ENERGY CONVERSION

ENERGY CONVERSIONDC GENERATOR

General Voltage Equation of DC Generator:let E instantaneous voltage per conductor voltageEc are value of the DC generated EMF per conductor, volt are effective flux per pole, weberN speed in rev. per min (rpm)P poleT - total flux count in one revolution per conductorec d/dtT weber /revEc =r/t = weber /rev x N rev/min x 1 min/60secEc = PN/ 60 voltLet a =number of parallel path in the armature winding = total of number of conductor to the generator E = are DC generated conductor per parallel path = /a x PN/ 60Weber/sec = Vdt E= x PN/ 60a

Note : number of parallel path, a= number of group of coils Low voltages for lap winding a=mP High voltages for wave winding a= 2m M-plex of the windingExample:A 6 pole DC generator has an armature winding with504 conductors connected in 6 parallel path. Calculate the generated voltage in the machine if each pole produces 1.65 x 106 maxwells and the armature speed is 1800 rpm.

Solution:

= (1.65 x 106 Maxwells) * (1 Wb/108 lines) = 1.65 x 10-2Wb

E = PN/ 60a = [(504 cond) *(6 poles)*( 1.65 x 10-2Wb)*(1800rpm)]/(60*a) = 249.5 VoltsDYNAMO a rotating electrical machine which an energy transformation takes place

2 types of Dynamo:Generator Converts mechanical energy to electrical energyMotor Converts electrical energy to mechanical energy

*1 Volt = 1 Wb/sec*1 Wb = 108 Maxwells2 types of Producing EMF by Electromagnetic action:

By changing or varying the magnitudeof the flux acting n a coil or conductor.Faradays law of Electromagnetic Induction-the EMF induced in the terminal of a coil is directly proportional to the rate of change in flux.e = -N (d/dt)where: e = induced or transformed emfN = number of turns(d/dt) = change in flux per unit time (Wb/sec) By relative movement between the magnetic field and the conductor or coil.e = Lvwhere: e = generated emf = flux density (Wb/m2)L = eff. Length of the cord (m)v = component velocity having a direction L to the direction of flux

2 ways of Producing Relative Motion between the Conductor and Magnetic field:Field Stationary and Conductor moving DC generatorConductor stationary and Magnetic moving (AC Gen)

Example: If a single conductor is arranged such that 34.29 cm of its length passes through a form magnetic field of 8106 lines/cm2 and moves at the rate of 139.7cm in 1sec. Determine the generated or induced emf at any instant.

Solution:e = Lv ; lines=maxwellwhere: = (8106 lines/cm2) * (1Wb/108 lines)e = (8106 Wb/cm2) * (34.29 cm) * (139.7cm/sec)e = 38.83 x 10-3 Wb/sec or VoltTYPES OF DC GENERATORS:

1. Series DC Generator - it has a field winding called series field winding connected in series with the armature.

Formulas:Ia = ILEb = Vt + Ia(Ra+Rbc+Rs)Vbc = IaRbcPg = EgIaPi = Pa + SPLPi = Pg + total lossesPo = VtIL%= (Po/Pi) * 100% where:Ia = armature currentIf = field currentIL = load currentRf = shunt resistanceRbc = brush contact resistanceRs = series field resistancePi = input powerPo = output powerPg = generated powerSPL = stray power lossEg = generated voltageVt = terminal voltage2. Self-excited Shunt DC Generator - it has a field winding called shunt field winding which is connected across the armature use for constant voltage application.

Formulas:Ia = IL + If Eg = Vt + Ia(Ra+Rbc)Vt = IfRfVbc = IaRbcPg = EgIaPi = Pg + SPL = Po + total lossesPo = VtILIL = P/(Vt)%= (Po/Pi) * 100% where:Ia = armature currentIf = field currentIL = load currentRf = shunt resistanceRbc = brush contact resistanceRs = series field resistancePi = input powerPo = output powerPg = generated powerSPL = stray power lossEg = generated voltageVt = terminal voltageExample Problem:A 20kW 220V DC shunt generator has a brush resistance of 0.005 ohms, an armature resistance of 0.065 ohms and a shunt field resistance of 200 ohms. What power is develop in thee armature when it delivers its rated load.

