Utilisation of Electrical Energy Introduction

Last time … Induction motor introduction. Poles, slip & rpm. Wiring options. Equivalent circuit of induction motor. No-load power factor and branch current. Equivalent impedance, Zeq. Assignment 4 (induction motors) working time. 13/01/2019 UEE

Introduction - objectives Full-load power factor and branch current continued. Equivalent impedance, Zeq. Argand diagrams. Full-load power and efficiency. Locked rotor tests. Torque of induction motors. Assignment 4 (induction motors) working time. 13/01/2019 UEE

Full-load calculations recap These calculations ignore the magnetising branch current initially as Vs is the same. This simplifies the model of the motor and leaves only series resistances and reactances which form the total impedance, Zeq. It can be shown that the real part of the load Rr varies with the rotor speed but the reactance Xr remains constant. A variable resistor represents the rotor speed. 13/01/2019 UEE

Full-load induction motor model VS 13/01/2019 UEE

Complex notation for Zeq When adding the resistances and reactances, it is sometimes easier to use complex notation. Zeq is the resultant of series impedances. Ohmic part is the real part of the impedance. Reactive part is the imaginary part of the impedance. 13/01/2019 UEE

Zeq calculations Zeq = equivalent impedance. Zeq = (Rs + R`r /s) + j(Xs + Xr). Where s = slip. Remember for inductive loads, current lags the voltage. 13/01/2019 UEE

Example 5 (from last week) An induction motor has the following parameters: Rs = 0.6W, R`r = 0.4W, Xs = Xr = 0.6W, slip s = 0.04. Zeq (full load) = (0.6 + 0.4/0.04) + j(0.6 + 0.6) = 10.6 + j1.2 W. To be continued…. 13/01/2019 UEE

Argand diagrams Plots the real (Re) and imaginary (Im) parts of Z. Z = 10.6 + j1.2 13/01/2019 UEE

Calculations Zeq (full load) = 10.6 + j1.2 W. This has both a real and an imaginary component plotted on the Argand diagram. It can be expressed as the magnitude of Z, often seen as |Z| (r on the diagram) which is (10.62 + 1.22) followed by the phase angle which is found by tan-1(1.2 / 10.6). 10.6 + j1.2 W = 10.67 ∟6.8°. 13/01/2019 UEE

Example 1 The magnitude of Z is known as the Modulus (Mod) and often seen as |Z|. The angle q is known as the Argument (Arg). Find the modulus and argument of the complex number 5.5 + j3.3. |Z| = (5.52 + 3.32) = 6.41W. Arg(Z) = tan-1(3.3 / 5.5) = 30.96 °. Z =6.41∟ 31 ° W. 13/01/2019 UEE

Power factor Now you have the argument, it is easy to calculate the power factor (PF) for an impedance. PF = cos(Arg(z)) E.g. what is the power factor for Z =6.41∟ 30.96 ° W? PF = cos(30.96) = 0.86. 13/01/2019 UEE

Converting Mod Arg to complex impedance Sometimes it is necessary to convert quantities expressed as Mod∟Arg to the complex form. |Z| ∟q = |Z| cosq + j|Z| sinq W. 45.2 ∟75.1 ° = 45.2 cos(75.1) + j45.2 sin(75.1) = 11.62 + j43.68 W. Try worksheet questions 7 – 9 to practise this. Remember that sin and cos values can be negative. 13/01/2019 UEE

Worked example 01 (star wiring) 13/01/2019 UEE

First steps – branch current Io & PF Ic = Vs / Rc = 240 / 120 = 2.0 A. Im = Vs / Xm = 240 / 25 = 9.6 A. From the diagram, Io = Ic – jIm = 2.0 – j9.6 A. jIm is –ve because current lags voltage for inductors. Io = 9.81∟-78.23° A. Now we also have the PF = cos (-78.23) = 0.204. 13/01/2019 UEE

Now F/L stator & rotor impedance Zeq = (Rs + R`r /s) + j(Xs + X`r). At full load (F/L) slip s = 0.04. Zeq at full load = (Rs + R`r /0.04) + j(Xs + X`r). Zeq = (0.2 + 0.15 / 0.04) + j(0.6 + 0.7) = (0.2 + 3.75) + j(1.3) = 3.95 + j1.3 W = 4.16 ∟18.2° W. 13/01/2019 UEE

