Voltage Induced in a Rotating Loop Assumptions: Air gap flux density is radial. The flux density is uniform under magnet poles and vanishes midpoint between poles (Neutral plane). As the rotor moves at velocity vin a magnetic field B, the voltage induced in each segment is given by equation 1. Segment ab ind ( v B) e l vbl positive into page under pole face e ba 0 beyond the pole edge Figure 7-1 A simple rotating loop between curved pole faces. (a) erspective view; (b) field lines; (c) loop view; (d) front view. 1 2 2. Segment bc. e cb 0 Since v x B is into (or out of) page and thus perpendicular to l. 3. Segment cd. The total induced voltage on the loop is given by e 2vB under the pole faces e + e + e + e l 0 beyond th epole edges ind ba ad dc cb vbl positive out of page under pole face e dc 0 beyond the pole edges 4. Segment da e ad 0 3 Figure 7-3 The output voltage of the loop 4 1
The tangential velocity v of the loop can be expressed as v r ω m Getting DC Voltage Out of Rotating Loop: Commutator Hence, 2 0 Where, ( l) p φ π r B A B 5 6 The Induced Torque in the Rotating Loop Suppose a battery is connected to the loop The force on a current-carrying conductor placed in a magnetic field is given by 1. Segment ab. F i l ( x B) F Bl i tangent to direction of motion ab The torque on the rotor caused by the force is o rf sinθ r(ilb)sin rilb ab 90 CCW (Theta is angle between r and F) 2. Segment bc. Figure 7-6 (a) A DC motor with commutator. (b) Derivation of an equation for induced torque. 7 F bc 0 since l is parallel to B 8 2
3. Segment cd Fcd i ( l x B) ilb tangent to direction of motion The torque on the rotor caused by the force is 4. Segment da. F da 0 since is parallel to B l 2 rilb under the pole face + + + ind ab bc cd da 0 beyond the pole edge Since ( l) p φ π r B A B o rf sinθ ri l B sin90 ri l B CCW cd 2 φ i under the pole face ind π 0 beyond the pole edges 9 10 Given: r 0.5 m R 0.3 Ω V B 120 v l 1.0 m B 0.25 T What happens when the switch is closed? (1) Current flows in loop 400. 0, 0 3
What happens when the switch is closed? (2) The current produces a torque Combining the forces on each segment + + + What happens when the switch is closed? (3) The rotor begins to turn, an induced voltage develops ( ) Combining the voltages on each segment + + + +0+ +0 2 90 +0+ 90 +0 2 2 0.5 400 1 0.25 100 Nm What happens when the switch is closed? (4) Both the current iand torque Twill fall as increases since and steady state will be reached when, 0, 0 The steady state velocity will be 2 2 2 120 2 0.5 0.25 1 480 / What happens when a 10 Nm load torque is applied? The load torque will cause speed to fall. Then as 2 falls, the current increases since 2. and steady state will be reached when The steady state velocity will be 40 A... 2 108 432 / 2 0.5 0.25 1 4
How much power is supplied to the shaft? Tω (10)(432) 4320 w How much power is supplied by the battery? V B i (120)(40) 4800 w How much power is lost in the resistor? i 2 R (1600)(0.3) 480 w Suppose a torque of 7.5 Nm is applied in the direction of rotation. What is the new steady-state speed? The steady state velocity will be 30 A out of machine! Since.. the speed increase will cause >... 2 129 516 / 2 0.5 0.25 1 Induced Voltage Equations in DC Machine Z Z E e vb A l a a v rω Z E φω Where: 2π a Kis a machine s constant Z is total number of conductors is number of pole ais the number of current paths 19 m Zrω Bl m E A a Flux per pole BA B Therefore, φ Kφω A m m 2π rl Induced Torque Equations in DC Machine cond ri cond Bl IA i cond a rbli A Z ind a φ 2π rl BA B Therefore, Z φi KφI 2π a ind A A 20 5
LAB 5 RIOR REARATION: Complete the following at a time determined by the laboratory instructor. 1.Show that the mechanical power output in watts of a motor can be found from the equation mech n( rpm) T ( Nm) 2 π ( rad / revolution) n T Nm n T ( watts) n( rpm) T ( Nm) ( ) ( Watts) 9.55 60(sec/ min) 9.55 sec 9.55 LAB 5 RIOR REARATION: 2. A DC motor turns at a speed of 1460 rpm and produces an output torque of 3.0 Nm. The DC voltage applied to the motor is 100V and a current of 5.1A flows through the motor. a. What is the efficiency of the motor? η out in mech e 1460( rpm) 3.0Nm / 9.55 458.6 89.9% 100V 5.1A 510 b. How much are the power losses in the motor? (loss) in out 510 458.6 51.5 W Introduction to Rotating Machines Introduction to Rotating Machines 6