If = Vt/Rf = 220V/(200 ohms) = 1.1 A

IL = P/(Vt) = 20kW/220V = 90.9091 A

Ia = IL + If = 90.9091A + 1.1A = 92 A

Eg = Vt + Ia(Ra+Rbc) = 220V + 92A(0.005 ohms + 0.065 ohms) = 226.44 V

Pg = EgIa = 226.44V(92 A) = 20.8325 kW3. Compound DC Generator - it has a shunt field and series field winding either connected long shunt or short shunt.a. Short Shunt Compound DC Generator

Formulas:Ia = IL + IfEb = Vt + Ia(Ra+Rbc+Rs)Vt = IfRf - ILRfVbc = IaRbcPg = EgIaPi = Pa + SPLPi = Pg + total lossesPo = VtIL%= (Po/Pi) * 100%

b. Long Shunt Compound DC Generator

Formulas:Ia = IL + IfEb = Vt + Ia(Ra+Rbc+Rs)Vt = IfRfVbc = IaRbcPg = EgIaPi = Pa + SPLPi = Pg + total lossesPo = VtIL%= (Po/Pi) * 100%Example Problem: A 5kW 120v compound generator has an armature resistance of 0.23ohms, a series resistance of 0.04ohms and a short shunt resistance of 57.50ohms. Assuming a long shunt connection and a voltage drop at the brushes of 2v, Calculate the genereated EMF at full load.

Solution:

IL = Po /Vt = 5000w/120v = 41.67A

If = Vt /Rf = 120v/57.5ohms = 2.1A

Ia = IL + If = 41.67A + 2.1A = 43.77A

Eb = Vt + Ia(Ra+Rbc+Rs)

= Vt + IaRa + IaRbc + IaRs

= 120v + 43.77A(0.23ohms) + 2v + 43.77A(0.04ohms) Eb = 134vTRANSFORMERExact Circuit Diagram of Transformer:

Formulas:IN2 = IE2 + Io2 core loss = EPIE

where in:IN = no load current Io = magnetizing current = current that will produce mutual fluxIE = no load energy current = current that will supply core lossGU= conductanceB0 = susceptance 1) Equivalent Circuit Referred to the Primary Schematic:(no load when In is neglected)Schematic:Formulas:Re-p = rp + a2rs Xe-p = xp + a2xsZe-p = sqrt(Re-p2 + Xe-p2)

2) Equivalent Circuit Referred to the Secondary (no load current is neglected) Schematic:Formulas:Re-s = rs + rp /a2Xe-s = xs + xp/a2Ze-s = sqrt(Re-s2 + Xe-s2)

Example:1.)A 100kVA 2400v/240v 60Hz transformer has the following constants:rp = 0.42ohms rs = 0.0038ohms xp = 0.72ohmsxs = 0.0068ohmsCalculate the following:Re-p and Re-sXe-p and Xe-sZe-p and Ze-s

Solution:a = Vp/Vs = 2400/240 = 10

a. Re-p = rp + a2rs = 0.42ohms +102(0.0038ohms) = 0.8ohmsRe-s = rs + rp /a2 = 0.0038ohms + 0.42ohms/(102) = 0.008ohms b. Xe-p = xp + a2xs = 0.72ohms + 102(0.0068ohms) = 1.4ohmsXe-s = xs + xp/a2 = 0.0068ohms + 0.72ohms/(102) = 0.014ohms c. Ze-p = sqrt(Re-p2 + Xe-p2) = 1.61ohmsZe-s = sqrt(Re-s2 + Xe-s2) = 0.0161ohms2) Using the same data obtain, calculate the ff. voltage drop in primary and secondary turns.