Calculating I`r We have Vs and Zeq so I`r can be calculated. I`r = Vs / Zeq = 240 / (4.16 ∟18.2° ) = 57.7 ∟-18.2° (angle becomes negative because 1 / j = -j). 57.7 ∟-18.2° needs to be expressed as real and imaginary parts again. I`r = |Z| cosq + j|Z| sinq = 57.7 cos (-18.2) + j57.7 sin (-18.2) = 54.81 - j18.02 A. 13/01/2019 UEE

Calculating total current Is Is is the sum of Io and I`r – already calculated. Is = 2.0 – j9.6 A + 54.81 - j18.02 A. We can simply add the real and imaginary parts separately. Is = 56.81 – j27.62 A. Convert back to the Mod ∟ Arg form. Is = 63.17 ∟-25.9 °. 13/01/2019 UEE

Power factor at rated Full-load Is = 63.17 ∟-25.9 °. PF = cosq = cos -25.9° = 0.90. Remember for inductive loads, current lags the voltage. 13/01/2019 UEE

Calculating power of induction motors The load is represented by R`r / s. This varies with the speed of the motor represented by s. Full load slip s = 0.04 (given earlier). Power = I2R. Phase power = PS (stator to rotor). Rotor power Ps = (I`r)2(R`r / s). I`r = 57.7 ∟-18.2° but we only need to use the real part i.e. 57.7 A. 13/01/2019 UEE

Induction motor power contd. Ps = 57.72 x (0.15 / 0.04) = 3329 x 3.75 = 12,485 W. There are 3 phases and windings so total rotor power = 3Ps = 3 x 12,485 = 37,455 W = 37.46 kW. Mechanical power = Pm. Pm = (1 - s) Ps = (1 – 0.04) x 12,485 = 11,986 W. Gross mechanical power = 3Pm = 3 x 11,986 = 35,958 W = 36.0 kW. Shaft power = 3Pm – mechanical losses. 13/01/2019 UEE

Shaft power & efficiency Shaft power = 3Pm – mechanical losses. Assume mechanical losses of 2 kW. Shaft power = 36.0 - 2 = 34.0 kW. Efficiency h = power out / power in x 100%. Power in = 3VsIs cosq. (PF calculated previously). Pin = 3 x 240 x 63.17 x 0.90 = 40,934 W = 40.9 kW. Efficiency h = 34.0 / 40.9 x 100 = 83.1%. 13/01/2019 UEE

Locked rotor test The locked rotor test is carried out on an induction motor. Also known as short circuit test, locked rotor test or stalled torque test. From the locked rotor test, short circuit current at normal voltage, power factor on short circuit, total leakage reactance, and starting torque of a motor can be calculated. 13/01/2019 UEE

Torque of induction motor When calculating torque, the slip s must be taken into consideration. At start, slip s = maximum = 1. This changes the value of the load R`s to its minimum possible value. At rated speed, slip is the lowest value so the value of the load R`s is the maximum value. 13/01/2019 UEE

Key to abbreviations Supply frequency – f, Hz. p - number of pole pairs (2 poles = 1 pole pair). ns - synchronous speed, rev.s-1 or rpm. s – slip. Note: convert all speeds to rev.s-1. when calculating. 13/01/2019 UEE

Starting torque for ns = 15 Tstart = 3Vs2 /(2pns) x R`r / s[(Rs + R`r /s) + (Xs + X`r)] but s =1 in this case (start-up) so can be cancelled. Tstart = 3Vs2 /(2pns) x R`r / [(Rs + R`r ) + (Xs + X`r)]. Using the given values for the stator and rotor…. Tstart = 3 x 2402 /(2p x 15) x 0.15 / [(0.2 + 0.15 ) + (0.6 + 0.7)] = 1,833 x 0.09091 = 166.7 Nm. 13/01/2019 UEE

Slip at max torque = sTmax sTmax = R`r / (Rs2 + (Xs + Xr)2) = 0.15 / (0.22 + 1.32) = 0.15 / 1.315 = 0.11 When maximum torque is delivered, the slip is 0.11 or 11%. Rotor rpm at max torque = ns(1 - s) = ns(1 – 0.11) = ns x 0.89 = 15 x 0.89 = 13.35 rev.s-1 = 801 rpm. 13/01/2019 UEE

Next time Torque continued. Determining equivalent circuits. 13/01/2019 UEE

Summary Full-load power factor and branch current continued. Equivalent impedance, Zeq. Argand diagrams. Full-load power and efficiency. Locked rotor tests. Torque of induction motors. Assignment 4 (induction motors) working time. 13/01/2019 UEE