}}:cdot cdot cdot left ( mathrm{7} right )}}$$ In practice, transformers are connected to an electric supply of constant frequency and constant voltage, thus, both f and Bm are constant. Therefore, the core or iron loss is practically remains constant at all loads. We can reduce the hysteresis loss by using steel of high silicon content to construct the core of transformer while the eddy current loss can be minimized by using core of thin laminations instead of solid core. The open-circuit test is performed on a transformer to determine the iron or core loss. Copper Loss or I2R Loss Power loss in a transformer that occurs in both the primary and secondary windings due to their Ohmic resistance is called copper loss or I2R loss. We usually represent the copper loss by PC. Therefore, the total copper loss in a transformer is the sum of power loss in the primary winding and power loss in the secondary winding, i.e., $$mathrm{mathit{P_{c}}:=:mathrm{Copper:loss:in:primary:+:Copper:loss:in:secondary}}$$ $$mathrm{Rightarrow mathit{P_{c}}:=:mathit{I_{mathrm{1}}^{mathrm{2}}}mathit{R_{mathrm{1}}}:+:mathit{I_{mathrm{2}}^{mathrm{2}}}mathit{R_{mathrm{2}}}:cdot cdot cdot (8)}$$ Since, $$mathrm{mathit{I_{mathrm{1}}}mathit{N_{mathrm{1}}}:=:mathit{I_{mathrm{2}}}mathit{N_{mathrm{2}}}}$$ $$mathrm{Rightarrow mathit{I_{mathrm{1}}}:=:left ( frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}} right )mathit{I_{mathrm{2}}}:cdot cdot cdot (9)}$$ $$mathrm{therefore mathit{P_{c}}:=:left [ left ( frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}} right )I_{mathrm{2}} right ]^{mathrm{2}}:mathit{R_{mathrm{1}}}:+:mathit{I_{mathrm{2}}^{mathrm{2}}}mathit{R_{mathrm{2}}}:=:left [ left ( frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}} right )^{mathrm{2}}mathit{R_{mathrm{1}}}:+:mathit{R_{mathrm{2}}} right ]mathit{I_{mathrm{2}}^{mathrm{2}}}:cdot cdot cdot (10)}$$ From Equation (10), it is clear that the copper loss in a transformer varies as the square of the load current. For this reason, the copper loss is also referred as “variable loss” because in practice a transformer is subjected to variable load and hence has variable load current. We conduct the “short-circuit test” on the transformer to determine the value of its copper loss. In a practical transformer, the copper loss accounts for about 90% of the total power loss in the transformer. Stray Loss In practical transformer, a fraction of the total flux follows a path through air and this flux is called leakage flux. This leakage flux produces eddy currents in the conducting or metallic parts like tank of the transformer. These eddy currents cause power loss, which is known as stray loss. Dielectric Loss The power loss occurs in insulating materials like oil, solid insulation of the transformer, etc. is known as dielectric loss. The dielectric loss is significant only in transformers working on high voltages. Although, in practice, the stray loss and dielectric loss are very small, constant and may be neglected. From the above discussion, we found that a transformer has some losses which are constant and some other are variable. Thus, we may categorize losses in a transformer in two types namely constant losses and variable losses. Therefore, the total losses in a transformer are the sum of constant losses and variable losses, i.e., Total losses in transformer = Constant losses + Variable losses Learning working make money

Electrical Transformer In electrical and electronic systems, the electrical transformer is one of the most useful electrical machine. An electrical transformer can increase or decrease the magnitude of alternating voltage or current. It is the major reason behind the widespread use of alternating currents rather than direct current. A transformer does not have any moving part. Therefore, it has very high efficiency up to 99% and very strong and durable construction. Electrical Transformer A transformer or electrical transformer is a static AC electrical machine which changes the level of alternating voltage or alternating current without changing in the frequency of the supply. A typical transformer consists of two windings namely primary winding and secondary winding. These two windings are interlinked by a common magnetic circuit for transferring electrical energy between them. Principle of Transformer Operation The operation of the transformer is based on the principle of mutual inductance, which states that when a changing magnetic field of one coil links to another coil, an EMF is induced in the second coil. When an alternating voltage V1 is applied to the primary winding, an alternating current flows through it and produces an alternating magnetic flux. This changing magnetic flux flows through the core of the transformer and links to the secondary winding. According to Faraday’s law of electromagnetic induction, an EMF E2 is induced in the secondary winding due to the linkage of changing magnetic flux of the primary winding. If the secondary winding circuit is closed by connecting a load, then this induced EMF E2 in the secondary winding causes a secondary current I2 to flow through the load. Although the changing magnetic flux of primary winding is also linked with the primary winding itself. Hence, an EMF E1 is induced in the primary winding due to its own inductance effect. The value of E1 and E2 can be given by the following formulae, $$mathrm{mathit{E_{mathrm{1}}}:=:-mathit{N_{mathrm{1}}}frac{mathit{dphi }}{mathit{dt}}}$$ $$mathrm{mathit{E_{mathrm{2}}}:=:-mathit{N_{mathrm{2}}}frac{mathit{dphi }}{mathit{dt}}}$$ Where N1 and N2 are the number of turns in the primary winding and secondary winding respectively. On taking the ratio of E2 and E1, we get, $$mathrm{frac{mathit{E_{mathrm{2}}}}{mathit{E_{mathrm{1}}}}:=:frac{mathit{N_{mathrm{2}}}}{mathit{N_{mathrm{1}}}}}$$ This expression is known as transformation ratio of the transformer. The transformation ratio depends on the number of turns in primary and secondary windings. Which means the magnitude of output voltage depends on the relative number of turns in primary and secondary windings. If N2 > N1, then E2 > E1, i.e., the output voltage of the transformer is more than the input voltage, and such a transformer is known as set-up transformer. On the other hand, if N1 > N2, then E1 > E2 i.e., the output voltage is less than input voltage, such a transformer is called step-down transformer. From the circuit diagram of the transformer, we can see that there is no electrical connection between the primary and secondary instead they are linked with the help of a magnetic field. Thus, a transformer enables us to transfer AC electrical power magnetically from one circuit to another which a change in the voltage and current level. Important Points Note the following important points about transformers − The operation of transformer is based on the principle of electromagnetic induction. The transformer does not change the frequency, i.e. the frequency of input supply and output supply remains the same. Transformer is a static electrical machine, which means it does not have any moving part. Hence, it has very high efficiency. Transformer cannot work with direct current because it is an electromagnetic induction machine. There is no direct electrical connection between primary and secondary windings. The AC power is transferred from primary to secondary through magnetic flux. Learning working make money