Solution:

Ip = (kVA)/Vp = (100kVA)/2400V = 41.67A

Is = (kVA)/Vs = (100kVA)/240V = 416.7A

IpRe-p = (41.67A)(0.8 ohms) = 33.336V

IsRe-s = (416.7A)(0.008 ohms) = 3.3336V

IpXe-p = (41.67A)(1.4 ohms) = 58.338V

IsXe-s = (416.7A)(0.014 ohms) = 5.8338VTEST FOR TRANSFORMER1) Open Circuit TestSchematic:Formulas:VMR = Vs-ratedWMR = rated core loss

2) Short Circuit TestSchematic:Formulas:AMR = Ip-ratedWMR = rated copper loss

Example problem:1) A 5kVA 2300/230 60Hz transformer.short circuit testopen circuit testVMR = 142VVMR = 230VAMR = 2.17AAMR = 1.01AWMR = 110 WattsWMR = 40 WattsEstablished the parameters, use the short circuit test data.

solution:

Re-p = WMR/AMR2 = (110 Watts)/(2.17A)2 = 23.36 ohmsZe-p = VMR/AMR = (142V)/(2.17A) = 65.44 ohmsXe-p = sqrt.[(65.44)2-(23.36)2] = 61.13 ohmsRe-s = Re-p / a2 = 0.2336 ohmsZe-s = Ze-p / a2 = 0.6544 ohmsXe-s = Xe-p / a2 = 0.6113 ohmsTRANSFORMER EFFICIENCYALL DAY EFFICIENCY - if the transformer primary is connected to the source, whether loaded or not loaded, there is core loss in the transformer.- if the transformer is loaded, there is copper loss in the transformer. at no load, copper is negligible.

%d = (output energy/day * 100%)/(output energy/day + energy losses) Ex. Given 100kVA transformer connected 24 hrs. a day, daily load cycle: 6 hours 90kW load0.9 p.f.4 hours25kW load0.5 p.f.14 hours0kW load --Core loss at rated voltage = 1000 WattsCopper loss with full load current = 1680 WattsDet. the all day efficiency.

Solution:

(90kW/0.9) = 100 kVA

(25kW/0.5) = 50 kVA

Energy output/day = (90*1000*6) + (25*1000*4) = 640,000 Watt-hours

Energy losses/day = (1000*24 ) + (1680*6) + [(0.5)2 * (1680*4)] = 35,760 Watt-hours

%d = [640,000/(640,000+35,760)] * 100% = 94.71%DIRECT CURRENT MOTOR

>>>f = IL;fIb = ( ILin)/11,300,000 Where:f = force acting upon the conductor (Newton) = the flux density of electromagnetic field (Wb/m2)I = Current supplied by the conductor (Ampere)L = eff. length of the conductor (meter)>>>T = f * rWhere:T = Torque produced in the armature per conductor (N-m)f = force produced (N)r = torque arm or radius of armature (m)>>>T = * (Ia/a) * z * rWhere:Ia = Armature currenta = Number of parallel paths Types of DC Motor:I. According to Main Field Winding1) Series MotorFormulas:IL = IaVt = Eb + Ia(Ra+Rb+Rs)Vbc = IaRbcPa = EaIaPo = Pa - SPLPo = Pi - total accessPi = VtIL%= (Po/Pi) * 100%SPEED CHARACTERISTICS - Variable SpeedTOURQUE CHARACTERISTICS - High Stating TorqueTo reverse the direction of the rotation of this motor position of the brush must be interchanged.CAUTION! - Never operate this motor without load for it will race a runway.2) Shunt MotorFormulas:IL = Ia + IfVt = Eb + Ia(Ra+Rbc)Vt = IfRfVbc = IaRbcPa = EbIaPo = Pa - SPLPo = Pi - total accessPi = VtIL%= (Po/Pi) * 100%SPEED CHARACTERISTICS - Adjustable SpeedTOURQUE CHARACTERISTICS - Medium Stating TorqueTo reverse the direction of the rotation of this motor position of the brush must be interchanged or the connection of the field winding must be reversedbut not doing both. CAUTION! - Never open the field det of this motor while running for it will race or runaway.3) Compound Motor>Types of Self-Excited Compound DC Motor:DC Motor according to electrical connection of the series and shunt field windings with the armature.

