EECE 370 Lecture, Spring 2008

Chapter 4: AC Machinery Fundamentals 4-1. The simple loop is rotating in a uniform magnetic field shown in Figure 4-1 has the following char acteristics: B= 05 T to the right r=0lm ?=05m @= 103 rad/s (a) Calculate the voltage e,.,(£) induced in this rotating loop. (b) Suppose that a 5 Q resistor is connected as a load across the terminals of the loop. Calculate the current that would flow through the resistor. (c) Calculate the magnitude and direction of the induced torque on the loop for the conditions in (b). (d) Calculate the electric power being generated by the loop for the conditions in (bd). (e) Calculate the mechanical power being consumed by the loop for the conditions in (b). How does this number compare to the amount of electric power being generated by the loop? N 5 oo > B Bis a uniform magnetic field, aligned as shown. SOLUTION (a) The induced voltage on a simple rotating loop is given by Cina (zt) = 2roBi sin wt (48) Eig (t) = 2(0.1 m)(103 rad/s )(0.5 T)(0.5 m)sin103¢ pq (t)=5.15 sin103¢ V (b) Ifa5 Qresistor is connected as a load across the terminals of the loop, the current flow would be: i()=“a= 5.15 = V =1.08 sin 103¢ A (ce) The induced torque wauld be: Ting (t) =2rilB sin @ (4-17) Z,4(t)= 2(0.1 m)(1.03 sin a A)(0.5 m)(05 T) sin wt Figg (t) = 0.0515 sin? wt N-m, counterclockwise (a) The instantaneous power generated by the loop is: 103 4-2. 4-3. P(t)= egqi = (5.15 sin w V)(1.03 sin at A) =5.30 sin’ar W The average power generated by the loop is Py. = i 5.30 sin’wt dt= 2.65 W Tr ave (e) The mechanical power being consumed by the loop is: P = t= (0.0515 sin? wt V)(103 rad/s) =5.30 sin’ax W Note that the amount of mechanical power consumed by the loop is equal to the amount of electrical power created by the loop. This machine is acting as a generator, converting mechanical power into electrical power. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and 400 Hz. So uTION The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is 1, = 120, P The resulting table is Number of Poles f =50Hz ff, =60Hz ff, =400Hz 2 3000 r/min 3600 r/min 24000 r/min 4 1500 r/min 1800 r/min 12000 r/min 6 1000 r/min 1200 r/min 8000 r/min R 750 rémin 900 r/min 6000 r/min 10 600 r/min 720 r/min 4800 r/min 12 500 r/min 600 r/min 4000 r/min 14 428.6r/min 514.3 r/min 3429 r/min A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. All coils in each phase are connected in series, and the three phases are connected in A. The flux per pole in the machine is 0.060 Wb, and the speed of rotation of the magnetic field is 1800 r/min. (a) What is the frequency of the voltage produced in this winding? (b) What are the resultng phase and terminal voltages of this stator? SoLuTION (a) The frequency of the voltage produced in this winding is gh (1800 r/min )(4 poles) 7 = = 60 Hz 120 120 (b) There are 12 slots on this stator, with 40 turns of wire per slot. Since this is a four-pole machine, there are two sets of coils (4 slots) associated with each phase. The voltage in the coils in one pair of slots is E, -V2xN,.¢f —J2x(40 t)(0.060 Wb)(60 Hz) — 640 V There are two sets of coils per phase, since this is a four-pole machine, and they are connected in series, so the total phase voltage is 104 44. 45. V, = 2(640 V)= 1280 V Since the machine is A-connected, V, = Ys =1280V. A three-phase Y-connected 5U-Hz two-pole synchronous machine has a stator with 2000 turns of wire per phase. What rotor flux would be required to produce a terminal (linc-to-line) voltage of 6 KV? SoLuTION The phase voltage of this machine should be V, =V, / V3 =3454 V. The induced voltage per phase in this machine (which is equal to V, at no-load conditions) is given by the equation E,= V2x. N.OfF so o=—Ea 3464 V ~ J22N,f —Jtx(2000t)(50 Hz) =0.0078 Wb Modify the MATLAB program in Example 4-1 by swapping the currents flowing in any two phases. What happens to the resulting net magnetic field? SoLuTion This modification is very simple—just swap the currents supplied to two of the three phases. M-file: mag_field2.m M-file to calculate the net magetic field produced by a three-phase stator. dP dP de % Set up the basic conditions bnax = 1; % Normalize bmax to 1 freq = 60; % 60 Hz w= 2¥*pi*freg; % anyluar velocity (rad/s) %® First, cenerarte the three component magnetic fields ir t = 9:1/6c00:1/50; Baa sin(w*t) .