a) Short Shunt Compound Motor

b) Long Shunt Compound MotorExample: A 250 V series DC motor with compensating windings has a total series resistance RA + RS of 0.08 . The series field consists of 25 turns per pole and the magnetization curve is

Find the speed and induced torque of this motor when its armature current is 50 A.Calculate and plot its torque-speed characteristic.a) To analyze the behavior of a series motor with saturation, we pick points along the operating curve and find the torque and speed for each point. Since the magnetization curve is given in units of mmf (ampere-turns) vs. EA for a speed of 1200 rpm, calculated values of EA must be compared to equivalent values at 1200 rpm.For IA = 50 A

Since for a series motor IA = IF = 50 A, the mmf is

From the magnetization curve, at this mmf, the internal generated voltage is EA0 = 80 V. Since the motor has compensating windings, the correct speed of the motor will be

The resulting torque:

Example 5.5: A 100 hp, 250 V compounded DC motor with compensating windings has an internal resistance, including the series winding of 0.04 . There are 1000 turns per pole on the shunt field and 3 turns per pole on the series windings. The magnetization curve is shown below.

The field resistor has been adjusted for the motor speed of 1200 rpm. The mechanical, core, and stray losses may be neglected.Find the no-load shunt field current.Find the speed at IA = 200 A if the motor is b) cumulatively; c) differentially compoundedAt no load, the armature current is zero; therefore, the internal generated voltage equals VT = 250 V. From the magnetization curve, a field current of 5 A will produce a voltage EA = 250 V at 1200 rpm. Therefore, the shunt field current is 5 A.

When the armature current is 200 A, the internal generated voltage is

The effective field current of a cumulatively compounded motor will be

From the magnetization curve, EA0 = 262 V at speed n0 = 1200 rpm. The actual motor speed is

c) The effective field current of a differentially compounded motor will be

From the magnetization curve, EA0 = 236 V at speed n0 = 1200 rpm. The actual motor speed is

II. According to Source of Excitation Current for the Field Winding1) Self-Excited DC Motor- The source of excitation current for the field winding is from the supplied armature voltage.

2) Separately Excited DC Motor- The source of excitation current for the field winding is an independent external emf source.

Some Motor Applications:a) Pumpsb) Fanblocksc) Grindersd) Elevatorse) Food Mixersf) ConveyorsMOTOR COUNTER EMFEc - emf generated in the armature conductors whose direction or polarity is opposite to that of the applied emf or current to the armature conductors.Formulas:Ec = (PNZ)/60aVa = Ec + IaRaVa' = Va + Vb = Ec + IaRa + Vb Vb = IbRb Pd = EcIa = [(PNZ)/60a] * IaExample 5-1The armature of a permanent-magnet dc generator has a resistance of 1 and generates a voltage of 50 V when the speed is 500 r/min. If the armature is connected to a source of 150 V, calculate the following:

a. The starting currentb. The counter-emf when the motor runs at 1000 r/min. At 1460 r/min.c. The armature current at 1000 r/min. At 1460 r/min.

Figure 5.3 See Example 5.1.Solutiona. At the moment of start-up, the armature is stationary, so Eo=0 V (Fig. 5.3a). The starting current is limited only by the armature resistance:

/ =Es/R= 150 V/1 = 150 A

b. Because the generator voltage is 50 V at 500 r/min, the cemf of the motor will be 100 V at 1000 r/min and 146 V at 1460 r/min.

c. The net voltage in the armature circuit at 1000 r/min is

Es-Eo= 150 - 100 = 50 V

The corresponding armature current is

I= (Es-Eo)/R = 50/1 = 50 A (Fig.5.3b)

When the motor speed reaches 1460 r/min, the cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is only

/ = (Es-Eo)/R =(150 - 146)/1 = 4A

and the corresponding motor torque is much smaller than before (Fig. 5.3c).