* (ccs(0) + j*sin(G)): Bbb = sin(w*t4+2*pi/3) .* (cos(2*pi/3) + j*sin(2*pi/3)); Bee = sin(w*t-2*pi/3) .* (cos(-Z*pi/3) + j*sin(-2*pi/3)); % Calculate Bnez Bnet = Baa + Bbb + Bec; % Calculate a circle representing the expectec maximum %® value of Bnet circle = 1.5 * (cos(w*t) + j*sin(w*t)?}; 6 Plot the magnitude and direction ot the resulting magnetic % fields. Note that Baa is black, Bbb is blue, Bee is % magneta, and Bnet is red. for ii = 1:lengzh(t) % Plot the reference circls plot(circ_e,'k'); hold or; % Plot the four magnetic fields plot([C real(Baa(ii))],[0 imag(Baa(ii))],'k','!LineWidth!' ,2); plot([C real(Bbb(ii))],[0 imag(Bbb(ii))],'b','LineWidth' ,2); 105 4-0. 4-7, plotilO real(Bee(ii))!],1O imag(Bee(ii})!],'m','LineWidth',2) ; PloLi[O real (Buel (it))],[O inag (Buel (lid) )],'o', "LineWidbi', 3); axis square; axisi|-2 2 -2 2]); drawnow; hold off; ene When this program executes, the net magnetic field rotates clackwige, instead of counterclockwise. Tf an ac machine has the rotor and stator magnetic fielc’ shown in Figure P4-1, what is the direction of the induced torque in the machine? Is the machine acting as a moter of generator? Sarsrion Since t,,, =&B, XB... the induced torque is clockwise, opposite the direction of motion. The machine is acting, as a generator. The flux density distribution over the surface of a two-pole stator of radius 7 and length /is given by B=B,, cos (@,2-) (4.376) Prove that the total flux under each pole face is é= 2riB,, Nore B Air-gap flux density: ir gap B (a) = By cos (@,,1— a) Ba (c) Voltage 1s really into the page, since B is negative here. SoLution The total flux under a pole face is given by the equation o= B-dA Under a pole face, the flux density B is always parallel to the vector dA, since the flux density is always perpendicular to the surface of the rotor and stator in the air gap. Therefore, é= BdA A differential area on Lhe surface of a cylinder is given by the differential length along, the cylinder (2) times the differential width around the radius of the cylinder (rd@ ). dA =(di)(rda' where r is the radius of the cylinder Therefore, the flux under the pole face is g=- Ral rd@ Since 7 is constant and B is constant with respect to /, this equation reduces to g=rl Bae Now, B=B,, cos (ext a) = 8,, cos @ (when we substitute 9= wt a), so g=rl Bde xi2 . IZ gar By, cs 6d0= "1B, [sing] =rB, 1 ( 1) O= 2rlB,, 107 4-8. Tn the early days of ac motor development, machine designers had great difficult y controlling the core losses (hysteresis and eddy currents) m machines. They had not yet developed steels with low hysteresis, and were not making laminations as thin as the ones used today. To help control these losses, early ac motors in the USA were run from a 25 Hz ac power supply, while lighting systems were run from a separate 60 Hz ac power supply. (a) Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at 25 Hz. What was the fastest rotational speed available to these early motors? (>) Por a given motor operating at a constant flux density B, how would the core losses of the motor running at 25 Hz compare to the core losses of the motor running at 60 Hz? (c) Why did the early engineers provide a separate 60 Hz power system for lighting? SOLUTION (a) The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is _ 120 f, P The resulting table is Number of Poles =f, = 25 Hz 2 1500 r/min 4 750 r/min 6 500 r/min 8 375 r/min 10 300 r/min 12 250 r/min 14 214.3 r/min The highest possible rotational speed was 1500 r/min. (b) Core losses scale according to the 1.5" power of the speed of rotation, so the ratio of the core losses at 25 Hz to the core losses at 60 Hz (for a given machine) would be: Ls ratio = 1500 =0.269 or 26.9% 3600 fc) At 25 Hz, the light from incandescent lamps would visibly flicker in a very annoying way.